The Forcing Conundrum

Guest Post by Willis Eschenbach.

For all of its faults, the IPCC (Intergovernmental Panel on Climate Change) lays out their idea of the climate paradigm pretty clearly. A fundamental part of this paradigm is that the long-term change in global average surface temperature is a linear function of the long-term change in what is called the “radiative forcing”. Today I found myself contemplating the concept of radiative forcing, usually referred to just as “forcing”.

So … what is radiative forcing when it’s at home? Well, that gets a bit complex … in the history chapter of the Fourth Assessment Report (AR4), the IPCC says of the origination of the concept (emphasis mine):

The concept of radiative forcing (RF) as the radiative imbalance (W m–2) in the climate system at the top of the atmosphere caused by the addition of a greenhouse gas (or other change) was established at the time and summarised in Chapter 2 of the WGI FAR [First Assessment Report].

tropopause temperature by latitude

Figure 1. A graph of temperature versus altitude, showing how the tropopause is higher in the tropics and lower at the poles. The tropopause marks the boundary between the troposphere (the lowest atmospheric layer) and the stratosphere. SOURCE 

The concept of radiative forcing was clearly stated in the Third Assessment Report (TAR), which defined radiative forcing as follows:

 The radiative forcing of the surface-troposphere system due to the perturbation in or the introduction of an agent (say, a change in greenhouse gas concentrations) is the change in net (down minus up) irradiance (solar plus long-wave; in Wm-2) at the tropopause AFTER allowing for stratospheric temperatures to readjust to radiative equilibrium, but with surface and tropospheric temperatures and state held fixed at the unperturbed values.

In the context of climate change, the term forcing is restricted to changes in the radiation balance of the surface-troposphere system imposed by external factors, with no changes in stratospheric dynamics, without any surface and tropospheric feedbacks in operation (i.e., no secondary effects induced because of changes in tropospheric motions or its thermodynamic state), and with no dynamically-induced changes in the amount and distribution of atmospheric water (vapour, liquid, and solid forms).

So what’s not to like about that definition of forcing?

Well, the main thing that I don’t like about the definition is that it is not a definition of a measurable physical quantity.

We can measure the average surface temperature, or at least estimate it in a consistent fashion from a number of measurements. But we can never measure the change in the radiation balance at the troposphere AFTER the stratosphere has readjusted, but with the surface and tropospheric temperatures held fixed. You can’t hold any part of the climate fixed. It simply can not be done. This means that the IPCC vision of radiative forcing is a purely imaginary value, forever incapable of experimental confirmation or measurement.

The problem is that the surface and tropospheric temperatures respond to changes in radiation with a time scale on the order of seconds. The instant that the sun hits the surface, it starts affecting the surface temperature. Even hourly measurements of radiative imbalances reflect the changing temperatures of the surface and the troposphere during that hour. There is no way that we can have the “surface and tropospheric temperatures and state held fixed at the unperturbed values” as is required by the IPCC formulation.

There is a second difficulty with the IPCC definition of radiative forcing, a practical problem. This is that the forcing is defined by the IPCC as being measured at the tropopause. The tropopause is the boundary between the troposphere (the lowest atmospheric layer, where weather occurs), and the stratosphere above it. Unfortunately, the tropopause varies in height from the tropics to the poles, from day to night, and from summer to winter. The tropopause is a most vaguely located, vagrant, and ill-mannered creature that is neither stratosphere nor troposphere. One authority defines it as:

The boundary between the troposphere and the stratosphere, where an abrupt change in lapse rate usually occurs. It is defined as the lowest level at which the lapse rate decreases to 2 °C/km or less, provided that the average lapse rate between this level and all higher levels within 2 km does not exceed 2 °C/km.

This is an interesting definition. It highlights that there can be two or more layers that look like the tropopause (little temperature change with altitude), and if there is more than one, this definition always chooses the one at the higher altitude.

In any case, the issue arises because under the IPCC definition the radiation balance is measured at the tropopause. But it is very difficult to measure the radiation, either upwelling or downwelling, at the tropopause. You can’t do it from the ground, and you can’t do it from a satellite. You have to do it from a balloon or an airplane, while taking continuous temperature measurements so you can identify the altitude of the tropopause at that particular place and time. As a result, we will never be able to measure it on a global basis.

So even if we were not already talking about an unmeasurable quantity (radiative change with stratosphere reacting and surface and tropospheric temperatures held fixed), because of practical difficulties we still wouldn’t be able to measure the radiation at the tropopause in any global, regional, or even local sense. All we have is scattered point measurements, far from enough to establish a global average.

This is very unfortunate. It means that “radiative forcing” as defined by the IPCC is not measurable for two separate reasons, one practical, the other that the definition involves an imaginary and physically impossible situation.

In my experience, this is unusual in theories of physical phenomena. I don’t know of other scientific fields that base fundamental concepts on an unmeasurable imaginary variable rather than a measurable physical variable. Climate science is already strange enough, because it studies averages rather than observations. But this definition of forcing pushes the field into unreality.

Here is the main problem. Under the IPCC’s definition, radiative forcing cannot ever be measured. This makes it impossible to falsify the central idea that the change in surface temperature is a linear function of the change in forcing. Since we cannot measure the forcing, how can that be falsified (or proven)?

It is for this reason that I use a slightly different definition of the forcing. This is the net radiative change, not at the troposphere, but at the TOA (top of atmosphere, often taken to mean 20 km for practical purposes).

And rather than some imaginary measurement after some but not all parts of the climate have reacted, I use the forcing AFTER all parts of the climate have readjusted to the change. Any measurement we can take already must include whatever readjustments of the surface and tropospheric temperatures that have taken place since the last measurement. It is this definition of “radiative forcing” that I used in my recent post, An Interim Look at Intermediate Sensitivity.

I don’t have any particular conclusions in this post, other than this is a heck of a way to run a railroad, using imaginary values that can never be measured or verified.

w.

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441 Responses to The Forcing Conundrum

  1. GlynnMhor says:

    “… imaginary values that can never be measured or verified.”

    That about says it all, doesn’t it?

  2. Truthseeker says:

    Willis, you conclude with …
    “I don’t have any particular conclusions in this post, other than this is a heck of a way to run a railroad, using imaginary values that can never be measured or verified.”
    I think that is precisely the point. Since you cannot measure of verify these critical factors, you cannot “prove” the IPCC wrong. If you cannot prove them wrong, they must be right (at least in IPCC logic world).
    A bureaucrat never lets real science get in the way of a policy for control over others.

  3. noaaprogrammer says:

    Imaginary by design or by accident?

  4. NZ Willy says:

    Radiative forcing has a familiar analog in albedo, which is reflectivity of sunlight. But albedo restricts sunlight downwards, “radiative forcing” restricts heat upwards. There are times when cloud cover does indeed restrict sunlight from reaching us and the sky is darkish — so, conversely, “radiative forcing” would make the Earth look cool from space, i.e. Earth would be retaining heat. But the fly in this ointment is that Earth, as a black box, must radiate more as it heats. There is no glass skin as with a greenhouse. When the clouds clear, the excess heat radiates away into space.
    My car’s windshield still freezes in wintertime when parked out in the open, while the side windows remain thawed. This is because the side windows are warmed by the ground’s infrared and the windshield, which points to the sky and does not see the ground, is not. My car’s windshield tells me that there is no radiative forcing to prevent it from freezing — it shows that heat is radiating into space just like it ever has, IPCC Chicken Littles notwithstanding.

  5. Bill H says:

    Just like the hidden agenda in the emails of climate-gate the “you cant prove me wrong” fantasies of the IPCC and Alarmists make it impossible to prove or disprove the theroy… Its kind of like hiding your work, methods, and answers then telling people you will just have to trust us..

  6. john robertson says:

    By design, every action of the UN points to this choice of the IPCC to use imaginary values that can neither be measured, verified or discarded.
    It leaves people who question the scientific claims of the IPCC in a fog, its like arguing with the mist.
    It turned out that critiquing the science, with all its flaws, was just a means of tying critic up in knots.
    Of course the science was settled, there is none and none required.
    And after Climate gate and Copenhagen they are exposed for what they be.
    Bandits and Con artists, they never had science to support the plan and never felt they needed any .
    This scheme needs to be rewarded, by sentences proportional to those of Bernie Madoff.
    Who, in hindsight, was just following the lead of his government.

  7. AndyG55 says:

    Willis, the tropical tropopause goes up and down.
    Does it at any time go above 20km?
    Also the mid lat tropopause looks like it has slight change point at 20km

    I ask this because using a point just at or only slightly above a change-point seems a strange thing to do.. ……. or did you mean 20 miles (30km) ?

  8. AndyG55 says:

    Darn it ,, I am referring to your TOA height. !

  9. Johnny says:

    It seems to me that climate science is not a science in the Newtonian sense. Results take 30 years to be disproven and even if all experimental evidence is contrary to the results of the science it doesn’t shake people’s beliefs. We have a science where the people who create the theory also get to play with the raw data and “adjust it” constantly to reflect their view of the perturbations they believe are most important so that the data conforms better to their model predictions. This is a science where repeatable experiments are impossible. Where the central physical laws proposed don’t have to be proven by any actual experiments or measurements to be “settled”. It sounds more like a middle ages religion than a science like physics or chemistry.

  10. My son, a surveyor, heard me mention watt per sq km at the troposphere and immediately interjected “At what altitude?”
    Unless light passing through a horizonal sq m is parallel, like some laser light, then because there are more sq m of conceptual surface at a higher altitude quasi-sphere around the globe, the flux through each is less when there is a steady source. How does one define the flux when the “number of sq m” on a surface is forever changing rapidly with altitude, but by definition requires measurement to be taken when steady state is approached?
    I have no idea about what models do about the variation in numbers of sq m shared by a watt as altitude is changed. Can someone assure me that the effect is built into the math?
    If it is not, then a watt per sq m is ‘more powerful’ at low altitude than one at high and it would be hard to construct a Willis fig 1.

  11. Sparks says:

    But, Willis did you back that up with some fancy talk and show a chart?

  12. Spector says:

    I have been assuming that they are assuming that the outgoing radiation at the top of the troposphere is presumed to be driving or ‘forcing’ the state of the troposphere as a whole.

    Radiative Forcing
    From The Wikipedia:
    “In climate science, radiative forcing is defined as the difference between radiant energy received by the earth and energy re-radiated to space. Typically, radiative forcing is quantified at the tropopause in units of watts per square meter of earth’s surface. A positive forcing (more incoming energy) warms the system, while negative forcing (more outgoing energy) cools it. Causes of radiative forcing include changes in insolation (incident solar radiation) and in concentrations of radiatively active gases and aerosols.”

    http://en.wikipedia.org/wiki/Radiative_forcing

    I think that ‘Top of Troposphere, Net Radiation Input’ might be a better term, being less presumptive. I see that the MODTRAN plots they have, show only thermal outgoing radiation as a positive value.

  13. So the thing is noting but a large fudge factor that is made up to meet the needs or desires of the modelers.

  14. E.M.Smith says:

    “Well, the main thing that I don’t like about the definition is that it is not a definition of a measurable physical quantity.”

    Hip Hip Hooooray!!! Willis Sees the Light!

    @AndyG55:

    See the chart at the bottom of here:

    http://chiefio.wordpress.com/2012/12/12/tropopause-rules/

    The one with density and temperature and height.

    http://chiefio.wordpress.com/2012/12/12/tropopause-rules/atmosphere-temp-press-profile-1076/

    Puts 20 km as top of the tropoPause and just at the bottom of the stratosphere, but not near the Top Of The Atmosphere at all, as the Mesosphere goes to 80 km…

    I ‘eyeball it’ at a bit under 1% of sea level pressure though, so well above the bulk of the air mass.

    It also ought to be about at the point where molecules are far enough apart that a radiative regime can begin. (Part of what I find funny about the IPCC definition is that they define a radiative based metric at the point where the air below it is, by definition, in a non-radiative driven mode. It is only above the tropopause that radiation matters to IR outbound.)

  15. Robert Clemenzi says:

    At equilibrium, for every altitude, the flux in equals the flux out. It does not matter if it is the tropopause or not. (That is the definition of *equilibrium*.) That is why the IPCC definition has to be so strange.

    As for using a linear equation – what ever the shape of the actual function is, the first derivative will provide the slope of a straight line that can be used to estimate values over some range. Since CO2 is such a minor player, it is highly likely that the error between the straight line approximation and the actual function is less than we can measure.

  16. Larry in Texas says:

    Thanks very much Willis, this is one of the most informative pieces you have ever written (not that the others weren’t informative, but this one was well written for a layman like me). Now, if you could define for me what the “lapse rate” is, I might get a better handle on a very complicated subject for me.

  17. Willis Eschenbach says:

    AndyG55 says:
    December 12, 2012 at 9:32 pm

    Willis, the tropical tropopause goes up and down.
    Does it at any time go above 20km?
    Also the mid lat tropopause looks like it has slight change point at 20km

    I ask this because using a point just at or only slightly above a change-point seems a strange thing to do.. ……. or did you mean 20 miles (30km) ?

    Here’s the definition from NASA:

    Top of Atmosphere:
    a given altitude where air becomes so thin that atmospheric pressure or mass becomes negligible. TOA is mainly used to help mathematically quantify Earth science parameters because it serves as an upper limit on where physical and chemical interactions may occur with molecules in the atmosphere. The actual altitude used for calculations varies depending on what parameter or specification is being analyzed. For example, in radiation budget, TOA is considered 20 km because above that altitude the optical mass of the atmosphere is negligible. SOURCE

    w.

  18. ntesdorf says:

    “This means that the IPCC vision of radiative forcing is a purely imaginary value, forever incapable of experimental confirmation or measurement” This is quite consistent with all other aspects of CAGW theory. If the theory is threatened, the theory shifts ground to new unmeasureables grounds, until in turn these are threatened and the Caravan is forced to move on to Climate Change, Climate Disrution, Extreme Climate or whatever ground is still available. Slowly, very slowly, people are noticing this and quietly distancing themselves to solid ground..

  19. Willis Eschenbach says:

    Geoff Sherrington says:
    December 12, 2012 at 10:28 pm

    My son, a surveyor, heard me mention watt per sq km at the troposphere and immediately interjected “At what altitude?”
    Unless light passing through a horizonal sq m is parallel, like some laser light, then because there are more sq m of conceptual surface at a higher altitude quasi-sphere around the globe, the flux through each is less when there is a steady source. How does one define the flux when the “number of sq m” on a surface is forever changing rapidly with altitude, but by definition requires measurement to be taken when steady state is approached?
    I have no idea about what models do about the variation in numbers of sq m shared by a watt as altitude is changed. Can someone assure me that the effect is built into the math?
    If it is not, then a watt per sq m is ‘more powerful’ at low altitude than one at high and it would be hard to construct a Willis fig 1.

    Geoff, your son is right, but it is a difference that doesn’t make a difference. Here’s why.

    Suppose we take the TOA as our altitude, about 20 km up from the surface. The radius of the earth is about 6378 km, and the radius of the TOA is 6398 km. The surface area scales as the square of the radius.

    This means that the area of the TOA layer is about 0.6% larger than the surface of the earth. As a result, for any but the most detailed of analyses, the difference can be (and usually is) safely ignored.

    w.

  20. Michel says:

    “Top Of the Atmosphere” and “Troposphere” are useful concepts to develop climate models. They are moving targets. No exact physical definition of such boundaries will ever be possible.
    “Forcing” before, during, or after allowance for thermal equilibrium to be re-established is also just a concept. Any one can chose a definition that will be adequate for his modelling purpose.

    Models are no scientific hypothesis. They are speculations based on more or less plausible scenarii and mathematical formulas about the behaviour of physical parameters within a complex non-linear multi-variable system.
    When model calculations get too complicated, then variable reduction and linearizations are sought to simplify them. This doesn’t make the model better but at least it becomes computable.
    A model delivering a fair or a wrong prediction (into the future or as reconstruction of the past) does not prove or falsify a scientific theory. It is just a successful or a failed speculation.

  21. rxc6422rxc says:

    Could this factor be similar to what we call a “heat transfer coefficient” in engineering? This is a factor that is used to calculate the flow of heat in, say, a heat exchanger where you have fluids on each side of a metallic barrier. The simplified equation is Q=UAdT, where Q= btu/hr, A=area of heat transfer surface, dT=temperature difference between the two fluids, and U=overall heat transfer coefficient. U can usually be broken down into different components to account for resistance from the hot fluid to the metal surface, resistance to conduction in the metal, and then heat transfer from the surface of the metal to the cooler fluid. And each of those heat transfer coefficients can be affected by he fluid state (liquid, gas, 2-phase), the degree of superheat of the fluid relative to the metal temperature, etc.

    These coefficients are calculated/inferred from temperature measurements of the fluid and the solid metallic surface, and once calculated, are used to determine the performance of the equipment. They can be very hard to measure/calculate, so engineers usually incorporate safety factors into their overall equipment performance calculations. One would normally include allowances for fouling and corrosion in U, and you might even include additional area to the heat exchanger to make sure that the equipment performs as expected.

  22. peterg says:

    I suppose the forcing is a control theory abstraction. Add a GHG gas. Model its impact as if it were an extra radiative input. Deduce the increase in surface temperature. So it only has value as an imaginary construct for a particular way of modelling the “control” system so as to get the answers you want your model to give.

  23. AlecM says:

    Agreed, it’s a stupid parameter. However. the level of stupidity is far worse than Willy claims.

    Radiative forcing is the temperature radiation field from GHG thermal emission. Except for some water vapour side bands, it annihilates the surface radiation field. There is no net ‘forcing’ so no CO2-AGW. The GHE is from surface temperature rising above the level set by lapse rate.

    This is basic heat transfer physics between two emitters at nearly the same temperature that all process engineers know but far too may physicists buy into ‘back radiation’, the artefact of a pyrgeometer measurement.

    This is the biggest scientific cock up in History compounded by the Houghton mistakes of assuming lapse rate is caused by GHGs and using the two – stream approximation when only net radiative flux can do thermodynamic work. A truly massive scam by con artists even now trying to justify non-science.

  24. Liberal Skeptic says:

    Untestable hypothesis is the reason people outside of the theory don’t take string theory seriously yet. This doesn’t look any different.

    On a related note, I pointed out an appeal to authority logical fallacy in someone famous’s tweet regarding climate change last night and now i’ve got a bunch of people tweeting me who clearly have no idea what a logical fallacy is. No wonder they can’t see that Catastrophic climate change is just alarmism. This is basic critical thinking.

  25. Spector says:

    RE: Willis Eschenbach says: (December 12, 2012 at 11:51 pm)

    “… For example, in radiation budget, TOA is considered 20 km because above that altitude the optical mass of the atmosphere is negligible.

    Again, I see a standard definition is not quite what I would expect it to be, as this definition neatly avoids the strong CO2 radiation from the mesosphere. The default MODTRAN setting of 70 km would make much more sense for me. 20 km may be good enough for optical calculations, but I think it’s too low for good LWIR estimates.

  26. richardscourtney says:

    Willis:

    I write to make two observations.

    Firstly, Vincent Grey made your point about the non-physical nature of ‘radiative forcing’ in his peer review comments of the first IPCC Report. And he has made it in his review comments on each of the subsequent IPCC Reports, too.

    I suspect it may be to your mutual benefit if the two of you were to converse on this matter.

    Secondly, you rightly say that an important issue with radiative forcing is that its calculation holds surface temperature constant. However, some changes to global temperature result from surface effects which spread from the surface through the climate system; for example, Lindzen has pointed out that redistribution of heat by ocean currents could account for all twentieth century global temperature rise.

    Considering radiative forcing to be THE driver of global temperature change is nonsense when there are effects whereby some changes to global temperature drive radiative forcing.

    Richard

  27. Willis Eschenbach says:

    Robert Clemenzi says:
    December 12, 2012 at 11:36 pm

    At equilibrium, for every altitude, the flux in equals the flux out. It does not matter if it is the tropopause or not. (That is the definition of *equilibrium*.) That is why the IPCC definition has to be so strange.

    Thanks, Robert. Since the earth is almost never in equilibrium at any altitude, I’m not sure how this is relevant or what it means to the IPCC definition.

    As for using a linear equation – what ever the shape of the actual function is, the first derivative will provide the slope of a straight line that can be used to estimate values over some range. Since CO2 is such a minor player, it is highly likely that the error between the straight line approximation and the actual function is less than we can measure.

    This is not true for a governed system such as the climate, which is ruled by emergent phenomena that keep the temperature within bounds. See my post The Details Are In The Devil for a discussion of some of these issues.

    It’s like looking for a linear relationship between the temperature in my house and the outside temperature. The answer is, there is no correlation, linear or any other kind, between the outside temperature and the temperature of my house because it is a governed system. It has a thermostat that keeps it at about the same temperature regardless of the outside temperature.

    And that kind of relationship, Robert, you cannot approximate by a straight line, a curved line, or any other kind of line.

    w.

  28. Meynard François says:

    The physical problem of the coupling between radiation convection heat exchange is in general difficult to calculate exactly. Going back to the early papers of Manabe & Wetherhald, you see that an “ad hoc” algorithm has been used in order to calculate something. It consists in discoupling the calculation in two steps.

    1) pure radiative heat exchange due to a so called radiative forcing.
    2) then the corrective effect of the other heat flows convective.

    Apparently this “CO2 green house effect algorithm” is still used today in the large 3-D climate models. But, from a physical point of view this is incorrect, unless it is proven that this algorithm reflects the real coupled heat exchange. To my knowledge this has not been scientifically established yet. Does somebody know more about this point ?

  29. Stephen Richards says:

    i’ve got a bunch of people tweeting me who clearly have no idea what a logical fallacy is. No wonder they can’t see that Catastrophic climate change is just alarmism.

    You must have got one from my daughter ; )) She can talk me through her logical sequences for 10 mins and end it with me completely confused.

  30. Willis Eschenbach says:

    Spector says:
    December 13, 2012 at 12:57 am

    RE: Willis Eschenbach says: (December 12, 2012 at 11:51 pm)

    “… For example, in radiation budget, TOA is considered 20 km because above that altitude the optical mass of the atmosphere is negligible.

    Again, I see a standard definition is not quite what I would expect it to be, as this definition neatly avoids the strong CO2 radiation from the mesosphere. The default MODTRAN setting of 70 km would make much more sense for me. 20 km may be good enough for optical calculations, but I think it’s too low for good LWIR estimates.

    I’d agree. Note that the NASA definition says “The actual altitude used for calculations varies depending on what parameter or specification is being analyzed.”

    w.

  31. Roger Longstaff says:

    To me, the IPCC definition says it all: “….at the tropopause AFTER allowing for stratospheric temperatures to readjust to radiative equilibrium…”.

    There is never radiative equilibrium in the atmosphere as a consequence of planetary rotation, so the differences between equilibrium and non-equilibrium thermodynamics make the IPCC analysis fatally flawed. Add to this the incorrect use of mathematics (they do not use Holder’s Inequality, as far as I am aware) and the whole thing becomes completely meaningless.

  32. peter Miller says:

    This reminds me of a discussion I had many years ago with the late Prof. Krige, when he was supervising my Master’s thesis.

    The statistical method of kriging is well known in the mining world for resource and reserve calculation.

    I was telling him I could not duplicate the results of two companied’ official reserves from drill results.

    He looked at my workings and commented: “but you haven’t made the necessary adjustments.”

    So I asked: “how do I know what ‘adjustments’ to make?”

    His response was: “from observations.”

    Ever since that day, I have treated statistical models and interpretations with a sack of salt.

    So, in the case of radiative forcings, we need to make a huge number of observations to ascertain some form of average unrepresentative value and this, from a practical point of view, is virtually impossible. ,

    Makes perfect clumate science sense to me.

  33. E.M.Smith says:

    Oh what a Humpty Dumpty world… where even “Top Of Atmosphere” doesn’t mean the actual top of the actual atmosphere but varies based on what you want it to mean… Sigh…

  34. richard verney says:

    NZ Willy says:
    December 12, 2012 at 9:11 pm
    /////////////////////////////////////////////////////////////////////////
    Willy, I made a very similar observation about 6 or 8 montha ago on another article written by Willis. Save that I was talking about dew, and save that I would suggest that it is warm air being convected from the ground that keeps the car doors/door windows ice free.

    Is it not the case that dew would be a very rare event if DWLWIR truly had sensible energy? After all it is claimed that DWLWIR can warm the oceans (although I am very sceptical of that claim due to the absorption characteristics of water and since the top micron layer of the ocean is frequently little more than airborn windswept spray and spume such that it is divorced from the ocean below).

    Without checking, I believe that 40% of all DWLWIR is a absorbed within 1 micron of water and 60% within 2 or 3 microns. With that absorption and the DWLWIR figure given by Trenbeth, there is enough energy to evaporate water such that dew (which is often fractions of a millimetre)would quickly evaporate.

    Whilst the point you raise is slightly off topic, I think that you are right to look at the practical application of the theory to the real world as we experience every day in our lives.

  35. TimTheToolMan says:

    Willis notes “This means that the area of the TOA layer is about 0.6% larger than the surface of the earth. As a result, for any but the most detailed of analyses, the difference can be (and usually is) safely ignored.”

    Is it really that large? That would make the difference between the TOA and the surface averaged over the entire planet about 2 W/m2. Considering the net absorbed energy is thought to be less than 1 W/m2, I dont think that could be considered too small to worry about!

    http://spark.ucar.edu/longcontent/energy-budget

  36. Dumbed Down Science says:

    The IPCC are the true deniers, scientific charlatans degrading climate, one false hypothesis after another.

  37. AndyG55 says:

    E.M.Smith says:
    Puts 20 km as top of the tropoPause and just at the bottom of the stratosphere, but not near the Top Of The Atmosphere at all,

    Seems it is a NASA definition.. seems pretty silly to me to use a point so close to the change of the troposphere. but i guess if it makes it easier for them !!!!

    Just saying ;-)

  38. AndyG55 says:

    What I really am trying to say is that its always better to do your calculation “away” from areas where things could be changing. and at 20km , we are very close the point where deltaT changes from +ve to -ve. Not clever.

  39. Sparks says:

    E.M.Smith says:
    December 13, 2012 at 1:56 am

    Shut up and obey you’re masters

  40. AndyG55 says:

    Hey Willis, in figure 1, what lapse rate is being used ??

    I’m really tired, but for say the polar regions, a drop of 50C in just under 9km doesn’t make sense.

  41. Mack says:

    Tim the Tool Man,
    Good show putting up that link to the Earth’s Energy budget…..none of the figures you see in that cartoon are actual measurements but may well have been pulled from Trenberth’s ass.

  42. Ken Coffman says:

    I was convinced Willis’ logic, integrity and intellect would eventually lead him to discard the added-CO2-causes-warming theme. Anyone with engineering skill would know what happens when you add additional agents of conduction or convection to a system. Agents of conduction are agents of cooling. Anyone puzzled by what radiation can do should simply imagine the radiative path replaced by a conductive path. Conduction, convection and radiation always work in the same direction. The rate and effectiveness are different, but the outcome is always in the same direction.

  43. Bob Ryan says:

    Willis: the problem you point too has been a long running issue in the philosophy of science. You may be interested in the work of David Papineau and in particular his book ‘Theory and Meaning’. He would argue that much of modern science is made up of theoretical language and constructs which are not directly observable or measurable. He has a very interesting talk on this issue at http://philosophybites.com/2009/01/david-papineau-on-scientific-realism.html.

  44. Bob Ryan says:

    Sorry, hasty read through. My comment at 4.01am: in the first sentence it should be ‘to’ not ‘too’.

  45. Jim Cripwell says:

    The thing I find most troubling about these discussions, is that the warmists claim that they can accurately estimate the change in radiative forcing which occurs when CO2 concentrations change, by using radiative transfer models. Now my understanding is that radiative transfer models are engineering type models which have been fully validated. I cannot understand why radiative transfer models are suitable to estimate radiative forcing. They were never designed to estimate radiative forcing, and no-one seems to have shown that they are capable of estimating radiative forcing. Myhr et al 1998 merely states that they have used three such radiative transfer models, with not a word as to why they are suitable to do the job. Any comments, Willis?

  46. Dirck says:

    “A fundamental part of this paradigm is that the long-term change in global average surface temperature is a linear function of the long-term change in what is called the “radiative forcing”. ”

    That is not a fundamental part of anything except your catastrophic misunderstandings of climate science.

  47. richardscourtney says:

    Dirck:

    Your post at December 13, 2012 at 4:12 am says in total

    “A fundamental part of this paradigm is that the long-term change in global average surface temperature is a linear function of the long-term change in what is called the “radiative forcing”. ”

    That is not a fundamental part of anything except your catastrophic misunderstandings of climate science.

    You are wrong.

    See? I have just repeated your silly game. It’s not helpful, is it?

    And, incidentally, if pressed I can show that the IPCC says you are wrong, but I won’t until you have justified your untrue assertion that Willis’ statement is wrong.

    Richard

  48. phi says:

    Willis Eschenbach,

    “I don’t have any particular conclusions in this post, other than this is a heck of a way to run a railroad, using imaginary values that can never be measured or verified.”

    It’s even worse than that. GHG forcing is in fact undefined and indefinable. The need to calculate an instantaneous balance just after the addition of material created ex nihilo does not allow this material to adapt his temperature to the environment. As the temperature of material created ex nihilo is not defined, the GHG forcing is not defined either.

  49. Ryan says:

    I have been thinking of a way to demonstrate to the layman the basic problems with the greenhouse theory and I came up with this. It assumes you have a so-called “radiator” in your house and it is working hard this winter.

    Put your hand on the “radiator”. Is it hot? Well that’s because it has hot water inside (about 60Celsius so not THAT hot but still pretty warm) and the well-known phenomena of conduction is carrying that heat to the surface. Now put your face in front of the radiator – feel the heat? No? Well what your feeling (or rather not feeling) is the radiation coming off the “radiator”. Now put your face just above the hot “radiator”. Do you feel the heat now? Sure you do. You can actually feel a slight breeze of warm air past your face due to the well-known phenomena of convection. The fact is the phenomena of conduction and convection are far more powerful means of removing heat energy from warm surfaces than radiation is. The hotter the surface the more conduction and convection you will get.

    For the atmosphere we get the same thing. During the day the land and sea gets warmed by the sun. Replacing oyxgen by CO2 has almost no impact on this so peak daytime temperatures of land and sea cannot be impacted by greenhouse gases. The air gets warm too – partly from the suns rays directly but in great part due to the land and sea causing convection currents in the air. These convection currents go up a long way – right up to the troposphere at about 30,000 ft where the jet planes and the high cirrus clouds are. The closer you are to the ground the stronger the convection currents are – on a hot day you can often see the effect of strong convection as the heating effect is enough to bend the light through the air due to changes in density. Close to the ground the air temperature is determined almost entirely by these strong convection currents.

    The ground and sea itself maybe at a completely different temperature to the air just above it. Why is that? During the day the suns rays penetrate the atmosphere. It doesn’t matter if we think of the visible sunlight or infra-red – if it didn’t penetrate the atmosphere the Earth would be a cold dark place. It isn’t. Those photons ejected by the sun come hurtling across space, through the air and then meet the immovable object known as the land. Those photons might on occassion give up their energy to the air but most simply crash into land or sea where they are completely absorbed. The suns rays hitting the atmosphere may cause it to heat up quite readily because the air isn’t very dense – it is a different story for land and sea because the high density of atoms in both makes heating the land and sea hard work for those incoming photons. CO2 plays absolutely no part in this at all.

    It is during the night that CO2 has a part to play. It acts like a nightime insulating blanket has appeared over the Earth, but it is lightly distributed throughout the atmosphere at low density. However, it isn’t a proper insulating blanket because in such a blanket the air is in pockets and is not free to move – in the lower atmosphere there are strong air currents always conducting heat away from the warm land and sea. At ground level CO2 cannot stop much of the heat radiating from planet Earth (and indeed if it did then we would already be at saturation point and adding more CO2 wouldn’t make any difference) and as far as conduction and convection is concerned CO2 is just as capable of conducting heat away as oxygen. Frankly CO2 can make no difference at ground level. Even just above ground level the air itself may be warmer but there is little energy in warm air itself because it has low density and that extra energy is distributed throughout the atmosphere. You cannot melt the ice of the Arctic by simply making the nightime air above it very slightly warmer – it takes far too much energy to melt water and there just isn’t ever enough trapped in the air. You can only melt the ice by making the Arctic oceans warmer – and that cannot be done by making the air a slightly better insulator.

    In summary the greenhouse effect largely ignores the fact that where life exists on Earth the temperature is almost entirely dependent on absorption, conduction and convection.

  50. BillC says:

    Willis,

    Apologies for a shorter than necessary comment to explain what I’m thinking but I’d like your reaction if any. Isaac Held has blogged about a radiative forcing term defined where you let not only the stratospheric temperatures adjust, but also the immediate surface and atmosphere. This forcing is used to constrain the heat flux into the ocean. See the last (long) paragraph of this post.

    While this doesn’t satisfy several of your criteria, it at least provides a break point on the flux/temperature diagram that is more physically meaningful, and in the absence of feedbacks should yield a fairly linear flux-to-temperature relationship at a given, single location on the globe.

    thoughts?

  51. JP Miller says:

    I wonder what would happen if Willis, E.M. Smith, and Stephen Wilde were to put their ideas, data, and analytical approaches together and present a new formulation for climate processes including the likely impact of CO2 and water vapor. I imagine them getting it vetted by thoughtful professional climate heavy-weights and it getting published in Nature. Nah, never happen. The world’s poorer for it…..

  52. phi says:

    François Meynard,
    (December 13, 2012 at 1:10 am)

    “Apparently this “CO2 green house effect algorithm” is still used today in the large 3-D climate models.”

    I ask myself this question for a long time. The fact is that the narrative always considers convection as feedback. This is obviously unacceptable in a potentially unstable system.

  53. Chris B says:

    AlecM says:
    December 13, 2012 at 12:37 am

    ======================

    Here’s what I found in “Lapse Rate” on Wiki (I don’t pretend to understand it):

    Thermodynamic SS/Radiative GHG lapse rate
    Robert H. Essenhigh developed a comprehensive thermodynamic model of the lapse rate based on the Schuster-Schwarzschild integral (S-S) Equations of Transfer that govern radiation through the atmosphere including absorption and radiation by greenhouse gases.,.[11][12] “The solution predicts, in agreement with the Standard Atmosphere experimental data, a linear decline of the fourth power of the temperature, T^4, with pressure, P, and, at a first approximation, a linear decline of T with altitude, h, up to the tropopause at about 10 km (the lower atmosphere).” The predicted normalized density ratio and pressure ratio differ and fit the experimental data well. Sreekanth Kolan extended Essenhigh’s model to include the energy balance for the lower and upper atmospheres.[13]

  54. As a son of a rocket engineer, I find all this talk about “top of the atmosphere” at 20 km to be rather misguided.

    If a space vehicle comes within 120 to 160 km of the Earth’s surface, atmospheric drag will bring it down in a few days, with final disintegration occurring at an altitude of about 80 km. http://www.braeunig.us/space/orbmech.htm

    OK, difference science, different criteria.
    But why is 20 km considered to be TOA for climate science?
    20 km is only 65,617 feet. The anvils of thunderheads reach that high. The U-2 and SR-71 are air-breathing planes that aerodynamically fly above that altitude.

    20 km might be “Top of the Weather”, but it is misguided to think of it as a real Top of the Atmosphere.

  55. Jeff Alberts says:

    My car’s windshield still freezes in wintertime when parked out in the open, while the side windows remain thawed. This is because the side windows are warmed by the ground’s infrared and the windshield, which points to the sky and does not see the ground, is not. My car’s windshield tells me that there is no radiative forcing to prevent it from freezing — it shows that heat is radiating into space just like it ever has, IPCC Chicken Littles notwithstanding.

    My experience is different. My car’s windows won’t frost over when exposed to sunlight, but any in the shade will, whether the windshield or side windows.

  56. Jpatrick says:

    Yes. When you do “science” without operational definitions the results can be interpreted as the beholders wishe.

  57. Roger Longstaff says:

    Stephen Rasey says: December 13, 2012 at 8:12 am:

    “As a son of a rocket engineer, I find all this talk about “top of the atmosphere” at 20 km to be rather misguided”

    I completely agree. The system that I work on (SKYLON) breathes air up to 26 km, in a lifting, climbing, accelerating trajectory. To define TOA @ 20 km is moronic.

    In my view, TOA should be defined as a notional boundary at which only radiative energy transfer is possible – 300km, anyone?

  58. john robertson says:

    @JP Miller 6:58am. Just stay tuned we are watching that happen on WUWT, the future of honest science? And a forum of empirically based speculation.As science is never absolute, informed speculation is the next best thing.
    With so much specialization in the sciences now, it is necessary for outside points of view.
    The field description of expert(Drip under pressure) has never been so accurate.
    Open access and source can provide exposure of the “minor error” inherent in those crucial misconceptions that seem to ooze out of our universities.
    As for climate heavy weights… probably more on the blogs than present in the IPCC team .

  59. mkelly says:

    http://climaterealists.com/index.php?id=9004

    The link above shows what power CO2 has. The photo is of someone back yard. It shows what NZWILLY spoke of concerning his wind shield. CO2 has no power to warm the earth as it cannot even melt alittle frost.

  60. Willis Eschenbach says:

    AndyG55 says:
    December 13, 2012 at 3:20 am

    Hey Willis, in figure 1, what lapse rate is being used ??

    I’m really tired, but for say the polar regions, a drop of 50C in just under 9km doesn’t make sense.

    Hi, Andy, thanks for the question, but I’m not clear what you mean. There is no lapse rate involved in the calculations in Figure 1. They are the R^2 values of the correlation between the temperature and the TOA radiation imbalance.

    w.

    [UPDATE: Sorry, wrong thread. The real answer to your question is that the lines are obviously some kind of average or approximation of the actual lapse rate, which is often called the "environmental" lapse rate. By inspection, the lapse rate is about 63°C from the surface up to 9 km (see the "Polar" line). That gives us a lapse rate of about 0.7°C per 100 metres.

    The theoretical dry lapse rate is about 1°C per hundred metres, and the theoretical wet lapse rate is about 0.55°C per hundred metres. As we would expect, the average environmental rate is somewhere between the two ...

    w.]

  61. Diogenes says:

    Jeff

    Now I wish I had taken a photo of neighbour’s car at 3pm, just before sunset. Both the windscreen and all the side windows were still frosted over. They were like that at dawn. We had sunshine but a temperature of around freezing did not let the ice melt. Also, my other neighbour has a garage with a flat roof. Some rain fell on it a few weeks ago and did not clear before temps fell below freezing. The glacier is still there and does not appear to be receding. I live close to London.

    And it is brass monkeys in my barrel.

  62. Willis Eschenbach says:

    Dirck says:
    December 13, 2012 at 4:12 am

    “A fundamental part of this paradigm is that the long-term change in global average surface temperature is a linear function of the long-term change in what is called the “radiative forcing”. ”

    That is not a fundamental part of anything except your catastrophic misunderstandings of climate science.

    As Richard Courtney mentioned above, the kind of statement you made goes nowhere. All that you are saying is “nuh uh, does not”, which is valueless in a scientific discussion.

    If you want to get traction on a scientific site like this, that is far from enough. You need to say what and why and how. Back up your assertions with facts and citations. Give references to show that you are correct. Let me show you how it works. For example, I can say that my statement above, which you object to for unknown reasons, is fully supported by the work of Stephen Schwartz, as I discussed in my post called The Cold Equations.

    Then, after you read those two references to find out what I’m talking about, it’s your move.

    w.

  63. Larry in Texas asks:
    December 12, 2012 at 11:44 pm

    could define for me what the “lapse rate” is?

    In this context, it is the change in temperature with a change in height.

    AndyG55 says:
    December 13, 2012 at 3:20 am

    in figure 1, what lapse rate is being used ??

    I’m really tired, but for say the polar regions, a drop of 50C in just under 9km doesn’t make sense.

    That drawing is obviously wrong. The polar lapse rate is drawn as about 7.2 K/km, the correct value is about 6.5 K/km. Also, the polar tropopause is not drawn correctly. Over Antarctica, in winter, the tropopause is higher and colder than the tropical tropopause. Figure 1 shows a shape more like what is found over Alaska, not over the pole.

    All the books say that the tropopause is about 17 km at the equator and 7 km at the poles. However, on 08-16-2008, the Antarctic polar tropopause was over 20 km.

    If you want to see what the tropopause really looks like, you can plot real radiosonde data. I have a program that will show a full year’s data for several different locations.

    http://mc-computing.com/Science_Facts/Lapse_Rate/Lapse_Rate_Animations.html

  64. Willis Eschenbach says:

    BillC says:
    December 13, 2012 at 6:53 am

    Willis,

    Apologies for a shorter than necessary comment to explain what I’m thinking but I’d like your reaction if any. Isaac Held has blogged about a radiative forcing term defined where you let not only the stratospheric temperatures adjust, but also the immediate surface and atmosphere. This forcing is used to constrain the heat flux into the ocean. See the last (long) paragraph of this post.

    Thanks, Bill. There are a variety of radiative forcings used in the field. Here’s the distinctions as explained by James Hansen et al. in “Efficacy of Climate Forcings

    w.

  65. NZ Willy says:

    richard verney said: “Willy, I made a very similar observation … save that I would suggest that it is warm air being convected from the ground that keeps the car doors/door windows ice free.”

    Warm air cannot be the agent which keeps cars’ side windows ice free, because that would equally well keep the windshield ice free — but the windshield freezes. Furthermore, the grass tops in nearby fields, and rooftops, freeze too. This is a common phenomenon in New Zealand in winter. What they all have in common is that they do not see the ground, and so don’t get the infrared radiation from the ground.

    If there was “radiative forcing”, then these sky-facing surfaces would not freeze as they do because radiative forcing equates to infrared from the sky. But they do freeze, so there is no infrared from the sky, ergo, no radiative forcing. So my point is totally on-topic, despite your further demur.

  66. Willis Eschenbach says:

    Roger Longstaff says:
    December 13, 2012 at 8:43 am

    Stephen Rasey says: December 13, 2012 at 8:12 am:

    “As a son of a rocket engineer, I find all this talk about “top of the atmosphere” at 20 km to be rather misguided”

    I completely agree. The system that I work on (SKYLON) breathes air up to 26 km, in a lifting, climbing, accelerating trajectory. To define TOA @ 20 km is moronic.

    Let me say again what the NASA quote is careful to point out.

    The actual altitude used for calculations varies depending on what parameter or specification is being analyzed.

    So for some purposes, 20 km is perfectly OK, while for others it is far too low.

    For my purposes, I take the measurements of the satellites as reflecting the true TOA radiation, and I make no estimate of what actual altitude that might reflect … although the 70 km used by MODTRAN seems like a reasonable number.

    w.

  67. Willis Eschenbach says:

    I was totally off base in an answer above, viz:

    Willis Eschenbach says:
    December 13, 2012 at 8:53 am

    AndyG55 says:
    December 13, 2012 at 3:20 am

    Hey Willis, in figure 1, what lapse rate is being used ??

    I’m really tired, but for say the polar regions, a drop of 50C in just under 9km doesn’t make sense.

    Hi, Andy, thanks for the question, but I’m not clear what you mean. There is no lapse rate involved in the calculations in Figure 1. They are the R^2 values of the correlation between the temperature and the TOA radiation imbalance.

    w.

    I have updated my answer above to add:

    UPDATE: Sorry, wrong thread. The real answer to your question is that the lines are obviously some kind of average or approximation of the actual lapse rate, which is often called the “environmental” lapse rate. By inspection, the lapse rate is about 63°C from the surface up to 9 km (see the “Polar” line). That gives us a lapse rate of about 0.7°C per 100 metres.

    The theoretical dry lapse rate is about 1°C per hundred metres, and the theoretical wet lapse rate is about 0.55°C per hundred metres. As we would expect, the average environmental rate is somewhere between the two …

    w.

  68. Curious George says:

    The argument seems to be based on average temperatures. However the total (blackbody) radiation is proportional to the fourth power of the absolute temperature, so we are “averaging” a highly nonlinear relationship. For example, the total blackbody radiation more than doubles when temperature rises from -23C (250K) to 37C (310K).

    A related question: how good a blackbody the Earth surface is? Do oceans, deserts, forests, cloud tops behave the same way? Do IPCC calculations involve the heat capacity, which varies wildly in these examples? Do they consider daily or seasonal variations?

  69. Robert Clemenzi says:

    NZ Willy says:
    December 13, 2012 at 10:16 am

    If there was “radiative forcing”, then these sky-facing surfaces would not freeze as they do because radiative forcing equates to infrared from the sky. But they do freeze, so there is no infrared from the sky, ergo, no radiative forcing.

    It is my understanding that a freezer makes ice even though the inside walls of the freezer are emitting infrared radiation. This is because everything above absolute zero emits infrared radiation. Therefore, the fact that something freezes does NOT prove that there is “no infrared from the sky“. It only demonstrates that the amount of IR from the sky is less than the amount of IR released in forming frost.

  70. Gail Combs says:

    Johnny says:
    December 12, 2012 at 10:04 pm

    …. It sounds more like a middle ages religion than a science like physics or chemistry.
    >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
    Actually my husband pointed out that IPCC ‘science’ most closely resembles Alchemy.

    It looks like the problem does not reside in just climate science either.

    This … interview with Eric S. Raymond I think summarizes the Alchemist position prevalent in much of clinical medicine with regard to its software today: ‘Page 8:

    The occult school of alchemy didn’t turn into the science of chemistry until alchemists abandoned the practice of secrecy and instead started sharing results with each other and checking each other’s experiments. And that was a very early stage in the development of modern science and engineering. It happened about 400 years ago. And the thing that we’ve discovered over the last 400 years, as we’ve pursued experimental science and developed engineering from an art into a craft into a repeatable discipline, is that human beings doing complex, creative work, doing design work, make mistakes. There is no way to mechanically check the results of creative work. If you could do that, it wouldn’t be creative work; it would be something you could do with a machine. So the only way to check complex, creative work for correctness is by the critical judgment of peer experts…’
    http://linuxmednews.com/1078854224/addPostingForm

    So it looks like we are taking a GIANT STEP BACKWARDS 400 yrs to the middle ages. Of course this is where the Neo-luddites want to drag us back to in all technologies.

  71. Willis Eschenbach says:

    mkelly says:
    December 13, 2012 at 8:50 am

    http://climaterealists.com/index.php?id=9004

    The link above shows what power CO2 has. The photo is of someone back yard. It shows what NZWILLY spoke of concerning his wind shield. CO2 has no power to warm the earth as it cannot even melt alittle frost.

    Thanks, mkelly. I see you and other folks above making the argument that since the downwelling longwave radiation doesn’t do what y’all expect it to do, it must not exist, or it must not have any power to warm things … neither one is correct.

    Let me start by pointing out that the downwelling longwave radiation (DLR) is a measurable quantity. It is not a fantasy, you can take a radiation measuring device out anywhere on the earth and measure its instantaneous value. It is real, verifiable radiation, which of course contains energy. Since the radiation contains energy, and since energy cannot be destroyed, the radiation adds that energy to whatever it strikes. This is fundamental physics and is beyond question in a scientific sense.

    What people seem to forget, however is that the radiation is coming from a cold place. Globally, the average amount of downwelling energy is on the order of about 320 watts per metre squared (W m-2). Sounds like a lot, huh? People say “Why doesn’t that heat things”.

    The answer is, the blackbody temperature corresponding to that radiation is right around freezing. Of course, that temperature corresponding to the DLR is lower in winter and higher in summer, and higher at the tropics than at the poles.

    So the question you pose is, how can radiation coming off of an atmosphere at a temperature of about freezing warm the earth? And that is a reasonable question.

    The answer to that is “warm the earth compared to what?”.

    Right now, what we have is the downwelling longwave radiation from the atmosphere at 0°C. The alternative to that is what we would have if there were no greenhouse gases (GHGs). In that case, we would only have the downwelling radiation from outer space, which is at about -270°C … which one would you think would leave the planet warmer?

    So you are right, MKelly, that wintertime downwelling radiation, which is coming from the atmosphere at well below freezing, is not enough to melt frost. Nor would we expect it to.

    But that doesn’t mean that the radiation doesn’t exist.

    And more to the point, what it does mean is that the earth with downwelling longwave radiation from an atmosphere a few degrees below freezing is much warmer than the earth would be in the absence of GHGs and exposed directly to the background temperature of space. This why clear winter nights are the coldest. With little water vapor in the winter, on a clear night you can feel the cold of space sucking the heat out of the landscape.

    But when a cloud passes over in that winter night, it is immediately much warmer. That is the effect of the downwelling longwave radiation from the cloud. It doesn’t melt the ice on the winter night, how could it? In that case the downwelling radiation is coming from the cloud, which is colder than the surface and is well below freezing.

    So what DLR can do is keep the earth from being much colder than it would be in the absence of GHGs. Does that mean that DLR “warms” the earth? People object to that terminology, for reasons that escape me, so I put it this way:

    The earth with GHGs is warmer than it would be if a lack of GHGs exposed us directly to the bitter cold of outer space.

    In other words, DLR doesn’t melt the ice … but it keeps the surface getting much colder, and that leaves the earth a warmer place than it would be in the absence of CO2 and other GHGs.

    w.

  72. Stephen Wilde says:

    “But when a cloud passes over in that winter night, it is immediately much warmer. That is the effect of the downwelling longwave radiation from the cloud.”

    I don’t think so.

    Descending air is adiabatically warmed by conversion of PE to KE as air descends from above,

    With a clear sky at night, radiation from the surface is faster than the warming effect of descending air and an inversion forms.

    If the descending air is moist, clouds then form at the inversion height because the surface cooling reduces the temperature of the descending air to below its dew point.

    Those clouds are then at the temperature of the descending, adiabatically warmed, air from above and thus warmer than the air at the ground so they start to conduct energy downward and that offsets radiative cooling from the surface, often warming the lower layer in the process.

    So, why should you believe that?

    Consider the difference between a low cloud base and a medium or high cloud base.

    The nearer the cloud base to the surface the greater the warming effect which points to conduction as the cause.

    If downward IR were the cause it would make no difference whether the cloud base were high or low.

    We know full well from observations that low clouds break down an inversion more effectively than high clouds.

    So, conduction it must be.

  73. Frank says:

    Willis: Here are some points your presentation missed:

    We can measure the radiative imbalance (not radiative forcing) from space by starting with incoming solar and then subtracting reflected solar and OLR. The errors in these measurements apparently are too large to make a useful comparison between observed radiative imbalance and theoretical radiative forcing.

    The upward and downward radiative flux through the atmosphere can only be calculated if one specifies the composition and temperature (which controls emission) everywhere along the path. In instantaneous change in composition will produce and instantaneous change in flux. In locations where convection is not important (at the tropopause and higher), one can use the power absorbed and emitted by the changed flux to calculate how the temperature will change WITH TIME when assumptions about the composition of the atmosphere are changed. The radiation flux is then recalculated with the new temperature information and a new equilibrium state will be reached. This is what the IPCC does with its defines radiative forcing after the tropopause and the stratosphere reach radiative equilibrium. Since temperature below the tropopause is controlled mostly by convection and the lapse rate, calculations based on radiative equilibrium make no sense. Instead, the IPCC assumes that (to a first approximation) the lapse rate remains constant so that a 1 degC temperature increase at the tropopause will be linked with a 1 degC rise at the surface. The tropopause is special because it allows one to realistically calculate temperature change by radiative forcing/equilibrium and have some rational – a fixed lapse rate – for expecting that temperature change to be relevant to what happens at the surface.

    One can calculate radiative forcing at the surface from 2X CO2 (about 1 W/m2). Since the average photon reaching the earth will have been emitted from GHG’s close surface where it is warmer, 2X CO2 will increasing DLR. At the surface, however, you can’t predict whether the additional energy will leave the surface by convection or increased OLR (after surface warming).

    The IPCC’s definition of radiation forcing has the limitations you describe. However, it is the best system we have for coming up with one number that compares and integrates the effects of changing atmospheric gases, solar output, aerosols, surface albedo, clouds, etc. When small percent changes in radiative forcing and temperature are involved, a linear relationship is a reasonable approximation.

    We sent radiosondes into the atmosphere twice a day from about a hundred locations on the planet. We have excellent information about what is happening in the vicinity of the tropopause.

  74. anticlimactic says:

    We are expected to believe without question that GHGs warm the Earth in various ways, but the more I think about the less I can see how they can.

    All atoms in the universe are constantly trying to shed their heat and fall to absolute zero [-273C] and the only reason they don’t is that they are usually receiving about as much energy as they are losing.

    Assuming a GHG molecule absorbs some heat from the Earth, it will be immediately reradiated, roughly 50% back to Earth and 50% out in to space. When the heat returns to Earth it will again be reradiated and again only 50% of any captured heat will be returned – and so on.

    The warming effect of GHGs will approach zero in less than a minute!

  75. richard verney says:

    NZ Willy says:
    December 13, 2012 at 10:16 am
    ///////////////////////////////////////////////////////////////////////////////////////
    I agree with the general thrust of your final paragraph, although never freeze may be a bit too strong.
    As regards the second para, you overlook that radiative heating is the least efficient form of heating. I have gas central heating with radiators. Ther gas boiler heats water which is circulated throgh the radiators. One has to place one’s hand within about 1/2 inch of the horizontal panel which sits parallel with the wall before one can feel noticeable heat. On the other hand, one can feel noticable heat when one’s hand is three feet above the top of the radiator. This is not withstanding that the surface area of the top is a fraction of the surface ares of the horizontal panel. Although they are called radoators, it is convection that performs the bulk of the heating. The same is so with a BBQ. One cannot cook meat 4 inches from the side of a BBQ whereas one can cook meat a foot above it. Radiative heating soon gets swamped by convection.

    As regards the car, the windscreen, the bonnet and the roof are in the shaddow of the convective heat being given off by the ground on top of which the car is parked, and the adjacenet ground surrounding the car. Unless there is a lot of wind, the convective heat cannot heat the windshield because it is rising vertically.

    But consider my point with dew. Consider Spring or Autumn. You have a hollow. When the sun rises, it heats one side of the hollow, the other side of the hollow remains in shaddow. Relatively quickly the dew is burnt off on the sunny side. However, dew on the shaddow side can linger most of the day.

    Now Trenbeth suggests that DWLWIR has almost double the power of solar irradiance. In my example, with early morning sun the angle of incidence is low so solar energy in this example is very weak (much below the Trenbeth average). Not withstanding that it is very weak, it can relatively quickly burn off the dew on the sunny side of the hollow. Yet even in hours, DWLWIR cannot burn of the dew in the shaddow part of the hollow. Why is that if DWLWIR has real/sensible energy?

    One may be able to measure a signal from DWLWIR (Willis I do not doubt that one can measure a signal). But the issue is, can it do any work? Can it actually heat anything? (Willis is there any published paper showing that DWLWIR actually heats something, rather than mere speculation?) If DWLWIR cannot burn off dew in the shaddow area of a hollow notwithstanding exposure of the dew to DWLWIR of many many hours, I would suggest that it appears that DWLWIR lacks significant sensible energy.

    PS> My example is a deep hollow, or perhaps a valley, one side is sunny and the other is not.

  76. mkelly says:

    Willis says: “Thanks, mkelly. I see you and other folks above making the argument that since the downwelling longwave radiation doesn’t do what y’all expect it to do, it must not exist,…”

    Willis you often get annoyed with folks who don’t quote you or put words in your mouth. I have never said that it does not exist I have said and maintain that it has not power to do anything. ie work. So please don’t think or say that I said it does not exist. The question is can it do anything and I say no.

  77. phi says:

    Frank,

    “…the IPCC assumes that (to a first approximation) the lapse rate remains constant so that…”

    Yes. But the added CO2 is not constrained by the lapse rate because this gas is created ex nihilo. Emulate its temperature on the lapse rate is an arbitrary operation. The value obtained (radiative forcing) is therefore also arbitrary.

  78. richard verney says:

    Stephen Wilde says:
    December 13, 2012 at 11:35 am
    ////////////////////////////////////////////////////////////////////
    The cloud example is a difficult example and proves next to nothing because one cannot separate the various mechanisms that may be at play, and thereby correctly ascribe cause and effect.
    The cloud is an example of the classic greenhouse, ie., it really is akin to the glass roof in that it hinders convection thereby trapping warm(er) air. So whilst one may feel less cold on a cloudy night, one does not know whether this is due to the points you raise to which I would add the obstruction of convection (which would otherwise be more efficiently cooling the body of air immediately above the ground), or whether as Willis argues there is more DWLWIR.

  79. Willis Eschenbach says:

    Curious George says:
    December 13, 2012 at 10:40 am

    The argument seems to be based on average temperatures.

    What argument are you talking about? Details, friends, the devil is in the details.

    w.

  80. Willis Eschenbach says:

    Frank says:
    December 13, 2012 at 11:36 am


    One can calculate radiative forcing at the surface from 2X CO2 (about 1 W/m2).

    Sure, you can calculate it … but is it correct? That’s the recurring question, we have no answer yet. As Shakespeare put it:

    GLENDOWER. I can call spirits from the vasty deep.
    HOTSPUR. Why, so can I, or so can any man; But will they come when you do call for them?

    You go on to say:

    Since the average photon reaching the earth will have been emitted from GHG’s close surface where it is warmer, 2X CO2 will increasing DLR. At the surface, however, you can’t predict whether the additional energy will leave the surface by convection or increased OLR (after surface warming).

    The IPCC’s definition of radiation forcing has the limitations you describe. However, it is the best system we have for coming up with one number that compares and integrates the effects of changing atmospheric gases, solar output, aerosols, surface albedo, clouds, etc.

    It’s the “best system we have”? Sez who? You? Color me unimpressed. We need to start with something we can actually measure. If you want to develop theoretical values from that, fine. But:

    • If you can’t measure it, you can’t falsify it.

    • If you can’t falsify it, it’s not science.

    • If it’s not science, how can it be the best system we have?

    When small percent changes in radiative forcing and temperature are involved, a linear relationship is a reasonable approximation.

    I hear people say this all the time. The temperature at the surface of the earth ranges from about 134°C to – 90°C, a swing of about 200°C. The solar forcing goes from zero to a thousand watts per square metre in the tropics every day.

    What part of that says “small percent changes in forcing” to you?

    There is a further problem with this, which is that in some cases the averages are totally deceptive. I discussed this in “The Details Are In The Devil“. In some cases, there may be no relationship between forcing and temperature. See my previous post, “An Interim Look At Intermediate Sensitivity“. In that case, not only is the linear approximation incorrect. Any putative relationship is incorrect.

    w.

  81. george e. smith says:

    Well nowhere in the earth climate system, is anything IN EQUILIBRIUM.

    “””””…..The radiative forcing of the surface-troposphere system due to the perturbation in or the introduction of an agent (say, a change in greenhouse gas concentrations) is the change in net (down minus up) irradiance (solar plus long-wave; in Wm-2) at the tropopause AFTER allowing for stratospheric temperatures to readjust to radiative equilibrium, but with surface and tropospheric temperatures and state held fixed at the unperturbed values……””””

    In particular, given that the earth rotates, so that no point on earth ever is in thermal equilibrium. or radiative equilibrium, so this definition is BS.

    Moreover, there is no known way to hold either surface or tropospheric temperatures or “state” fixed at any value let alone any unperturbed value.

    Who are these people who think it is ok for them to simply make up their own physical laws, and make comlex systems obey their silly ideas ?

  82. Willis Eschenbach says:

    mkelly says:
    December 13, 2012 at 11:50 am

    Willis says:

    “Thanks, mkelly. I see you and other folks above making the argument that since the downwelling longwave radiation doesn’t do what y’all expect it to do, it must not exist,…”

    Willis you often get annoyed with folks who don’t quote you or put words in your mouth. I have never said that it does not exist I have said and maintain that it has not power to do anything. ie work. So please don’t think or say that I said it does not exist. The question is can it do anything and I say no.

    mkelly, If you would quote my entire sentence, rather than selectively quoting me as you have done, you would note that I actually said (emphasis mine):

    I see you and other folks above making the argument that since the downwelling longwave radiation doesn’t do what y’all expect it to do, it must not exist, or it must not have any power to warm things … neither one is correct.

    I said that you and others were making one of two arguments, EITHER that DLR doesn’t exist, OR making the argument that it does not have any power to warm things.

    Obviously, you are included in the OR clause … so I did not misquote or misrepresent you in the slightest. As you say, you claim that measurable downwelling longwave radiation, which obviously contains energy (that’s how we measure it), has no power to do anything at all. Which is exactly how I described your argument.

    Now, this is obviously some new kind of radiation, because every other kind of radiation that I’ve ever heard of has the power to do something. Generally speaking, for example, all radiation warms whatever absorbs the radiation.

    But you have a brand new kind of radiation, one that contains energy and yet does not warm what it is absorbed by, it has no power to do anything …

    So where does the energy go when such radiation hits the earth? It can’t be destroyed. Your claim is that it is not turned into heat as everyone else might think, including myself … so where does it go? It’s not turned into light, or mechanical motion, and according to you it’s not turned into heat … so where is the energy that was contained in the downwelling radiation?

    Regards,

    w.

  83. AndyG55 says:

    Its morning here now, and even though I haven’t had my coffee yet…

    It appears Figure 1 is based on wet lapse rate.

    If you take the tropical (pink) line.. its surface value is 20C (ignore where the numbers are, they are badly printed) That line crosses the grid again at -30C at 9km so the slope is 5.5C/km

    Same slope with the polar (blue line) but isn’t polar air mostly dry ?

    Still.. the point sits that using 20km as TOA is really pretty ridiculous.

  84. Willis Eschenbach says:

    richard verney says:
    December 13, 2012 at 12:05 pm

    So whilst one may feel less cold on a cloudy night, one does not know whether this is due to the points you raise to which I would add the obstruction of convection (which would otherwise be more efficiently cooling the body of air immediately above the ground), or whether as Willis argues there is more DWLWIR.

    I am not “arguing” that there is more downwelling longwave radiation under clouds than under a clear sky. That is a measurable fact, and has been measured many times, both during the day and the night. It is not due, as you postulate, to the “obstruction of convection”. Heck, many clouds are created by convection, they are areas of increased rather than decreased convection, and despite that, they are sources of increased downwelling radiation … go figure.

    Finally, we know it is not just interrupted convection because of the speed of the phenomenon. Just like when a cloud covers the sun it immediately becomes cooler (radiation), the same is true of night-time winter clouds. As soon as a cloud covers the night sky, the surface immediately becomes warmer. This is the sign of radiation, not convection.

    Look, guys, this stuff is not theory. This part of climate science is measurable, testable, and is even detectable by our own senses. There are good arguments against the conventional consensus theory. The idea that clouds don’t warm winter nights through increasing DLR is not one of them. Nor is the idea that DLR somehow magically does nothing when it is absorbed by the surface, the idea that the energy in the downwelling radiation just disappears. All of that is just wishful thinking.

    w.

  85. Mack says:

    Comment from Ryan Dec 13th 2012 6.40am …..Good comment Ryan . Well said.

  86. Stephen Wilde says:

    richard verney said:

    “So whilst one may feel less cold on a cloudy night, one does not know whether this is due to the points you raise to which I would add the obstruction of convection”

    I agree, Richard.

    Suppression of convection facilitates downward conduction.

    It cannot be anything to do with DLR because if it were the effect would be the same from both high and low clouds yet we know it is not.

    I don’t see it as anything to do with DLR in any event because the solar throughput is unhindered at equilibrium. At equilibrium solar shortwave in must equal longwave out at top of atmosphere.

    Solar shortwave comes in and planetary longwave goes out and at equilibrium they are always equal.

    What matters is how the atmosphere intervenes and that is a constant related to mass. gravity and insolation and nothing else.

    PE + KE = Constant.

    Only mass, gravity and insolation set that constant.

    The atmosphere shuffles KE and PE through the speed of adiabatic processes as necessary to maintain that constant.

    If the radiative characteristics of individual atmospheric components vary so as to disrupt the balance between solar shortwave in and longwave out then the atmospheric heights change to redistribute KE and PE to negate the imbalance.

    The constant being set by mass, gravity and insolation, any other factor such as the radiative characteristics of individual constituents of the atmosphere can only affect the proportion of KE relative to PE.

    The more the atmosphere expands from more GHGs the more PE there can be and the less KE (cooling) and the more the atmosphere contracts from less GHGs the less PE and the more KE (warming).

    GHGs might cause the atmosphere to expand but if they do that then PE increases relative to KE and there is an equal and opposite cooling effect due to the reduction in KE.

    If they cause the atmosphere to contract then KE increases relative to PE and there is an equal and opposite warming effect due to the increase in KE.

    That is the true Thermostat Hypothesis and Willis’s version is correct as far as it goes but it does not extend to the entire globe which is where it fails.

  87. Stephen Wilde says:

    “Heck, many clouds are created by convection, they are areas of increased rather than decreased convection, and despite that, they are sources of increased downwelling radiation ”

    Really ?

    Clouds created by convection as compared to clouds formed beneath an inversion layer result in air flowing in from surrounding sunlit regions to feed the uplift.

    No increase in DLR there.

  88. Curious George says:

    The argument seems to be based on average temperatures.

    Willis: What argument are you talking about? Details, friends, the devil is in the detail.

    The red, blue, and green lines in Fig. 1 are all averages – I may be wrong. Has IPCC ever considered temperature variations in space or time? I thought Willis might know.

  89. mkelly says:

    Willis Eschenbach says:
    December 13, 2012 at 12:52 pm
    “So where does the energy go when such radiation hits the earth?”

    First if you did not indend for me to be included in the group of folks that say the radiation for CO2 does not exist then I with draw my comment.

    It is called reflection. You assume it is absorbed.

    Just because radiation strikes a body does not mean it is absorbed and turned into thermal energy (heat).

    I have said before if you or others can write a radiative heat transfer equation showing how CO2 warms the surface of the earth I’ll happily accept that. ie makes the surfaced go from 255K to 288K.

  90. AndyG55 says:

    Ummm.. clouds are where the convective flow stops.
    If you stop the convective flow, the convective flow below is forced to slow down.
    Upward energy transfer is thus slowed down.

    Think of a big heavily laden truck going up a hill on a one lane road, with many cars behind that want to go faster.

  91. richard verney says:

    (since heat cannot be effectively conducted down

  92. richard verney says:

    Some posts have been a little off topic, but that is no doubt because one can hardly add to Willis’s closing comment. It is only too true that one can’t have any particular conclusions given the use of imaginary values that can never be measured or verified.
    I am in the mkelly camp that the issue is not whether DWLWIR exists, but rather it is one of whether DWLWIR has sensible energy/the ability to do real work in the Earth environment.
    The key to AGW is the behavoir of DWLWIR and water. The reason for this is threefold. First, it is the oceans that control the weather. Second (and this is related),the latent heat content of the oceans dwarfs the latent heat content of the atmosphere. Third, the absorption of DWLWIR in water is very different to that in dry air (such that the behavoir of DWLWIR over the oceans is different to that over land). IR cannot penetrate water to any significant degree such that water acts as an IR block. Almost all IR is absorbed within 10 microns, with about 40% of all DWLWIR being absorbed within 1 micron and 60% within 3 microns. Sunlight can penetrate water to great depths. This is fundamental.
    The power from sunlight can warm water; it does not immediately cause the top micron layers to burn off/evaporate but instead the energy penetrates to depth slowly heating a vast volume of water at depth. IR is very different since approximately 60% of all IR is absorbed within 3 microns. This means that energy in IR does not heat a vast volume of water at depth, but instead it is concentrated and heats a relatively small volume at the very top 1 to 3 micron layer. Since the energy is not dissipated, but concentrated, the very top microns surely must burn off.
    If one considers the amount of DWLWIR that is, according to Trenbeth/AGW proponents, hitting the oceans, as a matter of the absorption characteristics of water, the energy in this (assuming that DWLWIR can perform sensible work) would lead to the top micron layer quickly evaporating. If this was happening, then there would be 10s of metres of rainfall annually, which there is not. The fact that there is not such an amount of rainfall suggests: (i) the characteristics of absorption of IR in water is wrong (extremely unlikely) such that IR is not completely absorbed in the first microns but can infact penetrate many metres, (ii) DWLWIR does not exist or not in the quantity postulated by AGW proponents, or (iii) DWLWIR lacks sensible energy. It is a signal but cannot perform real work in the conditions encountered in Earth’s environ.
    Until one can witness DWLWIR performing real work such as burning off dew, or melting ice, questions quite obviously arise as to whether it does (or does not) possess sensible energy.
    PS. If one looks at the temperature profile of the oceans, the top few microns are cooler than the top 15 microns such that the direction of heat flow is up towards the surface, not down towards the deeper ocean.

  93. richard verney says:

    PS. In my last post where I use IR, I mean LWIR. The wavelength is very important to the characteristics of absorption.

  94. One can generally state the Forcings Theory as,

    The Earth’s climate warms/cools from changes in the net radiation balance at a hypothetical contiguous shell that surrounds the climate.

    TOA would fit the contiguous shell surrounding the climate criteria. And the tropopause would, as you point out, seem to fail the contiguous criteria.

    Ignoring heat from the Earth’s interior, the Forcings Theory, as stated above, would appear to be a priori true. But being a priori true, doesn’t make it science.

    Firstly, we cannot measure (at our hypothetical shell) the radiative source for most radiative forcing changes. Therefore cannot measure the change in a particular forcing.

    Secondly, the theory gives us no criteria to determine whether something is a radiative forcing or not. Which means there may be yet undiscovered forcings, as GCRs were a few years ago.

    All in all, I have trouble considering the Forcings Theory as properly scientific.

  95. Mike Borgelt says:

    The graphic showing a smooth variation in troposphere height from equator to poles is wrong.
    There are discontinuities between the equatorial, mid latitude and polar tropopauses. The jet streams occur there.

  96. Willis Eschenbach says:

    mkelly says:
    December 13, 2012 at 1:32 pm

    Willis Eschenbach says:
    December 13, 2012 at 12:52 pm

    “So where does the energy go when such radiation hits the earth?”

    First if you did not indend for me to be included in the group of folks that say the radiation for CO2 does not exist then I with draw my comment.

    It is called reflection. You assume it is absorbed.

    I assume nothing of the sort. I know it is absorbed. The average absorptivity of the earth’s surface at the relevant frequencies is greater than 95%. In a previous post, I said:

    NOTE 3: My bible for many things climatish, including the emissivity (which is equal to the absorptivity) of common substances, is Geiger’s The Climate Near The Ground, first published sometime around the fifties when people still measured things instead of modeling them. He gives the following figures for IR absorptivity/emissivity at 9 to 12 microns:

    Water, 0.96
    Fresh snow, 0.99
    Dry sand, 0.95
    Wet sand, 0.96
    Forest, deciduous, 0.95
    Forest, conifer, 0.97
    Leaves Corn, Beans, 0.94

    and so on down to things like:

    Mouse fur, 0.94
    Glass, 0.94

    You can see why the error from considering the earth as a blackbody in the IR is quite small.

    I must admit, though, that I do greatly enjoy the idea of some boffin at midnight in his laboratory measuring the emissivity of common substances when he hears the snap of the mousetrap he set earlier, and he thinks, hmmm …

    So we know for a fact that only a bit of it is reflected, a few percent. Thus, your claim that it is reflected is falsified. Again, these are measurements, not theory.

    Just because radiation strikes a body does not mean it is absorbed and turned into thermal energy (heat).

    True, and in the case of the earth more than 95% of it is absorbed and turned into thermal energy.

    I have said before if you or others can write a radiative heat transfer equation showing how CO2 warms the surface of the earth I’ll happily accept that. ie makes the surfaced go from 255K to 288K.

    Sure, be glad to. See my previous posts, The Steel Greenhouse, and People Living In Glass Planets.

    All the best,

    w.

  97. Willis Eschenbach says:

    richard verney says:
    December 13, 2012 at 1:40 pm


    I am in the mkelly camp that the issue is not whether DWLWIR exists, but rather it is one of whether DWLWIR has sensible energy/the ability to do real work in the Earth environment.

    The power from sunlight can warm water; it does not immediately cause the top micron layers to burn off/evaporate but instead the energy penetrates to depth slowly heating a vast volume of water at depth. IR is very different since approximately 60% of all IR is absorbed within 3 microns. This means that energy in IR does not heat a vast volume of water at depth, but instead it is concentrated and heats a relatively small volume at the very top 1 to 3 micron layer. Since the energy is not dissipated, but concentrated, the very top microns surely must burn off.

    If one considers the amount of DWLWIR that is, according to Trenbeth/AGW proponents, hitting the oceans, as a matter of the absorption characteristics of water, the energy in this (assuming that DWLWIR can perform sensible work) would lead to the top micron layer quickly evaporating. If this was happening, then there would be 10s of metres of rainfall annually, which there is not. The fact that there is not such an amount of rainfall suggests: (i) the characteristics of absorption of IR in water is wrong (extremely unlikely) such that IR is not completely absorbed in the first microns but can infact penetrate many metres, (ii) DWLWIR does not exist or not in the quantity postulated by AGW proponents, or (iii) DWLWIR lacks sensible energy. It is a signal but cannot perform real work in the conditions encountered in Earth’s environ.

    Richard, like mkelly you are postulating an entirely new class of radiation, radiation which contains no energy. Until you have some evidence that such a bizarre creature exists, I suspect that most folks will have the same reaction as mine … I’ll believe it when you show me such radiation, or any serious scientific claim that such energy-free radiation exists.

    Until then, infrared radiation contains energy, get used to it. Your job is to explain what happens to that energy if it is not absorbed by the ocean. You can’t get out of it by simply making up a fantasy about energy-free radiation.

    Until one can witness DWLWIR performing real work such as burning off dew, or melting ice, questions quite obviously arise as to whether it does (or does not) possess sensible energy.

    See my post above for an explanation about why DLR won’t melt ice, and despite that, it contains energy, and why the world is warmer with DLR than without DLR.

    Richard, in my post “Radiating the Ocean“, I listed four reasons why people’s theory that DLR can’t warm the ocean is wrong. I closed by saying:

    Note that each of these arguments against the idea that DLR can’t warm the ocean stands on its own. None of them depends on any of the others to be valid. So if you still think DLR can’t warm the ocean, you have to refute not one, but all four of those arguments.

    Look, folks, there’s lot’s of good, valid scientific objections against the AGW claims, but the idea that DLR can’t heat the ocean is nonsense. Go buy an infrared lamp, put it over a pan of water, and see what happens. It only hurts the general skeptical arguments when people believe and espouse impossible things …

    So there is the hurdle you need to get over in order to sustain your claim that DLR can’t warm the ocean. You need to refute all four of them if you want to get any traction.

    w.

  98. Matthew R Marler says:

    I don’t know of other scientific fields that base fundamental concepts on an unmeasurable imaginary variable rather than a measurable physical variable.

    Historically, most of them have done that aplenty. Conceptual and definitional clarity have usually followed decades of confusion. Consider the fundamental concept of “ergodic”, described in Lawrence Sklar’s book “Physics and Chance”, or just about any history of atomic theory.

    Possibly the most famous imaginary is the “body in uniform motion”.

  99. TimTheToolMan says:

    Willis writes a strawman argument “Go buy an infrared lamp, put it over a pan of water, and see what happens.”

    You have just made DLR > ULR which isn’t the case for the ocean. You’ve fallen into the precise trap that you’ve argued all along doesn’t matter regarding whether something is cooling more slowly or actually “warming”.

  100. Mack says:

    Willis,
    Maybe I can get a bit of traction over DLR “can’t warm the ocean” I suppose anything in nature is possible but what you are contending here is that a trace gas wafts around in the air above the ocean waves and warms the water beneath more than otherwise by the sun. Call me a stickler for not introducing amounts and figures but when thinking about all this I get a feeling of insignificance and impossibility, and so should you.

  101. Jeff Alberts says:

    Diogenes says:
    December 13, 2012 at 8:54 am

    Jeff

    Now I wish I had taken a photo of neighbour’s car at 3pm, just before sunset. Both the windscreen and all the side windows were still frosted over. They were like that at dawn. We had sunshine but a temperature of around freezing did not let the ice melt. Also, my other neighbour has a garage with a flat roof. Some rain fell on it a few weeks ago and did not clear before temps fell below freezing. The glacier is still there and does not appear to be receding. I live close to London.

    And it is brass monkeys in my barrel.

    I’m not doubting what you saw, just doubting that it’s due to radiation from the ground.

    I live north of Seattle, on the US northwest Pacific coast. When the ground frosts over here, any areas not hit by sun will usually stay frosted over throughout the day if the ambient temp doesn’t get too far above freezing. One road I drive to/from work has about a mile stretch that never gets direct sun. It will stay frosty all day long, even though all the other roads are dry and clear. It’s pretty dangerous.

    I can often also see frost and non-frost areas side by side where the shade and sunlight meet.

    Sorry, not getting the brass monkeys reference :(

  102. Willis Eschenbach says:

    Matthew R Marler says:
    December 13, 2012 at 5:00 pm

    I don’t know of other scientific fields that base fundamental concepts on an unmeasurable imaginary variable rather than a measurable physical variable.

    Historically, most of them have done that aplenty. Conceptual and definitional clarity have usually followed decades of confusion. Consider the fundamental concept of “ergodic”, described in Lawrence Sklar’s book “Physics and Chance”, or just about any history of atomic theory.

    Possibly the most famous imaginary is the “body in uniform motion”.

    Thanks, Matthew. You are conflating an unmeasurable variable (the radiation imbalance with the troposphere and surface fixed) and an abstract concept (“body in uniform motion”). The difference is that the radiation imbalance is a number, while a body in motion is not a number.

    w.

  103. Willis Eschenbach says:

    TimTheToolMan says:
    December 13, 2012 at 5:08 pm

    Willis writes a strawman argument

    “Go buy an infrared lamp, put it over a pan of water, and see what happens.”

    You have just made DLR > ULR which isn’t the case for the ocean. You’ve fallen into the precise trap that you’ve argued all along doesn’t matter regarding whether something is cooling more slowly or actually “warming”.

    Not sure what your point is here, Tim. Do you think that the ocean is warmer with DLR than without? I do. Do you think that the pan of water will be warmer with the DLR from the lamp than without? I do.

    If you don’t, as I mentioned above, you should read “Radiating the Ocean“. There, I advance four arguments that each individually falsifies the idea that the ocean does not absorb downwelling longwave radiation.

    w.

  104. TimTheToolMan says:

    Willis write “Not sure what your point is here, Tim.”

    I know. Its because you dont actually understand the issue yet otherwise you wouldn’t have put up the strawman argument.

    Willis writes “If you don’t, as I mentioned above, you should read “Radiating the Ocean“. ”

    Actually I think it is you who should read that thread. Specifically my comments within it ;-)

  105. Willis Eschenbach says:

    Mack says:
    December 13, 2012 at 6:07 pm

    Willis,
    Maybe I can get a bit of traction over DLR “can’t warm the ocean” I suppose anything in nature is possible but what you are contending here is that a trace gas wafts around in the air above the ocean waves and warms the water beneath more than otherwise by the sun. Call me a stickler for not introducing amounts and figures but when thinking about all this I get a feeling of insignificance and impossibility, and so should you.

    Mack, welcome to the discussion. There are lots of things in nature which exist only in trace amounts and yet are very important. Consider ricin, a few milligrams of which can fell an adult human. Or more to the point, consider the ozone layer, which makes up only a very trivial part of the earth’s atmosphere, less than one percent. But without it, ultraviolet would cause lots of cancer.

    So the argument that CO2 is too small to have an effect just won’t wash.

    w.

    PS—Bear in mind that CO2 is not what is doing the heavy lifting as far as downwelling longwave radiation goes. That would be water, both in the form of water vapor and in the form of clouds.

  106. Willis Eschenbach says:

    Jeff Alberts says:
    December 13, 2012 at 6:33 pm

    Sorry, not getting the brass monkeys reference :(

    It’s from a saying about the temperature, “Colder than the testicles on a brass monkey”, expressed in earthier language of course …

    w.

  107. Willis Eschenbach says:

    TimTheToolMan says:
    December 13, 2012 at 7:03 pm

    Willis write “Not sure what your point is here, Tim.”

    I know. Its because you dont actually understand the issue yet otherwise you wouldn’t have put up the strawman argument.

    Dang, that was a short conversation. Of all the available options, including actually explaining what your point was, you decided to attack me.

    Perhaps people play like that where you come from, Tim. To me, it just highlights the fact that you would go to great lengths rather than answer a civil question.

    I have no time for people who don’t play well with others, Tim, especially jerks that attack me rather than answer a question. I have zero tolerance for your nasty snark. Please take your accusations elsewhere, I’m done responding to you.

    w.

  108. jae says:

    The problem, Willis, is that there is absolutely no empirical evidence that the backradiation does anything, as some of the commenters say. Of course the backradiation exists, but It is simply a fact of physics, a property of matter. It does no necessarily have anything to do with the temperature of the atmomsphere. The construct of “backradiation” looks right, because the backraidiation does exist. But IT IS ONLY A PROPERTY OF MATTER. Only the STORAGE OF ENERGY matters. Again, again, again, again, I keep comparing Atlanta and Phoenix, same latitude and elevation——AND Phoenix on a clear day ALWAYS has a higher temperature, day and night, than Atlanta, DESPITE THE FACT THAT ATLANTA HAS AT LEAST 4 TIMES AS MUCH GREEHNOUSE GASSES AS PHOENIX! OH, ORACLE, PLEASE EXPLAIN THIS PHENOMENON! THERE IS SIMPLY NO PROOF OF THE PUTATIVE GHE! AND WITHOUT EMPIRICAL PROOF, THERE IS NO SCIENCE THERE. WILLIS, SORRY, BUT THEORY IS NOT SCIENCE!

  109. TimTheToolMan says:

    Willis attacks me with “I have no time for people who don’t play well with others, Tim, especially jerks that attack me rather than answer a question.”

    But didn’t read posts in the other thread like I asked him to. I didn’t attack you Willis, I accused you of not understanding the issue and I stand by that statement because you made a strawman argument on warming water with an IR lamp vs understanding how DLR interacts with the ocean’s cool skin.

  110. Willis Eschenbach says:

    jae says:
    December 13, 2012 at 7:41 pm

    The problem, Willis, is that there is absolutely no empirical evidence that the backradiation does anything, as some of the commenters say.

    Oh, please. There is plenty of evidence.

    For example, there is clear empirical evidence that downwelling radiation keeps the oceans from freezing. The oceans emit about 390 W/m2. They absorb on the order of 170 W/m2 from the sun.

    If there were no energy coming from DLR, then the ocean would be constantly losing about 220 W/m2 … so what keeps it from freezing, jae?

    I say that the fact that the oceans are not frozen is conclusive evidence that the ocean is getting about 220 W/m2 from somewhere, otherwise it would be frozen. I say it is getting it from downwelling longwave radiation.

    Where do you think the energy to keep the ocean from freezing is coming from?

    w.

    PS—You say:

    But IT IS ONLY A PROPERTY OF MATTER. Only the STORAGE OF ENERGY matters. Again, again, again, again, I keep comparing Atlanta and Phoenix, same latitude and elevation——AND Phoenix on a clear day ALWAYS has a higher temperature, day and night, than Atlanta, DESPITE THE FACT THAT ATLANTA HAS AT LEAST 4 TIMES AS MUCH GREEHNOUSE GASSES AS PHOENIX! OH, ORACLE, PLEASE EXPLAIN THIS PHENOMENON! THERE IS SIMPLY NO PROOF OF THE PUTATIVE GHE! AND WITHOUT EMPIRICAL PROOF, THERE IS NO SCIENCE THERE. WILLIS, SORRY, BUT THEORY IS NOT SCIENCE!

    A word to the wise. Capital letters are the equivalent of shouting, and in the amount you used them, they are the infallible mark of someone with lots of passion and not much science. You should avoid that kind of shouting if you wish to be taken seriously.

    PPS—Why on earth would you expect Atlanta and Phoenix to have the same temperatures, or for Atlanta to be warmer?

  111. TimTheToolMan says:

    Willis writes in a reply to tallbloke the other thread “You’re looking at this backwards. In a system in which the surface is maintained at a slightly lower temperature than the bulk … what happens if you forcibly warm the surface? Think it through all the way, Roger. The very surface warms … until it’s slightly warmer than the bulk … which then heats up slightly, until the surface is slightly cooler than the bulk, and the previous condition (cooler surface) is restored.
    Unless your claim is that such a system can’t be heated from the top … but I don’t think that’s your claim.”

    And this indicates to me that Willis gets confused by DLR “warming”. At no time is the surface temperature warmer than the bulk, heating it up. This would be a new result that Willis would have to justify by reference. Willis doesn’t understand the process of warming yet. Thats not an attack, its an observation.

    The ocean is always losing energy. The relation is DLR < ULR. It doesn't matter if clouds increase the DLR by a seemingly whopping 100 W/m2 if the ocean surface is radiating 150 W/m2…its losing energy and the surface isn't heating and transferring its energy to the bulk.

    When Willis writes about heat lamps its a totally different situation. In that case there is sufficient energy to actually heat the water and there is no cool skin. The water is warming and not cooling. Hence there is no reason why the energy cant conduct down into the bulk.

  112. Willis Eschenbach says:

    TimTheToolMan says:
    December 13, 2012 at 7:58 pm

    … I didn’t attack you Willis, I accused you of not understanding the issue and I stand by that statement because you made a strawman argument on warming water with an IR lamp vs understanding how DLR interacts with the ocean’s cool skin.

    Ask me if I care …

    w.

  113. Matthew R Marler says:

    Willis, just chatting. I admire your work.

    You are conflating an unmeasurable variable (the radiation imbalance with the troposphere and surface fixed) and an abstract concept (“body in uniform motion”). The difference is that the radiation imbalance is a number, while a body in motion is not a number.

    Hmm. How about “black body radiation”? Is that an unmeasurable variable or an imaginary concept? How would you classify Avogadro’s number?

  114. Mack says:

    Willis,
    Medical conditions caused by small amounts of anything be it ricin or ozone will not wash with me either. We are talking about heat and the physics of gases and their interaction with the sea, not the cellular biology of humans.
    However, addressing your strawman arguement in the medical line……….as an example of ignorance and worry over small amounts, lets take the example of the flouridation of a towns water supply. There are people,(shall we call them ignorant cranks),who believe that increasing the F’ ion in a towns water supply from a naturally occuring say 0.02ppm.to about 0.8ppm will cause all sorts of disease,Alzhiemers, you name it, Yet there are towns in the US where there is (or was?) naturally occuring F’ in the water of 100s of ppm and the only ill effect is a mottling of the teeth.
    I hope you are not among the “ignorant cranks” Willis.
    But yes, you’re right about the “heavy lifting” of water,be it in the clouds,water vapour, fog, snow, ice,ocean, ie water everywhere. The Earth’s temperature is purely hydrological.
    Well the heavy lifting could be hydraulic Willis :) whatever.

  115. Bernie McCune says:

    I am not sure this is the place to post some observations that have been made by a small group of high altitude balloon flyers but I will make an attempt to say something about it which might be of value in trying to understand the strange environment of the troposphere and stratosphere. I was involved with a team that supported the NASA high altitude balloon program and also briefly the Air Force high altitude balloon program. Figure 1 is certainly an idealized view of this environment. These balloons fly for hours and sometimes days in this very strange environment. These very large balloons have been launched from all over the world with a very active Antarctic flight schedule as part of the program. Thermal effects relative to the balloon payloads (mostly radiative) are unusual since there is very little air and convective thermal exchange at altitude. Our experience of this environment can only be considered the blink of an eye in terms of months and years that go on when we are not there to observe these high altitude phenomenon. It will be very difficult to probe this dynamic environment because we just cannot position instrumentation there for very long. In situ measurement is, at this time, the only way to gather this data. The use of white paint and aluminized foil coverings can keep the balloon instruments at an acceptable temperature rather than freezing or cooking them. The “trop” is generally very cold (often -60 to -80 C) but as you pass up above it, temperatures can “warm up” to a toasty -30 C. Those boundaries are often dramatic. The short 90 minute flight of a radiosonde balloon can quickly probe this environment but most of these small “weather” balloon flights do not ever get above 100,000 feet. High altitude balloon flights are much rarer than sonde flights but they have reached altitudes of 160Kft and regularly fly at 120 to 140 Kft. The balloons fly at all hours and night flights have a really big problem of staying warm enough to stay alive. When internal temperatures in electronic packages drop much below -10 C there is a good chance they will stop functioning. The same group I worked with developed Space Shuttle and satellite payloads and we found that the thermal design for the orbiting space payloads was easy compared to balloon payloads. Being completely in the vacuum of space rather than in the last remnants of the atmosphere makes things much easier thermally. Both have some strange radiative thermal characteristics but remnants of gases complicates things for the high altitude balloon flights. Proximity to the earth’s surface for balloon flights is a thermal factor. Daytime albedo radiative effects on balloon payloads from clouds we were flying over can rapidly heat the electronics packages if we hadn’t designed the thermal shell properly. There are direct sun effects and then the reflected cloud effects from below. Antarctic flights occur when the sun never sets and there are serious snow and ice albedo effects from below. Those Antarctic flights have lasted for weeks and in at least one case the balloon did a double spiral twice around the continent (may have been about a 6 week flight)

    The winds at these high altitudes give an indication that there is still plenty of “gas” up there. Winds on the way up can vary in speed and direction with speeds of 0 to 50 knots being common. Contrary to some popular ideas, winds at 120 to 140Kft can reach these same speeds (50 to 60 knots). Winds if we happen to get into the jet stream as we climb into the “trop” can reach well over 100 knots (I have seen cases of 150 knots). Winds as we reach above the trop can be as low as 5 knots and sometimes drop even lower to almost 0. Mid latitude northern hemispheric winds at altitude tend to travel from west to east during the fall months and then there is a “turn around” that occurs over a period of days and weeks in the spring where winds at altitude travel from east to west until late summer where another “turn around” occurs. This turn around effect can be seen occurring over several months through a whole series of high to lower altitudes and latitudes. It is a very complex pattern but does come in a very seasonal clockwork fashion. Some interesting box winds occur during this changeable period. A balloon traveling to altitude can “see” east west winds through a several thousand foot climb and then get into a west east wind and so forth. The balloon can travel 10s of thousands of feet and remain mostly overhead during these unusual conditions. High altitude winds in the southern hemisphere are a mirror image of the NH winds and follow the mirror image seasonal pattern.

    I wanted to post these observations so that I could show you that we must be very careful about making assumptions about the environment that we think is there. There are some very dramatic differences from our surface environment that will make it hard for us to visualize what is happening there. Things are occurring up there mostly beyond our ability to “see” and since there are no manned vehicles ever flying through that area on a regular basis we really have no idea what is really happening over long periods of time except from some of these tiny snapshots of that near space overhead environment.

    Bernie

  116. Cliff Huston says:

    TimTheToolMan says:

    “And this indicates to me that Willis gets confused by DLR “warming”. At no time is the surface temperature warmer than the bulk, heating it up. This would be a new result that Willis would have to justify by reference. Willis doesn’t understand the process of warming yet. Thats not an attack, its an observation.”

    When a water molecule is impacted by DLR it becomes warmer than other surface molecules and the bulk molecules below. The fact that the surface water molecules on average are cooler than the bulk water molecules below, has nothing to do with the fact that DLR caused the water to be heated. The warmed molecule does not know it belongs to the cool layer and will be happy to transfer its energy to a cooler molecule, that belongs to the warm group below it.

    Feynman: Jiggling Atoms

    Think about it.

    Cliff

  117. phlogiston says:

    Bernie McCune says:
    December 13, 2012 at 11:31 pm

    Important observations, the turbulent and chaotic nature of high atmosphere winds will present a much larger surface area for thermal convective exchange and could profoundly affect vertical thermal dynamics in the atmosphere. As modellers of supernovae discovered, incorporating turbulent gas mixing can mean the difference between success and failure of effective simulation.

  118. Frank says:

    Willis: Thanks for your reply. I generally like your posts, but rebel when I think you overstated the skeptical case.

    About the surface radiative forcing, you wrote: “Sure, you can calculate it … but is it correct?” Later, about radiative forcing in general, you wrote: “If you can’t measure it, you can’t falsify it. If you can’t falsify it, it’s not science.” These are noble sentiments, but it think they are misapplied here. Like you, I’d like to see an experiment done with the atmosphere: Forget about MODTRAN or HITRAN; just point a giant variable-wavelength IR laser at a spaceship with a photodetector and take the absorption spectrum of the whole %*&^* atmosphere! Unfortunately, we can’t instantaneously double the amount of CO2 and take a second set of measurements. We also need to know about emission.

    I think our current information is a reasonable substitute for impractical experiments on the whole atmosphere. The absorption spectra of gases have been carefully studied by scientists since at least the 1930’s, long before politicization of climate science. The width of absorption lines of gases is determined by broadened by Doppler broadening (solid theoretical basis) and collision/pressure broadening (characterized imperfectly by experiment). Each GHG has been studied carefully in the laboratory and parameters characterizing the central frequency, intensity and pressure- and temperature-dependent half-width are entered in the HITRAN and MODTRAN databases (originally created for the aerospace industry). It’s trivial to pick a temperature, pressure and frequency range for a gas like CO2 and experimentally determine if the calculated and observed spectra agree. One can determine if the absorption spectrum of pure CO2 is changed by dilution with a mixture of nitrogen and oxygen. Characterizing the optical properties of GHG’s is clearly science and it probably has been done with a high degree of accuracy and reproducibility because the measurements are made in a laboratory.

    The second step is to use parameters obtained from laboratory studies to calculate what should happen as radiation passes upward or downward through pre-defined atmospheres meant to be representation of various locations on earth. The radiative forcing for 2X CO2 at the tropopause has been calculated from such models independently from MODTRAN or HITRAN data by several groups and by GCM’s (which don’t use line-by-line methods). The least reliable part of this process is dealing with clouds, because we don’t have a very good handle of the temperature and emission by cloud tops. We can’t measure radiative forcing in the atmosphere itself, but the results for dozens of models have been compared. While these calculations are not “falsifiable by experiment”, they aren’t equivalent to Shakespeare’s Glendower calling the spirits from the deep. Radiative forcing isn’t anywhere near as bad as the unfalsifiable temperature predictions made for 2100 by GCMs that can’t properly model convection and clouds and that contain a dozen or more adjustable parameters. The no-feedbacks climate sensitivity of about 1 degC for 2X CO2 appears far more reliable than any estimates of its amplification by various feedbacks; another creation of climate science that haven’t been confirmed by observation. There’s so much poor science to criticize that I personally wouldn’t invest my time complaining about radiative forcing. 3.7 W/m2 translates to a temperature rise of only 1 degK (without feedbacks), so I think radiative forcing is a good issue for skeptics.

    Radiative forcing is calculated at the tropopause because this is the lowest spot in the atmosphere where the temperature is controlled by radiative equilibrium. Where temperature is controlled by radiative equilibrium, radiative forcing can be converted to temperature change. Radiative forcing calculated at the surface can’t be converted to a temperature change, because some of the energy delivered by increased DLR can escape by convection rather than being used to warm the surface until radiative balance at the surface has been restored. If most of the energy delivered by increased DLR leaves by increased convection, 2X CO2 won’t produce much warming.

    As for your comments about large and small changes, an average of 239 W/m2 of solar radiation reaches the surface and troposphere and an average of 333 W/m2 of DLR reaches the surface. A radiative forcing of 3.7 W/m2 for 2X CO2 is small compared with these values. You correctly point out that there can be a 1000 W/m2 difference in solar radiation between day and night, but no one is worried about the temperature difference between day and night. Even if they were, DLR is still present during the night, and radiative forcing is only a small fraction of DLR. As for temperature change, you need to work in terms of degrees Kelvin (W = oT^4), not Celsius. Global warming and climate change aren’t concerned with the temperature the difference between the Sahara and Antarctic; just a rise of a few degK on a planet where the mean temperature is 288 degK. That is a small change.

    If you take the derivation of W = oT^4 with respect to temperature and do a little algebra, you get dW/W = 4*dT/T. For small changes (with no feedbacks), the percent change in T is one-fourth the percent change in W. The IPCC’s feedbacks allegedly increase this by a factor of 2 to 4.5.

  119. TimTheToolMan says:

    Cliff writes ” The warmed molecule does not know it belongs to the cool layer and will be happy to transfer its energy to a cooler molecule, that belongs to the warm group below it.”

    That would be true if there were cooler molecules below to transfer the heat to. But that’s not the case. Here is a diagram of how the very surface of the ocean looks

    http://en.wikipedia.org/wiki/File:MODIS_and_AIRS_SST_comp_fig2.i.jpg

    Its a logarithmic scale. Take notice of where the DLR interacts with the ocean. Thats from the dot at the very top down to about the second dot. Then below that, the ocean continues to warm for one mm or so. Thermodynamics tells us that if molecules are heated at the surface then that energy wont travel downwards because its going from a colder place to a warmer one.

    Its a very interesting area and is far from intuitive.

  120. richardscourtney says:

    Jeff Alberts:

    At December 13, 2012 at 6:33 pm you say to Diogenes

    Sorry, not getting the brass monkeys reference :(

    I reply with this off-topic explanation to assist non-UK readers of WUWT who may also not understand the phrase but may be amused by its explanation.

    Cold weather in the UK is often said to be,
    “Cold enough to freeze the balls off a brass monkey”.
    And this is often shortened to become a phrase such as “It’s brass monkeys outside” when describing cold weather.

    The origin of “brass monkeys” is not documented but is often said to have a naval history. Many similar phrases do have a naval origin (e.g. ‘show a leg’). And these phrases derive from the British humour of using an innocuous phrase in a manner which could be thought to be rude (i.e. double entendre).

    The story about “brass monkeys” goes like this.

    Warships prepared for battle by – among other things – stacking cannon balls alongside their cannons. The lower layer of balls needed to be constrained or the entire stack would collapse and roll away. Initially, this was achieved by putting the balls in flat trays with low walls around their edges, but lifting balls from the bottom layer of the trays was difficult (fingers could not get under a heavy iron ball). This was solved by replacing each tray with a thick, wooden board which had round depressions in its top surface. The iron balls would key into the depressions which were called ‘moons’ because their appearance resembled the Moon when they were seen in the dim side-lighting within a ship.

    The Royal Navy liked to place cannons beside the entrances to its shore installations and to stack pyramids of cannon balls beside the cannons. The wooden moon-keys for the balls soon degraded, so they were replaced by brass ones (and sailors were required to polish them daily).

    However, brass has a much higher coefficient of thermal expansion than iron. So, as temperature drops the brass moons contract relative to the iron balls. This squeezes the balls up so they are less well keyed into place. Indeed, on very cold nights it could become cold enough to ‘freeze the balls off a brass moon-key’ and the balls would roll down the street.

    Slurred pronunciation changes ‘moon-key’ to ‘monkey’.

    Richard

  121. phi says:

    Frank,

    “There’s so much poor science to criticize that I personally wouldn’t invest my time complaining about radiative forcing. 3.7 W/m2 translates to a temperature rise of only 1 degK (without feedbacks), so I think radiative forcing is a good issue for skeptics.”

    Well, science is not the stock market! Either radiative forcing has a meaning, or it does not.

    Adding CO2 to the atmosphere modifies the system and in particular the lapse rate. The surface Warming expected is a result of a change in lapse rate and not an effect of extra energy. There is no reason to think that we can characterize an increase of CO2 in the unit used to describe a change in albedo or solar activity.

  122. Ryan says:

    Willis, you are quite wrong to compare clouds to GHGs. That is an entriely different mechanism. Clouds are made up of dispersed droplets of liquid water (not gases) and they absorb a great deal of incoming radiation during the day as water has a very high specifi heat capacity. We can see that because not only do they make the ground darker, but they do so evenly across the entire visible spectrum. They do so into the infra-red as well. Consequently clouds contain a lot of heat energy. During the night they will emit that heat energy as radiation. That is like heating up a tank of hot water during the day and sitting it just above someones head at night. Yes, doing that will make you feel slightly warmer (but actually only slightly warmer because the effect due to emitting photons is actually tiny). This has nothing to do with the greenhouse effect, however, which claims that heat emitted from the ground will be absorbed and re-emitted by CO2. Even the much stronger cloud effect you have mentioned will noyl have a very small impact.

    It is very difficult to get atoms to emit photons. Every method we have tried in the lab tends to end up with massive amounts of conducted heat energy. A lightbulb emits photons which are visible – but you can’t feel the heat of them when they hit your face – the energy they have is tiny (thankfully our eyes are sensitive to this tiny amount of energy). You can feel the heat of the highly energetic atoms in the filament as they cause the glass envelope of the bulb to heat and the air above it by conduction. Getting photons out of a wire is like getting blood out of a stone. Much easier to get conducted heat out.

    The sun heats the Earth only because it emits such huge numbers of photons that travel to the equator of the Earth more or less completely unimpeded. Here they warm the oceans which then keep the northern lattitudes warm by a series of rather convenient ocean currents (the people of Britain should thank the people of Panama because their country keeps Britain warm). The sea temperature around the UK is about 10Celsius in the Winter and 16Celsius in the Summer. Doesn’t bear much relation to the air temperature above it (which can vary from -12 to +38) so why should CO2 make a difference?

  123. denniswingo says:

    Wills

    I find all of this quite interesting and am throwing my 0.5 cents in.

    The amount of radiative forcing delta between the 1950’s and today should be easily measurable. The U.S. Air Force through their “upper atmospheric research” program pushed the state of the art in IR spectrometers to the limit of measurement, or a fraction of an individual wavelength. These measurements were done in order to design the detectors for the Sidewinder and other IR heat seeking missiles. It was done at all altitudes up through at least 50-70,000 feet.

    These instruments measured the extinction coefficients for CO2, N2O, H2O, CH4, and other trace gas absorbers in the atmosphere. In theory the additional CO2 and CH4 and N2O in the atmosphere should show up as a broadening of the spectrum of absorption at ALL altitudes up to an including the altitudes where the lines desaturate due to reduced pressure and temperature.

    In theory additional CO2 (to use this as the example) should do two things. It should broaden the absorption spectrum (just compare the 1950’s values to today, easily measurable) AND it should increase the altitude where the lines desaturate (the altitude where the absorption lines are no longer fully saturated i.e absorbing all the radiation at that particular wavelength).

    Both of these quantities are measurable and then you can back out that the effects on local temperature. Everything I have ever seen on this subject relates to classical physics and this is not a classical physics phenomenon but a Quantum Mechanical phenomenon. I could blast a terawatt per square meter at 0.7 microns but since nothing at this wavelength absorbs that energy it is as if it does not exist. Energy is only transferred between radiation to thermal motion of a molecule through absorption and emission and the time constant between the two. That does not change. If the absorption spectrum is wider, more energy is absorbed (and emitted) but during the time between the two the molecular excitation can lead to a temperature rise. These quantities are strictly governed by QM rules and the equations state that absorption and emission are themselves variable, dependent on temperature and pressure of the ENTIRE atmosphere, not just the relative increase in concentration. This is what is called collision broadening of the absorption lines, which is a gaussian to lorentz transformation of the QM spectrum.

    Where is the basic QM physics in all of this? I rarely if ever see it discussed and these quantities are calcuable and measurable AND we have a huge baseline from the USAF work stretching from the early 1950’s on the subject.

  124. AlecM says:

    denniswingo: you won’t get any sensible discussion of the IR physics from IPCC luminaries such as Pierrehumbert because their house of cards’ science is based on fake physics. Thus they assume that a pyrometer measures energy flux when all it does is to measure an assembly of Poynting Vectors in the viewing angle of the instrument. If you made that viewing angle 2pi, in zero temperature gradient. the signal would be zero. So, ‘back radiation’ isn’t real and the positive feedback doesn’t exist.

    As for IR absorption, this article clearly states that it is by dribs and drabs in many collisions.:google “PhysTodayRT2011.pdf”. This is the false physics to which you refer: quantum exclusion means it can’t happen. The need for LTE means the excess energy, if there is any, pseudo-diffuses to be thermalised at heterogeneities, mainly clouds.

    The caveat about excess energy is that there isn’t any because thermal emission from the atmosphere annihilates the same emission from the surface. If this didn’t happen at equal temperature, we’d be an expanding ball of plasma. These people have no shame for their actions.

  125. Bernie McCune says:

    @ denniswingo

    I wonder what the payload weight on modern day experiment of this sort might be? Balloon payloads continue to shrink due to all the recent (past 15 years) miniaturization process (includes lower power requirements and of course much smaller and lighter [modern Li technology] battery packs). Large latex balloons could easily probe to 80 or 90 Kft and they are relatively cheap. Latex versus polyethylene balloons rise to a burst altitude and drop the payload. Probably a couple hour flight but data can be taken going up and coming down. They also make small polyethylene balloons though. Poly balloons, however, reach a float altitude and remain there until evening and night cooling lower the float altitude a little each day/night cycle (or a slow loss of He). A large balloon can carry several tons to altitude and NASA is presently doing work on long duration balloons. This type of platform could drop sondes on a very regular basis and probe the upper atmosphere clear to the ground. There could be thousands of these probes aboard a single balloon flight. Balloon flights tend to be in the $1M range so this experiment could be done for almost nothing by today’s standards. Time to get real about this stuff and stop trying to do it with computers!

    Bernie

    Bernie

  126. mkelly says:

    AlecM says:
    December 14, 2012 at 7:30 am
    “The caveat about excess energy is that there isn’t any because thermal emission from the atmosphere annihilates the same emission from the surface. If this didn’t happen at equal temperature, we’d be an expanding ball of plasma.”

    AlecM that is roughly my point. A photon that left the surface and returned via a CO2 emission cannot make the surface warmer only take it back at most to where it was. (yes I know it is not the same photon.)

    The simple heat transfer equation q/A= e * SB * (T1^4- T2^4) tells you that if T1 and T2 are equal then NO thermal energy is transfered so how can it be that if T2 is less than T1 there is heat transfer sufficient to raise the temperature of the surface 33K.

  127. denniswingo says:

    Bernie

    The answer is that the payload you speak of is easily doable and NASA does do some of this with their jet flight program where they fly spectrometers. I wish that I had the time to fully delve into this subject but to do it right is a full time job and no one is paying me to do so!

  128. Ryan says:

    Just been looking at some YouTube clips of firefighters using thermal imaging cameras. Plenty of evidence of convection currents carrying very hot air up to the ceiling (1200F)but the warming of walls and floors due to direct radiation is pretty minimal (200F). Most of this is probably direct radiation from the fire itself rather than heat radiated by hot air. Anyone that has stood in front of a bonfire will know you need to stand pretty close to feel the radiant heat (wouldn’t want to be standing right on top of one though!).

  129. Bernie McCune says:

    Sorry to go a bit off topic but how big and what pointing requirements (or other reqmts) do these instruments have? I suspect this might be a project that could easily be funded and could make several careers if there is a pressing need for it. I have an inkling that there might be niche available for this – no?

    Bernie

  130. Willis Eschenbach says:

    For TimTheToolMan and the others who think that downwelling longwave radiation can’t warm the ocean, not one single one of you has answered my simple question.

    What keeps the ocean from freezing, if not the downwelling longwave radiation (DLR)?

    We know from measurements that the sun puts about 170 W/m2 of energy into the ocean on a 24/7 average basis.

    We know from measurements that the oceans radiate about 390 W/m2 of energy on the same 24/7 basis.

    So … what is providing the addition 220 W/m2 or so of energy that is keeping the ocean from freezing? I say it is the DLR. I keep waiting and waiting for you folks that say it is not DLR to explain why the oceans are not frozen.

    Instead you give me explanations of why it is impossible, descriptions of mechanisms and the like … I don’t care about the mechanisms. I’m asking about the results of the claims. If it’s not the DLR that is keeping the oceans liquid, as y’all are claiming, then where is the energy coming from to keep them unfrozen?

    w.

  131. denniswingo says:

    We know from measurements that the sun puts about 170 W/m2 of energy into the ocean on a 24/7 average basis.

    Willis, first of all, we need to disentangle the 24/7 average basis because that term is a fiction.

    The oceans absorb energy from the sun during the day and radiate it at night, depending on the temperature of the atmosphere. This is basic physics.

    The temperature of the ocean at the water/air interface is a function of the temperature of the atmosphere at that interface. If the air temperature is less than the ocean temperature then the ocean radiates energy into the atmosphere. If the temperature of the atmosphere is greater than the temperature of the ocean then the ocean absorbs energy through molecular collisions (conduction) at the air/water interface.

    That is basic physics.

    This can be a weak effect (glassy ocean) or a strong effect (turbulent ocean).

    If there are clouds then there is a significant component of reflected IR radiation, but that IR radiation heats the atmosphere before it heats the water.

    To find out what the radiative absorption of the ocean is you have to know what the absorption spectrum of water in the ocean is and what the skin depth is to determine how deep that radiation penetrates. However, this only works if the atmosphere is warmer than the water. If there is a clear sky and no humidity, the ocean or any body of water loses heat rapidly. I used to live on a lake on the Tennessee river and had a temperature monitor near the water. On a clear, cold, low humidity night the temperature would show a drop after dark but about 2-3 hours after dark the temperature would climb a couple of degrees from the conduction/radiation of heat from the water.

    This is not that hard to figure out.

  132. richardscourtney says:

    Willis:

    Please keep going. You are winning, and you are winning for science.

    Richard

  133. phi says:
    December 14, 2012 at 4:35 am

    Adding CO2 to the atmosphere modifies the system and in particular the lapse rate. The surface Warming expected is a result of a change in lapse rate and not an effect of extra energy.

    That is not how I understand it. Please provide a reference.

  134. phi says:

    Robert Clemenzi,

    Adition of GHG in the atmosphere reduces radiation losses. It is these losses with the adiabatic cooling that determine the lapse rate. Add GHG therefore causes the change of the lapse rate. The proportion in which the surface will be heated depends on the proportion of energy flux reported on convection.

  135. phi says:

    Robert Clemenzi,
    You asked me for a reference, I do not have. It is a constant in climatology : the simplest things and the most obvious are generally not discussed but they are often denied.

  136. phi says:
    December 14, 2012 at 12:58 pm

    Actually, adding GHGs to the atmosphere increases radiation losses from the atmosphere. Everyone thinks it just increases to amount captured, but they actually cause more to be emitted than they capture.

    Adiabatic cooling determines the maximum lapse rate, not the actual (measured / typical) lapse rate.

    To see actual data, you can try my application.
    http://mc-computing.com/Science_Facts/Lapse_Rate/Lapse_Rate_Animations.html

  137. richardscourtney said @ December 14, 2012 at 11:15 am

    Willis:

    Please keep going. You are winning, and you are winning for science.

    Richard

    Willis, I’m sure Richard didn’t mean that, Please don’t go. We want you to stay! For the science.

  138. phi says:

    Robert Clemenzi,

    “Actually, adding GHGs to the atmosphere increases radiation losses from the atmosphere.”

    What you say is indeed true, but you talk about the global level. Less energy is radiated directly from the ground. The flux through the atmosphere is increased by the same and therefore more energy is lost by the atmosphere.

    But I was not talking about this global phenomenon. The potential of linear radiative losses in the column is reduced by the increase in opacity. The actual lapse rate is a consequence of both adiabatic (moist) cooling and radiative cooling. Increased GHG cause an increase in opacity, thus decreasing the potential of linear radiatives losses and therefore a decrease in lapse rate.

  139. phi says:

    Robert Clemenzi,

    I would add that the amount of energy that the ground can no longer evacuate directly due to the increase in greenhouse gases can be effectively modeled by a forcing. But this part is certainly not dominant in what is supposed to represent the radiative forcing.

  140. Brian H says:

    typo: “often taken to me 20 km” [Thanks, fixed. -w.]

    This handy-dandy ‘forcing’ power of the preferred variable is a limb of the “hidden variable fraud” written of on WUWT.

  141. richardscourtney says:

    The Pompous Git:

    re your post at December 14, 2012 at 11:14 pm.

    Thankyou for that. Yes, my choice of wording was unfortunate: of course I want more here from Willis.
    And I did laugh. 8-)

    Richard

  142. Mack says:

    Willis,
    You say “we know from measurement that the sun puts about 170w/sq.m into the ocean”
    How is that measurement obtained Willis?

  143. AlecM says:

    phi: your ‘consensus’ view is incorrect.

    Radiative equilibrium between equal temperature [100 m] lower atmosphere and Earth’s surface is by the annihilation of most surface IR in GHG bands by thermal IR. Otherwise we would be an expanding ball of gas. Because there can be no CO2-AGW, there can be no effect of increasing CO2 concentration on water vapour concentration so the Houghton/Hansen moist lapse rate GHE is disproved.

  144. phi says:

    AlecM,

    I’m sorry but I do not quite understand the meaning of your message. That said, I do not think my views are consensual, unfortunately!

  145. AlecM says:

    phi; let me explain.

    You are claiming the GHE is ~33 K, the level claimed by Hansen et al 1981. This is adapted from Houghton 1977 who made some serious mistakes.

    The climate models use the two stream approximation to claim IR warming of the atmosphere is ~7 times reality. That warming is mostly water vapour side bands. There is no CO2 IR absorption of surface IR. There is no positive feedback.

    This is because that CO2 band surface emission is annihilated by the thermal GHG IR from the atmosphere, standard radiative physics.

    So, there is no mechanism by which increase of CO2 can cause increase of water vapour. Therefore the Houghton/Hansen idea of the GHE being lapse rate warming is debunked.

    The real GHE is ~9 K with 24 K from lapse rate/gravitational potential energy.

    There is no CAGW risk, ever, not even from methane. The windmills and carbon trading are from evil corporations on the inside tack manipulating bent politicians like Obama.

  146. Willis Eschenbach says:

    Mack says:
    December 15, 2012 at 3:37 am
    Willis,

    You say “we know from measurement that the sun puts about 170w/sq.m into the ocean”
    How is that measurement obtained Willis?

    Good question, Mack. All such global numbers are best estimates of the global situation from a variety of measurements. See Trenberth 2008 for details.

    Note that the accuracy of these numbers is perhaps ± 10% or so. Note also that 10% inaccuracy doesn’t matter in the slightest regarding the question of what is warming the ocean. I just pulled up my copy of Trenberth 2008 to get the updated figures. The relevant values are:

    ENERGY LOSSES FROM THE OCEAN

    Radiation, ~ 400 W/m2

    Conduction/convection of sensible heat, ~ 20 W/m2

    Evaporative loss of latent heat, ~ 80 W/m2

    TOTAL LOSSES, ~ 500 W/m2

    So the ocean is constantly losing about half a kilowatt of energy per square metre on a 24/7 average basis.

    Now lets look at the other side of the ledger:

    ENERGY GAINS INTO THE OCEAN

    Downwelling Solar Radiation (DSR), 160 W/m2

    Downwelling Longwave Radiation (DLR), 340 W/m2

    TOTAL GAINS, ~ 500 W/m2

    Note that the ocean is in general energetic balance, in that the gains and losses are about equal. This is a necessary condition, otherwise long ago the ocean would have either frozen solid or boiled away.

    Note also that the existence of DLR constantly adding energy to the ocean is the current explanation of why the ocean is not frozen. I happen to think it is true.

    So for folks that say that there is no energy added to the ocean from DLR, then they have a huge, huge problem. With this problem, you can’t just peanut butter over the cracks by noting the uncertainties in the measurements. This is a problem involving a large amount of energy:

    If the ocean cannot absorb energy from DLR, 
    then where is the approximately 340 W/m2 coming from 
    to keep the ocean from freezing?

    As I said, nobody has been able to answer that question to date, nor do I expect an answer.

    All the best,

    w.

  147. phi says:

    AlecM,

    Your reflection is quantitative, I do not have the means to speak about it. I made ​​a qualitative critique. When I say that GHG reduce the amount of energy evacuated directly from the ground, I do not know how much it is, it could be zero, I don’t know.

  148. Stephen Wilde says:

    “If the ocean cannot absorb energy from DLR,
    then where is the approximately 340 W/m2 coming from
    to keep the ocean from freezing?”

    Descending air converting potential energy back to kinetic energy with the maximum effect at the surface.

    http://climaterealists.com/index.php?id=10775

    ” The Ignoring Of Adiabatic Processes – Big Mistake”

  149. denniswingo says:

    As I said, nobody has been able to answer that question to date, nor do I expect an answer.

    Except that you ignore when people do answer.

  150. AlecM says:

    Willis Eschenbach: ‘Radiation, ~ 400 W/m2

    Conduction/convection of sensible heat, ~ 20 W/m2

    Evaporative loss of latent heat, ~ 80 W/m2

    TOTAL LOSSES, ~ 500 W/m2′

    Oh where did you get your physics? 400 W.m^2 is the S-B prediction for an isolated black body in a vacuum at ~ 19.62 deg C. Assuming the adjacent ~100 m atmosphere is at the same temperature – 1 K, and its emissivity is 0.83, it radiates ~341 W/m^2 back meaning net IR = ~59 W.m^2..

    This is split 2:1 into atmospheric window IR which does no work and water vapour sidebands which end up causing more convection.

    Total losses ~ 159 W.m^2, about the same as the DSR. There is no DLR – it’s the artefact of the shielding around a pyrgeometer sensor. The DLR and the ULR are combined as net IR, the only way that IR can do work. There can be no two streams each heating the atmosphere.

    For further info see Poynting’s Theorem 1884.

  151. Curious George says:

    Errors large and small: Global warming enthusiasts predict a temperature rise of about 3 degrees Celsius, which is almost exactly 1% rise in the absolute temperature. The error from considering the earth as a blackbody in the IR is 3%-4% – not “quite small” in the circumstances.

  152. Mack says:

    Scroll up to my comments to Tim the Toolman to see what I think of Trenberth and his Earth’s Energy budget figures Willis.

  153. Mack says:

    Willis,
    The only things that are measured is an incoming solar radiation of about 1360w/sq.m.( a “solar constant”) which is a yearly global average which cannot be divided but remains simply as that, and that simply attenuates to a yearly global average at the Earth’s surface of about 340 w/sq.m. which corresponds to measured land based readings .

  154. jae says:

    Willis: you ask:

    “PPS—Why on earth would you expect Atlanta and Phoenix to have the same temperatures, or for Atlanta to be warmer?”

    One of the common ways to study physical phenomena is to try to keep all variables constant, except for the one you are investigatging. The Phoenix/Atlanta comparison is an attempt to hold (almost) everything constant EXCEPT the amount of GHG. Same latitude, same elevation; therefore, same solar insolation at TOA on a given day. But the amount of GHG in Atlanta is somewhere around 4 times as much as in Phoenix. Now, would not one wonder, if the amount of downwelling radiation is proportional to the amount of GHG and if downwelling radiation has ANYTHING to do with temperature, why it is not hotter in Atlanta on some clear day in July. I suspect that it is because there is so much more water available for evaporation in Atlanta, causing a cooling effect. HOWEVER, doesn’t this also show that the temperature at any location on Earth is not very dependant upon the amount of GHGs? I think it’s probably dependant only upon insolation and the amount of water available. Which is maybe why your thermostat hypothesis works, also.

    Relative to your kind advice:

    “A word to the wise. Capital letters are the equivalent of shouting, and in the amount you used them, they are the infallible mark of someone with lots of passion and not much science. You should avoid that kind of shouting if you wish to be taken seriously.”

    I’m sorry that you read the caps as shouting. I mean them to be only for emphasis. I’m just too lazy to underline, italicize, or bold.

  155. richard verney says:

    Willis
    Sorry for the late response but I have been without internet access.

    I recall your article (radiating the oceans); it was thought provoking (as is usually the case with your articles) and led to some interesting comments. I always enjoy your articles, because they are often insightful (sometimes very) and almost always give rise to lively debate.

    I am not seeking to create a new form of radiation. Any electrician will know that some measurements do not always tell the full story. A common example is a flat car battery. If you strap a voltmeter across a flat battery, it will often read 12 volts suggesting that everything is fine. It is only when you load it (say with the starter motor) that the voltage drops to say less than half. Sometimes, one can measure things but that does not necessarily mean that they have the ability to do work. I am not saying my example of a voltmeter and a flat car battery is completely analogous to the measurement of W/m2

    I am suggesting that perhaps (note I use the word perhaps) the 255K DWLWIR which is being measured is incapable of imparting sensible energy/work on 277K (or 288K) water. Why do I say this? Simples, because it does not appear to cause water to evaporate in the way that one would expect given the absorption characteristics of LWIR in water (60% of all LWIR being absorbed within the first 4 microns). This leads me to consider that there is a potential problem with the DWLWIR theory that requires an explanation.

    May be 255K DWLWIR can do all sorts of wonderful things on a 200K rock, or another gas molecule (such as an oxygen or nitrogen molecule) at 244K, but it appears that 255K DWLWIR has little (if any) impact on water. I have previously pointed out this problem with oceans, but one can also see the problem on land with dew and frost.

    It may well be the case that UWLWIR emitted at TOA into space may have a significant effect on the Earth facing side of orbiting satellites, but, if so, that merely demonstrates that one has to look at the environment in which the radiation is operating to see whether it possess sensible energy capable of performing working in that environ. I emphasise that I am only commenting upon DWLWIR in relation to water (in its many phases on planet Earth).

    Consider more carefully my example of dew (or a light film of frost) in a shady hollow. Let’s just do some maths. Let’s assume that an entire 1sqm piece of ground in a shady hollow is entirely covered by dew to a thickness of 1mm (of course, it will be less thick than this in many places). 1m by 1m by 1mm is 1 litre of water, weighing 1kg. DWLWIR, global ave is 333 W/m2. Latent Heat of Evaporation is 2260 KJ/Kg. So theoretically, DWLWIR has enough energy to evaporate water over a 1 sqm area at the rate of 0.14 gram/sec. Accordingly all the dew (1kg worth) in the shadow area would theoretically be evaporated by DWLWIR in less than 2 hours if DWLWIR can perform sensible work.

    In addition one has to bear in mind that once the sun went down the night before and temperatures began to drop, the dew as it was forming was receiving the DWLWIR all night long. So why is dew around at the break of daylight? After all, it has already had the benefit of more than 7 hours of DWLWIR if the dew formed sometime before midnight. If DWLWIR could do sensible work, as the dew is forming the DWLWIR would be burning it off such that one might reasonably not expect to see morning dew at all, certainly one would not expect to see dew hang around all day (or most of the day). Dew in the shady hollow that disappears in the early afternoon is probably burnt off by the rise in ambient air temperature taking place from sun up, not form DWLWIR.

    In contrast, we know that sunlight can do sensible work. Once the sun gets up, the dew (or frost) on the sunny side of the hollow can be burnt off quite quickly, maybe within 30 mins to 1 hour of sun up. This is notwithstanding that morning sun is very weak. Global ave is about 170 W/m2 (ie., only about half that of the DWLWIR) and due to the low incidence of morning sunlight, we could be talking of no more than 20% of that figure and yet this is sufficient to burn off the dew (or melt the frost unless the ambient air temp is very cold). You do not need many W/m2 because the film is very thin and there is little mass of water to be burnt off.

    So you need to explain why on a given day (with its then prevailing ambient temperature and humidity), low incident sunlight (which on any basis is weak) can do something within less than an hour that DWLWIR (which is claimed to be about twice as strong and even more in relation to low incident sunlight) cannot achieve in the whole day.

    Given the importance of the water cycle, the latent heat content of water (and its phase changes), the thermal capacity of the oceans in comparison to that of air, the different absorption characteristics of LWIR and Solar Radiance in water etc, if climate science was a real science there would be thousands of experiments conducted on the effect of DWLWIR revealing proper empirical data, and yet there is none!

    Willis, I am not saying that I am right and you are wrong. Like most readers of this blog, I have an open (but sceptical mind) and I am here to learn. I am merely pointing out that dew is a potential problem for the concept of DWLWIR (in the figures claimed for it) and I have yet to see any reasonable explanation why low incident sunlight can do something measurable in matter of minutes which DWLWIR cannot achieve in hours.

    Willis, before responding with generalities such as the oceans would freeze but for DWLWIR so we know that DWLWIR must be real and must be doing something, just take a little time and do the maths. Please address my specific point with the dew in a shady hollow which can be burnt off by direct low incident morning sun light, but cannot be burnt off in the shadow area by the DWLWIR.

  156. richard verney says:

    Mack says:
    December 15, 2012 at 3:37 am
    //////////////////////////////////////////////////////////////////

    You are correct to point this out. This is one of the many problems caused by the AGW proponents love for using averages. The use of averages prevents one from seeing what is truly happening in the real world, since the devil is in the detail.

    The figure of 170 W/m2 is a calculated global average for solar. However, we know that the sun is imputting more than that figure into the oceans since the oceans are not equally distributed over the globe.

    Once you take out the Arctic, Antartic, the land masses of the Northern Hemishpere, it is quite apparent that the bulk of the oceans occupy a more equitorial position and therefore receive more than the global average..

    It is of course, the equitorial ocean that acts as the heat pump for the planet. There is plenty of solar to keep this ocean warm (and not freeze). Indeed, Willis wrote an article on ARGO and the cap of temperature of 30degC. In that article, he accepted that theoretically, the tropical oceans would be far warmer if one considers the available energy from the strong equitorial sun.

    The null hypothesis is that the oceans on average radiate only about 70 W/m2 and therefore they would not freeze.

  157. george e. smith says:

    “””””…..denniswingo says:

    December 14, 2012 at 6:22 am
    Wills

    I find all of this quite interesting and am throwing my 0.5 cents in.
    ………………………….

    I could blast a terawatt per square meter at 0.7 microns but since nothing at this wavelength absorbs that energy it is as if it does not exist……”””””

    Dennis, why don’t you try blasting your TW/m^2 laser of 0.7 micron wavelength radiation into a bucket of water. Don’t worry about standing close to it; since nothing absorbs at that wavelength then of course nothing will happen.

    For the legal disclaimer, I have not personally tried this, so I recommend that nobody try it, until they have given you their signed release from liability, or alternatively, you have given them a written gaurantee of liability for any untoward result.

  158. richard verney says:

    Willis
    You seem to consider that the oceans would freeze but for DWLWIR. Perhaps you should ask yourself, why the oceans freeze if DWLWIR is keeping them warm.
    My case is simple, the tropical ocean receives plenty of solar such that it will not freeze (even without DWLWIR). Theoretically, the tropical ocean could get up to a temperature of well over 40degC (may be even about 50degC). It does not reach this theoretical temperature for a variety of reasons, one important one being the conveyor belt currents which take away heat and distribute that heat elsewhere. It is this heat pump that that prevents the oceans nearer the poles from being frozen all year round.
    As I have mentioned to you many times before, there is a problem with DWLWIR and the oceans. The problem is this.
    • The absorption characteristics of LWIR in water is such that 60% of all LWIR is absorbed within 4 microns. LWIR (unlike solar) does not penetrate to any significant depth. All the energy is concentrated in the first 12 or so microns, and 60% of it within the first 4 microns.
    • The very top surface skin layer of the ocean is cooler than the bulk ocean below. The heat flow at this layer of the ocean is therefore upwards. Because of the direction of heat flow, any energy absorbed in the top 4 microns (as would be the bulk of the DWLWIR) cannot be conducted downwards to the bulk ocean below. Willis I have shown you this profile before, (but unfortunately, I do not have the link to hand) look it up if you disagree.
    • The top micron layer is being heated very quickly. Photons travel at the speed of light, and the energy being inputted is in watts per second. Any heat in the top micron layer cannot therefore be effectively overturned as part of the ocean overturning because that process is a mechanical process that takes hours (not seconds).
    • LWDWIR (global average) is said to be about 333 W/m2. Latent Heat of Evaporation is 2260 KJ/Kg. So theoretically, DWLWIR has enough energy to evaporate water (over a 1 sqm area) at the rate of 0.14 gram/sec.
    • We know that the water ‘must’ be evaporating at the 0.14 gram/sec rate, because the heat that resulted from the DWLWIR being absorbed in the top micron layer cannot be conducted downwards (which if it could be would slow down the rate of evaporation) because the temperature profile in this layer of the ocean is upward directing (not downward directing and heat cannot travel against the flow). We know that the 0.14 gram/sec evaporation cannot be nullified by ocean overturning because that is a mechanical process that takes about ½ day, not seconds.
    So herein lies the problem. If DWLWIR has sensible energy and has the power claimed, it would give rise to water being evaporated at the rate of 0.14grams per second. Annually, this is equivalent to rainfall of about 4.5metres. But we know that annual rainfall of this magnitude is not occurring. The oceans are clearly not evaporating at this rate (which they would be if DWLWIR ‘heats’ the oceans since due to its absorption characteristics, if it does anything it can only heat the top microns of the ocean leading to quick evaporation). The fact that we are not observing this magnitude of rainfall suggests that DWLWIR lacks sensible energy.
    You need to explain why the oceans are not evaporating at the rate equivalent to 4.5 metres of rainfall annually. Of course, if you can put forward a physical mechanism explaining how heat concentrated in the top microns layer can be conducted downwards even though this is against the temperature profile of the top microns/millimetres of the ocean, and/or how ocean over turning can turn over the top micron layer faster than the rate of evaporation (0.14 grams per second), then I would be interested in considering this physical process.
    PS. We are fortunate that the absorption characteristics of solar is so very different, because if it was the same as LWIR, the oceans would have boiled off eons ago.
    PPS. If you read the comments to your earlier post on radiating the oceans, you will note the many commentators who consider that the oceans are only losing heat via radiating at the rate of 70 W/m2. The oceans are in perfect balance without the need for the double accounting that you apply.

  159. Mack says:

    Richard Verney says…
    “The figure of 170w/sq.m is a calculated global average for solar” The key word you used here is “calculated” . It is not measured but “calculated” . I only take heed of true measurements Richard, because the moment you start calculating you move away from reality and start drawing pictures of the sun and earth on the blackboard, so to speak .

  160. Michael Moon says:

    Excuse me. Willis Eschenback is a massage therapist with a degree in psychology. Transport of Heat and Mass does not appear in his curriculum. The Second Law also does not appear. Just to clarify, the Sun is at over 5500 degrees Kelvin. The atmosphere is virtually always slightly cooler than the surface of the Earth. Heat goes from hot to cold, not the other way. Pyrometers and so-called pyrgeometers measure the temperature of the atmosphere.

    News flash, if the atmosphere is cooler than the Earth surface, the atmosphere CANNOT warm the Earth’s surface. Carnot, anyone, Bueller? No that was not a miss-spelling, the guy’s name is Carnot, as in Carnot Cycle.

    Really, massage? Dude, you belong here why? Carnot is rather important in the engneering world, the engineers who built those power plants that keep the lights on, the heat on in your home, the roads, the cars, the trains, the Internet, the CPU, in your computer, the power grid, and on and on.

    Willis, the thing is, radiation can only TRANSFER heat from WARM things to COLDER things. And, by the way, when HEAT is TRANSFERRED, things get HOTTER. Accept it, consider it, make it an important part of your life, and never try to contradict it again. Yes cooler things still radiate. If the warmer things to which they radiate got HOTTER and began to radiate MORE, what stops this runaway????????

    Good Lord, no one one here new that???

  161. Michael Moon says:

    Knew that, the CPU in your computer, sorry sorry

  162. denniswingo says:

    Dennis, why don’t you try blasting your TW/m^2 laser of 0.7 micron wavelength radiation into a bucket of water. Don’t worry about standing close to it; since nothing absorbs at that wavelength then of course nothing will happen.

    Lets go there. Since water does not absorb radiation at 0.7 microns it is unlikely that anything would happen. Beyond that I did not say that nothing would absorb I said the atmosphere would not. If it did then the planet would have fried long ago as the insolation at that wavelength is in the several hundred watts per square meter. Solar power satellite designs use lasers at wavelengths near this to transfer megawatts per square meter with little atmospheric effects. Green lasers at about 550 nanometers are proposed for laser weapons with gigawatts/m2 for this same purpose. Thanks for helping me prove my point. Unless you understand the couplings between radiation sources and sinks you can’t understand this subject and damn few people do understand it. Here is a nice website on the absorption spectrum for water.

    http://www.lsbu.ac.uk/water/vibrat.html

  163. Michael Moon says:

    Or the First Law, or the Third Law, or the Zero’th Law. Really? Do we all remember all four??? Can we recite them? I can….

  164. Robert Clemenzi says:

    richard verney says:
    December 15, 2012 at 5:51 pm

    Your arguments about DWLWIR and dew are spot on except for one thing – the ground is also emitting radiation. Assuming that the ground and the atmosphere are at the exact same temperature, then the upwelling radiation will always be greater than the downwelling radiation because the ground emits with an almost continuous blackbody spectrum and the atmosphere has a significant range of frequencies where it is transparent (the spectral window). As a result, at night the ground cools until the atmosphere is warmer and the downwelling and upwelling radiation are equal.

    When the humidity is high enough, dew, fog, or frost will form. Each of these also emits upwelling radiation. Limiting the discussion to dew, the downwelling radiation keeps it from freezing. In the case of the poles, during the 6 month winter, without the downwelling radiation, the nitrogen in the atmosphere would condense into a liquid.

    In the morning, before sun up, the upwelling and downwelling radiation are about equal. In this case, the additional heat from the Sun makes the difference and burns off the dew.

    Since you understand electronics, think of W/m2 as current and T^4 as voltage. The surface and atmosphere are capacitors, the Sun is a current source, and deep space is ground. Add resistors as necessary to produce the correct currents.

    By the way, solar radiation is UV plus visible plus SWIR, and about 50% of the energy is in the SWIR spectrum. Most of that is absorbed more than 2cm below the surface of the ocean.

    http://omlc.ogi.edu/spectra/water/gif/hale73.gif
    http://omlc.ogi.edu/spectra/water/data/hale73.dat

  165. Sparks says:

    Black body Energy in -> Energy out-> a blanket <-Energy back in Energy out -> SPACE

  166. george e. smith says:

    “””””…..Willis Eschenbach says:

    December 13, 2012 at 12:00 am

    Geoff Sherrington says:
    December 12, 2012 at 10:28 pm

    My son, a surveyor, heard me mention watt per sq km at the troposphere and immediately interjected “At what altitude?”
    Unless light passing through a horizonal sq m is parallel, like some laser light, then because there are more sq m of conceptual surface at a higher altitude quasi-sphere around the globe, the flux through each is less when there is a steady source. How does one define the flux when the “number of sq m” on a surface is forever changing rapidly with altitude, but by definition requires measurement to be taken when steady state is approached?
    I have no idea about what models do about the variation in numbers of sq m shared by a watt as altitude is changed. Can someone assure me that the effect is built into the math?
    If it is not, then a watt per sq m is ‘more powerful’ at low altitude than one at high and it would be hard to construct a Willis fig 1.

    Geoff, your son is right, but it is a difference that doesn’t make a difference. Here’s why……”””””

    Geoff, does your son the surveyor, know that the radius of the earth orbit (mean) is 93 million miles. Compared to that, the 15 or so miles of atmosphere thickness, makes no difference to W/m^2 coming from the sun.

  167. Mack says:

    Willis,
    Consider the fact that in equatorial regions a sq.m metal plate (inches thick) will get hot enough to fry eggs on. Think of the equivalent electrical wattage required to achieve this and you are more likely to agree that the Earth’s surface receives a solar incoming radiation yearly average of 340odd w/sq.m. instead of the IPCC’s 161w/sq.m.
    Unless you say that some of this is caused by DLR? In which I would have to say to you in the nicest possible fashion…It’s the sun stupid. ;)

  168. richard verney says:

    Mack says:
    December 15, 2012 at 7:28 pm
    //////////////////////////////////////////////////////////
    I agree with your comment. I have many issues with the K&T energy budget, but I am merely trying to point out one issue that arises IF that budget were correct. IF DWLWIR possesses sensible energy (in the Earth environ) and IF it is as powerful as claimed, why is it not causing dew to evaporate, or the oceans to evaporate?

    The optical absorption of LWIR in water would suggest that there should be significant evaporation since the ‘theoretical’ energy being imparted cannot be spread throughout the bulk volume of the liquid. With Solar the energy penetrates to great depth such that Solar results in the very gradual and relatively slight heating of the bulk liquid. DWLWIR by comparison would concentrate all the energy that it possesses in just a few microns, rapidly raising the temperature of those few microns sufficiently to cause rapid evaporation.

    We are not seeing this rapid evaporation of dew, or, for that matter, of the oceans. So why not?

    All I am asking is what is the explanation as to why we are not witnessing this in the real natural world? That is what I want Willis to address.

  169. I’d like to add to Mr. Clemenzi’s electrical analogy…

    Since you understand electronics, think of W/m2 as current and T^4 as voltage. The surface and atmosphere are capacitors, the Sun is a current source, and deep space is ground. Add resistors as necessary to produce the correct currents.

    Sea water and dense things like concrete and bricks are big integrating capacitors. Comparatively, the atmosphere represents a tiny, irrelevant capacitor…comparatively to the day/night cycle and thermal capacity of water, of course.

    I wonder about Willis’ analytic skill if he really can’t figure out how tropical surface water, directly heated to an average temperature of greater than 20C by the sun, can be liquid without atmospheric backradiation. As usual, if you let your enemy frame the argument, you will lose. When integrating, it’s important that the dt be appropriate. Water is not heated by average insolation. It’s heated by peak insolation (1KW/m^2) and has thermal mass, so it’s able to store the peak energy it’s exposed to. Of course, it’s possible for a small thermal mass to heat a large thermal mass, by being a lot frigging hotter. To extend the electrical analogy, a small capacitor can add charge to a large capacitor, but only if the the voltage is huge. 1/2 C V^2, you know? Otherwise, the big capacitor charges the small capacitor, right? There’s no need to resort to nonphysical gyrations (ha, epicycles) of radiation to figure this out.

  170. richard verney says:

    Robert Clemenzi says:
    December 15, 2012 at 10:36 pm
    ////////////////////////////////////////////////////////////////////////

    My analogy with the car battery, was not intended as a direct analogy (as I mentioned). It was intended to illustrate a conceptual point, namely there are times when we measure something and not fully comprehend/appreciate the significance (or lack thereof) of what we have measured.

    I am well aware that Solar has a wide spectrum and because of it’s spectrum, and the absorption characteristics of water, it penetrates deep into water. Much more than the couple of centremetres mentioned by you. That is one of my points. That is why Solar does not burn of the oceans. It results in the very gradual heating of the bulk, which is to be contrasted with the position that arises with DWLWIR which if it does anything must rapidly heat just the first few microns of the ocean thereby leading to rapid evaporation of the top micron layer.

    You omit conduction. The ground also conducts heat to the atmosphere above. In my example of morning dew, for all practical purposes, Solar does not heat the dew at all. If the thin film of dew is 1mm thick, due to the absorption characteristics of water (to the various wavelengths of Solar), for all practical purposes, non of it is absorbed by the dew and it all goes to heat the ground below. The dew is burnt off by conduction brought about by Solar heating the ground below the dew.

    To evaporate water you need to consider energy over time. Just before sun up, whatever amount of energy is being conducted and radiated from the ground in the hollow, is for practical purposes the same all over the hollow. Likewise the ambient temperature and humidity of the air above the hollow, At sun up, the only additional factor is the input of Solar on the sunny side of the hollow. This additional energy could be measured, but we know it is slight since there is not much energy in low incident sunlight especially in the late autumn, winter, early spring. This slight additional energy is sufficient to burn off dew within 30 mins (may be a bit more or a bit less – depends upon the prevailing ambient conditions which prevailing ambient conditions are the same in the shady side of the hollow).

    As the day warms, the sunny side of the hollow conducts (and you would say radiates) more heat and the atmosphere gradually warms. At this stage of the day DWWIR is increasing. It is now more than was the case when the sun was first up. DWLWIR is omnidirectional so we know that the shady side of the hollow is now receiving some of this increased DWLWIR. The shady side of the hollow may experience this increase in the DWLWIR for hours and yet the dew is not burnt off from the shady side of the hollow. Given the length of time that the dew in the shady side of the hollow is exposed to this increase in DWLWIR, we know that the total energy received from this increased package must be more than the additional energy that was inputted to the sunny side of the hollow by Solar (alone) within the first 30 mins (or so) of daybreak.

    One can visualise what is going on without knowing the precise figures. But if the relation between Solar and DWLWIR is in the order suggested by the K&T energy budget, it would not take long for DWLWIR to burn off a thin film of dew especially given the absorption characteristics of water to DWLWIR. When considering this, one must bear in mind the thermal inertia of the ground. The ground is being heated by Solar but heating of the ground could be slow, whereas water theortectically absorbs LWIR almost instantaneously and nearly all the energy from DWLWIR is contrated in the first few microns such that one would expect rapid evaporation to occur. .

  171. mkelly says:

    Mr. Verney and Mr. Clemenzi your analogy to electrical circuits is quite proper. My heat transfer book uses circuit diagrams as analogy for resistance to heat flow.

  172. Willis Eschenbach says:

    Michael Moon says:
    December 15, 2012 at 9:39 pm

    Excuse me. Willis Eschenback is a massage therapist with a degree in psychology. Transport of Heat and Mass does not appear in his curriculum. The Second Law also does not appear. Just to clarify, the Sun is at over 5500 degrees Kelvin. The atmosphere is virtually always slightly cooler than the surface of the Earth. Heat goes from hot to cold, not the other way. Pyrometers and so-called pyrgeometers measure the temperature of the atmosphere.

    News flash, if the atmosphere is cooler than the Earth surface, the atmosphere CANNOT warm the Earth’s surface. Carnot, anyone, Bueller? No that was not a miss-spelling, the guy’s name is Carnot, as in Carnot Cycle.

    Really, massage? Dude, you belong here why?

    Egads, sir, why the vitriol?

    I belong here because I am a scientist. I have worked as a massage therapist, it is true, along with dozens and dozens of other jobs. I’ve been the guy who sweeps the floor at one end of the scale, and I’ve been the Chief Financial Officer of a company with forty million dollars in annual sales at the other end of the scale. I’ve worked as a musician, and a commercial fisherman. I’ve run an boatbuilding and machine shop on a tiny island in the South Pacific There is a link to my CV here for those interested. If you think that any of my jobs disqualify me as a scientist, you don’t understand how science works.

    More to the point, Michael, my climate science is good enough to be published in Nature magazine and other scientific journals. They had no trouble with it, they didn’t care that I have worked as a massage therapist … you see, it’s kinda crazy, but they were interested in my ideas, not my work history. It’s called “science”, you should look it up …

    And you? Why do you belong here? Have you published in Nature magazine?

    Regarding heat flow, you are missing the boat entirely. While NET heat flow always goes from hot to cold as you claim, with radiation there are always two flows of energy going on. There is a flow of energy from the warmer to the cooler area, and there is also a flow of energy from the cooler to the warmer area. It is only the NET flow that is constrained to always go in one direction. Get a college-level beginner’s physics book, they’ll cover it in there. Here’s an example illustration, from the University of Sydney.

    Note the flow of energy going both directions …

    So your claim about the atmosphere being unable to add energy to the surface is, well, not to put too fine a point on it, a common newbie mistake. Curiously, this mistake is usually made by people who are arrogant, nasty folk like you, puffed up and full of your self-importance, and eager to display your lack of knowledge.

    Finally, my scientific credentials are not the point. The important questions is, are my statements and claims true? It doesn’t matter who makes the scientific claim. The only issue is whether that claim is valid.

    w.

  173. Willis Eschenbach says:

    richard verney says:
    December 16, 2012 at 3:53 am

    Mack says:
    December 15, 2012 at 7:28 pm
    //////////////////////////////////////////////////////////

    I agree with your comment. I have many issues with the K&T energy budget, but I am merely trying to point out one issue that arises IF that budget were correct. IF DWLWIR possesses sensible energy (in the Earth environ) and IF it is as powerful as claimed, why is it not causing dew to evaporate, or the oceans to evaporate?

    The optical absorption of LWIR in water would suggest that there should be significant evaporation since the ‘theoretical’ energy being imparted cannot be spread throughout the bulk volume of the liquid.

    There is a huge incorrect assumption here, that DLR, once absorbed, cannot spread throughout the body of the ocean. The claim is made that this is so because it is said that the surface of the water is always and forever cooler than the bulk.

    But as you point out, Richard, this absorption of DLR energy does NOT cause huge evaporation to occur at the surface.

    Your conclusion from this is that the energy somehow is not being absorbed.

    My conclusion from this, on the other hand, is that:

    a) as soon as the DLR warms the surface of the water very slightly, the surface will be warmer than the bulk and energy will flow into the bulk, and then the surface will be cooler than the bulk again.

    b) waves, wavelets, and wind energy are constantly stirring the surface and mixing the energy into the bulk.

    Finally, guys, you are missing the question. I don’t care about the mechanisms. I care about the end result. Let me give you an example.

    Suppose we are looking at a mountain. You say “The mountain can’t be climbed”.

    I say “Take a look with the binoculars, there’s climbers nearing the top right now”.

    You say “But I have proven it is impossible.”

    I say “People are doing it right now as we speak”.

    You say “But there is no mechanism, I have proven to my own satisfaction that it is impossible. Therefore there cannot be people up there”

    Do you understand the metaphor? I say that DLR is warming the oceans, AS EVIDENCED BY THE FACT THAT THEY ARE LIQUID.

    You say that’s not possible, you advance 500 perfectly reasonable explanations why it can’t happen from DLR, because you can show that the mechanism won’t work.

    OK, fine, sez I, then what is keeping the oceans liquid?

    Crickets …

    I don’t care if you have proven 500 ways that it is impossible. I don’t care about the MECHANISM. I am discussing the real world results, which are a liquid ocean. If the DLR is not heating it … what is? So far, I have no plausible answer from anyone. Someone upthread suggested that it comes from the energy of descending air being compressed and warming up … I fear I didn’t dignify that one with a response, although I should have suggested that they run the numbers and look at the actual amounts of energy involved …

    So, I still await an answer—if DLR is not warming the ocean, what is?

    w.

  174. Pamela Gray says:

    A case in point. Decades ago, most psychiatrists were firm in the scientific conclusion that autism was caused by distant and cold mothering. Mothers called that notion bunk and balderdash. Many used foul language in their description of the consensus and worse in their description of psychiatry.

    So no, it does not take a card-carrying scientist to refute a consensus and eventually be proven correct. If you were one of those people who discredited people who’s job description was to raise their children, you were eventually labeled a fool.

    My advice to AGWers, do NOT walk in those shoes.

  175. Robert Clemenzi says:

    richard verney says:
    December 16, 2012 at 5:01 am
    ////////////////////////////////////////////////////////////////////////

    At night, the ground radiates energy and cools much faster than the air above it. Via conduction, this pulls energy (heat) from the air immediately above it. At this point, “hollows” are a bit more complicated than the Great Plains. In a hollow, this cold air runs downhill and collects in the valley. As a result, the cold layer above a plain, or calm ocean, may be only a few centimeters thick, but in a hollow it will be several meters thick. This is easy to see in the Smokey’s (an Eastern US mountain chain) where the hollows contain fog long after the peaks are clear.

    I agree, the Sun can not heat the dew directly. It heats the leaf, grain of sand, etc. and that heats the dew. Because the conduction from the leaf to the water is much greater than from the leaf to the ground (direct contact verses an air gap), the thermal mass of the bulk surface has almost no effect until after the dew is evaporated.

    When evaluating a transient response, the Solar radiation suggested by the K&T energy budget is misleading. 170 W/m2 is averaged over a full day and a full year. The peak radiation to burn off the dew is more like 1,361 W/m2, the TOA value. As for heating the leaf, remember that dew drops are small lenses and that means that the “temperature” can be much higher than predicted by SB. However, for the DWLWIR the dew is opaque, meaning – no lens effect. So comparing an increase of 40 W/m2 to an increase of 1,361 W/m2 plus a lens might explain the difference.

    The angle of the sun light is an interesting problem. In general, the amount of energy available is related to the cosine of the angle. However, because leaves are at all angles, and because of the lensing effect, a much higher than expected amount of energy is available to evaporate dew. As long as the sun shines, the amount of energy available is very significant. Passive (non-focusing) collectors are able to boil water in Canada during the winter. The cold air has more to do with where it comes from and the number of daylight hours than with the angle of the Sun.

    With respect to dew, in the very early morning, refraction is also an issue. This, plus the extra thickness of the atmosphere (6 miles tropical noon, 154 miles in the morning), is why the morning and evening suns feel so cool.

  176. Pamela Gray says:

    Ocean heating: The Coriolis affect combined with surface obstacles and currents define the major components of Earth’s unique oceanic fluid dynamics. Ocean layers are heated, especially around the equatorial belt, with IR heating (not long wave) which itself is filtered to various levels at the surface because of cloud variation. The heat absorbed by the oceans is then mixed and sloshed around or allowed to sit and layer calmly via trade wind variations.

    Oceanic heat loss: My educated guess is that in various ways, heat is largely released back into the atmosphere, sometimes because of storms, and usually because of its eventual trip to the poles where it evaporates and eventually escapes the confines of our planet. Some years and decades (long and short) the oceans lose a lot of heat, and some years and decades that cooling system doesn’t work so well. This gain/loss see-saw is anything but predictable.

    I think longwave radiation oceanic heating (the kind that is re-emitted from the ocean and re-emitted back into the ocean from the atmosphere) is a tiny player, buried in the noise, in the above description, and certainly not at all in measurable trends.

  177. Frank says:

    Phi replied to me: “Well, science is not the stock market! Either radiative forcing has a meaning, or it does not.”

    The IPCC’s definition of radiative forcing provides a very clear meaning. Radiative forcing is the result of a calculation that determines how the net flux of radiation through the atmosphere is expected to change as the atmosphere and/or the radiation entering the atmosphere changes. The IPCC has specified where (the tropopause) and how (allow stratosphere to reach radiative equilibrium) that calculation is to be done, and their choices make sense. To do the calculations, one needs carefully measured parameters from many reproducible laboratory experiments, atmospheric conditions and composition, and how much radiation enters the atmosphere from the sun, the surface and cloud tops. (All but the parameters change from location to location, so there is no single correct answer.) At least some of the software for doing these calculations is available on the Internet. If anything in climate science desires to be called “settled science”, it might be radiative forcing.

    Unfortunately, “radiative forcing” reported in W/m2 doesn’t mean anything to non-specialists. The IPCC chapters that I have read about GHG forcing don’t tell the public how to interpret radiative forcing in terms of a temperature increase. Instead they wait until they have amplified radiative forcing with poorly understood feedbacks to produce climate sensitivity or processed through climate models containing parameters that have been tuned so that 20th century warming is attributed to increasing GHGs.

    Here’s a reasonable interpretation of radiative forcing in terms of warming, generally called the no-feedbacks climate sensitivity. W = eoT^4, so W+dW = eo(T+dT)^4 where dW and dT are small changes, deltaW and deltaT. With a little algebra and ignoring terms with higher powers of dT, one can derive an equation I rarely encounter: dW/W = 4*(dT/T). It says that the percent change in radiation will be four times bigger than the associated percent equilibrium change in temperature (in degK). To compensate for a radiative forcing of 3.7 W/m2 (1.54% change) we need a 0.38% increase in temperature. Since the surface and lower troposphere receive about 240 W/m2 of solar radiation (after account for albedo), the equivalent blackbody temperature (255 degK) appears to be the most relevant temperature to use, affording a 1.0 degK warming. Where will this warming take place? Calculations can’t tell us, but 90% of the photons escaping to space are emitted from the upper troposphere, so a reasonable answer is that upper troposphere (where the temperature is near 255 degK) will warm 1.0 degK. If the lapse rate remains unchanged, the surface will warm by 1.0 degC too.

  178. Mack says:

    Willis says……..Dec15th 201 7.28pm
    /////////////////////////////////////////////////////////////////
    Everything you say is IF there Willis. Did Micheal Moon say you worked as a massage therapist? Probably explains why you you were asking everybody for a bit of traction further up .

  179. phi says:

    Frank,

    “If anything in climate science desires to be called “settled science”, it might be radiative forcing.”

    How is computed the CO2 forcing? By adding with a thought experiment some CO2 created ex nihilo. What is the temperature of this CO2? Obviously not defined. We uses the temperature profile of the atmosphere before the add but it’s a perfectly arbitrary operation. It is absolutely necessary to leave the new CO2 to adapt to its environment before making any calculation. Unfortunately, the principle of radiative forcing requires an instantaneous balance. This calculation is actually under-determined and the value obtained arbitrary.

    We can being aware of that also considering the schematic equation of heat transfer through the atmosphere: Ts = L * P + k * P ^ 0.25 where Ts is the surface temperature, P the heating power, L and K synthetics factors. The addition of CO2 occurs in L and can not be modeled by an alteration of P. There is an exception with the energy evacuated directly from the ground and now intercepted by the added CO2 but it is a small amount.

    “If the lapse rate remains unchanged, the surface will warm by 1.0 degC too.”

    1 ° C may be correct as an average, by cons, we know that it’s not valid for surface because the main effect of the added CO2 is to change the lapse rate by reducing the potential radiative losses along the column.

    This warming being not valid at the surface is critical because it is the surface temperature that initiates the feedback on water vapor.

  180. Greg House says:

    Willis Eschenbach says, December 16, 2012 at 9:55 am: “While NET heat flow always goes from hot to cold as you claim, with radiation there are always two flows of energy going on. There is a flow of energy from the warmer to the cooler area, and there is also a flow of energy from the cooler to the warmer area. It is only the NET flow that is constrained to always go in one direction. Get a college-level beginner’s physics book, they’ll cover it in there. Here’s an example illustration, from the University of Sydney.”
    ============================================================

    Willis, your illustration does illustrate an IDEA, but like any other illustration it does not prove the idea to be right or to be a scientific fact. You illustration only demonstrates that there are other people who share your idea about “NET”.

    Second, and this is important, this “NET” thing makes physical sense with regard to temperature only, if “a flow of energy from the cooler to the warmer area”, as you put it, has an effect on the temperature of the warmer area. If it does not, your “NET” does not make physical sense at all.

    Even if there is some effect on the temperature of the warmer area, but this effect is disproportionately or even vanishingly small, you can not just add or subtract energy and then derive temperature from the result of this arithmetic operation. This would be no science.

    Which means, until this “NET” thing is not proven experimentally, it remains a pure fiction.

    Please, Willis, present a clear reference to and a description of a REAL scientific experiment (not just a “thought experiment”) proving your idea about this “NET” effect.

  181. Willis Eschenbach says:

    Greg House says:
    December 16, 2012 at 1:40 pm

    … Please, Willis, present a clear reference to and a description of a REAL scientific experiment (not just a “thought experiment”) proving your idea about this “NET” effect.

    Greg, I doubt you can find what you are asking for. Such basic statements are rarely accompanied by experiments. The fact that solid objects radiate depending on their temperature is such basic science that it is presented as a statement of fact rather than with an experiment.

    Here is some information from MIT on exactly how to calculate the radiative heat exchange between two surfaces. They have no doubt that there are two separate radiative flows.

    Now, if you don’t believe what the folks at MIT say about radiation, I’m afraid you’ll have to take it up with them.

    w.

  182. TimTheToolMan says:

    Willis writes “For TimTheToolMan and the others who think that downwelling longwave radiation can’t warm the ocean, not one single one of you has answered my simple question.”

    I do think DLR slows the rate of cooling, Willis. I’ve told you this a number of times in the other thread. I’ve not said it doesn’t in this or any other thread so I dont know where that accusation keeps coming from.

    I am pointing out they so far I believe your understanding on how the ocean warms as a result is lacking. You’ve stated that there is no difference between warming and slower cooling but then you’ve made statements that clearly indicate you believe the DLR increases the SST. Directly!. It does not. It cant because the ocean is cooling Willis. The sun warms it at depth, not at the very surface when, for example, clouds come over.

    There IS a difference between slower cooling and warming because when you think “warming”, when you think more DLR means +100 W/m2 you can and do get it wrong.

  183. Willis Eschenbach says:

    Frank says:
    December 16, 2012 at 11:11 am

    Phi replied to me:

    “Well, science is not the stock market! Either radiative forcing has a meaning, or it does not.”

    The IPCC’s definition of radiative forcing provides a very clear meaning. Radiative forcing is the result of a calculation that determines how the net flux of radiation through the atmosphere is expected to change as the atmosphere and/or the radiation entering the atmosphere changes. The IPCC has specified where (the tropopause) and how (allow stratosphere to reach radiative equilibrium) that calculation is to be done, and their choices make sense. To do the calculations, one needs carefully measured parameters from many reproducible laboratory experiments, atmospheric conditions and composition, and how much radiation enters the atmosphere from the sun, the surface and cloud tops. (All but the parameters change from location to location, so there is no single correct answer.) At least some of the software for doing these calculations is available on the Internet. If anything in climate science desires to be called “settled science”, it might be radiative forcing.

    I am clear that you can “do the calculations”, as you point out, to give a number for the IPCC-defined “climate forcing”.

    The problem is, what you can’t do is measure, at any point in time and space, said climate forcing. So you can’t ever verify or validate your calculations, even in a rough sense. The thing being discussed (climate forcing under those conditions) doesn’t exist.

    This means that whatever numbers you might come up with as you calculated “climate forcing” CAN NEVER BE FALSIFIED.

    So yes, the IPCC definition has a clear meaning. The problem is that the clear meaning describes an impossible situation—they specify the climate forcing with the stratosphere responding and the troposphere and surface temperature held fixed, a situation which has never existed on the planet.

    And that means that no statements about climate forcing can be falsified, and thus they are not scientific statements.

    w.

  184. jae says:

    Willis notes:

    “Greg, I doubt you can find what you are asking for. Such basic statements are rarely accompanied by experiments. The fact that solid objects radiate depending on their temperature is such basic science that it is presented as a statement of fact rather than with an experiment.’

    BUT, sir, I think I just gave you THE experiment (Phoenix vs. Atlanta). The amount of radiation from greenhouse gases seems to matter very little, if at all. Water and water vapor can probably explain it all.

    I noticed that you have not responded to my last comment on this subject, and I am wondering why…

  185. I don’t know anyone who disputes the fact that things with mass and temperature radiate (except perhaps Myrhh, I’m not sure what he or she thinks). The Kent drawing shows two independent sources radiating toward each other. No problem. Going further, I don’t know anyone disagreeing with the idea of a passive receptor heating up and radiating. The problem is trying to do useful work with re-radiated energy…and in the presence of convection? Good luck. You can’t get an average temperature increase of 33C from backward atmospheric re-radiation. Something this blatant would be easy to instrument in a lab environment.

  186. Willis Eschenbach says:

    kencoffman (@kencoffman) says:
    December 16, 2012 at 4:36 am

    … I wonder about Willis’ analytic skill if he really can’t figure out how tropical surface water, directly heated to an average temperature of greater than 20C by the sun, can be liquid without atmospheric backradiation.

    And I, on the other hand, wonder exactly where you think the missing 340 W/m2 is coming from to keep the ocean liquid. Let me reprise the numbers for you:

    ENERGY LOSSES FROM THE OCEAN

    Radiation, ~ 400 W/m2

    Conduction/convection of sensible heat, ~ 20 W/m2

    Evaporative loss of latent heat, ~ 80 W/m2

    TOTAL LOSSES, ~ 500 W/m2

    So the ocean is constantly losing about half a kilowatt of energy per square metre on a 24/7 average basis.

    Now lets look at the other side of the ledger:

    ENERGY GAINS INTO THE OCEAN

    Downwelling Solar Radiation (DSR), 160 W/m2

    Downwelling Longwave Radiation (DLR), 340 W/m2

    TOTAL GAINS, ~ 500 W/m2

    Your claim that it is the sun sounds lovely, all full of sunshine and everything … but where are your numbers? You need to demonstrate mathematically, not just claim but demonstrate, that somehow the sun is delivering enough energy to keep not just the tropical ocean but the entire ocean liquid. If there is no DLR heating the ocean, that means that the sun must average half a kilowatt per square metre on a 24/7 basis …

    Those are my calculations. I await your calculations as to how the sun is putting out half a kilowatt as a global 24/7 average …

    And if you can’t do the calculations, I wonder about your analytical skill.

    w.

  187. Willis Eschenbach says:

    kencoffman (@kencoffman) says:
    December 16, 2012 at 4:36 am

    … I wonder about Willis’ analytic skill if he really can’t figure out how tropical surface water, directly heated to an average temperature of greater than 20C by the sun, can be liquid without atmospheric backradiation.

    Oh, yeah, Ken, one last point regarding your statement. If you just look at the tropics, it’s true there is more sun … but because the temperature is higher, there is also more evaporative, radiative, and sensible heat loss from the ocean. So the problem remains—the sun alone is not enough to keep it liquid.

    w.

  188. Willis Eschenbach says:

    jae, you pointed out I haven’t answered the following, thanks:

    jae says:
    December 15, 2012 at 4:36 pm

    Willis: you ask:

    “PPS—Why on earth would you expect Atlanta and Phoenix to have the same temperatures, or for Atlanta to be warmer?”

    One of the common ways to study physical phenomena is to try to keep all variables constant, except for the one you are investigatging. The Phoenix/Atlanta comparison is an attempt to hold (almost) everything constant EXCEPT the amount of GHG. Same latitude, same elevation; therefore, same solar insolation at TOA on a given day.

    Well, yeah … except that one of them is in the desert and one is not. As a result, they have very different responses to downwelling IR. There is an interesting discussion of these issues here.

    w.

  189. JazzyT says:

    There seems to be some confusion about the idea of GHGs emitting IR which is absorbed at the surface, and heating it. It might be better to say, “transferring heat to the surface” rather than “heating the surface” because, at night, in the absence of other effects, the surface will inevitably cool, but will cool more slowly than without the GHGs. During the day, the surface is warmed by the sun.

    The whole climate change issue involves questions of whether the Earth might warm, on the average, by 2 degrees C over the next century, or by 3 degrees, perhaps 5 degrees, or by 1 degree or less, with various consequences or lack thereof.. This projected temperature change, over a century, is superimposed on daily and seasonal changes. Daily changes of 10 degrees is quite ordinary, and seasonal changes of 30 degrees or more are quite common, as compared to a much smaller AVERAGE change over 100 years.

    There is NO need to worry about where the heat would come from for global warming; it comes in every day. If the Earth were to heat up, on the average, by 10 degrees C over the next century, most climate scientists think that truly cataclysmic results would follow–but NOBODY thinks that this is a remotely serious possibiltiy. The balance between incoming and outgoing energy is fine enough that one day’s sunlight changes the temperature much more than any average change that anyone seriously considers for the next century. (Heating the top layers of the oceans would take more than a day, but still much shorter than a century.)

    GHGs do indeed radiate IR to the surface and transport heat thereby. But the surface cannot actually warm from this unless the atmosphere is actually warmer (which could happen if, say, a warm front moves through). But the effect of GHGs is to slow the rate at which the surface cools through IR emission. The actual mechanism is complicated, but that’s the end result.

    We could try to calculate the temperature at the surface in terms of how much the atmosphere radiates to the ground, various heat capacities, etc. but that’s difficult to conceive of, and difficult to solve. There’s an easier way: despite the differences of night and day, seasons, weather, etc. the incoming and outgoing energy is always very close to balanced. Were this not so, the average temperature would go up, not over a century, but over a matter of weeks or days. It would probably be very exciting. But the main regulatory mechanism, the sharp dependence of IR emission on temperature, puts the brakes on quickly. As Frank mentioned above, emission goes as the fourth power of temperature. For every 1% increase in absolute temperature, which is about 3 degrees C, there’s about a 4% increase in IR emission. So, if nothing else has changed, that cools the surface down pretty quickly. (For a demo, wait until sunset.)

    So, to solve it the easy (or easier) way, we look at the percentage of IR that gets through (at each wavelength, it’s not quite simple) and then ask: what temperature would the surface have to be so that the heat that escapes through IR in 24 hours is equal to the heat received from the sun in one day? The more the GHGs “block” the IR the higher that temperature is. The word “block” is a gross oversimplirication, but in the end, that’s why the temperature goes up–more solar heat is retained, until the surface is warm enough to shove through an amount of energy that equals, on the average, what the sun puts in.

    Of course, after that, you look at the IR absorption spectra of the greenhouse gases, blackbody emission spectra, heat transport through air, currents, ocean currents, evaporation, precipitation, thermals caused by politicians emitting large amounts of hot air…lots of complications. But one thing you DON”T have to worry about is how the energy gets to the ground to heat it up. The sun takes care of that, every day.

  190. Willis Eschenbach says:

    TimTheToolMan says:
    December 16, 2012 at 5:35 pm

    Willis writes “For TimTheToolMan and the others who think that downwelling longwave radiation can’t warm the ocean, not one single one of you has answered my simple question.”

    I do think DLR slows the rate of cooling, Willis. I’ve told you this a number of times in the other thread. I’ve not said it doesn’t in this or any other thread so I dont know where that accusation keeps coming from.

    I am pointing out they so far I believe your understanding on how the ocean warms as a result is lacking. You’ve stated that there is no difference between warming and slower cooling but then you’ve made statements that clearly indicate you believe the DLR increases the SST. Directly!. It does not. It cant because the ocean is cooling Willis. The sun warms it at depth, not at the very surface when, for example, clouds come over.

    There IS a difference between slower cooling and warming because when you think “warming”, when you think more DLR means +100 W/m2 you can and do get it wrong.

    Tim, I find this insistence on some meaning of “warming” is semantical quibbling. For example, we say that putting on a jacket warms us up. Now, in fact all it does is slow down the cooling.

    So what? The result is the same, so what is the difference? We end up warmer when we wear a jacket than when we don’t. The ocean ends up warmer when there is DLR than when there is no DLR.

    You’ve stated that there is no difference between warming and slower cooling but then you’ve made statements that clearly indicate you believe the DLR increases the SST.

    Again, I find this to be semantic nonsense. The DLR slows the evaporation at the surface. As a result, the bulk of the mixed layer ends up warmer than it would be in the absence of DLR. So yes, DLR does indeed increase the SST down deep in the ocean.

    In fact, there is absolutely no difference between warming and slower cooling. Suppose I have a warm block of iron, losing energy at some rate X.

    I put a blanket on it, in an hour it may only experience half the temperature drop it experienced when it had no blanket.

    Then I run the experiment again. Instead of a blanket, I put propane torches on all sides of it to warm it perfectly evenly. As a result it only experiences half the temperature drop it experience when it had no blanket.

    Here’s the thing. There’s no difference in result between the blanket and the torches. If someone warmed it one way or the other in secret, you couldn’t tell the difference. The final temperature is the same, the amount of heat in the block of iron is the same.

    So it doesn’t matter to me in the slightest whether you call it “warming” or “slowing the cooling”. The end result is the same, the bulk of the mixed layer is warmer with DLR than without DLR.

    Finally, this again is like trying to prove that people can’t climb the mountain, when we can clearly see people who have climbed to the top … your problem is to figure out where the missing 340 W/m2 is coming from.

    w.

  191. Greg House says:

    Willis Eschenbach says:, December 16, 2012 at 5:23 pm: “Greg, [...] The fact that solid objects radiate depending on their temperature is such basic science that it is presented as a statement of fact rather than with an experiment.
    Here is some information from MIT on exactly how to calculate the radiative heat exchange between two surfaces. They have no doubt that there are two separate radiative flows.
    Now, if you don’t believe what the folks at MIT say about radiation, I’m afraid you’ll have to take it up with them.”

    =====================================================

    Willis, I am not questioning the fact that solid objects radiate depending on their temperature, or that there are separate radiative flows. You have probably misunderstood the point. I am questioning the alleged effect of radiation from colder bodies on TEMPERATURE of warmer bodies, and logically the calculations of what you call “NET” transfer. I hope you have understood the point now.

    Your references to “folks at MIT” are not a scientific argument unfortunately, anyway not when the very essence of your point is questioned. It is a wide spread logical fallacy called “appeal to authority”. I am very sure that “folks at MIT” can give references to other “folks” and so on. Such a circle of references is, however, not a scientific argumentation.

    You are apparently a knowledgeable man, so I guess, if you can not refer to a real scientific experiment confirming your (and other “folks’) assertion about that “NET” thing, you probably are capable of understanding that this assertion can not be considered a scientific fact, it is just a fiction.

  192. gymnosperm says:

    So here’s a stab at explaining why the oceans don’t freeze for lack of DLR:

    Take a water heater and remove the dip tube. Install a big pump that recycles water between the holes at the top of the tank. What you are doing is installing a flowmeter and thermometer on the outflow side and saying, “Wow, there is so much energy leaving this thing would freeze if it weren’t warmed by water coming back in.”…

  193. TimTheToolMan says:

    Willis writes “Here’s the thing. There’s no difference in result between the blanket and the torches.”

    You’ve made another strawman argument without even realising it… The thing is that DLR (and a blanket) redirects the object’s energy back at the object. Back radiation. And its just not possible for a blanket to make the block of iron warmer than when you started. Just because you can set a propane torch up to account for the same energy loss means nothing and doesn’t mean warming is the same as cooling.

    Do you stand by this statement that you made to tallbloke?

    “The very surface warms … until it’s slightly warmer than the bulk … which then heats up slightly, until the surface is slightly cooler than the bulk, and the previous condition (cooler surface) is restored.”

    Because this is very explicitly saying that DLR can warm the ocean warmer than it was before you started. This is like saying I can put a blanket on the iron or put coffee in a thermos and it will get warmer than when I started.

  194. AlecM says:

    Willis says this: ‘Regarding heat flow, you are missing the boat entirely. While NET heat flow always goes from hot to cold as you claim, with radiation there are always two flows of energy going on. There is a flow of energy from the warmer to the cooler area, and there is also a flow of energy from the cooler to the warmer area. It is only the NET flow that is constrained to always go in one direction. Get a college-level beginner’s physics book, they’ll cover it in there. Here’s an example illustration, from the University of Sydney.’

    This is plain wrong. Unfortunately most people are taught it as if it’s correct. The reason is they are taught the S-B equation and the difference, net energy flux is the difference between the two S-B predictions. This is incredibly misleading though and you along with most have fallen in to the trap.

    To prove this you go back two steps in the physics. The first is the Planck Irradiation Function which when integrated over the wavelengths or the wave numbers is the S-B equation. The second is to understand the the PIF at any wavelength/wave number is for a collimated beam identical to the Poynting Vector. This transfers to physics to Maxwell’s Equations.

    The wonder of this is Poynting’s Theorem which is that the net PV at any point in space is the vector sum of all the arriving PVs. In other words there can never be two streams doing work. Only the vector sum at a point can do work.

    This is why the climate models are total, absolute bunkum. You can’t compute absorption of energy from the two streams because it exaggerates warming by a factor of ~7. What’s more as there is zero net CO2 IR emitted from the surface [basic radiation physics at thermal equilibrium means net PV is zero where the two PVs are near black body iamplitude] there can never be any CO2-AGW or positive feedback.

    Please Willis, learn the real physics and tell these IPCC charlatans they have got it wrong. It goes back to Houghton using the two-stream approximation to calculate energy absorption. If he had had the scientific wit to work out that the DOWN warming is negative to offsets most of the UP warming we might have avoided this incredible mess of fake science.

  195. AlecM says:

    I shall also add a point I haw made in the past which is that the belief in DLR is entirely baseless because a pyrgeometer measurement is the artefact of the instrument’s shielding. The proof it to have two pyrgeometers back to back in zero temperature gradient – net signal = 0. Take one away and the signal jumps to a measure of the S-B flux for that body at its particular temperature and emissivity.

    However, until that electromagnetic wave combines destructively with the wave from the opposite direction, it cannot do any thermodynamic work. ‘Back radiation’ is an imaginary energy flow all of which is destroyed at the Earth’s surface. The only work that can be done from the Earth’s surface is the 63 W/m^2 net IR flux, the 17 W/m^2 convection and the 80 W/m^2 evapo-transpiration.

    Anyone who claims differently has to justify it from Maxwell’s Equations and they can’t.

  196. Willis Eschenbach says:

    Greg House says:
    December 16, 2012 at 10:22 pm

    Willis Eschenbach says:, December 16, 2012 at 5:23 pm: “Greg, [...] The fact that solid objects radiate depending on their temperature is such basic science that it is presented as a statement of fact rather than with an experiment.
    Here is some information from MIT on exactly how to calculate the radiative heat exchange between two surfaces. They have no doubt that there are two separate radiative flows.
    Now, if you don’t believe what the folks at MIT say about radiation, I’m afraid you’ll have to take it up with them.”
    =====================================================

    Willis, I am not questioning the fact that solid objects radiate depending on their temperature, or that there are separate radiative flows. You have probably misunderstood the point. I am questioning the alleged effect of radiation from colder bodies on TEMPERATURE of warmer bodies, and logically the calculations of what you call “NET” transfer. I hope you have understood the point now.

    Your references to “folks at MIT” are not a scientific argument unfortunately, anyway not when the very essence of your point is questioned. It is a wide spread logical fallacy called “appeal to authority”. I am very sure that “folks at MIT” can give references to other “folks” and so on. Such a circle of references is, however, not a scientific argumentation.

    You are apparently a knowledgeable man, so I guess, if you can not refer to a real scientific experiment confirming your (and other “folks’) assertion about that “NET” thing, you probably are capable of understanding that this assertion can not be considered a scientific fact, it is just a fiction.

    Greg, thank you for your explanation. Now, I see that in fact you agree that everything radiates.

    It appears that you believe, however, that when radiation from something cooler is absorbed by something warmer than the spot where the radiation originated, no energy of any kind is transferred to the warmer object.

    Is this in fact your belief? And if so … where does the energy go that was contained in the radiation? It can’t vanish. It’s not reflected. I say that energy is converted to heat. If I understand you, you say it is not converted to heat … so what is the energy converted into?

    As I said above, there’s plenty of things in science for which you won’t find published experimental results, because they are too basic. If you are dead-set on an experiment, here’s one for you. You can do it as a thought experiment, or as a real experiment if you wish.

    Put a 100-watt incandescent bulb in a fixture, turn it on and let it equilibrate, and measure the temperature.

    Put a 40-watt incandescent bulb in a fixture, turn it on and let it equilibrate, and measure the temperature.

    Clearly, at equilibrium the 100-watt bulb will be hotter than the 40-watt bulb.

    Now, for the final test, put the 100-watt and the 40-watt bulbs in full view of each other (but not so near as to allow heating by conduction/convection), turn them on, and let the whole setup equilibrate.

    So here’s the point of the experiment. What do you think the final temperatures of the bulbs will be when both are lit?

    a) Both bulbs will be the same temperature as when they were lit individually.

    b) Both bulbs will be cooler than when they were lit individually.

    c) The 100-watt bulb will be warmer, and the 40-watt bulb will be the same.

    d) The 40-watt bulb will be warmer, and the 100-watt bulb will be the same.

    e) Both bulbs will be warmer.

    Me, I can do it as a thought experiment, I don’t need to build the apparatus to be clear about what will happen. YMMV.

    All the best,

    w.

    PS—Were I in your shoes, I’d study what the folks at MIT say in the citation I referenced. I’d study it until I understood what they were saying, inside and out. Then I’d see if I could find theoretical or practical problems with their explanation before I advanced my own theory on how the world works … but hey, that’s just me.

    And no, this is not an “appeal to authority”. Instead, it is an appeal to investigate what the authorities have to say before striking out on your own. Some very bright people have given these questions some very profound thought, and it behooves us to investigate their contributions thoroughly in order to have a firm base for our understanding.

    And perhaps eventually, to expand or even overthrow their profound thoughts … but to do that, first you have to understand them.

    But like I said, that’s just me …

  197. Willis Eschenbach says:

    Folks, I’ll say it again. “Proving” that the DLR (downwelling longwave radiation) can’t heat the ocean is meaningless in this discussion. All of your convoluted explanations are wasted on me.

    I have asked a simple question.

    If DLR is not keeping the ocean from freezing, what is?

    Simple energy calculations show that the incoming sunlight at the surface (global 24/7 average ≈ 160 W/m2) is not enough to balance the known losses from radiation, conduction/convection, and latent heat loss (global 24/7 average ≈ 500 W/m2). The calculation is out of balance by about 340 W/m2. I (and most every scientist I know of) say that this 340 W/m2 is the energy that the ocean is absorbing from the DLR.

    Until someone can come up with an alternate source for the missing energy (about 340 W/m2), I’m going to just keep on saying it is supplied by the DLR. I’m sorry, but I have to deal with the reality that the oceans are not frozen. The geothermal heat flux is about three orders of magnitude too small to supply the missing energy necessary to balance the oceanic energy budget. I know of no other source of energy that would keep it liquid. Nor has anyone else identified one.

    I add to that the fact that downwelling longwave radiation is known to heat what it is absorbed by; the fact that DLR is absorbed by the ocean; the fact that DLR has been experimentally measured all around the world in a variety of seasons, times of day, and locations; and the fact that the energy contained in the DLR is of the right size (about 340 W/m2) to supply the missing energy to the ocean.

    Now, y’all can believe in rainbow energy or pyramid energy or whatever, I don’t care. But until someone can come up with an alternate source of 340 W/m2 of energy to keep the ocean liquid, I’m calling BS on the claim that the ocean is not warmer with DLR than it would be without it.

  198. AlecM says:

    Willis: The citation you reference, about radiative heat transfer calculations, is exactly what I was taught as a Metallurgical Engineer. Until 3 years’ ago when I set out to study exactly what the IPCC claimed, I have never thought beyond the S-B analysis. However, once I realised the climate models have in them a perpetual motion machine, creating much more IR energy to be absorbed than real net IR, I decided to find out why.

    The two-stream approximation cannot be used to calculate heat absorption unless one stream is defined as negative, the cooler to the warmer. You cannot have a perpetual motion machine. Unfortunately, the grant money would disappear because the CAGW scare vanishes, and we can’t have real science interfering with capitalism, can we?

  199. Mack says:

    Dec 14th @10.34 am “So what is providing…….. energy that is keeping the ocean from freezing? I keep waiting and waiting for you folks to say it is not DLR to explain why the oceans are not frozen” (then later)… “where is the energy coming from to keep them unfrozen?”
    So then he ,Willis (hopefully) reads what I said to him about Trenberth’s unreal geometrically derived IPPC crap of 160w/sq.m. reaching the Earths surface. But no he persists..2 days later..
    Dec16th 2012 @6.38pm “So the problem remains- the sun alone is not enough to keep it liquid”.

    You know how you might be reading something right in front of you but cannot believe what your eyes are seeing. This is exactly my case here. How can this guy Willis who calls himself a scientist (maybe that’s his problem) say such stuff.
    Willis…
    If you were to take the earth and for hypothesis sake freeze the seas. Then introduce a sun which will apply a continuous and timeless input of energy to the surface of 340w/sq.m. I’m sure that before too long we would approach what we have today in reality. I’m going to say this again Willis without any smilies or winkies . It’s the sun stupid.

  200. AlecM says:

    Willis: ‘If DLR is not keeping the ocean from freezing, what is?’

    The answer is very simple. The oceans do not emit 400 W/m^2 ULR. The only IR they emit is the net 60 W/m^2. This is an inescapable fact – only net energy flows are real.

  201. TimTheToolMan says:

    Willis writes “But until someone can come up with an alternate source of 340 W/m2 of energy to keep the ocean liquid, I’m calling BS on the claim that the ocean is not warmer with DLR than it would be without it.”

    Its not about the warming, its about the process. And mostly about the “warming is the same as reduced cooling” At least try to understand this. I’ve done you the courtesy of understanding your posts.

    This….
    http://wattsupwiththat.com/2011/08/15/radiating-the-ocean/#comment-721248

    And then this…
    http://wattsupwiththat.com/2011/08/15/radiating-the-ocean/#comment-723255

    And I think the only missing bit between those posts is that when the DLR increases and if there is still DSR to keep warming, then the slowing conduction in the skin will cause the skin’s temperature gradient to reduce (less is being lost from below now) and then the hook along with the SST will increase. At no point does the very surface warm beyond the temperature below due to the DLR.

  202. D Böehm says:

    I have to agree with Willis. The DLR is the reason the oceans are not frozen. If that is wrong, please provide a credible alternate explanation.

  203. AlecM says:

    D Boehm: ‘DLR’ is an artefact, Poynting Vectors representing the temperature of the emitter(s) in the viewing angle of the pyrgeometer sensor. It’s not real otherwise at twice average solar flux, you’d be able to feel it on the back of your hand by the heat sensors!

    It cannot do any thermodynamic work until it combines vectorially with information from the hotter body, This is why to measure real energy flux you need two pyrgeometers back to back and take the difference signal.

    The Trenberth ‘Energy Budget’ is total bunkum unless you give negative warming power for PVs from the cooler to warmer bodies. It’s far better to work with net energy flows!

  204. Willis Eschenbach says:

    AlecM says:
    December 17, 2012 at 12:47 am

    Willis: The citation you reference, about radiative heat transfer calculations, is exactly what I was taught as a Metallurgical Engineer. Until 3 years’ ago when I set out to study exactly what the IPCC claimed, I have never thought beyond the S-B analysis. However, once I realised the climate models have in them a perpetual motion machine, creating much more IR energy to be absorbed than real net IR, I decided to find out why.

    The two-stream approximation cannot be used to calculate heat absorption unless one stream is defined as negative, the cooler to the warmer. You cannot have a perpetual motion machine. Unfortunately, the grant money would disappear because the CAGW scare vanishes, and we can’t have real science interfering with capitalism, can we?

    I totally agree that you can’t have a perpetual motion machine. A greenhouse effect has nothing to do with perpetual motion. Actually, it doesn’t have anything to do with CO2 either, see my posts called The Steel Greenhouse and People Living in Glass Planets.

    w.

  205. Willis Eschenbach says:

    Mack says:
    December 17, 2012 at 12:51 am

    … So then he ,Willis (hopefully) reads what I said to him about Trenberth’s unreal geometrically derived IPPC crap of 160w/sq.m. reaching the Earths surface. But no he persists..2 days later…

    Mack, the lack of a citation to your words about Trenberth make it hard to figure what you are talking about. What you said about Trenberth in this thread, and then followed with a recommendation that I should read it again, was:

    … none of the figures you see in that cartoon are actual measurements but may well have been pulled from Trenberth’s ass.

    Now you want to bust me for not replying to that vague and unpleasantly scatological claim? Really? Trenberth goes to great lengths to explain exactly how he develops and derives each of the figures involved in his global energy budget. Have you even read the two papers involved (Earth’s Annual Global Mean Energy Budget and Earth’s global energy budget)? If you have a beef with how he is deriving one or more of the values, you should spell it out.

    Because around here, objecting to a scientific work by claiming that the scientist obtained the numbers rectally won’t get a reply from me, or from most folks. That’s just content-free blather, without a scrap of scientific value, devoid of fact, lacking details, claims, citations, and indeed lacking everything but braggadocio.

    Sorry to say, Mack, but I chose not to answer your post because there was nothing there to answer. For you to try to bust me now for not answering such vapidity is risible.

    w.

  206. Edim says:

    “Simple energy calculations show that the incoming sunlight at the surface (global 24/7 average ≈ 160 W/m2) is not enough to balance the known losses from radiation, conduction/convection, and latent heat loss (global 24/7 average ≈ 500 W/m2). The calculation is out of balance by about 340 W/m2. I (and most every scientist I know of) say that this 340 W/m2 is the energy that the ocean is absorbing from the DLR.”

    There’s no missing energy – simple Earth’s energy budget show that the surface absorbs 51% of the incoming solar energy and loses 23% by evaporation, 21% by IR radiation (6% radiated diractly to space) and 7% by convection.

    51 = 23 + 21 + 7

  207. Willis Eschenbach says:

    AlecM says:
    December 17, 2012 at 1:00 am

    Willis: ‘If DLR is not keeping the ocean from freezing, what is?’

    The answer is very simple. The oceans do not emit 400 W/m^2 ULR. The only IR they emit is the net 60 W/m^2. This is an inescapable fact – only net energy flows are real.

    Why is it that people think that thermal IR cannot be measured? Scientists routinely use instruments to measure it. What they measure is real. When they point their instruments at the ocean, they do not get 60 W/m2, or anything near that. To be radiating at that rate the ocean would be at -90°C (-135°F). Depending on the temperature of the ocean, the instruments say that the ocean is radiating IR constantly at somewhere on the order of 400 W/m, more where it’s warmer, less where it’s cooler. But not 60 W/m2.

    You get to have your own opinions about things, Alec, but you don’t get to make up how much the ocean radiates in the thermal infrared. We can measure that, and many people have done so.

    Finally, you have the stick by the wrong end when you say that “only net energy flows are real”. In fact, for radiation, net energy flows are a mathematical construct. The only thing that is real is the radiation coming from the two objects. Here’s my drawing of the thought experiment again, I see nobody answered my question about the thought experiment above … ah, well.

    What is real is the radiation (both light and thermal IR) coming from each of the lights. We can use a variety of instruments to measure the radiation coming off of each of the lights. That is what is real.

    Now, there is indeed a net flow of heat from the hotter bulb to the colder bulb, but it cannot be measured directly because it is a mathematical abstraction. In a system in which both lights are constantly both absorbing and radiating energy, the net flow is the difference between the two actual flows.

    Note that we cannot directly measure the net flow. All we can do is measure each individual flow and subtract one from the other.

    w.

  208. Willis Eschenbach says:

    TimTheToolMan says:
    December 17, 2012 at 1:28 am

    Willis writes

    “But until someone can come up with an alternate source of 340 W/m2 of energy to keep the ocean liquid, I’m calling BS on the claim that the ocean is not warmer with DLR than it would be without it.”

    Its not about the warming, its about the process. And mostly about the “warming is the same as reduced cooling” At least try to understand this. I’ve done you the courtesy of understanding your posts.

    Tim, once again you try to explain to me why DLR is not adding 340 W/m2 to the ocean.

    Look, Tim, I truly don’t care about the process. I’ll assume that your process is correct, and that for some strange reason the ocean is not absorbing 340 W/m2 from the DLR.

    My question is, if your process is correct and the ocean is NOT absorbing 340 W/m2 from the DLR … then where is it absorbing that amount of energy from?

    And to echo D. Boehm’s comment above, to date, I have heard no credible alternative explanation.

    w.

  209. AlecM says:

    Willis:I’m sorry to state this so bluntly, but pyrometers do not measure energy flux, they measure temperature. OK, a pyrgeometer is calibrated against a cavity black body in Power Units but that is the S-B power that would be emitted from the isolated emitter in a vacuum. The artefact is because the shield at the back of the detector stops EM information coming the other way.

    The only real energy flow is the difference signal between back-to back instruments. This is what the manufacturers state, see the bottom of this page: http://www.kippzonen.com/?product/16132/CGR+3.aspx

    These people have sold 1000s of pyrgeometers to climate research knowing full well that they are being misused because Meteorologists and Climate Scientists are taught ‘back radiation’ must be real because the single pyrgeometer is calibrated in W/m^2. This is the most blatant and longest [50 years] misunderstanding in science I have ever come across. and it must be stopped by forcing these academics to teach correct physics. [The other big mistakes are wrong IR physics and wrong cloud physics].

    So, there is no opposing 400 W/m*2 and 340 W/m^2 power fluxes. These are instrumental artefacts, a temperature signal expressed in the EM continuum. There is no energy transfer until the information in those two streams combines vectorially as the net energy flux and even then other emitters are involved, e.g. the cosmic microwave background for the atmospheric window radiation.

    ‘Back Radiation’ does not exist. For normal temperature gradients, it all disappears at the Earth’ surface.

  210. Ryan says:

    Willis: “Simple energy calculations show that the incoming sunlight at the surface (global 24/7 average ≈ 160 W/m2) is not enough to balance the known losses from radiation, conduction/convection, and latent heat loss”

    Well, I for one don’t see why I need to justify myself any further. The incoming heat at the equator is not 160W/sqm. It is 1000W/sqm. The water at the equator is forced to flow E-W and then has nowhere to go but northwards until it rotates around the poles and back down south like some enormous central heating system. That is why the oceans are not frozen. The specific heat capacity of the oceans is such that they can easily abosrb enough energy at the equator to remain unfrozen across the rest of the globe. This also explains why the Arctic ocean is not frozen all the way to the sea bed in winter despite having 6 months with no solar radiation let alone no DLR and with an opaque layer of ice on top. It also explains why the seas around Britain can be 10Celsius always in the winter even though the air temperature can persist below 0Celsius for days on end. You can’t use an average because the areas with less daytime heating will also have less cooling – i.e. the maths is different for the radiators of a central heating system than for the heat exchanger.

    If you want to prove to yourself that Trenberth’s figures must be nonsense, go and get a big mirror and put it in the garden in a cold night over your head facing down – see how much warmer it is? No? But surely it must be – after all now you are getting 100% DLR at close range, you don’t have to rely on the weak effect of unsatured CO2 spread throughout the atmosphere. Sit in a shed however and you might well feel a bit warmer because now you have reduced the conduction/convection effect.

    Radiation balances are only relevant when seen from space. Below the tropopause only the conduction/convection/evaporation is relevant. If it wasn’t for the cooling effect of the troposphere the oceans would be a lot WARMER, because vibrating atoms conduct heat between them in a far more effective manner than the emission of photons. Bear that in mind – if there was no air above the oceans they would only be able to lose heat by re-radiating photons – a much slower process than by radiating photons. They would be much hotter and would boil into space.

  211. Willis Eschenbach says:

    AlecM says:
    December 17, 2012 at 2:39 am

    D Boehm: ‘DLR’ is an artefact, Poynting Vectors representing the temperature of the emitter(s) in the viewing angle of the pyrgeometer sensor. It’s not real otherwise at twice average solar flux, you’d be able to feel it on the back of your hand by the heat sensors!

    It cannot do any thermodynamic work until it combines vectorially with information from the hotter body, This is why to measure real energy flux you need two pyrgeometers back to back and take the difference signal.

    Man, this is not even wrong. Where to start. First, DLR is not an artifact by any spelling. It is real, measurable thermal radiation. It is exactly the same kind of thermal radiation emitted by the objects around us.

    How much thermal radiation is there? Well, it depends on the temperature and nature of the object. For example, my wooden desk, with a temperature of say 20°F (~70°F) and an emissivity of say 0.95 or so is emitting thermal radiation at the rate of about 400 W/m2.

    Now, you can complain all you want that this is more than the average 24/7 solar insolation, and it is, but that doesn’t change the facts. If you were to measure the infrared radiation coming off of my desk, that’s about the number you’d get, 400 W/m2. That’s how we see things with infrared night-vision. That’s how we measure temperature with infrared thermometers. Things radiate, and how much they radiate is a function of their temperature.

    And indeed this is real radiation containing real energy. All that stuff about having to combine with other energy to do any work? Doubletalk. Look at the illustration of the two lamps. Radiation goes in all directions. You are claiming that the energy from one of those lamps individually cannot do work … why on earth not? You can get work out of the temperature difference between the two lamps as you point out, but you can also get work from either of them individually. Heck, if nothing else stick a Crookes radiometer in the light from one of them and shade it from the light of the other … you’re getting work out of just one of them.

    w.

  212. AlecM says:

    Willis; Firstly your warm desk. Put two pyrgeometers back to back between the desk and the sofa. The net signal will be near zero. This is a fact. A single pyrgeometer measures temperature expressed in power units. It is not a real energy flow.

    That only takes place when the two information streams combine. A rider to this is that the common idea that such streams are strings of photons is also very wrong. This is because the photons only exist at the instant energy quanta are transferred, and that can only happen once the net energy flux is made.

    As for the two lamps, solving that problem mathematically is very complex. The 40 W will rise in temperature more than the 100 W because its cooler glass will be a sink for the warmer 100 W envelope.

  213. AlecM says:

    Right, a better explanation of the two lamps.

    7% of the energy is emitted as SW light. The mechanism by which it does work is that EM vectors from the cooler body combine vectorially with the vectors from the hotter body and the net energy then passes from the hotter body to the cooler. Because the cooler body is at much lower temperature, the net flux is the same as the emitted flux.

    93% of the energy is IR. Some is absorbed by the glass bulb and this then radiates and convects. Concerning the former, what happens is that radiation from cooler bodies combines vectorially and because we are dealing with a greater proportion of the energy from the cooler body, the radiative heat loss is a smaller proportion than for the SW case. As a result, convection increases.

    Now consider the oceans. All the GHG thermal energy received from the atmosphere disappears, removing the same from the 400 W/m^2 you would get for an isolated ocean in a vacuum! So, the energy flow is one way, 60 W/m^2.

    Thinks of the opposing Poynting Vectors as an impedance to real IR energy flux. Higher amplitude at any given wavelength reduces that flux from the surface, and it does so unevenly depending on the spectral nature of the radiation from the other direction.

  214. TimTheToolMan says:

    Willis writes “My question is, if your process is correct and the ocean is NOT absorbing 340 W/m2 from the DLR … then where is it absorbing that amount of energy from?”

    The surface of the ocean IS absorbing the energy in the top 10um or so…and radiating it back up at about the same rate (some may be used in evaporation) for no net positive effect in the bulk. That is the vital bit. There is no energy gain in the bulk or in the top 1 mm or anywhere due to the DLR.

    The ocean is cooling at the surface and the DLR is not warming it. Not even a little bit. Warming comes from sunlight absorbed below and the SST is driven by the energy from below, not from the DLR although the DLR is a factor in determining where the balance is.

    Mathematically it accounts for some of the energy the ocean must radiate according to S-B.

    Do you understand why I think your statement to tallbloke is wrong?

  215. Mack says:

    Willis,
    It’s quite difficult for me (layman) to explain, but here goes…Trenberth and academics and AGW people seem not to be content with just accepting that the TOA receives a real measured ( since 1902 by waterflow and now by sattilite) reading of an incoming 1360w/sq.m. but they take this figure and like all true teachers get some idea that this is the amount emitted by the sun and immediatly go to the blackboard and draw a picture of two little circles of the sun and earth. At this point they have stepped out of reality. They then draw rays from the sun to the earth and through a process of maths and geometry work out what THEY think is the incoming TOA solar radiation. A sort of buggering up of a real measurement .Sort of one foot in reality and the other in an AGW fantasy. As I’ve told you before Willis this 1360w/sq.m. is a yearly global average which cannot be divided. It is the real deal. This is what we get. However you and the academics arrive at this CALCULATED figure of 340 w/sq.m incoming at the TOA, which then transforms into figures tearing around in your hysterical imaginations. The neat and paradoxical thing for you AGW deceiving liars is that what the earth receives in reality at the surface ie 340w/sq.m ,happens to coincide with the 340 w/sq.m. you lot work out for the TOA. In this way it easily confuses students because they are unaware of which surface you are referring to …. the actual earths surface or the “surface” of the TOA. When drawing your little pictures of the sun and earth on the blackboard what student is going to pick up on the little distance of the width of the atmosphere eh Willis. This is probably one reason your crap science has done so well. So get real Willis, The reality is that this earth receives 340 oddw/sq.m. at its surface (the sq,m metal plate in equatorial regions getting hot enough to fry eggs on ,remember?) get used to it and you can take Trenberths and your figures a shove em where there ain’t no w/sq.m.

  216. AlecM says:

    TimTheRToolMan, the operational IR emissivity of the oceans is negatively dependent on temperature so highest IR emission is from temperate seas. specifically off New England and Japan.

  217. I have no problem with modeling and simulation and superposition…these things are commonly used in engineering. However, I do have a problem with creating nonphysical influences and using them in the model. If you want to spread insolation across a disk, that’s fine if there is a physical mechanism that actually does that. I’m not going to spread insolation across a disk when I’m modeling a rotating sphere. If you want to average incoming energy into an average representing a daily dose, that’s fine if you can point to the physical mechanism that does that. I’m not going to average daily insolation across a 24-hour period if there is no physical mechanism. If you want to store and integrate thermal energy, then store it in something with thermal mass like water and dense masses and use those storage mechanisms in your integrating storage/delay/feedback model.

    I like the example of an independent 40W lamp radiating into a 100W lamp. Now replace the 40W lamp with a snowglobe and imagine how how much you must increase the 100W to get 40W back from the passive snowglobe…with unconstrained convection. This will give you a sense of proportion for the work done via “back radiation”.

    I like the idea of replacing an insulating blanket with a torch to create an equivalent (slower) cooling profile (that’s clever and wise). Now imagine what the torch must do to increase the average temperature of the warm block by 33C. Good luck, my friends.

  218. AlecM says:

    kencoffman: as an engineer, what do you think of the modelling when it uses the two stream approximation yet only net IR can do thermodynamic work!

    Including a TOA error, the result is to increase the IR energy absorbed in the atmosphere from 23 W/m^2 to 134.5 W/m^2, a factor of 5.84. this is the cause of the imaginary positive feedback. They offset higher temperature by exaggerating cloud albedo.

  219. Ryan says:

    The light bulbs. You are both wrong. If 7% of the energy from a bulb is visible light, it is not true that the remaining 93% is infra-red. Furthermore even if it was it is irrelevant because the laws of entropy make it clear that if photons are obsorbed it doesn’t matter what the frequency they will both decay to heat in the end – i.e. visible light photons cause just as much heating as infra-red.

    The reason the hot bulb will not measurably heat the coller bulb is because the bulb has a filament which is probably at 3000Celsius. It is so hot that it is actually boiling single electrons from its surface. The gas in the bulb is inert, so it doesn’t capture these free electrons, but the gas does get seriously hot. The envelope of the bulb does get extremely hot but radiation has almost nothing to do with this because most of the photons are by definition going straight through. Because the envelope of the bulb gets so hot (but not as hot as the filament because the area is much bigger) there will be considerable heating of the air around the bulb and this will take away almost all the heat from the hotter bulb and hence most of the energy. With all this going on around both bulbs the relatively small amount of radiation transfer would be difficult to measure reliably. However, if you ramped down the current in the cooler bulb so it was no longer emitting in the visible spectrum but was still a little warm, you of course would see it was illuminated by the hotter bulb so clearly it is receiving (photon) radiation albeit at a difference frequency. But clearly it isn’t getting much energy through that route because if you put your hand in between the two you won’t feel the heat very much (you will if you put it just above the bulb).

    If you wanted to do the experiment properly you would want the bulbs to be in a vacuum like the old bulbs because then there is little conduction as such – but then the tungsten gets so hot it actually starts to boil off whole atoms onto the inside of the glass (so there is conduction in reality and the bulbs can’t run so hot anyway).

    What is happening on the microscopic scale in the conduction case is that whole atoms are violently vibrating in a solid and impacting the atoms in the less dense and free-flowing gas around it. In the case of radiation massless photons are being absorbed by electrons causing them to move to a slightly higher orbit around the atom – you can imagine that this latter mechanism really isn’t as effective as conduction in taking heat away from a cooling object.

  220. Ryan says:

    Willis: The reason you are getting confused is that black-body radiation is measured relative to absolute zero. So, in your example the desk is radiating at 400W. Does this mean putting another desk in front of it will cause 800W of radiation? No. The two desks are at the same temperature so there can be no heat transfer between them. Both are radiating at 400W which seems mighty energetic, but since they are both at that temperature there really isn’t any energy transfer between them (oh, and you won’t easily be able to distinguish between them using a thermal imaging camera either). They would be receiving photons from each other for sure (i.e. the 1st desk can see the photons of the brown second desk just as you can) but the net energy transfer is zero (the 2nd desk is emitting just as much visible light to the first desk)

    Bearing in mind what I said about conduction being a much stronger effect than radiation, what happens to happens to still air above a large body of water? The conduction of the heat from the water causes the air to eventually reach the same temperature as the water (note this cannot happens the other way around as heat in free-flowing gases and liquids rises and the water is much desner than the air). If they are at the same temperature due to conduction, how much energy transfer can there be by radiation? The answer is surely zero.

    That, Willis, is why the AGW crowd have got it all wrong. Because conduction occurs much more strongly at ground level than radiation, the air is much the same temperature as the land and sea due to conduction. If the air is much the same temperature as land and sea, there can be no net radiative transfer from land or sea to the air. If there is no net radiative transfer from land or sea to air near ground level, then CO2 in the air cannot make a difference to the radiative transfer. Radiation only exceeds conduction above the tropopause when the low density of the air means conduction cannot readily occur and then radiation becomes the only viable way of losing the heat to the dead cold of space.

  221. gymnosperm says:

    Well of course DLR warms the skin, but that’s about all it warms. If it didn’t the ocean WOULD freeze. What we have is an enormous eddy of energy (amounting to nearly 60% of the total incoming) cycling between the skin and water vapor in the lower atmosphere. This is precisely the part of the atmosphere where a few extra molecules of CO2 will be lost in the vapor equivalent of 13000 cubic kilometers of water the atmosphere contains on average.

  222. AlecM says:

    Ryan: what’s so incredibly wrong about the two-stream approximation is that the IPCC consensus claims there is 333 W/m^2 real energy flow from the lower atmosphere to the surface and 396 W.m^2 heading upwards, heating the air both ways as the IR is absorbed by GHGs.

    These data are obtained by pyrgeometers and the meteorologists have always believed them to be real energy flows. However, only 63 W/m^2 real energy is radiated and these people can’t imagine that a body at 16°C, the surface, does not emit 396 W/m^2.

    Yet this is only the case for an isolated body in a vacuum. in reality, as you show, the lower atmosphere is the same temperature and for the specific case of the 15 micron CO2 band, there is zero net radiation from the surface to be absorbed by the CO2 in the air.

    In other words there can never be any CO2-AGW. Yet these people go on to claim fantastic warming by AGW oblivious to the realities of heat transfer, as this article shows.

    What’s worse., they get the IR wrong: There is no possibility of any thermalisation even if there were CO2 IR to be absorbed. The IR that is absorbed is indirectly thermalised at clouds mainly, and much of that goes into grey body radiation, 22% of which from the top escapes directly to space.

    In short they have made 7 mistakes in basic physics, 3 of which are embarrassing to professionals like me because they have had to have been really dumb, or fraudulent. Everything that could go wrong has gone wrong, not least 1000s of pyrgeometers measuring the temperature of the atmosphere which is then put into the Energy Budget as real energy. You could not make up this level of incompetence nor the blind persistence in the face of real science in insisting there are no errors.

  223. Phil. says:

    Regarding light bulbs there are now higher efficiency incandescent bulbs where the glass envelope is coated by a dichroic layer which reflects IR back towards the filament. The dichroic passes visible however. This back reflected light heats the filament thus producing more light emission. The result of this is to be able to run the light bulb on lower current to produce the same visible output. It’s not practical to heat the filament to a higher temperature because it will melt but by recycling the IR output you get more visible out in lumens/watt.

  224. Willis Eschenbach says:

    AlecM says:
    December 17, 2012 at 4:18 am

    Willis; Firstly your warm desk. Put two pyrgeometers back to back between the desk and the sofa. The net signal will be near zero. This is a fact. A single pyrgeometer measures temperature expressed in power units. It is not a real energy flow.

    This is too good. You use a physical measuring device to measure one physically real and measurable signal in the thermal infrared band.

    You measure another physically real and measurable signal in the same band. They are about equal. As you would imagine, the difference between them is about zero.

    Your conclusion?

    Your conclusion from the fact that they are equal is that the signals that you measured are imaginary. They are not real. They do not contain energy.

    You realize how crazy you are sounding?

    If the signals are not real and they contain no energy, then what the hell were you measuring?

    Finally, again you are monomoniacally focussed on the mechanism, on the process. I couldn’t care less about that. I care about where the 340 W/m2 is going to come from to keep the ocean from freezing … and in that regard, you aren’t helping in the slightest.

    But don’t feel bad, neither is anyone else.

    w.

  225. Willis Eschenbach says:

    Mack says:
    December 17, 2012 at 5:35 am

    … When drawing your little pictures of the sun and earth on the blackboard what student is going to pick up on the little distance of the width of the atmosphere eh Willis. This is probably one reason your crap science has done so well. So get real Willis, The reality is that this earth receives 340 oddw/sq.m. at its surface (the sq,m metal plate in equatorial regions getting hot enough to fry eggs on ,remember?) get used to it and you can take Trenberths and your figures a shove em where there ain’t no w/sq.m.

    Mack, I’m sorry, but that abusive rant just cancelled your vote with me. I have no time to deal with jerkwagons.

    Buh-bye, don’t let the door hit you on the way out. You certainly may talk with others here, but your conversation with me is cancelled because of your sudden storm of extreme dickishness.

    w.

  226. Willis Eschenbach says:

    Ryan says:
    December 17, 2012 at 3:58 am

    Willis:

    “Simple energy calculations show that the incoming sunlight at the surface (global 24/7 average ≈ 160 W/m2) is not enough to balance the known losses from radiation, conduction/convection, and latent heat loss”

    Well, I for one don’t see why I need to justify myself any further. The incoming heat at the equator is not 160W/sqm. It is 1000W/sqm.

    What part of “global 24/7 average” are you not following? We’re not talking about instantaneous sun at the equator, we’re talking about the global oceanic energy balance.

    w.

  227. Ryan says:

    Willis: Sorry you are not right. When you are measuring the IR you are not measuring the radiated energy from the desk, say, to the thermal imaging camera. You are measuring the infra-red energy that WOULD flow if the thermal imaging camera was at -273Celsius (i.e. absolute zero). So no, it is not an energy you can sense. There is no net energy transferred from the desk to you, for instance, even if the measurement tell you that it is emitting 400W/sqm. There is a small amount of net energy transferred from you to the desk because you are ever so slightly warmer, but it would really be negligible. Similarly there can be no net transfer of photonic energy from the ocean to the air if the air is already at the same temperature, otherwise this would violate the laws of entropy.

  228. Willis Eschenbach says:

    TimTheToolMan says:
    December 17, 2012 at 5:11 am

    Willis writes

    “My question is, if your process is correct and the ocean is NOT absorbing 340 W/m2 from the DLR … then where is it absorbing that amount of energy from?”

    The surface of the ocean IS absorbing the energy in the top 10um or so…and radiating it back up at about the same rate (some may be used in evaporation) for no net positive effect in the bulk. That is the vital bit. There is no energy gain in the bulk or in the top 1 mm or anywhere due to the DLR.

    OK, so you agree that the ocean is in fact absorbing the energy, and that the energy is in fact subsequently being lost again in various forms … but then you go on to say:

    The ocean is cooling at the surface and the DLR is not warming it. Not even a little bit.

    Look, you have agreed that the ocean is absorbing the energy. As soon as it is absorbed it turns into heat … but now you claim that all that added heat doesn’t change the temperature at all.

    Energy can’t be destroyed. Every bit of the DLR that you agree is absorbed by the ocean is turned into heat.

    You seem to think that that heat does nothing because at equilibrium, an equivalent amount of heat is being lost … but if we took the DLR away, the ocean would cool rapidly and the mixed layer would end up at a much cooler temperature.

    So to summarize, it seems that you agree with me that:

    1. The ocean absorbs DLR.

    2. The ocean ends up warmer with the DLR than it would be without it.

    w.

    PS—Now that you have come out of the closet and joined the dark sidy by agreeing that the ocean indeed does absorb DLR, watch and see how fast your friends evaporate …

    PPS—Your contention. that the amount of energy added to the ocean is rapidly being lost, doesn’t meant that the DLR doesn’t warm the ocean. To understand why, consider, for example, a rock sitting in the sun. The rock is basically at thermal equilibrium.

    Note that this is identical to the situation you describe with the DLR in the ocean. The amount of energy added to the rock by the sun is being constantly radiated away. So just as with DLR and the ocean, the rock doesn’t warm.

    Where your logic goes off the rails is where you interpret that situation (the amount of energy entering the system is equal to the amount leaving, rock and ocean not warming) as meaning that the sun doesn’t warm the rock, or that the DLR doesn’t warm the ocean. It doesn’t mean that at all. Yes, amount of energy leaving the rock equals the amount entering, just like with DLR and the ocean. But the sun is still warming the rock, and DLR is still warming the ocean.

    To see that, just remove the sun. The rock cools. Same is true of DLR and the ocean.

  229. gymnosperm says:

    The 340 comes from DLW but it doesn’t go any deeper than the skin. It is cycling so fast that in a quantum sense it is the same energy. Like cycling money between your checking to savings accounts and wondering why you’re still not broke. The small (@ 10%) losses from the eddy have been more than compensated for during the satellite era by incoming SW.

  230. Ryan says:

    1. The ocean absorbs DLR.
    Correct.

    2. The ocean ends up warmer with the DLR than it would be without it.
    Wrong. Because the ocean is emitting at the same rate, being as it is at the same temperature as the air from which it is receiving radiated heat. There is no net flow of radiated heat from ocean to air at the boundary between the two. They are at the same temperature due to conduction. Conduction drives the process to the ocean and air until they are at the same temperature, and then the net radiation in both directions cancels out. In entropy terms they are at the same levels of disorder already due to conduction, so further changes in disorder due to radiation cannot occur.

  231. Ryan says:

    “To see that, just remove the sun. The rock cools.”

    The sun is much hotter than the rock, so it can heat the rock. The air is not much hotter than the rock, so it cannot heat the rock. In any case, the hot rock is busily heating the air by conduction until the two are the same temperature. If the rock is colder than the air it will remove heat from the nearby air until they are at the same temperature. Once this happens radiation cannot play a part at all.

  232. Ryan says:

    Willis: Last comment for the evening. Do you appreciate that just because the atmosphere might be emitting 340W/sqm this only is relevant for a black body at -273Celsius? I.e. only a black body at -273Celsius can absorb this power? The Earth is not black body at -273Celsius, therefore it cannot absorb 340W/sqm. The figure of 340W/sqm is almost completely irrelevant. Deep space is at -273Celsius so losing the heat to space it is relevant, but not for the Earth.

    Watts are a rate of heat transfer. If I have two bodies at the same temperature I don’t get a heat transfer so I don’t get any Watts to measure. Period. Doesn’t matter if it is conduction or radiation.

  233. Willis Eschenbach says:

    Ryan says:
    December 17, 2012 at 7:16 am

    Willis: The reason you are getting confused is that black-body radiation is measured relative to absolute zero. So, in your example the desk is radiating at 400W. Does this mean putting another desk in front of it will cause 800W of radiation? No. The two desks are at the same temperature so there can be no heat transfer between them. Both are radiating at 400W

    No, Ryan, the desk is not “radiating at 400W”. That statement is scientifically meaningless. Come back when you have figured out why, and we can continue the discussion.

    w.

  234. jae says:

    Willis:
    At 6:47 on 12/16 you wrote:
    “Well, yeah … except that one of them is in the desert and one is not. As a result, they have very different responses to downwelling IR. There is an interesting discussion of these issues here.”

    Well, as you say, “well, yeah…” The link doesn’t work so I don’t know what you are getting at. My point is exactly that one place is a desert and the other one is not!! Why is the desert so much hotter than the humid area, despite having only 1/4 the amount of the magic GHGs??

    Please check link as I would certainly like to go there. I’ve been bringing this same issue up for about 5 years now, and still have not had any success in reconciling this very simple empirical observation with the postulated radiative GHE.

    Thanks!

    [Link fixed, thanks. -w]

  235. AlecM says:

    Willis: ‘You measure another physically real and measurable signal in the same band. They are about equal. As you would imagine, the difference between them is about zero.

    Your conclusion?

    Your conclusion from the fact that they are equal is that the signals that you measured are imaginary. They are not real. They do not contain energy.

    You realize how crazy you are sounding?’

    I am not mad, just a rather better physicist than most..Climate science is full of people who look at Stefan-Boltzmann and imagine they know everything. So let me explain things in a different way.

    Your desk and you sofa are in radiative equilibrium. If you were to measure the energy arriving at a pyrgeometer sensor between them over 360°, it would be zero.

    Now put a metal shield between the sensor and the desk. The pyrgeometer signal immediately jumps to the temperature signal for the sofa, the S-B level in a vacuum. But there is still no radiative transfer between the desk and the sofa so there is no energy flow from the pyrgeometer sensor to the sofa despite the output meter for a pyrgeometer being in units of power.

    Do you get it now? Just because the pyrgeometer displays W/m^2, that signal is only true for transmission to absolute zero. The only reason you detect it is because the device shields the signal from the other direction. <It is an artefact which does not exist without the shield in place.

    This most basic failure to understand what a pyrgeometer measures has meant climate science is based on an entirely false premise and has been so for 50 years because the Meteorologists and now climate scientists, even mainstream physics in poor institutions are taught false science.

    Most of the IR energy in Trenberth’s Energy budget does not exist.

    If anyone else cares to comment on this please do because I must ram this message home; climate science has fallen at the first hurdle and the alchemists in it are beavering away repeating the same mistake over and over again.. That is the real definition of madness – at least I vary my teaching approach!

  236. Willis Eschenbach says:

    Ryan says:
    December 17, 2012 at 9:25 am

    Willis: Sorry you are not right. When you are measuring the IR you are not measuring the radiated energy from the desk, say, to the thermal imaging camera. You are measuring the infra-red energy that WOULD flow if the thermal imaging camera was at -273Celsius (i.e. absolute zero). So no, it is not an energy you can sense.

    Man, you are a regular font of misinformation.

    When you are measuring IR, you are measuring IR, nothing else. You can re-baseline your IR measurements all you want, you can convert them into temperature in C or in F, but you are indeed measuring the radiated IR energy from the desk.

    I also note that you don’t seem to understand that the temperature of the camera doesn’t change the flow of IR from the desk. The desk doesn’t know what the temperature of the camera is, it radiates the same amount of IR regardless of the temperature of the camera.

    In other words, what you call “the infra-red energy that WOULD flow if the thermal imaging camera was at -273Celsius” is exactly the same as the amount of energy that WOULD flow from the desk if the thermal imaging camera was at room temperature.

    w.

  237. Willis Eschenbach says:

    Ryan says:
    December 17, 2012 at 10:02 am

    Willis: Last comment for the evening. Do you appreciate that just because the atmosphere might be emitting 340W/sqm this only is relevant for a black body at -273Celsius? I.e. only a black body at -273Celsius can absorb this power? The Earth is not black body at -273Celsius, therefore it cannot absorb 340W/sqm. The figure of 340W/sqm is almost completely irrelevant. Deep space is at -273Celsius so losing the heat to space it is relevant, but not for the Earth.

    Watts are a rate of heat transfer. If I have two bodies at the same temperature I don’t get a heat transfer so I don’t get any Watts to measure. Period. Doesn’t matter if it is conduction or radiation.

    Ryan, please, I beg of you, quit while you are behind. You are making a fool of yourself. Almost none of the statements in those two paragraphs are true. Stop posting, put down the mouse, and nobody gets hurt … then buy yourself a thermodynamics text and learn some science before you post again. It’s painful watching you flail around.

    I gotta admit, though, I doubt that you could top this particular piece of misinformation:

    The Earth is not black body at -273Celsius, therefore it cannot absorb 340W/sqm.

    I can only shake my head in awe at the massive stock of scientific misunderstanding that you’ve managed to pack into that one short sentence.

    w.

  238. gymnosperm says:

    One must also account for the observation that the oceans continued to warm after 1997 without wayming the atmosphere. DLR by definition is absorbed by the atmosphere and would warm it equally. Further, one must account for the crazy quilt distribution of ocean surface warming and cooling Bob Tisdale has shown.

  239. AlecM, I am perfectly comfortable with superposition where you isolate the effect of multiple independent sources and sum them. It’s a little more complex when the sources are dependent, but we have a strategy for that too…small signal analysis with preset initial conditions. Weather and climate are state machines where the next state is determined by input signals and the recorded history. I wish, like an academic, I could simply invent forcings and transfer functions to suit my pet theory, but sadly, as an engineer, I’m bound to physical reality. Greenhouse gases increase the Earth’s surface temperature by an average of 33C. Don’t worry about how that’s thermo-mechanically possible, just take their word for it.

  240. AlecM says:

    kencoffman: greenhouse gases do not cause the Earths temperature to rise by 33K. That was the first big scam of Hansenkoism/Houghtonism!

    This is because adiabatic dry LR, Cp/g, is independent of the gas and because there is no mechanism for CO2 to increase water vapour concentration, there is no link with moist LR.

    The real GHE is ~ 9 K, the remaining 24 K from gravitational potential energy. The 9 K is the heating of the surface because emissivity falls as GHG thermal emission switches of specific wavelength bands!

  241. JazzyT says:

    AlecM,

    You’ve really managed to confuse yourself on the whole issue of Poynting vectors and Poynting’s theorem. It’s a perfectly good way of analyzing electromagnetic waves to find the net flow of energy through a location, and the work done on charges in that location, and the change in energy density at that location. Energy can be radiated by an object while it simultaneously absorbs radiant energy from elsewhere; add these up and you’ll get the net energy gained or lost. If you work with energy, that’s a scalar. If you want to work with Poynting vectors, you can do so; it’s not that useful. But, recognize that for for IR, the emitted and absorbed radiation do NOT cancel each other to leave only their vector sum. The vector sum will give the amount and direction of the NET energy flow at some point, including any work done on electric charges. But that doesn’t negate the components. By the same token, if you deposit your paycheck and pay your rent, and then balance your checkbook, that does not mean that your rent is suddenly unpaid and your paycheck was cut; it just tells you how much is left over for other things.

    Waves can add up and cancel; that’s destructive interference. They can also add up and reinforce, with constructive interference. Vector sums of electric fields are useful for this; but if you take the cross product of electric and magnetic fields to get the Poynting vector, you lose polarization information, and can’t analyze interference. Also, you’d need phase information, given by path length differences. Besides that, the Earth’s surface and atmosphere are not going to demonstrate interference anyway; as extended, thermal sources, they don’t exhibit the necessary spatial and temporal coherence.

    Poynting vectors are great for anyzing the interactions of EM waves and conductors. But for heat transfer by IR, you really need to start with an outlook based on quantum physics. When IR is emitted or absorbed, this is the interaction between molecules and photons. Only rarely is it helpful to consider the IR as an EM wave interacting with wires or other configurations of conductors. So, the Poynting vector approach isn’t all that useful for heat transfer by IR. It’s still valid, as long as you don’t go tripping yourself up by confusing the vector sum of Poyning vectors at a point with the actual propigation of waves beyond that point, or confusing time-dependent vs. time-independent Poynting vectors, or similar confusions.

    Here’s somthing to sum it up:

    AlecM says:
    December 17, 2012 at 1:00 am

    Willis: ‘If DLR is not keeping the ocean from freezing, what is?’

    The answer is very simple. The oceans do not emit 400 W/m^2 ULR. The only IR they emit is the net 60 W/m^2. This is an inescapable fact – only net energy flows are real.

    No, both energy flows are real. IR from the surface can transfer energy to the atmosphere; IR from the atmosphere can transfer energy to the surface. Note, I said “transfer energy,” not “warm.” It’s only a warming process if the net heat transfer into an object is positive. But any object that radiates without absorbing at least an equal amount of radiation, will cool.

    All energy flows can transfer heat, but only NET energy flows can give NET heat transfer. So, in the above, the net energy flow represents the net heat loss by the ocean, which is given by the difference between IR emitted upward and downward IR absorbed.

    If you say that “the problem with this whole analysis is that they physicists don’t understand the physics,” then you’d better be a world-class physicist yourself. Otherwise, you probably missed something.

  242. AlecM says:

    JazzyT: for a collimated beam, the Poynting Vector is identical to the Planck Irradiance Function at that wavelength.

    The vector sum of the PVs UP and Down is simply a wavelength dependent version of the S-B1 – S-B2 calculation………..

  243. D.I. says:

    JazzyT says:
    “If you say that “the problem with this whole analysis is that they physicists don’t understand the physics,” then you’d better be a world-class physicist yourself. Otherwise, you probably missed something”.
    So what is a ‘World Class Physicist’ ? Is It some-one who has discovered a ‘New Law’ or some-one who is Popular for reciting ‘Old Laws’?
    What is your definition of ‘World Class’?.
    Just curious.

  244. denniswingo says:

    I also note that you don’t seem to understand that the temperature of the camera doesn’t change the flow of IR from the desk. The desk doesn’t know what the temperature of the camera is, it radiates the same amount of IR regardless of the temperature of the camera.

    In other words, what you call “the infra-red energy that WOULD flow if the thermal imaging camera was at -273Celsius” is exactly the same as the amount of energy that WOULD flow from the desk if the thermal imaging camera was at room temperature.

    If the camera was at a temperature lower than the desk energy would flow in the direction of the camera. This would be an extremely small amount of energy as the desk is radiating in other directions proportional to the area that the desk “sees”.

    If the camera were hotter than the desk then energy would flow in the direction of the desk, proportional to the area that the camera sees, also providing that all other energy sources are at a lower temperature than the camera.

    This is just basic thermodynamics.

    Thuswise, if the ocean is warmer than the sky/atmosphere it is radiating toward, it will radiate energy. It does not matter how much downwelling IR there is, from CO2 or a cloud, if the temperature of the ocean is greater than what it is looking at it will radiate energy and thus cool. It will also conduct energy away through molecular interaction with the atmosphere. There will also be convection that will speed the process along. However, to answer your question is that in order for the ocean to freeze, enough energy at the surface has to radiate away so as to bring the temperature at the surface down to freezing at -2c. Even then 80 more calories per cubic cm of energy has to radiate away in order to remove enough energy to support the phase change from water to ice.

    The ocean, if its temperature is less than the sky/atmosphere, will absorb energy from the atmosphere/sky. This is just basic thermodynamics.

    In the Arctic and Antarctic there is a phenomenon called flash freezing whereby the rough water is continually radiating energy away due to the greater surface area of the water. The temperature can go well below -2c at the surface of the water. Then when the wind dies, tens of square kilometers of water will freeze (phase change to ice) in seconds. Google this, there are videos of it out there some where.

  245. Greg House says:

    Willis Eschenbach says, December 17, 2012 at 12:24 am:
    Greg, [...] I see that in fact you agree that everything radiates.
    It appears that you believe, however, that when radiation from something cooler is absorbed by something warmer than the spot where the radiation originated, no energy of any kind is transferred to the warmer object.
    Is this in fact your belief? And if so …
    [...]Put a 100-watt incandescent bulb in a fixture, [...] Me, I can do it as a thought experiment, I don’t need to build the apparatus to be clear about what will happen.

    ========================================================

    No, Willis, there was absolutely nothing in my comments that suggested what you said about my “belief”. But nevertheless thank you for this nice example of a straw-man argument.

    You “thought experiment” is only an illustration of how you think or (want to make people to believe) how things work. It is not real, it is a product of fantasy and not a scientific argument at all. Your assertion about “NET” you made previously is not science, Willis, it is a fiction, because unproven experimentally. At best you can call it an “experimentally unproven hypothesis”, but not a scientific fact.

    The fact that a colder body radiates in a warmer body’s direction ALONE does not prove any particular effect on this warmer body including effect on the warmer body’s TEMPERATURE.

    Apparently, neither you nor warmists I talked to on this and other blogs can present a REAL experiment confirming that notion about “NET”. Which inevitably leads to the conclusion that there has never been one and the whole notion is just a fiction.

  246. Mack says:

    Yeah, Willis calls it a day with me because he can’t handle the kitchen heat.
    A slight improvement in the wording on my last posting to him Dec.17th 2012 @ 5.35 am……..
    “Sort of one foot in reality and the other in an AGW fantasy” better read as…….
    “Sort of one foot in reality and the other in a model of AGW fantasy”
    These academic teaching twats drawing pictures of the sun and earth on the blackboard fail to realise they are already modelling. They fail to realise they are in a classroom and not in the real world making real observations and taking real measurements.
    Science thus becomes abused …and Willis here is still abusing it.

  247. Willis Eschenbach says:

    Mack says:
    December 17, 2012 at 4:58 pm (Edit)

    Yeah, Willis calls it a day with me because he can’t handle the kitchen heat.

    Oh, please. I’m unable to make a dent in your lack of knowledge, so I don’t plan to try any further. Your assertions are not simply un-physical, they are a-physical. Please re-read what JazzyT says above. Truly, you don’t understand what you are talking about.

    w.

  248. Willis Eschenbach says:

    denniswingo says:
    December 17, 2012 at 1:57

    [Willis said:]

    I also note that you don’t seem to understand that the temperature of the camera doesn’t change the flow of IR from the desk. The desk doesn’t know what the temperature of the camera is, it radiates the same amount of IR regardless of the temperature of the camera.

    In other words, what you call “the infra-red energy that WOULD flow if the thermal imaging camera was at -273Celsius” is exactly the same as the amount of energy that WOULD flow from the desk if the thermal imaging camera was at room temperature.

    If the camera was at a temperature lower than the desk energy would flow in the direction of the camera. This would be an extremely small amount of energy as the desk is radiating in other directions proportional to the area that the desk “sees”.

    If the camera were hotter than the desk then energy would flow in the direction of the desk, proportional to the area that the camera sees, also providing that all other energy sources are at a lower temperature than the camera.

    This is just basic thermodynamics.

    Thanks, dennis. I’m not sure if we disagree or not. I think the difference is you are referring to the net energy flow while I am also referring to the individual energy flows that are measured in order to determine the net energy flow. Let me try again.

    The desk does not know the temperature of the camera. It radiates in all directions. It radiates about the same amount whether the camera is cold or hot.

    You are correct that if the camera is colder than the desk, there will be a net flow of energy from the desk to the camera. But this is NOT because of an increase in the energy flow from the desk to the camera. That stays the same, because the temperature of the desk stays the same.

    To the contrary, it is because of a decrease in the flow of energy from the camera to the desk. Because the camera is now cold, it radiates less energy.

    Because the flow of radiated energy from the camera to the desk is now smaller than the flow of radiated energy from the desk to the camera, the net flow of energy (Q1 minus Q2) is from the desk to the camera. As you point out, net flows of energy always go from warm to cold.

    It’s not just a good idea.

    It’s the Law.

    w.

  249. Mack says:

    I thought you were not talking to me Willis,
    Trenberth btw is a kiwi and “works” in the USA. I too am a kiwi an as far as I’m concerned you can b— well keep him.

  250. jae says:

    Willis: Did you forget me again? Waiting for The Link….

  251. Mack says:

    Also it might interest you to know Willis that Ernest Rutherford (google him up if you don’t know) was a kiwi born in this very town I live. So what has the world become when a Science Nobel Prize has become as worthless as an Oscar ,Willis. How do I compare Trenberth to Rutherford ? Don’t bother to answer because it was too easy even without the clue that one has (wittingly or unwittingly) denigrated science.

  252. JazzyT says:

    AlecM says:
    December 17, 2012 at 12:54 pm

    JazzyT: for a collimated beam, the Poynting Vector is identical to the Planck Irradiance Function at that wavelength.

    The vector sum of the PVs UP and Down is simply a wavelength dependent version of the S-B1 – S-B2 calculation………..

    These two statements are true. But, you seemed to be saying that once you had the vector sum of upward and downward PVs, that the resulting PV would represent a beam that was left after summing. That’s not true. You still have two beams (or really, some complicated wavefronts) at every point, including where the photons are emitted: the surface of the Earth, and the atmosphere. The resultant Poynting vectors simply give you the difference between upward and downward waves. This tells you about the net energy flow. This will be large, since it has to equal, on the average, the energy brought in by sunlight. This resultant PV can also be included in calculations about how part of the atmosphere is heated or cooled. But, it still represents energy flowing in different directions; it’s magnitude tells you about imbalances in incoming energy from different directons, and which way the NET energy flow is directed. It doesn’t mean that all the energy ends up actually going in that direction. The individual vectors partially cancel, but the photons, and wavefronts, themselves don’t annihilate.

  253. denniswingo says:

    You are correct that if the camera is colder than the desk, there will be a net flow of energy from the desk to the camera. But this is NOT because of an increase in the energy flow from the desk to the camera. That stays the same, because the temperature of the desk stays the same.

    To the contrary, it is because of a decrease in the flow of energy from the camera to the desk. Because the camera is now cold, it radiates less energy.

    Nope, the camera which was defined as colder is still radiating toward the desk, it is just that the net flow will aways be in the direction of the warmer to the colder object.

    Lets get away from simplistic examples.

    You have two stars in a double star system. one star is a red giant with a surface temperature of 5000 degrees K. The diameter of the star is 1 Astronomical unit (140 million kilometers)

    The second star is a main sequence star like the sun with a diameter of 0.06 AU with a surface temperature of 10,000 degrees K.

    The net energy flow between the two will ALWAYS be from the main sequence star to the red giant star due to the difference in temperature. This is true even though the total energy radiated by the red giant is orders of magnitude higher than from the main sequence star.

    Lets go to the satellite world.

    I have a flat spinning disk in orbit around the Earth with its flat face pointing toward the Earth (nadir), with its opposite space pointing toward Deep space (zenith). The altitude is 150 statute miles so that the nadir face sees only the Earth. The temperature of the Earth as it radiates energy to space is about 20 degrees C (this is a well known quantity in spacecraft design). The temperature toward Zenith (deep space) is the 4 degrees kelvin of the background of the universe. The disk is black on the nadir and the zenith face and for this example we ignore the edges.

    What happens to the temperature of the disk?

    Answer that question. The answers will surprise you.

    Next consider the case of the disk made from metal that is a perfect thermal conductor and the opposite case of a disk made of aerosol, a perfect insulator.

    If you can answer these questions correctly you understand the subject.

  254. JazzyT says:

    D.I. says:
    December 17, 2012 at 1:25 pm

    JazzyT says:
    “If you say that “the problem with this whole analysis is that they physicists don’t understand the physics,” then you’d better be a world-class physicist yourself. Otherwise, you probably missed something”.
    So what is a ‘World Class Physicist’ ? Is It some-one who has discovered a ‘New Law’ or some-one who is Popular for reciting ‘Old Laws’?
    What is your definition of ‘World Class’?.

    First, I’d say that if you’re going to buck the conventional wisdom and tell the physicists who buy into it that they’re wrong, you’d better be absolutely, completely familiar with the current state of the art, and also its history, so you don’t end up misinterpreting something or bringing up an apparent paradox that’s already been hashed out. “Discovering a ‘New Law'” is one thing that a physicist might do by bucking the trend, although invalidating an old law, or established theory, would be enough. It’s a lot easier to get people to buy in to the idea that the old law was wrong if you have a new law with greater explanatory power, e.g., relativity, quantum mechanics, QCD. Or, you could come up with a law explaining things that nobody really thought about before, predicting effects which then show up in observations. Thomas Young did this in the early 1800s, with his wave theories, followed by his famous two-slit experiment that showed the wave nature of light. Still, if one could simply point out a major flaw in accepted theory, and then, armed with a deep and detailed knowledge of that theory, convince everyone that this really was a fundamental flaw and not just some manageable complication, then that would be a way of “telling the physicists that they don’t understand the physics.” Anyone who adds a new law, or convincingly undermines an old one, has made a major contribution, and that would qualify them as a “world class physicist.” But it’s very unlikely that one could do that without having a deep and detailed understanding of the known physics, which would at least qualify as “highly competent,” and would fit a broader definition of “world class” (those who haven’t made that major contribution but could, given the opportunity).

    Physicists who are “popular for reciting ‘old laws'” might very well have that level of knowledge, though that might not be necessary, depending on what exactly they “recite.” But if that’s all they do, then, if they “recite” these laws correctly, i.e., the way other physicists understand them, then, by definition, they won’t be telling other physicists that they are wrong.

    Kind of a long answer, I’m afraid, but I liked the question, since it made me think it through.

  255. Ryan says:

    OK Willis, I’m getting tired of your snappy attitude. If you are so sure that the DLR is what prevents the oceans from freezing here is a simple experiment for you:

    Take a mirror about 1sqm. You will accept that a mirror is a higly efficient reflector of photonic radiation, correct? Put a red object in front of it and it will show up just as red in the mirror, right? So it isn’t a big stretch to assume that a fair amount of infra-red gets reflected in much the same way, right? And you’d be right, because the metal coating on a highly polished mirrored surface has a sea of densely packed electrons at its surface that frankly isn’t interested in absorbing photons and simply throws them straight back at you. We use this fact in vacuum flasks, patio heaters, electric fires etc.

    OK, so now take a bucket of water and put put it out in the garden on a warm night with the mirror on top, the mirrored surface pointing out to space. According to you, Willis, the bucket of water will freeze, because it is now not getting any heat from the sun or from this DLR which you claim is giving out almost as much heat as the sun during the night, because the mirror will reflect it all back into space with almost 100% efficiency.

    I am pretty confident that the bucket of water won’t freeze. No amount of your curt answers will change that. But go ahead, knock yourself out and prove otherwise.

  256. AlecM says:

    JazzyT: I notice that you imagine photons are part of the initial Poynting Vectors. I suspect this is not the case and here’s my reasoning.

    Planck invented the photon as a throwaway concept. He never liked it. It represents the transient when the quantum of energy changes state between the mechanical vibrational and the EM worlds.I don;t believe that the individual PVs have an connection to photons until the net vector forms!

  257. Ryan says:

    JazzyT: I for one am not claiming the physics is wrong. Willis isn’t actually arguing what the physicists are arguing. He seems to be arguing for the over-simplistic model created by Kevin Trenberth (presumably never reviewed by a physicist).

    The physicists make it clear that at the surface the atmosphere is effectively opaque to thermal radiation. Heat transfer is by conduction and convection. It is only above the tropopause where thermal radiation makes a difference. In effect, what the physicists are saying is that because the rate at which the planet tries to cool into the stratosphere is slowed down by adding more CO2 then the stratosphere will get warmer. Then, because the temperature gradient between the warmer stratosphere and the surface would be smaller, the processes of convection in the troposphere are slowed down as well so the troposphere warms up too.

    No physicist is claiming that the CO2 causes photons to zap from the nightime warm atmosphere directly into the ground (or ocean) heating it up. Willis has got that completely wrong but it isn’t his fault because Team AGW are passing around so many bad overly simplistic models. There is a fairly good explanation on Wikipedia, although it still has some errors due to lax language.

  258. Mack says:

    Ryan,
    I’m noticing Trenberth is washing his hands of the latest Earth’s Energy diagrams and leaving younger brainwashed minions to take over. Probably having an edgy feeling about being associated with the latest Train-drivers Guide To Science.

  259. Ryan says:

    Mack: Talking of Energy diagrams the Wikipedia page on the Greenhouse Effect has quite a good explanation of the physics under “Mechanism” I wouldn’t take issue with too much (I would only take issue with the impact at ground level due to a piddling amount of heating in the stratosphere). However, the energy diagram in the top right of the page is hilarious – it shows twice as much heating energy at ground level as is coming from the sun in the first place! I’m guessing that diagram is a Trenberth special.

  260. Greg House says:

    Ryan says, December 18, 2012 at 2:02 am: “[...]OK, so now take a bucket of water and put put it out in the garden on a warm night with the mirror on top, the mirrored surface pointing out to space. According to you, Willis, the bucket of water will freeze, because it is now not getting any heat from the sun or from this DLR which you claim is giving out almost as much heat as the sun during the night, because the mirror will reflect it all back into space with almost 100% efficiency.
    I am pretty confident that the bucket of water won’t freeze. No amount of your curt answers will change that. But go ahead, knock yourself out and prove otherwise.”

    =======================================================

    Ryan, if I was a crazy warmist, I would say: “of course, the bucket of water won’t freeze because of back radiation from the other side of the mirror pointing out to the bucket”. Got you.

  261. Robert Clemenzi says:

    Ryan says:
    December 18, 2012 at 2:02 am

    OK, so now take a bucket of water and put put it out in the garden on a warm night with the mirror on top, the mirrored surface pointing out to space.

    I am pretty confident that the bucket of water won’t freeze.

    Instead of a bucket, try a Styrofoam container. And use two of them, one with a mirror and one without. In the morning, see if the temperatures are the same. Be sure to use a clear night. No clouds. The effect will not be very large on a warm night with high humidity. I suggest a night between 35F and 40F.

    The main problem with your experiment is that both sides of an optical mirror are very good IR mirrors. Actually, the metal coating will have no effect since the glass itself is IR opaque. (Unless you are using a first surface mirror, of course.)

    In orbit, when a satellite enters the Earth’s shadow, its outside temperature drops below freezing in about one minute. Something causes it to take a bit longer on the Earth’s surface. Perhaps you can explain why.

  262. Mack says:

    Ryan,
    I was wondering if I could get through to Willis in a more simple and direct way,
    See this picture here………
    http://www.ipcc.ch/graphics/ar4-wg1/jpg/faq-1-1-fig-1.jpg
    The incoming solar radiation should be about 1360w/sq.m. NOT about 342w/sq.m.
    How wrong can numbers be in science to be wronger than this.

  263. Mack says:

    Maybe “wronger” should be “more wrong” :)

  264. Robert Clemenzi says:

    Mack says:
    December 18, 2012 at 10:59 am

    The incoming solar radiation should be about 1360w/sq.m. NOT about 342w/sq.m.

    342 * 4 = 1368

    The area of a circle is πR^2
    The area of a sphere is 4πR^2

    The Earth absorbs radiation as a circle and emits it over the area of a sphere. As a result, when averaged over a period of 24 hours, the average energy is one fourth the peak.

  265. Greg House says:

    Robert Clemenzi says, December 18, 2012 at 11:37 am: “The Earth absorbs radiation as a circle and emits it over the area of a sphere.”
    =======================================================

    A circle is flat, Robert. Last time I checked the Earth was not flat.

    Why do you need to reduce our mother Earth to a circle?

  266. D Böehm says:

    Greg House,

    You’re nitpicking. I’m sure you know Robert was referring to a disk.

  267. Greg House says:

    D Böehm says, December 18, 2012 at 1:36 pm: “I’m sure you know Robert was referring to a disk.”
    =====================================================

    The same goes for a disk. Last time I checked the Earth was not a disk.

  268. richardscourtney says:

    Greg House:

    Don’t be silly. The Earth occludes a disc. This thread is too important for you to waste space on it with nonsense.

    Richard

  269. Gary Pearse says:

    Greg, the earth is not a disk but for reception of the sun’s rays it essentially presents a disk. The square metres of the earth’s surface receive diminishing radiation from the sun as you move away in all directions from the point (centre of the disk) where the sun’s rays meet the earth at right angles. Clearly you can see that at the outside edges of the disk, there is very little energy impinging on the square meterage. Indeed, if you go around to the dark side, their is no sun’s energy impinging on that side of the sphere – as far as the sun is concerned the dark side is the other side of the disk. Robert’s observation is an accurate and cogent one that diserves repeting:

    “The Earth absorbs radiation as a circle and emits it over the area of a sphere.”
    Robert Clemenzi says: December 18, 2012 at 11:37 am

  270. farmerbraun says:

    Yes it would be nice to see a little more elucidation and less of the . . . whatever.
    We seemed to have sorted the semantic problem with “NET”: perhaps we can now return to the argument.

  271. Greg House says:

    richardscourtney says, December 18, 2012 at 1:53 pm: “Don’t be silly. … to waste space on it with nonsense”
    ========================================================

    I see a sign of desperation on your side, Richard.

    Coming back to Robert Clemenzi’s formulation, it was simply wrong. According to reality, the right formulation would be “the Earth absorbs radiation as a sphere and emits it as a sphere”. If we consider the earth to be a sphere, of course, what we do. And if we remember that this sphere is rotating.

    This might make some people unhappy, I understand that. But a sphere is a sphere and not a disk.

  272. Greg House says:

    Gary Pearse says, December 18, 2012 at 2:06 pm: “Greg, the earth is not a disk but for reception of the sun’s rays it essentially presents a disk. The square metres of the earth’s surface receive diminishing radiation from the sun as you move away in all directions from the point (centre of the disk) where the sun’s rays meet the earth at right angles. Clearly you can see that at the outside edges of the disk, there is very little energy impinging on the square meterage. Indeed, if you go around to the dark side, their is no sun’s energy impinging on that side of the sphere…”
    ==================================================

    Gary, I am well aware of this sort of argumentation and take it seriously. I hope you and others would take the counter-argumentation equally seriously.

    The first problem you encounter is the handling of the premiss “the incoming solar radiation should be about 1360w/sq.m.”. A “sq.m.” is always a “sq.m.”. Hence a hemisphere would receive 2 times as much Watt as the disk of the same radius. Sorry, but this is a simple math and a simple logic.

  273. richardscourtney says:

    Greg House:

    re you post to me at December 18, 2012 at 2:20 pm.

    There is no “desperation” in pointing out that you are wrong. You could have thanked me (and others) for pointing it out.

    I will not reply to any more nonsense you address to me because your nonsense is not worth the bother. And if you think me rude for not answering further posts you address to me then I could not care less.

    Richard

  274. Robert Clemenzi says:

    Greg, assume that there is spherical shell around the Sun with the same radius as the distance between the Sun and the Earth. The inside of that shell receives about 1368 W/m2 of electromagnetic radiation.

    It is fairly obvious that the side of the Earth not facing the Sun receives none of that. It is also obvious that when the Sun is low in the sky, the light is spread out over a larger area than when it directly over head. The idea is to come up with a formula that takes all this into account and determines the average amount of energy being absorbed by the planet as a whole.

    When looking at the Earth from the Sun, the amount of energy available to be absorbed is equal to 1368 W/m2 times the area of a disk with the same radius as the spherical Earth. However, when you consider a period of 24 hours, that energy is distributed over the entire surface of a sphere. That is what the equations produce, they average the amount of energy available on one side of a sphere over the entire surface of the same sphere.

  275. JazzyT says:

    AlecM says:
    December 18, 2012 at 3:24 am

    JazzyT: I notice that you imagine photons are part of the initial Poynting Vectors. I suspect this is not the case and here’s my reasoning.

    Planck invented the photon as a throwaway concept. He never liked it. It represents the transient when the quantum of energy changes state between the mechanical vibrational and the EM worlds.I don;t believe that the individual PVs have an connection to photons until the net vector forms!

    Planck didn’t even use photons; he just found that if he assumed that the actual emission and absorption had to take place in little packets, and then the math worked out. This avoided the so-called “ultraviolet catastrophe” wherein the emission spectrum, in theory, increased without bound at shorter wavelengths.

    In 1905, Einstein expanded on this idea to explain the photoelectric effect: by postulating that light itself was quantized, with the energy of each photon proportional to its frequency. High frequency photons could eject electrons from a piece of metal, perhaps with energy left over. Below some threshold, lower-frequency photons could not, even if you had a lot of them. Maxwell’s equations couldn’t explain this. Over time, Einstein’s hypothesis gained support, including a Nobel prize in 1921. (Relativity was part ot that too, but soft-pedaled because it still lacked empirical evidence.) Once Arthur Compton bounced photons off of electrons like billiard balls, and demonstrated the corresponding energy and momentum transfer in both, the photon idea was pretty much universally accepted. Compton was awarded a Nobel prize for this, too. At higher energies, it’s relatively easy to see the effects of single photons. In some gamma rayimaging equipment, you can watch photons arrive one at a time; or you can hear them as they’re detected in a Geiger counter.

    The Poynting vector of a wavefront is an abstraction, and in constructing it, you lose information about the electric and magnetic fields of that wavefront. The resultant Poynting vector added from several others doesn’t necessarily give the direction of any wavefront. To think otherwise would suggest that you could combine two flashlight beams, or laser beams, at 90 degrees, and get a single beam coming off at 45 degrees from each of the original ones. Or, that you could park two cars facing one another, shine the headlights of each right into the headlights of the other, and make them both go dark, or at least dim. This seems to be like what you are suggesting for IR; if it would work for IR, why not for visible light?

    You can stick with a Maxwell/Planck/Poynting interpretation if you like. But modern physicists tend to think that emission and absorption of photons is the best way to describe radiative energy transfer. Pyoynting vectors have their place, but that place is limited. If you have a comprehensive refutation of Einstein and Compton, not to mention Heisenberg and many others, that will convince all or most physicists working with radiation transport, I’ll look forward to seeing it.

  276. JazzyT says:

    in
    December 18, 2012 at 3:24 pm
    “best way to describe energy transfer” should be “best way to describe radiative energy transfer.”

    [FIXED. -w]

  277. Greg House says:

    Robert Clemenzi says, December 18, 2012 at 2:56 pm: “Greg, assume that there is spherical shell around the Sun with the same radius as the distance between the Sun and the Earth. [...] When looking at the Earth from the Sun, the amount of energy available to be absorbed is equal to 1368 W/m2 times the area of a disk with the same radius as the spherical Earth. However, when you consider a period of 24 hours, that energy is distributed over the entire surface of a sphere.”
    ========================================================

    Robert, about “W/m2″, please, go back and read my previous comment (December 18, 2012 at 2:35 pm). The argument there is actually sufficient. But I do not mind to talk about other details.

    As for 24 hours, the sphere makes one full rotation, hence it is not that only a hemisphere (or only one side of an imaginable disk, for debate’s sake) receives radiation, it is the whole sphere.

  278. Willis Eschenbach says:

    Greg House says:
    December 18, 2012 at 3:53 pm

    Robert, about “W/m2″, please, go back and read my previous comment (December 18, 2012 at 2:35 pm). The argument there is actually sufficient. But I do not mind to talk about other details.

    OK, the argument there was:

    The first problem you encounter is the handling of the premiss “the incoming solar radiation should be about 1360w/sq.m.”. A “sq.m.” is always a “sq.m.”. Hence a hemisphere would receive 2 times as much Watt as the disk of the same radius. Sorry, but this is a simple math and a simple logic.

    You go on to say:

    As for 24 hours, the sphere makes one full rotation, hence it is not that only a hemisphere (or only one side of an imaginable disk, for debate’s sake) receives radiation, it is the whole sphere.

    I’m not sure what the fuss is about here.

    The amount of solar energy available at the top of the atmosphere is called the “TSI”, the total solar irradiance. It is expressed in units of watts per square metre perpendicular to the sun’s rays.

    It is usually taken to be about 1366 W/m2.

    How much of that hits the earth? Well, that would be calculated as the total solar irradiation, TSI, times the area of the sun’s rays intersepted by the earth, which is Pi r^2.

    TSI (Watts/m2) * Pi r^2 (m2) = total intercepted energy (watts)

    where “r” is the radius of the earth. The units of square metres cancel out, so the answer is in watts. This gives us the total amount of solar energy intercepted by the earth.

    Now, what does that work out to be as a per-square-metre average over the entire earth’s surface? To figure that out, we divide total intercepted energy by the area of the earth’s surface, which is 4 Pi r^2

    intercepted energy (watts)
    —————————————- = average energy (W/m2)
    surface area (square metres)

    Substituting for the numerator and the denominator, we get

    TSI (Watts/m2) * Pi r^2 (m2)
    —————————————- = average energy (W/m2)
    4 Pi r^2 (square metres)

    The Pi r^2 cancels out, leaving

    TSI (Watts/m2)
    —————————————- = average energy (W/m2) = 1366/4 ≈ 342 W/m2
    4 (m2)

    So as a global average, it is common to say that the solar irradiation is 342 W/m2. This is not the total sun available at a particular time and place. Heck, at the equator at noon, you get over a kilowatt per square metre at the surface. But it is the global average sun available at the top of the atmosphere.

    Is this global average a useful number? Sure, for things that it is useful for … but you don’t want to use it in situations where it is not useful.

    Am I missing something? Like I said, not sure what the fuss is about.

    Best to all,

    w.

  279. Ryan says:

    Hmmm, Take the average of the lengths of a rectangle 1cm x 9cm and square the average to find the area, then you’ll realise just why playing with averages is not a good idea in non-linear systems.

  280. Ryan says:

    Robert Clememzi: Use whatever mirror you like Robert. Make it just a polished metal surface if you want. Put a nice absorbent black backing on the other side and allow the heat to conduct out from the bucket. No problem. Entirely up to you. Not sure glass is quite so opaque to IR as you imagine – perhaps you were thinking of UV?

    Spacecraft? Perhaps we should wrap them in bubble-wrap filled with CO2? Just a thought? Or maybe they only need 200km of atmosphere around them to keep them warm. Anyway, I’m off to take the radiator out of the car because someone told me that painting the engine white would be all the heat dissipation I need and then I’m going to write to Intel to tell them to take the fans of their processors and put them in clear glass packages to let the IR out.

  281. Greg House says:

    Willis Eschenbach says, December 18, 2012 at 4:42 pm: “The amount of solar energy available at the top of the atmosphere is called the “TSI”, the total solar irradiance. It is expressed in units of watts per square metre perpendicular to the sun’s rays. It is usually taken to be about 1366 W/m2. [...] Am I missing something?”
    ============================================================

    Yes, you are: my point.

    As I said, a square meter is always a square meter, regardless the form of the surface. Given the premiss of 1366 W/m2, the surface of a hemisphere would have 2 times as much square meter as the surface of a disk of the same radius, hence a hemisphere would receive 2 times as much Watt as the disk of the same radius.

    To your “perpendicular”, this is not scientific at all, Willis, because in units like “watts per square metre” or “whatever per square metre” the “square metre” is not a geometrical figure and can not be perpendicular to anything.

    Hence, based on he premiss of 1366 W/m2, a hemisphere would receive 2 times as much Watt as a disk of the same radius. Sorry again, but this is a simple math and a simple logic.

    If your calculation ignores that, and it does, it is simply wrong.

  282. Willis Eschenbach says:

    Greg, I’m sorry, but I couldn’t follow that at all.

    I’ve given it to you in simple math. If you can find a problem with my math, let me know. It’s not complex math, it’s very straightforward.

    If you can’t find any problems with my simple math (average solar energy per square metre equals total solar energy available divided by total square metres of Earth surface area, it’s not rocket science), please don’t send further explanations and imaginations of this type about square metres and the like, I can’t make sense of them.

    Regarding your curious aversion to the concept of “perpendicular”, I just looked up what NASA says on the subject:

    In addition, the total solar irradiance is the maximum power the Sun can deliver to a surface that is perpendicular to the path of incoming light.

    See that part about “perpendicular to the path”? That’s all I was saying. Sorry for the confusion.

    w.

  283. Mack says:

    It’s amazing how everybody reverts to hypothetical unreal geometrical calculations? Newsflash. The sun is a big ball of fire in the heavens . It is real and the measured flux,power whatever (its irrelevant because it’s timeless) is 1360oddw/sq.m

  284. Willis Eschenbach says:

    Mack says:
    December 18, 2012 at 5:54 pm

    It’s amazing how everybody reverts to hypothetical unreal geometrical calculations? Newsflash. The sun is a big ball of fire in the heavens . It is real and the measured flux,power whatever (its irrelevant because it’s timeless) is 1360oddw/sq.m

    My goodness. Simple arithmetic is now “hypothetical unreal geometrical calculations”.

    Mack, these kinds of calculations were developed by the Egyptians for very practical reasons.

    Yes, the sun provides 1360-odd watts, as you say.

    This is different from the average amount of solar energy received by a square metre of the surface of the Earth.

    That is calculated as total energy divided by total surface area.

    If you truly think that something as simple as energy divided by surface area is a “hypothetical unreal geometrical calculation”, I fear I don’t know how to assist you, other than to suggest that while the condition persists you stay out of scientific discussions, and that for your own cranial safety you avoid subjects like “normal uniform rational B-spline curves”.

    w.

  285. Greg House says:

    Willis Eschenbach says, December 18, 2012 at 5:45 pm: “If you can find a problem with my math, let me know. [...]
    Regarding your curious aversion to the concept of “perpendicular”, I just looked up what NASA says on the subject: “In addition, the total solar irradiance is the maximum power the Sun can deliver to a surface that is perpendicular to the path of incoming light.”

    =============================================================

    Right, theY say that; thus contradicting what they also say on the same page above: “At Earth’s average distance from the Sun (about 150 million kilometers), the average intensity of solar energy reaching the top of the atmosphere directly facing the Sun is about 1,360 watts per square meter, according to measurements made by the most recent NASA satellite missions. This amount of power is known as the total solar irradiance”.

    Note, by the way: one time the total solar irradiance (TSI) is an average, then it is a maximum. An average can not be a maximum at the same time.

    I guess, what they measured was indeed an average, and this average includes the whole hemisphere. Hence, this 1360 W/m2 has nothing to do with a perpendicular. I see a sign of distortion (deliberate or not is another story) there, and their calculation (like yours as well) is therefore wrong.

    Now to the problem with your math, it is obvious: your Pi r^2 (m2) (area of a disk) is wrong, 2Pi r^2 (m2) (area of a hemisphere) would be right. OMG, THE “EARTH WITHOUT ATMOSPHERE” IS NOT SO F—ING COLD ANY LONGER! (sarc, sorry)

    Of course, there is another general problem with operating with averages, Ryan has already said that (December 18, 2012 at 4:57 pm).

  286. jae says:

    Willis, did I leave you speechless? That WOULD be a FIRST!

    I hope the few CAPS did not bug you…

  287. jae says:

    Well, I guess that should read: “Willis, did I leave you RESPONSELESS?” Just where oh where is some physical evidence of a “radiative greenhouse effect?” Nice construct, but no data. No data, no science. Period.

  288. Robert Clemenzi says:

    Greg House says:
    December 18, 2012 at 6:28 pm

    Note, by the way: one time the total solar irradiance (TSI) is an average, then it is a maximum. An average can not be a maximum at the same time.

    The Earth’s orbit around the Sun is an ellipse. As a result, the amount of energy available varies over the course of a year. Integrated over that year, and then divided by the elapsed time, provides an average TSI for the year.

    The Earth rotates on its axis once per day. Integrating over that time, and then dividing by the elapsed time, provides an average value received at the surface which is one fourth the TSI. The surface receives a maximum amount of energy when it is perpendicular to the path of incoming light. At any other angle, the amount of energy “per unit area” is reduced.

    Try this. Point a flash light directly at a wall. Compare the area covered with the area covered when the same flash light is pointed at the wall at a large angle. In both cases, the same amount of light hits the wall. However, the area of the two beams is quite different. This is a demonstration of why the watts per unit area are different depending on where it is measured perpendicular to the surface of a sphere (such as the Earth). As long as you measure the power hitting a plane perpendicular to the energy source, the location on the sun lit side of a sphere won’t matter. But that energy is spread out over a larger area depending on the cosine of the angle between a line perpendicular to the surface and the direction of the Sun.

  289. Robert Clemenzi says:

    jae says:
    ///////////////////////////////

    The link was bad, but not that hard to debug. Either of these will work.

    http://www.warwickhughes.com/papers/idso98.htm
    http://www.warwickhughes.com/papers/Idso_CR_1998.pdf

  290. TimTheToolMan says:

    Willis writes “You seem to think that that heat does nothing because at equilibrium, an equivalent amount of heat is being lost … but if we took the DLR away, the ocean would cool rapidly and the mixed layer would end up at a much cooler temperature.”

    You seem to think I dont believe that DLR is what causes the oceans to be at the temperature they are. I have told you many times I do believe DLR is responsible for that and I am trying to discuss that process with you because your earlier comments show a lack of understanding what “cooling” really means.

    You mention a rock warming in the sun at equilibrium and then point out that adding more DLR by say adding GHGs makes the rock warmer. You think its because more DLR means the rock absorbs more energy from the DLR and heats up. This is incorrect. Or at least a less than optimal way of looking at it because you’re likely to make errors in reasoning like you did with tallbloke.

    Here is a simple thought experiment to show you. If the said rock is in equilibrium in the sun and you instantaneously do two things. Turn off the sun. And add as much DLR producing stuff you like. Have it wrapped in shiny tin foil if you like…

    What can you say about the DLR. Can you say anything about its increase or its limit?
    What immediately starts to happen to the temperature of the rock.?

    The answers to those questions I believe are relatively self evident and they show us that it is not the DLR that heats anything. Not even a little bit. The process is one of reduced cooling and the way it works is that the DLR absorbed by the rock (yes its absorbed) is radiated away again. Net energy flow in the IR range is away from the rock.

    You can say that mathematically you simply add the radiation and absorbtion to arrive at the net energy flow and note that the absorbtion component is towards the rock. You can even say that for a rock the absorbed energy can mix freely with any other energy and is completely indistinguishable from it.

    The issue with the ocean is that unlike a rock, this DLR cannot be absorbed any deeper than 10um (its really now a physical limitation due to the cool skin) and so the mechanism for warming is different to that of a rock. This is the process you’re getting wrong and I’ve been describing.

    I’ve said it before and I’ll say it again, the very suface does not and cannot become warmer than the bulk through increases in DLR and transfer that DLR supplied energy down. Warming is from the bottom up.

  291. Greg House says:

    Robert Clemenzi says, December 18, 2012 at 7:18 pm : [...]
    =========================================================

    Robert, thank you for the explanation. I agree, actually, with one exception, but this exception is not really relevant at the moment, so maybe later.

    The point is, however, at least as I see it, if you look closely at what I quoted from that NASA link Willis generously provided, you will see that those different angles/the whole hemisphere must have been included in their average. Look at their formulation again: “At Earth’s average distance from the Sun (about 150 million kilometers), the average intensity of solar energy reaching the top of the atmosphere directly facing the Sun is about 1,360 watts per square meter, according to measurements made by the most recent NASA satellite missions. This amount of power is known as the total solar irradiance”.

    What I suspect is this: one department measured/calculated the real average throughout the hemisphere and another (climate) department either deliberately distorted or simply misunderstood that and got as a result a much colder “Earth without atmosphere”.

  292. Willis Eschenbach says:

    jae says:
    December 18, 2012 at 6:31 pm

    Willis, did I leave you speechless? That WOULD be a FIRST!

    I hope the few CAPS did not bug you…

    Sorry, jae, I didn’t see your post. I’ve fixed the link.

    Thanks for persevering,

    w.

  293. Willis Eschenbach says:

    Greg House says:
    December 18, 2012 at 6:28 pm

    Willis Eschenbach says, December 18, 2012 at 5:45 pm:

    “If you can find a problem with my math, let me know. [...]
    Regarding your curious aversion to the concept of “perpendicular”, I just looked up what NASA says on the subject: “In addition, the total solar irradiance is the maximum power the Sun can deliver to a surface that is perpendicular to the path of incoming light.””

    =============================================================

    Right, they say that; thus contradicting what they also say on the same page above: “At Earth’s average distance from the Sun (about 150 million kilometers), the average intensity of solar energy reaching the top of the atmosphere directly facing the Sun is about 1,360 watts per square meter, according to measurements made by the most recent NASA satellite missions. This amount of power is known as the total solar irradiance”.

    Note, by the way: one time the total solar irradiance (TSI) is an average, then it is a maximum. An average can not be a maximum at the same time.

    I guess, what they measured was indeed an average, and this average includes the whole hemisphere.

    You might guess that, but you’d be wrong. Actually, what they mean by “average” is that the earth’s orbit is not a perfect circle, so it is closer to the sun during part of the year and further from the sun the other part of the year. As a result, you have to average over a year to get the average TSI …

    In other words, the average has nothing to do with what you fantasize it is about. Instead, it’s about the average distance from the sun. The clue is where they say “At Earth’s average distance from the sun, the average intensity …”

    w.

  294. RACookPE1978 says:

    Now, that number (1366) is “sort of” valid – after you follow all of the averages properly.

    But – that number is true ONLY for top of atmosphere.

    At the ground (where the radiation actually has to heat the ground/ice/water/trees/grass/desert/rock/tundra/etc), the CAGW crowd tends to keep playing the “average radiation per year” bit. At which point they ARE wrong. (Well, actually, they have to keep playing that “average” game, because the real numbers fail their approximations and averages.)
    At the ground, each watt of sunlight is diffused over a larger and larger area as the latitude increases towards the pole.
    Further, the area being irradiated decreases as you go further north: sunlight between the tropic of Cancer and tropics of Capricorn is higher in the sky, penetrates less air mass, and is more high;y concentrated that between 67.5 (north or south) and the pole.

    Yes, at the equator, at local solar noon, the air mass = 1.0 (on a clear day, no haze, dust,clouds, or aerosols, etc.) But at the 82 north, air mass may vary from as much 30.0 down to 3.0 … Does that change how much energy is available to heat the arctic ice or water? Certainly.
    The sun shines for 0, 3, 12, 15, 18 or 24 hours in a day – depending on latitude and time-of-year. Does that change how much energy is available to heat the surfaces? Certainly, but they use “averages” despite the meaningless “calculation” achieved.

    At 82 north latitude, the sun is only 8 degrees above the horizon – at noon! – at time of minimum ice extents. Run that calculation for how much energy is present and tell me that minimum ice extents in the arctic “heat” the exposed water.
    The heat rates from solar irradiation mean something further south, but do not heat exposed water in the Arctic at times of minimum sea ice extent, yet using an “average” value fools you into accepting such claims.

  295. gnomish says:

    ” the instruments say that the ocean is radiating IR constantly at somewhere on the order of 400 W/m”
    that’s enough energy to boil nearly a gallon of water in an hour…
    jeez…. if i lived in a boat, i’d cook!!!!

  296. Willis Eschenbach says:

    TimTheToolMan says:
    December 18, 2012 at 7:44 pm

    Willis writes

    “You seem to think that that heat does nothing because at equilibrium, an equivalent amount of heat is being lost … but if we took the DLR away, the ocean would cool rapidly and the mixed layer would end up at a much cooler temperature.”

    You seem to think I dont believe that DLR is what causes the oceans to be at the temperature they are. I have told you many times I do believe DLR is responsible for that and I am trying to discuss that process with you because your earlier comments show a lack of understanding what “cooling” really means.

    OK, my bad. I read your postings to mean that the ocean was not warmed by the DLR.

    So … where are our disagreements?

    I’ve said it before and I’ll say it again, the very suface does not and cannot become warmer than the bulk through increases in DLR and transfer that DLR supplied energy down. Warming is from the bottom up.

    I fear I don’t understand this part. Is it your contention that the DLR doesn’t cause the bulk of the upper ocean to end up warmer than if there were no DLR?

    Here’s how I see it. Let us suppose that on average the ocean is losing energy about half of a kilowatt per square metre. Of this loss, about 400 W/m2 is by radiation, and another 20 in sensible heat and 80 in latent heat. So in say 24 hours, it will radiate on the order of 12 kilowatt-hours of energy.

    This radiation is balanced by 160 W/m2 of solar energy, and 340 W/m2 of DLR.

    Now, if there is no solar energy or DLR being absorbed, all of that 12 kWhrs of energy comes out of the bulk of the mixed layer. The bulk of the mixed layer would end up much cooler than it is now.

    If there were only solar energy warming the ocean, then the ocean would be cooler, but not as much. Only about 8 kWhrs of energy would be coming out of the bulk of the mixed layer.

    With the DLR and the sun, we have energetic balance. The amount leaving is equal to the amount entering. As a result, there is no heat lost from the bulk of the mixed layer.

    And as a result, if there is DLR, the entire bulk of the mixed layer ends up warmer, despite the fact that the DLR is absorbed in the very skin itself.

    Two final comments. First, you are totally glossing over mechanical mixing. Wind, wave, and spray are constantly churning the topmost layers of the ocean. I’ve sailed over much of it, and calm times are very uncommon. As a result, your lovely picture of a permanent microscopically thin skin layer that is cooler than the rest of the ocean simply doesn’t obtain over wide areas of the surface—they are being tossed, turned, and churned.

    Finally, note that the situation you correctly describe, of a surface layer which is slightly cooler than the underlying layers, is thermally unstable. As a result, that layer is constantly sinking, and being replaced by warmer water from below.

    Let me take some numbers, which are not real numbers but will do for purposes of discussion. Let’s suppose that the surface is at 5°C and the underlying layer is at 7°C. I know that we are actually talking about very slight temperature difference, this is for illustration of a mechanism.

    The water at 5° is denser than the water at 7°. So it is constantly sinking, and is constantly being replaced by 7° water, which immediately starts cooling by radiation, conduction, and evaporation.

    Now, suppose the DLR is switched on. Lets say it warms the skin water to 6°. Note that it is still cooler than the underlying layer, so your condition is satisfied.

    What is the fate of the 6° water?

    Well, it is cooler than the underlying 7° water, so it will sink … which mixes the DLR energy down into the bulk … which you seem to think is impossible.

    People have the mistaken idea that the ocean is thermally stratified, with the warmest on top. And it is … but only during the day. I learned a lot about the thermal behavior of the ocean when I started night diving.

    At night, the ocean undergoes a curious transition. After sunset, the surface of the ocean starts cooling. At a certain point, when surface waters have cooled enough, columns of descending water form. These mix the cooler surface waters down into the mixed layer beneath. This circulation, while slow, moves massive amounts of water over the course of each night. At night, if you are swimming along slowly at a depth of say three metres (10′) under the surface, you can feel the descending columns when you encounter them. They are cooler than the surrounding water.

    Here’s the thing it took me a while to understand. Yes, during the day the ocean picks up a whole lot of energy. This comes from the sun, which penetrates tens of metres deep into the ocean. But at night, every night, it loses just about that same amount of energy. And unlike during the day, it has to lose it all at the surface. That means that each nocturnal overturning of the ocean has to move lots and lots of water up to the surface every night. At night, if you are swimming slowly say three metres (10′) under the surface, you can feel the descending water. it’s cooler than the surroundings, with a distinct thermocline dividing it from the slowly upwelling surroundings. If you hang in it with neutral buoyancy, you slowly drift downwards.

    Now, if there is DLR, it does several things to this nightly overturning. First, it slows the heat loss of the surface water. As a result, it delays the onset of nightly oceanic overturning. Finally, it slows the overturning once it is started.

    The net result of each of these changes is that the bulk of the mixed layer ends up warmer with the DLR than it would be without DLR. Note that curiously, the energy from the DLR doesn’t need to mix downward in order to affect the bulk temperature of the upper ocean. All it needs to do is slow the overturning and the concomitant heat loss, and the bulk of the upper mixed layer ends up warmer.

    So I agree with you that DLR is absorbed by the ocean. I agree that it is absorbed in the very skin. I think where we part company is that I say if there were no DLR the entire ocean would freeze, top to bottom, whereas you seem to think that DLR somehow only affects the very skin layer.

    But then, I may be totally misunderstanding your position once again. Anyhow, give me your thoughts.

    Regards,

    w.

  297. Willis Eschenbach says:

    gnomish says:
    December 18, 2012 at 11:17 pm

    ” the instruments say that the ocean is radiating IR constantly at somewhere on the order of 400 W/m”
    that’s enough energy to boil nearly a gallon of water in an hour…

    Quick reality check. Four hundred watts per square metre is about the amount of energy radiated constantly from any object at around 20°C (70°F). For example, that’s the rate at which my desk radiates at room temperature. If you have discovered some way to boil water by exposing it to my desk, please let me know.

    w.

  298. gnomish says:

    sure- let’s do have a reality check:

    Volume: 1 gallon
    Energy: 400W
    Start Temp: 0C
    End Temp: 100C
    Efficiency 100%
    Time to Temp: 66 minutes
    http://www.phpdoc.info/brew/boilcalc.html

    ‘those instruments’ are saying if i put a 4 ft diameter parabolic mirror over the water, it could direct enough energy to do my cooking, heat my house and power a steam turbine 24/7.
    forget nuclear, coal, solar and wind! energy supply is so solved by conundrum power!!

  299. Ryan says:

    This 400W/sqm concept is what is causing the problem. The 2nd Law of Thermodynamics states that heat can only flow from a hot object to a colder object. This is equally true of IR as it is of conduction/convection, otherwise the fundamental principle of entropy would be violated.

    So 400W/sqm cannot be true of the NET energy flow from two objects at the same temperature. It could only be true if the 2nd object is at a temperature of absolute zero (-273Celsius). At absolute zero a body emits no heat energy at all so we don’t have to subtract that power from the power emitted by the desk, for instance.

    Now for radiation from the sun the radiated power is relevant because the earth is, frankly, quite close to absolute zero compared to the temperature of the sun (5500Celsius). So when we talk about radiated IR power from the sun we can effectively use the measured radiated power DIRECTLY because the actual temperature of the Earth is sufficiently close to being absolute zero that the small error introduced is more or less negligible compared to other quantities related to climate. We cannot do the same for the radiated power from a desk to the floor because the desk and the floor are at much the same temperature so no net energy can flow between the two, i.e we would need to subtract the power emitted from the floor from the power emitted by the desk to leave us with a net power transfer of 0W/sqm.

    So you see this gives Team AGW an ideal opportunity to misdirect the layman. They can use the radiated power from the sun more-or-less directly, but then do the same with IR to suggest they can add together. This is not in fact the case. In both cases you need to subtract the Earth’s IR emitted power at the point of equilibrium to find the net power transfer. The situation is further complicated because from the Earth to the atmosphere you don’t actually have only IR to deal with – most of the cooling of the planet’s surface is done by conduction/convection which is why you have the rapid cooling with altitude that you see in the graphs above until you reach the tropopause.

    It is perhaps worth bearing in mind that the Sun to Earth temperature difference is far greater than the Earth to space temperature difference, with a subsequent impact on the point at which equilibrium is reached with respect to heat outflow being equal to heat inflow.

  300. gnomish says:

    “Why does the Porridge Bird lay his egg in the air?”
    UNHAPPY MACNAM. UNHAPPY MACNAM.
    (firesign theater, I Think We’re All Bozos on This Bus)

    conundrum has been reduced to absurdity.
    gnite.

  301. TimTheToolMan says:

    Willis writes : “Finally, note that the situation you correctly describe, of a surface layer which is slightly cooler than the underlying layers, is thermally unstable. As a result, that layer is constantly sinking, and being replaced by warmer water from below.”

    I’ll start here and…no. Below is one of my quotes from the earlier thread and can I say that I know its never easy with so many posts but if someone disagrees and provides a reference (sorry it wasn’t a link at the time though) then that deserves a look.

    In the skin layer, its conduction not convection that transmits energy.

    “Molecular transport is the only mechanism for the vertical diffusion of heat and momentum in the cool skin and viscous layer”

    …from “Cool-skin simulation by a one-column ocean model – Chia-Ying Tu and Ben-Jei Tsuang” http://researcher.nsc.gov.tw/public/tsuang/Data/722815193371.pdf

    The final mile so to speak is conduction through the viscous layer. And that makes some sort of sense because if convection were possible then there wouldn’t be a cool skin.

    Ok, so for your example “This radiation is balanced by 160 W/m2 of solar energy, and 340 W/m2 of DLR.”

    The way I see it is that the 160W non-IR + 340W of DLR is balanced by more than 340W of ULR such that the ULR (eg 340+some W) + sensible heat loss (eg 20W) + evaporation (eg 80W) + any residual mixed and conducted down longer term into the ocean (eg the rest) = the 500W supplied.

    So the very surface is radiating all the 340W DLR (because the ocean is radiating at least this much to be cooling at the surface) and then a bit more which is being supplied from the bulk.

    From your example, the “bit more” might be 59.4W because the amount of energy thought to be accumulating in the ocean is, I believe, about 0.6W. This is “the rest”. Personally I dont have a lot of faith in those figures.

    So…The very surface is radiating at the same rate its being absorbed. If DLR increases, then the very surface radiates that new larger amount because despite that increase the DLR still less than the ULR.

    There is no problem with energy needing to “go somewhere” in the case of more DLR. The ocean surface is always cooling. That is why its vital to keep “cooling” in mind when considering this.

    This post is already too long so you can see my posts in the other thread that describe the mechanism for warming that involves changes to the SST driven from below.

    One final comment though regarding overturning is… yes, overturning happens and changes in DLR leading to changes in overturning is possibly another mechanism to accumulate energy in the ocean.

  302. Heat transfer is by conduction and convection. It is only above the tropopause where thermal radiation makes a difference.
    –Ryan

    As an engineer, I am very interested in thermal mechanisms that can do useful work. I’m itching to discover and develop the technology, get the patents and trade secrets locked down and get ready for the IPO that will make me even more filthy stinking rich. However, when an academic climatologist tells me about radiation effects above the troposphere, the first thing I think is: how much is the air above the troposphere different from empty space? The answer? Not much, So, this nearly empty space won’t have much interaction with radiation–like the vacuum of space, it’s a more a place where radiation passes freely than a place where much thermodynamic interchange takes place.

    There is a central difference between engineering and academia. In academia (like politics), all that is needed is a complex (the more convoluted, the better) storyline…a plausible or semi-plausible path from data and theory to support the peer-reviewed paper and the grant/paycheck pipeline. The most absurd nonsense can pass through this system like grease through water fowl.

  303. Ryan says:

    Ken Coffman: As an engineer myself I certainly agree with your points. In fact it turns out to be even more bizarre than you might think. The top of the stratosphere is much warmer than the bottom. So in reality there isn’t any real cooling at all in the atmosphere as it stands at present until you get to 50km up, when the atmosphere starts to cool in the mesosphere, Even then it doesn’t cool very fast – at the top of the mesosphere at 85km up it is still at -85Celsius so quite warm.

    The real cooling is done at the thermosphere but then it becomes a bit more abstract because if you measure a temperature in the exosphere the air molecules are so far apart that the thermometer doesn’t get to meet many of them – so the molecules clearly have energy but you can’t measure it! So do you say that the top of the atmosphere is at absolute zero and take the temperature gradient from there, or do you admit there is still some energy even at the top of the atmosphere and in fact the top of the atmosphere never loses all its heat.That makes a big difference because the thermosphere can range from 350km to 800km deep.

    The theory is that radiative heat transfer occurs above the tropopause, but I’m not convinced. The temperature gradient through the stratosphere is so low that you wouldn’t expect any net radiative transfer. You would still expect conduction, just at such a low level of energy that it doesn’t cause much turbulence and hence is difficult to measure. As you say, once you get to the upper levels of the atmosphere where radiative heat transfer could occur, most of it probably just flies through the atmosphere without every seeing a molecule, let alone specifically a CO2 molecule. Lets be honest, most of the visible light reflected from the Earth’s surface passes through unchecked and looks the same from the ISS as from the ground. IR cameras on weather satellites see the land, sea and top of clouds – no real sign of obscuring radiation from the upper atmosphere even from water vapour. Where is the evidence of IR from the upper atmosphere? Has anybody tried deploying an IR camera on a satellite that is tuned to look for the signal blocked by CO2? This would give us a clearer picture of what the CO2 is really doing and where, surely?

  304. Layman Lurker says:

    Fascinating discussion. TTTL I haven’t read your other thread comments so forgive me if I’ve mangled your logic, but if I understand you correctly you are implying that an increase in DLR (say from a doubling of CO2) will increase the ocean skin temperature and therefore ULR will increase in an offsetting manner. Correct? If this is the case then would it be fair to say that the ocean energy imbalance created by perturbing DLR would be expressed by a disequilibrium in the ocean skin vs bulk temperature differential?

  305. Greg House says:

    Willis Eschenbach says:
    December 18, 2012 at 10:23 pm
    In other words, the average has nothing to do with what you fantasize it is about. Instead, it’s about the average distance from the sun. The clue is where they say “At Earth’s average distance from the sun, the average intensity …”
    ========================================================

    No, look at the formulation again: “At Earth’s average distance from the Sun (about 150 million kilometers), the average intensity of solar energy reaching the top of the atmosphere directly facing the Sun is about 1,360 watts per square meter, according to measurements made by the most recent NASA satellite missions. This amount of power is known as the total solar irradiance”.

    They use the word “average” twice in the same sentence with obviously different meanings: one time it is “the average distance from the sun” and the second time it is “the average intensity of solar energy reaching the top of the atmosphere directly facing the Sun”. That means that they average twice, and the second time it can be only over the whole hemisphere, directly facing the Sun, as they put it.

    And again, in this formulation there is nothing that suggests any “perpendicular” thing. So, who is fantasizing here? OK, you have not made it up personally, as your link demonstrates, but it does not change the essence of that distortion.

    I am sorry, but what we can see is a pattern, I am afraid:

    1)To cope with the problem with the 2nd Law of Thermodynamics, “climate science” invented the “NET” thing that is not to be found in the historical formulations of the 2nd Law.

    2)To somehow theoretically substantiate the experimentally unproven “warming via back radiation”, they cut the solar power in half.

    3)And the last not least they calculate a so called “global temperature” without any basis in science (“extracting” temperatures for large areas from single or few weather stations, “reconstructing” temperatures etc.) and apparently even without any scientifically established definition.

  306. I often wonder at the concept of “Fraunhofer” heaters as defined by Mosher and others. Sure, certain narrow wavelengths are blocked (actually converted from coherence to incoherence, but nevermind) as if there is a direct correlation between this “blocking” and heating. If this was an effective method of heating or insulating, it would be another miracle for engineering. Hot diggity, let’s make us some CO2-thermal insulators or stimulated CO2-thermal heaters and make the world a much more comfortable place. Academic knuckleheads don’t seem to realize how great it would be to heat something by 33C (or block this effect for cooling purpose) with a radiation-modulating machine built only from cold, thin air. It’s a good thing these guys aren’t designing iPhones…you know, things from the engineering domain that actually have to work and do useful stuff.

  307. gnomish says:
    December 19, 2012 at 12:34 am

    sure- let’s do have a reality check:

    Great link by the way
    http://www.phpdoc.info/brew/boilcalc.html

    You missed an important point, those equations only work inside a calorimeter. Insulated sides and top are required. But lets reverse the equation and see what happens if we remove 400 W/m2. Assuming a start temperature of 15C (a bit high).

    Volume: 1 gallon
    Energy: 400W
    Start Temp: 0C
    End Temp: 15C
    Efficiency 95%
    Time to Temp: 10 minutes

    Yes, I know that the amount of energy leaving the system will decrease as the temperature decreases. At 4C (maximum density), the system should loose 334W/m2. Assuming that is the typical value (since the upper skin cools almost immediately to that temperature), it takes about 12 minutes to cool that much. It should actually take a bit longer to account for mechanical mixing since only the surface is able to loose heat. After 15 minutes or so, ice begins to form. Given an 8 hour night, the gallon should freeze solid .. unless there is some other source of heat to replace that lost by radiation.

    To determine the amount of heat radiated at a given temperature use the SB equation. This is the page I use – just set the Albedo to zero to get an approximate value.
    http://mc-computing.com/Science_Facts/Temperature_Conversions.html

  308. Robert Clemenzi says:

    Ryan says:
    December 19, 2012 at 6:45 am

    IR cameras on weather satellites see the land, sea and top of clouds – no real sign of obscuring radiation from the upper atmosphere even from water vapor.

    That’s because the monitored frequency is in the water vapor/CO2 hole. Otherwise, the ground would not be visible.

  309. Layman Lurker says:

    Apologies “TTTL” should have been TTTM in my comment above.

  310. jae says:

    Willis:

    Thanks for the link. I’ve read that Idso paper several times, but not recently; so I read it again. Willis, the paper is mainly about SENSITIVITY, and I don’t see how it addresses my question. In fact, it highlights my quandry. If the temperature sensititivity to increases in radiation is about 0.1C/Watt m-2, per the Idso paper, that would indicate that it should be MUCH, MUCH hotter in Atlanta than in Poenix, since there would be MUCH more downwelling radiation from all the water vapor in the air in Atlanta. But it is much cooler, day and night.

    What I believe is going on is that the water in humid areas like Atlanta exerts a negative feedback (via evaporation and clouds) and acts as a thermostat, similar to what you see in the tropics. In fact, Idso says the following under the subtitle “Cooling the Global Greenhouse,” in the paper you linked (page 7):

    “Within this context, it has been shown that a 10% increase in the amount of low-level clouds could completely cancel the typically-predicted warming of a doubling of the air’s CO2 content by reflecting more solar radiation back to space (Webster & Stephens 1984). In addition, Ramanathan & Collins (1991), by the use of their own natural experiments, have shown how
    the warming-induced production of high-level clouds over the equatorial oceans totally nullifies the green-house effect of water vapor there, with high clouds dramatically increasing from close to 0% coverage at sea surface temperatures of 26°C to fully 30% coverage at 29°C (Kiehl 1994). And in describing the implications of this strong negative feedback mechanism,
    Ramanathan & Collins state that ‘it would take more than an order-of-magnitude increase in atmospheric CO2 to increase the maximum sea surface temperature by a few degrees,’ which they acknowledge is a considerable departure from the predictions of most general circulation models of the atmosphere.”

    If such negative feedback does, indeed, exist, then increases in any GHGs will affect only very dry areas (and in normally wet areas when they are in drought (it never goes above 100 F in humid areas like Atlanta, unless it is extremely dry).

    IOW, we probably cannot affect the temperature much by adding GHGs…

  311. gnomish says:

    Mr. Clemenzi:
    “You missed an important point, those equations only work inside a calorimeter. Insulated sides and top are required.”
    I didn’t miss that but it wasn’t the point at all.
    The point was 400W/sq.m. radiation is being claimed.
    IF 400W is being radiated, then it can be used to power a half horsepower generator,
    if that isn’t the case, then watts up with that?
    the answer is, logically, that there is no 400W of radiated power.
    Don’t tell me that radiation can’t be concentrated by a reflector, please.
    the only radiation that can not be concentrated by a reflector is nonexistent.
    That’s the conundrum reduced to absurdity.
    please feel free to address that point.

  312. Willis Eschenbach says:

    gnomish says:
    December 19, 2012 at 11:12 am

    … IF 400W is being radiated, then it can be used to power a half horsepower generator,
    if that isn’t the case, then watts up with that?

    I swear, to you guys, theory is far more important than measurements.

    There is, in fact, something on the order of 400 W/m2 being emitted by everything around the planet that is at around 20°C. You can measure it. You can look through night-vision glasses and see it. People have measured it around the world. Stefan-Boltzmann studied it at great length.

    And you want to stand there and say it doesn’t fit your theories, so it can’t be happening.

    A much more valuable question for you would be, what’s wrong with my theories, which obviously and blatantly disagree with observations? Because trying to claim that an object at around 20°C is NOT radiating around 400 W/m2, well, that’s flying in the face of centuries of evidence.

    w.

  313. Willis Eschenbach says:

    gnomish says:
    December 19, 2012 at 12:34 am

    sure- let’s do have a reality check:

    Volume: 1 gallon
    Energy: 400W
    Start Temp: 0C
    End Temp: 100C
    Efficiency 100%
    Time to Temp: 66 minutes
    http://www.phpdoc.info/brew/boilcalc.html

    ‘those instruments’ are saying if i put a 4 ft diameter parabolic mirror over the water, it could direct enough energy to do my cooking, heat my house and power a steam turbine 24/7.
    forget nuclear, coal, solar and wind! energy supply is so solved by conundrum power!!

    No, that’s not what “those instruments are saying”. Instruments don’t deal with theories about parabolic mirrors. That is what your flawed understanding is saying.

    Since the energy source is at ~ 20°C, how do you plan to make heat flow uphill to water at ~ 100°C?

    Yes, the energy source is radiating at 400 W/m2 … but let’s say the water starts out at 20°C like your energy source. The water (and the pot it is in) are both radiating at 400 W/m2 just like the energy source.

    So how much heat is going to flow?

    If you said “None, Willis”, then you are following the story.

    The problem is, radiation at 400 W/m2 won’t warm something radiating at 400 W/m2.

    Gnomish, when I come up with something that seems impossible, that seems to contradict the laws of nature, the first place I look for a problem is in my own understanding. It is by far the most likely explanation.

    In this case, you seem to have forgotten that if water is at 20°C, it is constantly losing energy at the rate of 400 W/m2. If you put it next to something radiating 400 W/m2, all that does is break even. It keeps the water from cooling. But it will never boil it.

    Look up or think about the difference between energy and heat. Heat is net energy flow. The water at 20°C is radiating at 400 W/m2, and the heat source is radiating the same. So while there is a constant exchange of energy between the two objects, there is no flow of heat.

    For example, if the pot of water started out frozen, then there would be much less radiation leaving the pot of water, only about 315 W/m2. In that case, putting it by my desk (at 20°C) will definitely warm it by radiation. Why? Because the net heat flow starts out at about 400 – 315 = 85 W/m2, so it warms the water. But it can never warm it to above 20°C, at that point the energy going each way is equal, so there is no net heat flow.

    You need to distinguish between a flow of energy and a flow of heat. They are very different creatures.

    All the best,

    w.

  314. DirkH says:

    gnomish says:
    December 19, 2012 at 11:12 am
    “the only radiation that can not be concentrated by a reflector is nonexistent.”

    Try that with diffuse radiation.

  315. gnomish says:

    so when are watts not heat? are you channeling myrhh now?
    are radiant watts going to refuse to flow when directed?
    everything you said re checking the premises when logic contradicts them is valid.
    try it, then.

    forget about parabolic mirrors which are used all the time to focus radiant energy- that’s just one of those immutable facts that passes every empirical test.
    talk to my CO2 laser that burns the neighbor’s ants 10 meters away.
    it’s your theory that must be made to fit that reality.
    something is obviously and absurdly wrong with your theory.
    otherwise, show me how to boil water with your magic desk.

  316. gnomish says:

    gotta tell you, too- my microwave oven doesn’t need to be 100C to boil water.
    reconcile that with your theory if you can.

  317. gnomish says:

    there are no downhill watts, willis.
    watts don’t have varying potentials.
    watts don’t have to climb up hills.
    watts is watts. that measure converts directly to joules/s, calories/s, btu/s
    but it does not convert to temperature, willis.

  318. DirkH says:

    gnomish says:
    December 19, 2012 at 12:31 pm
    “watts is watts. that measure converts directly to joules/s, calories/s, btu/s
    but it does not convert to temperature, willis.”

    It does:
    https://en.wikipedia.org/wiki/Stefan-Boltzmann_law

  319. TimTheToolMan says:

    LL writes “if I understand you correctly you are implying that an increase in DLR (say from a doubling of CO2) will increase the ocean skin temperature and therefore ULR will increase in an offsetting manner.”

    Yes, the ocean skin temperature, the SST, is set by the waters just below the surface. If DLR is higher then the “compensating” ULR as you put it is also higher and the energy required from below is less. The top of the ocean (the “hook”) accumulates energy (ie warms up) until the SST has increased to the point where S-B is again satisfied for equilibrium.

    http://en.wikipedia.org/wiki/File:MODIS_and_AIRS_SST_comp_fig2.i.jpg

    So to warm the ocean from increased DLR, it is the SST driven by the hook temperature that is a little higher and then it would indeed be fair to say

    “If this is the case then would it be fair to say that the ocean energy imbalance created by perturbing DLR would be expressed by a disequilibrium in the ocean skin vs bulk temperature differential?”

    So specifically to change the SST by changing the DLR means the temperature gradient in the skin changes as a result of changes in conduction. ie for more DLR, conduction reduces and the gradient reduces. But once the new equilibrium is established the temperature gradient in the skin changes back to what it was because that reflects the gradient required to radiate the incoming solar energy + sensible + latent heats.

    At RC they suggested the change in this temperature gradient would be permanent less energy leaves the ocean and the ocean accumulates energy and thats how AGW warms the ocean. They are wrong. IMO this (ocean energy accumulation) is the most important process for AGW and so it’d be nice if they understood it.

    http://www.realclimate.org/index.php/archives/2006/09/why-greenhouse-gases-heat-the-ocean/

    “Reducing the size of the temperature gradient through the skin layer reduces the flux. Thus, if the absorption of the infrared emission from atmospheric greenhouse gases reduces the gradient through the skin layer, the flow of heat from the ocean beneath will be reduced, leaving more of the heat introduced into the bulk of the upper oceanic layer by the absorption of sunlight to remain there to increase water temperature.”

  320. TimTheToolMan says:

    LL writes “if I understand you correctly you are implying that an increase in DLR (say from a doubling of CO2) will increase the ocean skin temperature and therefore ULR will increase in an offsetting manner.”

    Actually a slight alteration to my previous pos for clarificationt, its not the *temperature* that changes in the ocean skin due to the increased DLR. Not directly anyway. But the energies for ULR and from conduction through the skin change and the resulting hook changes alter the skin temperature

  321. Robert Clemenzi says:

    gnomish says:
    December 19, 2012 at 11:12 am

    Don’t tell me that radiation can’t be concentrated by a reflector

    It is easy to concentrate energy from a point source using a mirror or lens. However, DWIR is coming from a diffuse source (everywhere). As a result, it is difficult, but not impossible, to concentrate that energy with a mirror. Several papers claim 2x to 4x increases. For example,
    http://optoelectronics.eecs.berkeley.edu/ey1990sem2123.pdf

    It seems logical that a reverse design could cool things (freeze water).

    However, that does not change the fact that the calculator you are using computes the time for 400 W/m2 greater than what the surface is emitting. That is why the calorimeter is important. In that case, the water is not loosing any radiation. In the case of the oceans, you have to account for and replace the additional losses.

  322. Gail Combs says:

    TimTheToolMan says:
    December 19, 2012 at 1:51 pm

    LL writes “if I understand you correctly you are implying that an increase in DLR (say from a doubling of CO2) will increase the ocean skin temperature and therefore ULR will increase in an offsetting manner.”

    Yes, the ocean skin temperature, the SST, is set by the waters just below the surface. If DLR is higher then the “compensating” ULR as you put it is also higher and the energy required from below is less. The top of the ocean (the “hook”) accumulates energy (ie warms up) until the SST has increased to the point where S-B is again satisfied for equilibrium…..
    >>>>>>>>>>>>>>>>>>>>>>>>>>>
    How can that happen when Infra-red wavelengths do not penetrate the ocean beyond a few microns and also have very little actual energy? graph

    The downward long wave infrared radiation from the atmosphere has a lot of low energy spread over a large number of wavelengths and can not penetrate deeper than a fraction of a millimeter (10 micrometers). This is much less than the boundary layer where evaporation takes place (500 microns), and that is the only thing the DLR infrared radiation can do; help to evaporate water. This is a close up graph

    The wavelengths of CO2 are between 13.5 and 16.5 microns, and penetrates only 5 to 10 microns deep. That is a very loose translation of the information from this site in Dutch.

    The other missing piece is the amount of energy you are talking about when speaking of the earth’s black/grey body radiation. It is miniscule at each wavelength compared to the sun graph

    Update 12-05-2012

    Oceanograaf Dr. Robert E. Stevenson schrijft in zijn rapport op pagina 8:

    The atmosphere cannot warm until the underlying surface warms first. The lower atmosphere is transparent to direct solar radiation, preventing it from being significantly warmed by sunlight alone. The surface atmosphere thus gets its warmth in three ways: from direct contact with the oceans; from infrared radiation off the ocean surface; and, from the removal of latent heat from the ocean by evaporation. Consequently, the temperature of the lower atmosphere is largely determined by the temperature of the ocean.

    Warming the ocean is not a simple matter, not like heating a small glass of water. The first thing to remember is that the ocean is not warmed by the overlying air.

    Yes the CO2 down radiation can warm the ocean but so does a little boy who isn’t potty trained and I am not sure who actually has the most influence.

  323. TimTheToolMan says:

    Gail writes “How can that happen when Infra-red wavelengths do not penetrate the ocean beyond a few microns and also have very little actual energy?”

    You need to read the rest of my posts Gail. DLR doesn’t directly increase the energy of the ocean, DLR is absorbed into that top 10um and radiated again at the same rate. You’re right in that it does not and cannot enter the bulk of the ocean as “energy input”

  324. Gary Pearse says:

    Ryan says:
    December 19, 2012 at 1:00 am
    “This 400W/sqm concept is what is causing the problem. The 2nd Law of Thermodynamics states that heat can only flow from a hot object to a colder object.”

    Ryan, refrigeration flows the heat from a cold object to a hotter sink. Indeed it makes the condensing coils even hotter! Mind you, it requires energy input to do it. And of course, the “heat” from your frozen pork chops is “used” to effect a phase change in the refrigerant. Remember, heat is not a synonym for temperature. Don’t be sloppy with the 2nd law.

  325. gnomish says:

    Robert Clemenzi:
    srsly – there is a problem here.
    it looks like I’m going to have to baby step through the logic – care to do this with me and identify the contradiction?
    proposition #1: is the ocean radiating 400W / sq m constantly or not?
    this is true or not – no need to quibble over nanowatts.
    so, it’s yes or no.

  326. Robert Clemenzi says:

    400 is not the value I would use, but yes it is radiating energy.

    Do you agree that if something is radiating more energy than it receives, it will git colder?
    And conversely, if it receives more energy than it emits, it will get warmer?

  327. gnomish says:

    DirkH
    pardon the quibble, but my statement was absolutely true. to express your blackbody relationship of radiant energy proportional to an exponent of temperature requires an additional factor not present in any temperature measurement or power measurement.
    watts = amps * volts
    does a thermometer specify the emissivity or temperature of an amp or a volt? no.
    but for now, please consider what it would be the actual effect of having 3.6A of 110V incandescent bulbs burning on every square meter of your ceiling.
    that’s the focus of my issue, i.e., what seems completely absurd to me. it passeth not the sniff test.

  328. mkelly says:

    Gail Combs you have placed another excellent post. Enjoyed the links. Don’t read Dutch though. :-)

  329. gnomish says:

    R.Clemenzi:
    “Do you agree that if something is radiating more energy than it receives, it will git colder?
    And conversely, if it receives more energy than it emits, it will get warmer?”
    yes.
    proposition #2) is it true that 400W of power is enough to boil a gallon of water in about an hour?
    again – it’s true or it’s not. if it takes a calorimeter, so be it. a few minutes here or there is not a quibble.

  330. Gail Combs says:

    mkelly says:
    December 19, 2012 at 5:31 pm

    Gail Combs you have placed another excellent post. Enjoyed the links. Don’t read Dutch though. :-)
    _________________________________
    Google Translate is your friend. http://translate.google.com/?tl=nl&q=undefined

    I happen to read enough German (required of a chemist) that I can sort of wade through it, but google is a heck of a lot easier.

  331. jae says:

    Willis says to gnomish a few comments above here:

    “I swear, to you guys, theory is far more important than measurements.”

    Huh? Willis, what are you saying? Maybe YOU are the one who thinks theory is more important than empirical evidence, eh?? Just where are the measurements for downwelling long-wave IR for Atlanta and Phoenix, so I can compare them. Atlanta should have many times more W/m2 than Phoenix, since Atlanta has, on average, about 4 times as much GHGs doing the downward radiation gig. For some reason, I can easily access direct radiation from the Sun at these locations, but there appears to be no database for GHG radiation at night. WHY??? It should be very easy for someone to measure the downwelling radiation at night, so we can see just what those GHGs are doing. But I don’t know of any database for such radiation. Do you? Would that data not be an important to all the “climate scientists?” Or are they all just watching radiation cartoons?

  332. Greg House says:

    jae says, December 19, 2012 at 6:39 pm: “It should be very easy for someone to measure the downwelling radiation at night, so we can see just what those GHGs are doing.”
    ======================================================

    Jae, to learn that those “GHGs” can not be doing well, nowhere, it is sufficient to learn about the Wood’s experiment. The “climate science” must have been aware of it since 1909.

    That is the reason, I guess, for their bogus “otherwise much colder Earth”-calculation and the apparent absence of any experimental proof for their “back radiation warming”.

  333. Robert Clemenzi says:

    gnomish says:
    December 19, 2012 at 5:46 pm

    proposition #2) is it true that 400W of power is enough to boil a gallon of water in about an hour?

    Only if it is not losing energy – either by radiation or thru the walls of the container. Or any other method!

  334. jae says:
    December 19, 2012 at 6:39 pm

    Atlanta should have many times more W/m2 than Phoenix, since Atlanta has, on average, about 4 times as much GHGs

    Are you sure? Please provide data.
    CO2 should be the same at both.

    Assuming H20 at 30,000 ppm, that would translate to an RH of
    100.10 % at 76°F
    63.65 % at 90°F – Atlanta
    34.84 % at 110°F – Phoenix
    17.49 % at 135°F – desert
    The fact that the RH is different does not necessarily mean that the amount of water in the atmosphere is different. Water on the ground tends to cool things off and I think that that is the main difference. If want to see some actual data for Jacksonville and Tucson, you can try my program.

    http://mc-computing.com/Science_Facts/Lapse_Rate/Lapse_Rate_Animations.html

  335. jae says:

    Greg House says:

    “Jae, to learn that those “GHGs” can not be doing well, nowhere, it is sufficient to learn about the Wood’s experiment. The “climate science” must have been aware of it since 1909.”

    Greg, Wood’s experiments related to real greenhouses, not to an open atomsphere. Not relevant here, PROBABLY (but not FOR SURE, since convection is so extremely important that it might completely overwhelm any radiative GHE effect–another topic).

  336. Greg House says:

    jae says, December 19, 2012 at 8:04 pm: Greg, Wood’s experiments related to real greenhouses, not to an open atomsphere.
    =========================================================

    This is so wrong, Jae, I can not believe it.

    The Wood’s experiment demonstrates that “back radiation warming” is either zero or negligible.

    The back radiation in his experiment was produced by glass, not by so called “greenhouse gases”, but this does not matter at all, because his experiment debunks the very MECHANISM (“back radiation warming”), by which the “greenhouse gasses” are claimed to warm, anyway according to the IPCC version of the “greenhouse effect”.

    So, please, read it again, in this form this time: Wood’s experiment = back radiation does not work.

  337. gnomish says:

    thanks, R. Clemenzi.
    presently you may help me resolve my issue, for which i thank you for your time.
    proposition #3: is there any distinction between watts radiated by ocean water or watts radiated by an electrical element?

  338. gnomish says:
    December 19, 2012 at 8:41 pm

    proposition #3: is there any distinction between watts radiated by ocean water or watts radiated by an electrical element?

    They are the same :)

    Watt = Joules per second
    http://en.wikipedia.org/wiki/Watt
    It is simply “the rate of energy conversion or transfer”.

    The amount of energy transferred is the number of watts times an amount of time.

    “Heat is energy transferred from one body to another by thermal interactions” – wiki ref

  339. gnomish says:

    R. Clemenzi – thanks for your patience and persistence.
    i’m almost at the crux of the matter, now.
    proposition #4:
    if a person sits on a 100W electrical heating pad, will his 1/4 sq meter butt be noticeably warmed?

  340. Willis Eschenbach says:

    gnomish says:
    December 19, 2012 at 12:31 pm

    there are no downhill watts, willis.
    watts don’t have varying potentials.
    watts don’t have to climb up hills.
    watts is watts. that measure converts directly to joules/s, calories/s, btu/s
    but it does not convert to temperature, willis.

    gnomish, one of the brilliant discoveries of the early scientists was that solid things radiate in the infrared. A couple of guys named Stefan and Boltzmann discovered the relationship between the temperature, and how many watts are radiated at that temperature. That turns out to be

    W = σ e T4

    where W is watts, σ is the Stefan-Boltzmann constant ( 5.67E-8 ), e is the emissivity of the object, and T is the temperature in kelvins.

    So watts of radiated energy do indeed convert to temperature, gnomish.

    Again let me say, truly you do not understand this stuff. Get a textbook and catch up, your gyrations are painful to watch.

    w.

    PS—Regarding my used of “uphill” which you reference in your post, I had said:

    Since the energy source is at ~ 20°C, how do you plan to make heat flow uphill to water at ~ 100°C?

    When I asked above how you planned to make the energy flow “uphill”, the word “uphill” was slang for making the energy flow from cooler to warmer.

    Sorry for the confusion.

  341. Robert Clemenzi says:

    gnomish says:
    December 19, 2012 at 5:25 pm

    what it would be the actual effect of having 3.6A of 110V incandescent bulbs burning on every square meter of your ceiling.

    Assuming that the ceiling is about 75°F and already emitting 441 W/m2.
    396 + 441 = 837 W/m2
    yielding an expected floor temperature of 167°F. However, this this would also heat the ceiling to about the same value. Maybe higher if the walls can’t git rid of enough energy. A bit lower if the windows are open.

  342. gnomish says:

    hold off, Willis, if you please.
    emissivity is not a property of temperature nor of heat, so there is no direct conversion between degrees and watts, as i stated.
    please let Robert finish his attempt to help me resolve a point you are missing.

  343. Robert Clemenzi says:

    gnomish says:
    December 19, 2012 at 11:04 pm

    proposition #4: if a person sits on a 100W electrical heating pad, will his 1/4 sq meter butt be noticeably warmed?

    Yes, even if it is turned off. When you stand, the air cools you (assuming it is less than your body temperature). When you sit, the heat can no longer escape. As a result, what you are sitting on warms up to your internal body temperature. So does that part of your anatomy. Both objects heat up.

    When you turn on the heating pad, things get even hotter. Fortunately, many heating pads have a thermostat so they won’t get hot enough to burn you. After a few minutes, these pads will be putting out only a few watts, not their full rated values.

  344. gymnosperm says:

    “Now, if there is DLR, it does several things to this nightly overturning. First, it slows the heat loss of the surface water. As a result, it delays the onset of nightly oceanic overturning. Finally, it slows the overturning once it is started.”
    ====================================
    Why would it slow the heat loss? By warming the skin it will increase conduction to the air, increase radiation to the air, and increase evaporation to the air.

    Ok, so it would reduce your “overturning” which amounts to a sort of inverse convection, but your data set for this phenomenon seems to be personal sensation on a few night dives. Have you never felt these during the day? I have.They are very common. How far above the bottom were you diving?

  345. gnomish says:

    R. Clemenzi-
    to save Willis needless pain, I’ll cut to the chase.
    (i hope i’ve eliminated all extraneous variables to avoid gyratory discursion)
    if an electrical source of radiation emitting 400W (from a gram of nichrome wire isolated in a thermos bottle but for a one sq. meter exit in my direction) produces noticeable heat on my skin but a square meter of ocean (which does not significantly chill in the process) fails to do so, what is the explanation?
    thank you for being most considerate and obliging.

  346. Willis Eschenbach says:

    jae says:
    December 19, 2012 at 6:39 pm

    Willis says to gnomish a few comments above here:

    “I swear, to you guys, theory is far more important than measurements.”

    Huh? Willis, what are you saying? Maybe YOU are the one who thinks theory is more important than empirical evidence, eh?? Just where are the measurements for downwelling long-wave IR for Atlanta and Phoenix, so I can compare them. Atlanta should have many times more W/m2 than Phoenix, since Atlanta has, on average, about 4 times as much GHGs doing the downward radiation gig. For some reason, I can easily access direct radiation from the Sun at these locations, but there appears to be no database for GHG radiation at night. WHY??? It should be very easy for someone to measure the downwelling radiation at night, so we can see just what those GHGs are doing. But I don’t know of any database for such radiation. Do you? Would that data not be an important to all the “climate scientists?” Or are they all just watching radiation cartoons?

    So, just what is your contention here? That DLR isn’t measured? I discussed the DLR measurements at the TAO buoys here. There’s downwelling longwave radiation measurements for a half dozen sites here.

    w.

  347. Willis Eschenbach says:

    gnomish says:
    December 19, 2012 at 11:11 pm

    hold off, Willis, if you please.
    emissivity is not a property of temperature nor of heat, so there is no direct conversion between degrees and watts, as i stated.
    please let Robert finish his attempt to help me resolve a point you are missing.

    Look, my friend, you made a statement to me about how it is not possible to convert watts to temperature.

    I answered you, pointing out people have been doing that exact conversion for centuries. Now you are bitching and whining because I answered it? Well, excuuuuse me.

    Sorry, that’s how it works. It’s called a discussion. Of course, my pointing out ugly reality, that people have converted watts to temperature for a couple centuries, doesn’t faze you. You have your fantasy, and by God, neither Stefan nor Boltzmann are gonna get in your way.

    Now, you come back with the meaningless statement that since emissivity isn’t a function of temperature, there is no “direct conversion” of watts to temperature … what on earth are you on about? For a given object,

    W = C T4

    where “C” is a constant. How direct a relationship do you want?

    Finally, “hold off” and “let Robert finish”? What does “hold off” mean? I am not stopping Robert in any sense, how could I. Whether I “hold off” or not, Robert is free to try to assist you. You should check up on how this “blog” thing works, gnomish. Robert doesn’t have to stand in line and wait merely because I am posting.

    w.

  348. Willis Eschenbach says:

    Robert Clemenzi says:
    December 19, 2012 at 7:26 pm

    gnomish says:
    December 19, 2012 at 5:46 pm

    proposition #2) is it true that 400W of power is enough to boil a gallon of water in about an hour?

    Only if it is not losing energy – either by radiation or thru the walls of the container. Or any other method!

    Gnomish, let me suggest close attention to this answer from Robert. Absorbing radiant energy doesn’t mean that an object will warm up. It all depends on how much energy that the object is losing at the same time.

    If you could hold your thermal losses to zero, the constant addition of even 1 W would be enough to eventually boil water.

    But we don’t have that situation. Suppose we set your gallon of water in front of my desk. My desk is radiating, with the amount of radiation being proportional to T4. At 20°C (~ 70°F), it’s radiating about 400 W/m2.

    Assuming that the gallon of water is at 20°C as well, it is also radiating 400 W/m2, just like the desk.

    And as a result, the water is gaining energy at a rate of 400 W/m2, and at the same time is radiating energy away at the same rate, 400 W/m2 … so it doesn’t change temperature at all.

    HTH,

    w.

  349. Willis Eschenbach says:

    jae, you asked about records of downwelling longwave. Here is a day of downwelling and upwelling longwave and shortwave.

    Figure S1. Up and downwelling radiation. Local time is shown along the top. SOURCE.

    Note that the DLR is about 300 W/m2. This is because it is winter and the location is cold (Colorado). As a result, the DLR is coming from a colder atmosphere. Assuming an emissivity of 0.95, that converts to a few degrees below freezing.

    Of course, the upwelling longwave also reflects the winter. It is slightly warmer than the overlying atmosphere, as we would expect. You can see how it increases with the warmth of the day, and then decreases as night approaches. It reaches its coldest point in the early morning.

    w.

  350. Ryan says:

    The comments made by my fellow engineer have really got me thinking about this subject in a clear way.

    What we have is a system where the tropopause as shown in the diagram is at -60Celsius. It is about 10,000m high. The ocean might be at say +25Celsius. Between the two we therefore have a temperature of 8%Celsius over a distance of 10,000m which is a very low temperature gradient indeed!

    Now normally we think of air as being quite thermally conductive, because we use it as a coolant, but if you but enough of it in series you end up with a very good insulator. This is what is happening here. It is analogous to the electrical domain – copper is a very good conductor but if you put enough of it in series you end up with a very high resistance (a 7km loop of telephone wire has a resistance of about 2000ohms).

    Due to the fact we have so much atmosphere between us and the tropopause the atmosphere itself at any two points say a metre apart are very nearly at thermal equilibrium – this means there can be almost NO cooling of the Earth’s surface either by radiation or by conduction through the atmosphere within the troposphere.

    Above the tropopause the atmosphere actually starts to get warmer due to direct heating by incoming UV that reaches the ground. This tells us several things: The tropopause can only be cooling by being in direct thermal contact with space – i.e. it is radiating into space. It also tells us that the tropopause is still warmer than space largely because it finds it difficult to radiate the heat through the relatively warm layer of the upper stratosphere.

    The stratosphere was not discovered (or expected) until 1900. Arrhenius proposed the greenhouse effect in 1876. Arrhenius had now way of knowing that the Earth’s atmosphere is actually a lot warmer at the top than at the bottom, so the actual physical model he proposed and which stands to this day was originally based on a lack of knowledge of the complexity of the Earth’s atmosphere.

    Given that the troposphere has a very low temperature gradient I would suggest that very little of the Earth’s incoming solar energy can be dissipated through this route. I suggest it can only be dissipated via direct re-radiation to space. Measurements suggest this is not the case however. I propose that the measurements are wrong. That might seem like a bold claim but we have to remember that the Earth would be -18Celsius without an atmosphere and is +14Celsius with it – that’s about a 10% difference compared to absolute zero (-273Celsius). So you have to get those measurements spot on to be sure you have modelled the causes of the small impact of the atmosphere correctly. Considering that the heating of the Earth is likely to be greatest over particular parts of the land near the equator at noon, we must also consider that this is also the maximum time of cooling. Therefore you need to measure the cooling at the same time of maximum heating – a difficult task to accomplish accurately. Note also that using some kind of average won’t wash – you need to break up the surface of the Earth into small areas and make the measurements accurately for each one to be sure you understand the contribution that each makes to the heating effect at ground level and any subsequent cooling. I think that is this was done properly it would show that warming of the upper stratosphere by UV is what causes the Earth to be warmer than expected as it prevents significant cooling through the atmosphere. If this is the case then CO2 can be expected to have little impact even if raised by 2 orders of magnitude since the thermal gradient would only go from very low to even lower.

    ["8% Celsius rise" "85 (degrees) Celsius" rise ? Mod]

  351. Ryan says:

    Willis: once again be careful about your tendency to be overly assertive. Gnomish is essentially correct that there is no direct relation ship between watts of radiated power and temperature in any given real-world case. This is because the radiated power of IR, for instance, is an electro-magnetic wave, it is not a beam of “hot” (in that sense it is quite different from conduction/convection which are indeed flows of “hotness” with no need for conversion between energy types). The electro-magnetic wave needs to be converted to heat energy to get an increase in temperature.

    Stefan/Boltzmann theorised a perfect absorbtive object that converts the electro-magnetic energy to heat with 100% efficiency and therefore were able to develop a relationship between the electro-magnetic power reaching the object and the temperature. Any real-world application needs to take into account that the process is likely to be nowhere near 100% efficient. The process is nowhere near 100% efficient for the surface of the Earth.

  352. Gail Combs says:

    Willis Eschenbach says:
    December 20, 2012 at 12:19 am
    ….So, just what is your contention here? That DLR isn’t measured? I discussed the DLR measurements at the TAO buoys here.…..
    >>>>>>>>>>>>>>>>>>>>>
    Willis, I Just took a look at that information and I have a question.

    First clouds are composed of tiny water droplets or ice crystals that are suspended in the air. Those clouds would have a temperature T, and would radiate as a gray body. This radiation would be a percentage of the theoretical black body radiation which is proportional to the fourth power of the absolute temperature as expressed with Stefan-Boltzmann Law. Would this not be confounding the whole idea of DLR from greenhouse gases. Heck would not the entire atmosphere have DLR because it is a gray body radiator and therefore the DLR would again have nothing to do with the Greenhouse gas theory except as a confounding contributor?

    One site said the emissivity of the atmosphere is about 0.83 based on the gases but not the diatomic gases (Link will not connect so that is off the google blurb)

    For clouds (liquid and solid) the emissivity of water and of frost crystals is 0.98 link

    Now I am intrigued so I did more poking around and I found this which I think you might be very interested in, Willis.

    Parameterization of atmospheric long-wave emissivity in a mountainous
    site for all sky conditions
    J. Herrero and M. J. Polo
    Received: 14 February 2012 – Accepted: 11 March 2012 – Published: 21 March 2012

    ABSTRACT
    Long-wave radiation is an important component of the energy balance of the Earth’s surface. The downward component, emitted by the clouds and aerosols in the atmosphere, is rarely measured, and is still not well understood. In mountainous areas, the models existing for its estimation through the emissivity of the atmosphere do not give good results, and worse still in the presence of clouds….. This study analyzes separately three significant atmospheric states related to cloud cover, which were also deduced from the screen-level meteorological data. Clear and totally overcast skies are accurately represented by the new parametric expressions, while the intermediate situations corresponding to partly clouded skies, concentrate most of the dispersion in the measurements and, hence, the error in the simulation. Thus, the modeling of atmospheric emissivity is greatly improved thanks to the use of different equations for each atmospheric state.
    ——–
    Introduction Long-wave radiation has an outstanding role in most of the environmental processes that take place near the Earth’s surface (e.g., Philipona, 2004). Radiation exchanges at wavelengths longer than 4 μm between the Earth and the atmosphere above are due to the thermal emissivity of the surface and atmospheric objects, typically clouds, water vapor and carbon dioxide. This component of the radiation balance is responsible for the cooling of the Earth’s surface, as it closely equals the shortwave radiation absorbed from the sun. The modeling of the energy balance, and, hence, of the long-wave radiation balance at the surface, is necessary for many different meteorological and hydrological problems, e.g., forecast of frost and fog, estimation of heat budget
    from the sea (Dera, 1992), simulation of evaporation from soil and canopy, or simulation of the ice and snow cover melt (Armstrong and Brun, 2008)….

    Downward long-wave radiation is difficult to calculate with analytical methods, as they require detailed measurements of the atmospheric profiles of temperature, humidity, pressure, and the radiative properties of atmospheric constituents (Alados et al., 1986; Lhomme et al., 2007). To overcome this problem, atmospheric emissivity and temperature profile are usually parameterized from screen level values of meteorological variables. The use of near surface level data is justified since most incoming long-wave radiation comes from the lowest layers of the atmosphere (Ohmura, 2001).

    …. the effect of clouds and stratification on atmospheric emissivity is highly dependent on regional factors which may lead to the need for local expressions (e.g., Alados et al., 1986; Barbaro, et al., 2010). on environmental processes, especially if snow is present. As existing measurements are scarce (e.g., Iziomon et al., 2003; Sicart et al., 2006), a correct parameterization of downward long-wave irradiance under all sky conditions is essential for these areas….
    Conclusions
    The long-wave measurements recorded in a weather station at an altitude of 2500 m in a Mediterranean climate are not correctly estimated by the existing models and frequently used parameterizations. These measurements show a very low atmospheric emissivity for long-wave radiation values with clear skies (up to 0.5) and a great facility for reaching the theoretical maximum value of 1 with cloudy skies.….

    This seems to indicate that at least at an altitude of 2500 m the DLR doubles on cloudy days. The statement “…a great facility for reaching the theoretical maximum value of 1 with cloudy skies.” seems to indicate the emissivity of water and of frost crystals is 0.98 holds true for clouds.

    The bit of poking around that Sleepalot (comment) and I did (comment 1) and (comment 2) also seems to indicate the emissivity? GHG effect? of water vapor is much higher than that of dry air/CO2 even at sea level.

    What was interesting about the data from the Barcelos, Brazil was that the average humidity was 90% in the month of May. The average humidity was 80% for the ten clear days and the temperature was nearly the same day or night whether it was clear or cloudy for that month.

    Given the obvious effects of clouds, water vapor and the oceans I can not believe the IPCC pretty much ignores their effects on the earth’s climate. The radiative forcing attributed to water vapor in the IPCC Components of Radiative Forcing Chart makes absolutely no sense given the data from Barcelos, Brazil vs Adrar, Algeria. Heck they only list H2O in the statosphere! Where the heck is the rest of the radiative forcing from water ??? Dumped in with CO2 as a multiplier effect! (Grumble)

    So now that I have finished my ramble, I will clarify my question. Are we looking at gray body radiation from the atmosphere because the sun warms the atmosphere/water vapor/clouds or are we looking at IR emitted from the earth and captured and re-emitted by greenhouse gases?

  353. Now normally we think of air as being quite thermally conductive, because we use it as a coolant, but if you but enough of it in series you end up with a very good insulator.
    –Ryan

    Whenever talking about the conductive qualities of air, remember: whatever the atmosphere does, it is in series with the vacuum of space…which is the only reason we spend so much time considering the slippery effects of radiation. In the electrical analogy, the “resistance” of space is like a billion ohms, while the atmosphere would be much less and from radiative terms is mostly irrelevant…though the atmosphere can “smear” peak temperatures which has a great effect on peak radiation (keep in mind the t^4 relation). I think the real key to any comfort we enjoy on Earth is simply related to how ineffective radiation is at heating and cooling. Our comfort comes from the asymmetry of the hot ball of fire radiating very intensely and our cool ball of integrating/storing water radiating very weakly.

    As an engineer trying to cool hot CPUs, sometimes thermal radiation works in my favor and sometimes it works against me, but the bottom line is, compared to conduction and convection, radiation doesn’t do much…and its effects can often be ignored with little error. Radiation from weak radiators is a steady process, but it’s weak and slow…to do anything useful with radiation (like increasing the average temperature of something by 33C), the energy source must be very intense compared to the energy represented by the +33C I’m trying to get.

  354. Ryan says:

    Ken, generally I agree with you, but I would say that space (being absorbtive and at absolute zero ) is the electrical analogue of ground (0V). The sun is the electrical analog of a voltage source at 5500V (!) although we probably need to split it into three supplies at different frequencies for UV, IR and visible. The stratosphere has a high impedance to UV, the earth has a high impedance to all three and the troposphere has a low impedance to all three (but there is an awful lot of it). Seems to me you could do a mesh analysis like you would do for electrical currents and “solve” for the atmosphere!

    I don’t suppose climatologists are used to doing a mesh analysis. Maybe that is the crux of the problem. The science of multiple sources at different frequencies and multiple energy flows has been known for years in electrical/electronic engineering but is unknown outside this field.

  355. gnomish says:

    exhibiting melodramatic monologue, willis said:
    “You have your fantasy, and by God, neither Stefan nor Boltzmann are gonna get in your way.
    Now, you come back with the meaningless statement that since emissivity isn’t a function of temperature, there is no “direct conversion” of watts to temperature … what on earth are you on about? For a given object,
    W = C T4
    ———————————————————————————————————————-
    *sigh*… i reply:
    just as there is a relationship between volume and mass, the properties are not convertible without the additional factor of density.
    likewise, temperature and watts are not convertible without an additional factor.
    you know this, right? you also understand the meaning of ‘directly converted’, right?
    you removed that variable from the equation, thus misstating it. why so tricksy?

    i may be about to learn something. i asked you please to stand aside and not interfere.

    but nooo- you are desperate for attention. i can oblige for i am a connoisseur of schadenfreude and you are offering a feast.
    pay close attention – there will be a test:
    trolling your own thread doesn’t save WUWT from obscurity as you once claimed in a fever of unjustifiable self importance.
    you are not well skilled in the art as you clearly imagine yourself to be.
    whining about non-existent whining is an old trick of feeble old trolls that can be wrapped around their necks till they shriek.
    i wonder if you even know how to keep score, because you’re losing badly right from the starting line.

  356. Categorize this under: “The Truth is Out There”…

    The good news about natural convection at high altitude is that the heat transfer doesn’t degrade as badly as it does in forced convection. Because the flow is generated by the temperature difference, it tends to be self-correcting, in a manner of speaking. With a fan, you a stuck with a fixed volumetric flow, so the mass flow rate, and the heat transfer, goes down linearly with density.
    But in natural convection, if the density goes down, the mass flow goes down, and the temperature difference goes up. But that forces the flow rate to increase, and partially compensates for the loss of mass flow.

    Tony Kordyban
    http://www.coolingzone.com/library.php?read=515

  357. Willis Eschenbach says:

    Ryan says:
    December 20, 2012 at 6:17 am

    Willis: once again be careful about your tendency to be overly assertive. Gnomish is essentially correct that there is no direct relation ship between watts of radiated power and temperature in any given real-world case.

    Thanks, Ryan, but if you and Gnomish were “essentially correct” that there is “no direct relationship between watts of radiated power and temperature”, then there couldn’t possibly be such a creature as an infrared thermometer.

    But there are infrared thermometers, you can buy them at Home Depot. You point them at an object, they measure the IR of the object, and they tell you the temperature of the object.

    So … if there is no direct relationship between IR and temperature as you and Gnomish foolishly claim … how does an IR thermometer work?

    You also say:

    Stefan/Boltzmann theorised a perfect absorbtive object that converts the electro-magnetic energy to heat with 100% efficiency and therefore were able to develop a relationship between the electro-magnetic power reaching the object and the temperature. Any real-world application needs to take into account that the process is likely to be nowhere near 100% efficient. The process is nowhere near 100% efficient for the surface of the Earth.

    How far it deviates from the theoretical perfect given blackbody is calculated using the “emittance”. Your claim that the process is “nowhere near 100 percent efficient” indicates that you don’t understand what is going on, because the emittance of the earth in the IR region is generally greater than 95%, which is not far from 100% in my book … and it’s not “efficiency” that you are measuring in any case, it is emittance.

    Since all of the points that you made are incorrect, I’d have to say, Ryan … that once again you should be careful about your tendency to be overly assertive.

    Sorry, couldn’t resist …

    w.

  358. Willis Eschenbach says:

    gnomish says:
    December 20, 2012 at 11:14 am (Edit)

    exhibiting melodramatic monologue, willis said:

    “You have your fantasy, and by God, neither Stefan nor Boltzmann are gonna get in your way.
    Now, you come back with the meaningless statement that since emissivity isn’t a function of temperature, there is no “direct conversion” of watts to temperature … what on earth are you on about? For a given object,
    W = C T4”

    ———————————————————————————————————————-
    *sigh*… i reply:
    just as there is a relationship between volume and mass, the properties are not convertible without the additional factor of density.
    likewise, temperature and watts are not convertible without an additional factor.
    you know this, right? you also understand the meaning of ‘directly converted’, right?
    you removed that variable from the equation, thus misstating it. why so tricksy?

    Oh, OK, I see what you are on about. It seems that you think that something like

    W = T4

    is a “direct relationship” but

    W = C T4

    where “C” is a constant for any given object is not a “direct relationship”.

    Hmmm … I’m afraid I can’t see the difference. Perhaps you could explain it to the class.

    OK, gnomish, have it your way. For me a “direct relationship” means that one of the variables is directly PROPORTIONAL to the other variable in some sense. For example, I would say the nutrition of a child has a direct relationship with its health. Similarly, I would say that the amount of gunpowder in a cannon has a direct relationship with how far the cannonball goes.

    Now, to make the actual calculation for the cannonball’s trajectory you need other details—air temperature, weight of the ball, air density, and the like.

    But for me, none of that wipes away the direct relationship between gunpowder and distance.

    Obviously, you have some very different meaning for a “direct relationship”, my bad for not reading your mind and knowing that.

    w.

  359. Willis Eschenbach says:

    gnomish says:
    December 20, 2012 at 11:14 am

    … I may be about to learn something. i asked you please to stand aside and not interfere.

    Yes, and I pointed out that there is no possible way for me to interfere with your learning something here on the blog, whether I post or not.

    Either you learn or you don’t, gnomish, and your claim that I am somehow standing in the way of you learning something is a risible and physically impossible excuse.

    w.

  360. Willis Eschenbach says:

    Gail Combs says:
    December 20, 2012 at 6:18 am

    Willis Eschenbach says:
    December 20, 2012 at 12:19 am

    ….So, just what is your contention here? That DLR isn’t measured? I discussed the DLR measurements at the TAO buoys here.…..

    >>>>>>>>>>>>>>>>>>>>>
    Willis, I Just took a look at that information and I have a question.

    First clouds are composed of tiny water droplets or ice crystals that are suspended in the air. Those clouds would have a temperature T, and would radiate as a gray body. This radiation would be a percentage of the theoretical black body radiation which is proportional to the fourth power of the absolute temperature as expressed with Stefan-Boltzmann Law. Would this not be confounding the whole idea of DLR from greenhouse gases. Heck would not the entire atmosphere have DLR because it is a gray body radiator and therefore the DLR would again have nothing to do with the Greenhouse gas theory except as a confounding contributor?

    Thanks, Gail. The emissivity of clouds is generally taken as 1.0, they are so close to a theoretical black body.

    Would this “confound the whole idea of DLR from GHGs?

    Well, no. They are different phenomena. You can see it in records of downwelling longwave radiation (DLR). There is a background DLR level, due to water vapor and CO2 and other GHGs. But when a cloud comes over, the DLR (generally) increases, because the clouds such excellent absorber-emitters of longwave radiation.

    Take a look at the graphic I posted above. The background DLR is about 300 W/m2, and you can see the variations in that value due (mainly) to clouds.

    w.

  361. Greg House says:

    Willis Eschenbach says, December 20, 2012 at 12:07 pm:
    “Thanks, Ryan, but if you and Gnomish were “essentially correct” that there is “no direct relationship between watts of radiated power and temperature”, then there couldn’t possibly be such a creature as an infrared thermometer.
    But there are infrared thermometers, you can buy them at Home Depot. You point them at an object, they measure the IR of the object, and they tell you the temperature of the object.
    So … if there is no direct relationship between IR and temperature as you and Gnomish foolishly claim … how does an IR thermometer work?”

    ========================================================

    Anyway, not the way you imagine it works, dear Willis. Generally, do not just believe your eyes, dig deeper.

    From my humble knowledge, IR thermometer calculates temperature on the basis of assumed emissivity of the target. Guess, what happens, if emissivity of the target is significantly different from the assumed one?

    Hence, Willis, if an IR thermometer was not fed with that assumption (by the manufacturer), it would be unable to calculate the temperature of the target at all.

    As you can possibly see now, things are sometimes different from what they appear to be. I hope it will be sufficient reason for you to reduce a little bit the teaching connotation your comments sometimes probably unintentionally contain.

  362. willis Eschenbach says:

    Greg House says:
    December 20, 2012 at 1:54 pm

    Anyway, not the way you imagine it works, dear Willis. Generally, do not just believe your eyes, dig deeper.

    From my humble knowledge, IR thermometer calculates temperature on the basis of assumed emissivity of the target. Guess, what happens, if emissivity of the target is significantly different from the assumed one?

    Hence, Willis, if an IR thermometer was not fed with that assumption (by the manufacturer), it would be unable to calculate the temperature of the target at all.

    As you can possibly see now, things are sometimes different from what they appear to be. I hope it will be sufficient reason for you to reduce a little bit the teaching connotation your comments sometimes probably unintentionally contain.

    Say what? Is any of that supposed to be news?

    Of course the IR thermometers have to deal with the emissivity, Greg. Stefan-Bolzmann’s law contains emissivity, how could they not have to deal with that? The good IR thermometers have an emissivity adjustment, viz:

    Five Ways to Determine Emissivity
    There are five ways to determine the emissivity of the material, to ensure accurate temperature measurements:
    • Heat a sample of the material to a known temperature, using a precise sensor, and measure the temperature using the IR instrument. Then adjust the emissivity value to force the indicator to display the correct temperature.
    • For relatively low temperatures (up to 500°F), a piece of masking tape, with an emissivity of 0.95, can be measured. Then adjust the emissivity value to force the indicator to display the correct temperature of the material.
    • For high temperature measurements, a hole (depth of which is at least 6 times the diameter) can be drilled into the object. This hole acts as a blackbody with emissivity of 1.0. Measure the temperature in the hole, then adjust the emissivity to force the indicator to display the correct temperature of the material.
    • If the material, or a portion of it, can be coated, a dull black paint will have an emissivity of approx. 1.0. Measure the temperature of the paint, then adjust the emissivity to force the indicator to display the correct temperature.
    Standardized emissivity values for most materials are available (see pages 114-115). These can be entered into the instrument to estimate the material’s emissivity value. SOURCE

    I’m surprised you would assume I don’t know these things …

    My issue is, I’m not at all sure what your point is here. We seem to have a semantics problem. I call a relationship like

    W = C T4

    where C is a constant for any given body a “direct relationship”.

    You don’t.

    Having established that, what is the underlying point that you are trying to make? That objects don’t radiate in the IR? That their radiation is not a function of temperature and emissivity? I’m not clear on what you think I’ve done wrong.

    Let me see if I can explain this a different way:

    If I have a lump of steel, in any condition, and I double its temperature, I will get sixteen times the wattage of radiation from it that I got at the original temperature.

    If I have a lump of copper, and I double its temperature, I will get sixteen times the radiation.

    The same is true for every object that you can name. If you double its temperature, you get sixteen times the radiation. The is because for any given object that you can name, radiation (watt/m2) is directly proportional to the fourth power of the temperature (K).

    I call that a “direct relationship”. YMMV, you may wish to call it something else, but either way, what is the underlying error that you seem convinced that I am making?

    Regards,

    w.

  363. Willis Eschenbach says:

    One last comment on IR thermometers. While it is true that emissivity plays a part in the relationship between temperature and IR radiation, the difference is more apparent than real.

    This is because most objects have an emissivity of 0.9 or better. Let me quote from a previous post:

    My bible for many things climatish, including the emissivity (which is equal to the absorptivity) of common substances, is Geiger’s The Climate Near The Ground, first published sometime around the fifties when people still measured things instead of modeling them. He gives the following figures for IR emissivity at 9 to 12 microns:

    Water, 0.96
    Fresh snow, 0.99
    Dry sand, 0.95
    Wet sand, 0.96
    Forest, deciduous, 0.95
    Forest, conifer, 0.97
    Leaves Corn, Beans, 0.94

    and so on down to things like:

    Mouse fur, 0.94
    Glass, 0.94

    You can see why the error from considering the earth as a blackbody in the IR is quite small.

    I must admit, though, that I do greatly enjoy the idea of some boffin at midnight in his laboratory measuring the emissivity of common substances when he hears the snap of the mousetrap he set earlier, and he thinks, hmmm …

    As you can see, an IR thermometer with an assumed emissivity of 0.95 will only have an error of a couple of percent for everything from sand to glass to trees to mouse fur …

    w.

  364. gnomish says:

    a couple of percent Kelvin, you might point out, translates to @ 10 degrees F.
    that might be considered significant when the matter of global warming hinges on tenths and hundredths.
    wrt your misdirection: i said ‘direct conversion’ not ‘direct relationship’. but you knew that.
    presently, i hope R. Clemenzi may give a conclusive reply to my question.
    his comments are more digestible because he has no need to assuage a deep insecurity by fighting with ignorant nobodies like myself.

  365. Greg House says:

    Willis Eschenbach says, December 20, 2012 at 12:07 pm:
    “So … if there is no direct relationship between IR and temperature as you and Gnomish foolishly claim … how does an IR thermometer work?”
    Willis Eschenbach says, December 20, 2012 at 12:24 pm:
    “For me a “direct relationship” means that one of the variables is directly PROPORTIONAL to the other variable in some sense.”

    Willis Eschenbach says, December 20, 2012 at 2:33 pm:
    “My issue is, I’m not at all sure what your point is here. We seem to have a semantics problem.”
    ===========================================================

    No, Willis, it is that you present a (bad) semantic “solution” to a non-semantic problem.

    Your “directly PROPORTIONAL to the other variable in some sense” is a contradiction in itself.

    No IR thermometer can determine the temperature of the target from IR alone. Which means, it can receive more IR from an object having lower temperature and the same amount of IR from another object having higher temperature. This is not what people would call, in your words, “directly PROPORTIONAL”. It is just your semantic “solution” to your problem.

    This is the 4th “semantic” solution you have presented to the public recently. Let me remind you of the other 3 modern warmism is based on:

    1) The “semantic” solution to the problem with the 2nd Law of Thermodynamics: “climate science” invented the “NET” thing that is not to be found in the historical formulations of the 2nd Law, nor is it apparently proven experimentally.

    2) The “semantic” solution to the experimentally unproven “warming via back radiation”: to somehow theoretically substantiate it they cut the solar power in half.

    3) And last not least, the “semantic” solution to the problem with the calculations of so called “global temperature”: they just call what they calculate without any basis in science (“extracting” temperatures for large areas from single or few weather stations, “reconstructing” temperatures etc.) “global temperature”, apparently even without any scientifically established definition.

  366. Mack says:

    Well as a result of all this carry on here you won’t be blathering on any more about oceans freezing will you Willis. Here’s some Xmas holiday reading for you….. call it Xmas cheer. because the message should bring comfort and joy.
    http://jennifermarohasy.com/author/nasif-s-nahle/

  367. Willis Eschenbach says:

    Greg House says:
    December 20, 2012 at 3:36 pm

    Willis Eschenbach says, December 20, 2012 at 2:33 pm:

    “My issue is, I’m not at all sure what your point is here. We seem to have a semantics problem.”

    ===========================================================

    No, Willis, it is that you present a (bad) semantic “solution” to a non-semantic problem.

    Greg, I don’t care about the semantics. I don’t care what you call these things. I just want to know what your issue is. You seem very emphatic about something, but what?

    I assure you that statements like:

    The “semantic” solution to the experimentally unproven “warming via back radiation”: to somehow theoretically substantiate it they cut the solar power in half.

    do not assist me in trying to find out what it is that your are trying to say, or what error of mine you are pointing out. What on earth does that statement of yours mean? On second thought, never mind, I don’t want to know. What I do what to know is, what is it that you are on about?

    Please boil it down to as few and as clear words as possible, because I can’t make sense of the above.

    w.

  368. Gail Combs says:

    Willis Eschenbach says:
    December 20, 2012 at 1:26 pm
    …Well, no. They are different phenomena. You can see it in records of downwelling longwave radiation (DLR). There is a background DLR level, due to water vapor and CO2 and other GHGs. But when a cloud comes over, the DLR (generally) increases, because the clouds such excellent absorber-emitters of longwave radiation…..
    >>>>>>>>>>>>>>>>>>>>>>>>>>>
    I am still confused. Maybe it is just semantics. I interpret what you are saying to mean that without a Greenhouse Gas (GHG) Effect the atmosphere immediately after sunset would drop to absolute zero and therefore all DLR is due to the GHG and there is no DLR (IR) from the atmospheric gases being at a temperature above 0K.

    1. I think we all agree there is IR radiation from all matter that is not at absolute zero, correct?

    I did have a bit of difficulty finding what the gray body radiation for the atmosphere/gas was. (White screen of death problem) However see bottom of comment.

    2. Gases do emit IR radiation and the amount is dependent on the partial pressure and the temperature. (see bottom)

    3. Radiation from a gas not at absolute zero is due to temperature and has nothing whatever to do with the greenhouse gas effect if you are considering only gray/black body radiation.

    4. Pure nitrogen will warm to room temperature through conductivity and convection. Pure air (without GHGs) will also warm without the influence of GHG effect. Although this observational data seems to indicate sunlight has a heck of a lot to do with the warming of the atmosphere.

    5. The links I made in the last post indicate the emissivity of air, at least in those mountains, was 0.5 or less without clouds. The below articles indicate CO2 has an emissivity of 0.0027. Since you indicated in another comment you consider all emissivities ~ 0.95 I think this may be the key concept.

    6. I think we can at least agree that the emissivity of clouds is 0.98 to 1.0 and the emissivity of dry air is less, somewhere between 0.5 to 0.83.

    7. There is also the Specific Heat Capacity of Water water has to absorb 4.184 Joules of heat for the temperature of one gram of water to increase 1 degree celsius (°C). Nitrogen has a specific heat at constant pressure of 1.04 @ at 68F (20C) and 14.7 psia (1 atm) and for air ~ 1.00

    Could the difference in DLR between clear and cloudy days be due mostly to the difference in the emissivity and heat capacity of air vs clouds without considering IR energy bounced from the earth to the clouds and back again?

    From the table, the emissivity of carbon dioxide decreases with height and its partial pressure. In addition, the total emittance of CO2 with a partial pressure (Pp) of 1 atm*m would not exceed 0.9 W/m^2. At its current partial pressure, the CO2 has a total emittance of 0.423 (second line in blue characters) For this reason the value for the total emittance (€) given by some authors from the IPCC -5.35 W/m^2- is not the actual value, but an adaptation to make the numbers agree with pre-assumed and subjective numbers. The IPCC team of experts has changed the radiative forcing so many times that the IPCC team have had to admit that the numbers are not real. The real values for the emittance or “radiative forcing” have been provided by the heat transfer science and thermodynamics.

    Knowing the real values of emissivity, total emittance and absorptivity of the carbon dioxide obtained by other scientists by experimentation and observation of nature, the formula derived from the Stephan-Boltzmann’s equation adopts the following form, introducing values from the real world:

    ΔT = (0.423 W/m^2) [ln ([CO2] current / [CO2] standard)] / 4 (5.6697 x 10^-8 W/m^2*K^4) (300.15) ^3
    http://www.biocab.org/emissivity_co2.html

    Determination of Mean Free Path of Quantum/Waves and Total Emissivity of Carbon Dioxide Considering its Molecular Cross Section.

    …The formula to calculate the total emissivity of fluids, plasma and free electrons is as follows:

    ε [epsilon] = (1-e^ (t * (- (1/s))) / √π

    Where ε [epsilon] is the total emissivity of emitter particles and t is the crossing time lapse of a quantum/wave.

    Known values:

    t = 0.0049 s

    Introducing magnitudes:

    ε [epsilon] = (1-e^[(0.0049 s * (-1/s))]) / (√3.1415…) = 0.0027….

    That is emissivity for CO2 is 0.0027

  369. Willis Eschenbach says:

    gnomish says:
    December 20, 2012 at 3:34 pm

    a couple of percent Kelvin, you might point out, translates to @ 10 degrees F.
    that might be considered significant when the matter of global warming hinges on tenths and hundredths.
    wrt your misdirection: i said ‘direct conversion’ not ‘direct relationship’. but you knew that.
    presently, i hope R. Clemenzi may give a conclusive reply to my question.
    his comments are more digestible because he has no need to assuage a deep insecurity by fighting with ignorant nobodies like myself.

    I’m not fighting with you, gnomish, I’m trying to find out what you are so concerned about. Again, I have to ask, what is your point? Surely it is not that a couple of percent kelvin is around 6°C, we both knew that. But what is it that you find so important? It seems you think I’ve make some terrible error in saying that my desk radiates about 400 W/m2, or maybe it was something else that I said, but I haven’t a clue.

    Is your point really that you said “direct conversion” and I said “direct relationship”? Are you serious when you claim that this was somehow to “misdirect” you?

    Rather than spending your time accusing me of “deep insecurity” and “misdirection”, and complaining that I’m not as good as someone else at making “digestible” comments, how about you focus on the science instead? I freely confess that my failings are manifold, I was born yesterday, but my morals and motives have nothing to do with the science. Your continued insistence on attacking me is warping your thinking and causing you to lose focus on the science itself. It’s gotten so bad you think I’m attacking you. I’m not. I’m just saying, both I and a variety of knowledgeable people have commented upthread that you are badly mistaken in some of your views. You should at least consider the possibility that they might be right …

    My regards to you,

    w.

    PS—The error from a preset emissivity of 0.95 being in error by a couple of percent is not ten degrees F in any case:

    Calculated temperature, 390 W/m2, emissivity 0.95, 18.5°C
    Calculated temperature, 390 W/m2, emissivity 0.97, 17.0°C

    Difference, 1.5°C, or 2.7°F

    So the 2% error in emissivity translates to about a 1.5°C (2.7°F) error, not a 10°F error as you state.

  370. jae says:

    Willis:

    Thanks for the links to the radiation charts. I was unaware of them. I’m intrigued by the fact that the downwelling radiation appears to be little affected by virtually ANYTHING during warm days when it is clear. Pick a day in July, August, or September and plot downward solar radiation, downward IR, and temperature–and then look at all the sites. You get about 400 Wm-2, regardless of altitude, latitude, temperature, time of day, and amount of GHGs (water vapor). I find this weird. Speechless for now…

  371. TimTheToolMan says:

    Willis writes “There is a background DLR level, due to water vapor and CO2 and other GHGs. But when a cloud comes over, the DLR (generally) increases, because the clouds such excellent absorber-emitters of longwave radiation.”

    This IS semantic but IMO important. The DLR increases because the flow of IR energy away is reduced, not inherantly because clouds are good absorber-emitters. That happens miles away. The DLR that increases, the stuff you might measure at the ground, isn’t DLR that was emitted by the cloud.

    Its just the way you think about it, Willis and its cooling not warming.

  372. Willis Eschenbach says:

    jae says:
    December 20, 2012 at 5:42 pm

    Willis:

    Thanks for the links to the radiation charts. I was unaware of them. I’m intrigued by the fact that the downwelling radiation appears to be little affected by virtually ANYTHING during warm days when it is clear. Pick a day in July, August, or September and plot downward solar radiation, downward IR, and temperature–and then look at all the sites. You get about 400 Wm-2, regardless of altitude, latitude, temperature, time of day, and amount of GHGs (water vapor). I find this weird. Speechless for now…

    Every time I look at some of those charts, I come away mystified by something.

    Regarding variations in the DLR, what kind of variations were you expecting? All of them have diurnal variations on the order of 80-100 W/m2. It’s difficult to say what the average might be for any of them without downloading the data. I see some that vary maybe 320-420 W/m2, average maybe 370, and others that vary 290-370 or so on the same day, average maybe 330.

    The S-B temperatures corresponding to those average radiation levels with emissivity = 0.95 are about 56°F (13°C) for the larger average and about 44°F (7°C) for the smaller average of 330 W/m2.

    These were early summertime records I looked at, June 1st. The downwelling radiation represents the temperature of the atmosphere from which the radiation originated.

    It is not widely appreciated that almost three-quarters (72%) of the downwelling longwave radiation at the surface comes from the lowest 87 metres (285 feet) of the atmosphere. Or to look at it another way, 90% of the downwelling radiation hitting the ground comes from the lowest half kilometer (a third of a mile) of the atmosphere. SOURCE: My bible, Geiger’s classic text “The Climate Near The Ground”, sixth edition (2003). An early version (1950) is a free download here. Modern version is on Amazon, new $125, used $45.

    As a result, the intensity of the downwelling radiation is mostly representative of the air temperature of those lowest atmospheric layers. For these lowest levels, that range of 45-55°F in the summertime, seems reasonable. I usually figure that temperature drops about a degree C per 100 metres. At an altitude of 500 metres, that would be a drop of 5°C (9°F) from the surface air temp or so.

    Finally, I would expect the variations in lower atmospheric temperatures over hundreds of miles to be far less than those observed at ground stations, with their host of microclimates. I’d expect the atmosphere to average a whole lot of that out, and over large areas.

    … But of course with steep thermoclines where you least expect them, you gotta love Nature, she does pierced, patched, and plotted very well, but she doesn’t do gradual all that stunningly. Even in the atmosphere.

    Anyhow, that’s my thoughts on the matter. I hardly pretend to understand the nuances (and in some cases the broad strokes) of what I find in the actual observations … gotta love it, always more to learn.

    w.

  373. Greg House says:

    Willis Eschenbach says, December 20, 2012 at 4:57 pm: “I assure you that statements like: “The “semantic” solution to the experimentally unproven “warming via back radiation”: to somehow theoretically substantiate it they cut the solar power in half.” do not assist me in trying to find out what it is that your are trying to say, or what error of mine you are pointing out. What on earth does that statement of yours mean?
    ==========================================================

    OK, Willis, I will explain it again, because the point is very important. And it is not your error primarily, because it probably was not you who invented that bogus calculation about the -18C cold “Earth without atmosphere”. Let us do it point by point.

    1. The basic NASA’s statement about total solar irradiance (TSI) is this: “At Earth’s average distance from the Sun (about 150 million kilometers), the average intensity of solar energy reaching the top of the atmosphere directly facing the Sun is about 1,360 watts per square meter, according to measurements made by the most recent NASA satellite missions. This amount of power is known as the total solar irradiance”. There are at least 2 things in this statement indicating that TSI goes for the whole hemisphere and not for an imaginable circle/disk.

    First, they say reaching the top of the atmosphere directly facing the Sun. It is obvious that the top of the atmosphere directly facing the Sun is not a flat disk, but a hemisphere, exactly like the half of the Earth directly facing the Sun.

    Second, they call it “TOTAL solar irradiance”. The word TOTAL is another indication that the whole hemisphere is meant.

    Conclusion: in your calculation the real solar power is cut in half. This is one error.

    2. The other statement you quoted from the same NASA page contradicts this one, they replace in fact “average” and “total” with “maximum” for no scientific reason.

    This replacement is what I call a semantic “solution” to a problem. The problem is, as I said earlier, that there is apparently no physical experimental proof of warming via back radiation, and that “solution” helps a little bit.

    Another problem with you calculation is, even if we put aside that hemisphere/disk error, that your Earth does nor rotate at all. You consider only the half of the Earth to receive solar power. This is apparently another error, because the Earth does rotate. The WHOLE rotating Earth is receiving solar power and the part that temporarily is not, has not been proven to cool down to the absolute zero, before it faces the sun again. If it does not cool down to the absolute zero, the average temperature will be higher. In your calculation it is not taken into consideration, this is the second error.

    The third error is your way of operating with averages, Ryan already said that on this thread (December 18, 2012 at 4:57 pm) and gave you an easy example. Unfortunately, you failed to comment on that, as well as on the “rotating Earth” argument of mine.

    I hope this explanation will help.

  374. Willis Eschenbach says:

    TimTheToolMan says:
    December 20, 2012 at 6:39 pm

    Willis writes

    “There is a background DLR level, due to water vapor and CO2 and other GHGs. But when a cloud comes over, the DLR (generally) increases, because the clouds such excellent absorber-emitters of longwave radiation.”

    This IS semantic but IMO important. The DLR increases because the flow of IR energy away is reduced, not inherantly because clouds are good absorber-emitters. That happens miles away. The DLR that increases, the stuff you might measure at the ground, isn’t DLR that was emitted by the cloud.

    You are correct, the downwelling radiation from upper layers (clouds or not) will be absorbed at lower levels. Those levels will radiate in turn, upwards and downwards. Some of that eventually reaches the ground.

    When a cloud comes over, it intercepts 100% of the upwelling radiation. As you point out, this reduces the flow out to space, with half of the radiation being redirected back to the surface.

    Its just the way you think about it, Willis and its cooling not warming.

    Not clear what this means.

    w.

  375. TimTheToolMan says:

    Mathematically yes, practically no. Which way is the energy flow?

    Actually more the point how does the temperature gradient from ground to cloud look do you think?

  376. Robert Clemenzi says:

    Greg House says:
    December 20, 2012 at 9:11 pm

    It is obvious that the top of the atmosphere directly facing the Sun is not a flat disk

    That is exactly what the phrase reaching the top of the atmosphere directly facing the Sun means. You need to read a textbook instead of the high level summary on the NASA page. General science pages tend to not use phrases like “normal to the surface” because too many people won’t understand it. Consider a round disk in space and a line between the center of the disk and the Sun. If the disk is oriented at right angles to that line, we say that the line is “normal” to the plane of the disk. Now change the angle between the plane of the disk and the line. When you look at the circular disk from the Sun, it will now have the apparent shape of an ellipse. The area of that ellipse will be smaller than the area of the disk. The question is – Will the disk still get the same number of watts per unit area that it received when it was oriented normal to the line between the Sun and the disk? It should be obvious that the amount of sunlight per unit area decreases as the disk is tilted. Typically, the new value is computed by taking the amount of energy available, dividing by the area of the disk, and multiplying by the area of the ellipse seen from the Sun.

    This amount of power is known as the total solar irradiance

    That refers to the total amount of energy available to objects the same distance from the Sun as the Earth. The word “total” is usually meant with respect to what is available at the surface. In other words, the amount at the surface is the total available amount minus what is absorbed in the atmosphere and reduced by an amount related to the angle between the local surface normal and the position of the Sun. Sometimes, the word “maximum” is used to stress that nothing that occurs in the atmosphere can increase that amount of energy above that value.

    your Earth does nor rotate at all

    The reason the energy shown in most diagrams is one fourth the energy at the top of the atmosphere is two fold.
    * The Earth is a sphere
    * The Earth rotates

  377. Ryan says:

    Yesterday I had a mug of hot coffee sitting on my desk in one of those insulated flask mugs. Unfortunately I left it there too long and it had got cold. I guess it took about half an hour to get cold. It started off at about 70Celsius I guess and within 2000seconds the water got cold. I did some calculations in miyt head and came to the conclusion that a body of water 1sqm in area heated to the same temperature would be losing something like 10,000watts/sqm, of that order.

    Now that’s a high rate of loss but a coffee mug is nowhere near thermal equilibrium with the air. The heat from the mug is dispersing into the air and just a meter up you can’t feel the heat anymore. You could argue that it isn’t very well constrained to stay in a single column of air, but consider cars in a traffic jam – each engine is at about 70Celsius but if you are on a bridge over the traffic jam you can’t really feel the extra heat.

    What the previous paragraph tells me is that the daytime heating of land and sea by the sun doesn’t really heat up the atmosphere that much. The atmosphere really just smooths the heat out from day to night just as the ocean does. There’s just too much of it for the heating of just one day to make much difference to the temperature of the troposphere. When I go skiing in the Alps the valley can be at +25Celsius but at 3000m up you don’t see much melting of the ice due to all this extra heating – the lapse rate of the air just shrugs it off.

    Then I turned this idea on its head. I thought “if the hot coffee is losing heat at a rapid rate but gives me a steep temperature gradient through the air, what does that tell me about how much power is being lost through the air if the temperature gradient is very low?” Another calculation showed a very low rate of heat energy loss by this means – as you would expect if you have a low temperature gradient there just isn’t enough “ooomph” to get a lot of heat flow. A few watts/sqm.

    What this tells me is that actually the atmosphere on its own would boil us all alive if there was no other cooling path. This has nothing to do with greenhouse gases. Just the sheer quantity of atmosphere causes a temperature gradient to build across it that means the temperature gradient is very low and you can’t get much heat flow through it. Fortunately the heat from the Earth can simply flow directly to space. Essentially the mechanism for keeping the Earth warmer than expected is simply photonic energy reaching land and sea directly but then be impeded in its escape by a vast body of air where greenhouse gasses are actually pretty irrelevant because most of the “cooling” is actually due to convection and so the Earth needs to reach a hotter equilibrium point in order to emit most of its heat energy by direct IR to space.

    Trenberth’s “missing heat” from satellite measurements isn’t due to extra heat being stored in the deep oceans it is simply that the observations are difficult to do sufficiently accurately (especially if you are looking to find heat trapped by CO2 in a mechanism that doesn’t exist).

  378. jae says:

    Willis says at 12/20, 7:32:

    “Regarding variations in the DLR, what kind of variations were you expecting? All of them have diurnal variations on the order of 80-100 W/m2. It’s difficult to say what the average might be for any of them without downloading the data. I see some that vary maybe 320-420 W/m2, average maybe 370, and others that vary 290-370 or so on the same day, average maybe 330.”

    Not so, Willis; that’s what bugs me. Look at all sites on July 15, 2012, for example. ALL SITES show DW IR of about 400 +/- 25 w/m2 DAY AND NIGHT! That’s weird, considering the enormous variation between the sites in terms of amounts of GHG (water vapor), altitude, latitude, temperature, sun/no sun, etc.. I don’t understand it at all.

  379. Ken Coffman says:

    This would be an interesting experiment: heat up one-square-foot cubes of iron and balsa wood at 300C until the temperature of both is stable as measured with an IR thermometer. Let them sit for ten minutes, then I’ll put my bare hand on the balsa wood and an academic climatologist can put his or her hand on the iron block.

    Here’s a slight modification which will represent equal storage of thermal energy. Heat up 100-pound blocks of iron and balsa wood at 300C until the temperature of both are stable as measured with an IR thermometer. Let them sit for ten minutes, then I’ll put my bare hand on the 100-pound balsa wood block and an academic climatologist can put his or her hand on the hundred pound iron block.

    In both cases, which test subject is more likely to need to visit an emergency room?

    Interesting how the temperature of stuff does not tell a full thermodynamic story, eh? Wouldn’t we conclude from this that it’s very important to know what substance is doing the radiating? Further, might we not conclude that whatever radiates from air to land is irrelevant with regard to the surface temperature?

  380. Willis Eschenbach says:

    Greg House says:
    December 20, 2012 at 9:11 pm (Edit)

    Willis Eschenbach says, December 20, 2012 at 4:57 pm:

    “I assure you that statements like: “The “semantic” solution to the experimentally unproven “warming via back radiation”: to somehow theoretically substantiate it they cut the solar power in half.” do not assist me in trying to find out what it is that your are trying to say, or what error of mine you are pointing out. What on earth does that statement of yours mean?

    ==========================================================

    OK, Willis, I will explain it again, because the point is very important. And it is not your error primarily, because it probably was not you who invented that bogus calculation about the -18C cold “Earth without atmosphere”. Let us do it point by point.

    Many thanks for your explanation, Greg. Let me say in front that I am happy to answer your questions.

    1. The basic NASA’s statement about total solar irradiance (TSI) is this: “At Earth’s average distance from the Sun (about 150 million kilometers), the average intensity of solar energy reaching the top of the atmosphere directly facing the Sun is about 1,360 watts per square meter, according to measurements made by the most recent NASA satellite missions. This amount of power is known as the total solar irradiance”. There are at least 2 things in this statement indicating that TSI goes for the whole hemisphere and not for an imaginable circle/disk.

    First, they say reaching the top of the atmosphere directly facing the Sun. It is obvious that the top of the atmosphere directly facing the Sun is not a flat disk, but a hemisphere, exactly like the half of the Earth directly facing the Sun.

    Second, they call it “TOTAL solar irradiance”. The word TOTAL is another indication that the whole hemisphere is meant.

    Conclusion: in your calculation the real solar power is cut in half. This is one error.

    Rather than focus on the words of one person at NASA, lets develop the TSI from first principles.

    The amount of sun reaching the earth is related to a couple things—the distance from the sun to the earth and the luminosity (brightness) of the sun. This makes sense, because the TSI decreases with distance from the sun and increases with the luminosity of the sun.

    The formula to calculate TSI is simple.

    TSI = L / (4 Pi A2)

    where L is the luminosity of the sun, 3.84e+26 Watts, and A is the unit distance from the sun to the earth, 1.50e+11 metres. Since A gets squared, the answer is in watts per square metre.

    What we are doing here is just total solar energy divided by total area at the distance of the earth, to get the energy per unit area (square metre) at the orbit of the earth. The area is the area of a sphere with a radius equal to the distance from the earth to the sun, which is 4 Pi A2.

    When we substitute, we get

    TSI = 3.84e+26 W / (4 * 3.1416 * (1.5e+11)2)

    which simplifies to 1,364 W/m2. Note that this is perpendicular to the sun, as it is measured on a sun-centered sphere whose radius is the distance from the earth to the sun. In other words, your speculation about what NASA meant by “average” is incorrect, and a plain reading of the NASA text suffices. The TSI is about 1360 W/m2 measured on a surface perpendicular to the sun.

    2. The other statement you quoted from the same NASA page contradicts this one, they replace in fact “average” and “total” with “maximum” for no scientific reason.

    This replacement is what I call a semantic “solution” to a problem. The problem is, as I said earlier, that there is apparently no physical experimental proof of warming via back radiation, and that “solution” helps a little bit.

    Can’t say why NASA does what they do, you’ll have to ask them about that. I have shown you upthread the calculation to go from the 1,364 W/m2 TSI, to the TSI averaged over the whole surface of the earth. I note that no one has found any problems with that calculation, which is total energy intercepted by the earth, divided by the total area of the earth.

    If you don’t calculate it as energy divided by area, how else are you planning to calculate solar per unit area of the planet?

    As to “experimental proof of back radiation”, say what? I’ve been posting graphs of actual measurements of “back radiation” (DLR) over a 24 hour period. I am stunned that you still sit there and fatuously claim there is no “experimental proof” that DLR exists, when it is measured daily by scientists all over the world … you really are going a long ways off the path to try to establish that you are right, including totally ignoring scientific measurements of DLR made by lots and lots of folks.

    Another problem with you calculation is, even if we put aside that hemisphere/disk error, that your Earth does nor rotate at all. You consider only the half of the Earth to receive solar power. This is apparently another error, because the Earth does rotate. The WHOLE rotating Earth is receiving solar power and the part that temporarily is not, has not been proven to cool down to the absolute zero, before it faces the sun again. If it does not cool down to the absolute zero, the average temperature will be higher. In your calculation it is not taken into consideration, this is the second error.

    Go back and look at the calculation of the average power. I use this mysterious thing called “mathematics”. I take the total amount of solar energy intercepted by the planet, and divide it by the ENTIRE SURFACE AREA OF THE PLANET, using the following magical incantation:

    Average energy per unit area = total energy / total area.

    Study that staggeringly complex formula for a while, and you’ll see that I have indeed given you the average sunlight over the total surface of the planet, and that I have absolutely NOT done what you say, to “consider only half of the earth to receive solar power).

    The third error is your way of operating with averages, Ryan already said that on this thread (December 18, 2012 at 4:57 pm) and gave you an easy example. Unfortunately, you failed to comment on that, as well as on the “rotating Earth” argument of mine.

    Ryan said:
    December 18, 2012 at 4:57 pm

    Hmmm, Take the average of the lengths of a rectangle 1cm x 9cm and square the average to find the area, then you’ll realise just why playing with averages is not a good idea in non-linear systems.

    This is supposed to be news? The lengths of the two sides of a rectangle are A and B. The area is A * B = AB.

    The average of A and B is (A+B)/2. When we square that, we get (A2 + 2AB + B2)/2.

    And yes, (A2 + 2AB + B2)/2 is not equal to AB. We’ve definitively shown that bad math gives wrong answers. So what?

    Ryan has proven that to get the area of a rectangle, you can’t average the sides and square them. Whoa, Greg, that’s heavy, man … you also can’t take the square root of the sides and square them to get the area. And you can’t take the difference of the sides and square them to get the area, or the logarithms of the sides, or the sum of the sides.

    All of which proves, I suppose, that “playing with square roots and logarithms and subtraction and addition is not a good idea in non-linear systems”.

    Hey, playing with anything in non-linear systems if you are not a mathematician is always a mistake, step back from the equations and nobody gets hurt … but what does that have to do with averages and logarithms and addition?

    Actually, Greg, all Ryan has shown is that to get the area of a rectangle, the correct formula is A times B, and guess what? If you use any other formula, you get a wrong answer … whoa, shocking news, huh? He has conclusively proven is that if you use averages incorrectly, you will get an incorrect answer …

    That is the weakest argument against using averages that I’ve ever heard, by an order of magnitude, which is why I forbore commenting earlier, from a sense of mercy. There are valid strong arguments against using averages, particularly in responsive governed systems like the climate, but Ryan’s argument is not one of them.

    I hope this explanation will help.

    Thanks for being clear about your issues, Greg, it helps a lot.

    w.

  381. Willis Eschenbach says:

    Ryan says:
    December 21, 2012 at 2:18 am

    Yesterday I had a mug of hot coffee sitting on my desk in one of those insulated flask mugs. Unfortunately I left it there too long and it had got cold. I guess it took about half an hour to get cold. It started off at about 70Celsius I guess and within 2000seconds the water got cold. I did some calculations in miyt head and came to the conclusion that a body of water 1sqm in area heated to the same temperature would be losing something like 10,000watts/sqm, of that order.

    Now that’s a high rate of loss but a coffee mug is nowhere near thermal equilibrium with the air. The heat from the mug is dispersing into the air and just a meter up you can’t feel the heat anymore. You could argue that it isn’t very well constrained to stay in a single column of air, but consider cars in a traffic jam – each engine is at about 70Celsius but if you are on a bridge over the traffic jam you can’t really feel the extra heat.

    Ten thousand watts per square metre? That makes my bad number detector ring like crazy. Let’s run the calculations (details in the appendix).

    A cup of coffee is usually about 8 ounces, or half a pound, or about 226 grams.

    The specific heat of water is about 4.186 joules/gram/kelvin

    The change from hot to cold coffee is about 50 kelvin.

    That means that the coffee loses about 47,500 joules as it cools 50K in 2,000 seconds, or about 24 joules per second … which is also 24 watts, since a watt is a joule per second.

    Now, I figured the coffee to be in a cylinder about 10 cm tall and 10 cm across. This gives a surface area of about 0.047 square metres.

    Dividing that into the watts to give us watts per square metre, we find that the coffee cup is losing heat, not at the claimed rate of ten thousand watts per square metre, but about 500 watts per square metre. You are wrong by more than an order of magnitude.

    Ryan, I have to tell you that I have stopped reading your post at that point. Why?

    Because a man who wants to do the math in his head, doesn’t bother checking it with Excel, gets it horribly wrong, and then wants to lecture me on the startling conclusions revealed by his math, that man is not worth my time to read.

    I knew your numbers were bad on my first glance … and the fact that you did not notice that, didn’t bother to check, and then spun a whole story out of your results, tells me a lot. I don’t have a lot of time, so I want to concentrate on people who are paying attention to the details … and that’s clearly not you.

    One of the lessons I have learned (to my cost) is that when a number seems way high or way low, maybe it’s not, but I damn well better check it. The same is applicable on the other end of the scale, when a number seems too good to be true, it likely is. I’ve been bitten too often by both of those bugs.

    w.

    NOTE: Here’s the values for the calculation

    Item, Value, Units
    Specific heat of water, 4.186, J/g/K
    Weight of coffee, 0.5, lbm
    Weight of coffee, 226.80, g
    Start temp, 70, °C
    End temp, 20, °C
    ∆ temp, 50, ∆K
    Energy loss, 47468.44152, joules
    Time, 2000, seconds
    Wattage, 23.73422076, joules/sec
    Cup height, 10, cm
    Cup radius, 5, cm
    Surface area, 471.24, cm2
    Surface area, 0.047, m2
    Energy loss, 504, W/m2
  382. Willis Eschenbach says:

    jae says:
    December 21, 2012 at 6:28 am

    Willis says at 12/20, 7:32:

    “Regarding variations in the DLR, what kind of variations were you expecting? All of them have diurnal variations on the order of 80-100 W/m2. It’s difficult to say what the average might be for any of them without downloading the data. I see some that vary maybe 320-420 W/m2, average maybe 370, and others that vary 290-370 or so on the same day, average maybe 330.”

    Not so, Willis; that’s what bugs me. Look at all sites on July 15, 2012, for example. ALL SITES show DW IR of about 400 +/- 25 w/m2 DAY AND NIGHT! That’s weird, considering the enormous variation between the sites in terms of amounts of GHG (water vapor), altitude, latitude, temperature, sun/no sun, etc.. I don’t understand it at all.

    Jae, you might try this. Go to the setup page. Run the various locations for 15 July again, and this time, instead of just getting the graph for downwelling longwave radiation, get the air temperature as well (right column).

    You’ll see that the air temperature for all of the locations is roughly similar, they run from about 20°C to about 30°C, some a bit higher, some a bit lower. Remember that we can take an effective radiation altitude of half a km, and that the temperature drop with altitude is about 1°C per hundred metres. That means the air half a km up will be about 5°C cooler than at ground level. This gives air temps at altitude running from around 15°C to 25°C.

    S-B radiation for those temperatures assuming gray body (emissivity = 0.95) gives equivalent radiations of 371 W/m2 and 426 W/m2 …

    Which is about what you said for that day, 400 ± 25 W/m2.

    Finally, remember that even a doubling of CO2 is only supposed to add 4 W/m2 or so, and when the total longwave signal is 400 W/m2, that’s only 1%.

    w.

  383. Greg House says:

    Willis Eschenbach says, December 21, 2012 at 9:52 am: “As to “experimental proof of back radiation”, say what? I’ve been posting graphs of actual measurements of “back radiation”[...]
    ====================================================

    Willis, this is unfortunately another example of a “straw man argument”. I never said or asked anything about an “experimental proof of back radiation”.

    It was about apparent absence of experimental proof of “WARMING/(slowing down cooling) via back radiation”. I hope you will understand the difference.

  384. Willis Eschenbach says:

    Ken Coffman says:
    December 21, 2012 at 9:37 am

    This would be an interesting experiment: heat up one-square-foot cubes of iron and balsa wood at 300C until the temperature of both is stable as measured with an IR thermometer. Let them sit for ten minutes, then I’ll put my bare hand on the balsa wood and an academic climatologist can put his or her hand on the iron block.

    Here’s a slight modification which will represent equal storage of thermal energy. Heat up 100-pound blocks of iron and balsa wood at 300C until the temperature of both are stable as measured with an IR thermometer. Let them sit for ten minutes, then I’ll put my bare hand on the 100-pound balsa wood block and an academic climatologist can put his or her hand on the hundred pound iron block.

    In both cases, which test subject is more likely to need to visit an emergency room?

    Interesting how the temperature of stuff does not tell a full thermodynamic story, eh? Wouldn’t we conclude from this that it’s very important to know what substance is doing the radiating? Further, might we not conclude that whatever radiates from air to land is irrelevant with regard to the surface temperature?

    Ken, as a man who both utilizes and enjoys thought experiments, thank you for yours.

    You are correct that equal temperature or equal amounts of radiation doesn’t mean other things are equal. That’s no surprise, there are various things that we can measure regarding temperature. We can measure total heat in an object, for example. We can measure its thermal conductivity, which you have implicity measured by placing your hand on the two blocks. These are independent properties of the material.

    Now, thermal conductivity is a property of the material itself, one which is not affected by temperature. Steel conducts the heat to your hand rapidly. Balsa wood doesn’t. This is true whether they are hot or cold.

    So you are right, there’s a host of different things that you’d need to tell the “full thermodynamic story”. But here’s the part you are overlooking. All those things like the one you list have nothing to do with radiation.

    The only things that you need to know to determine the radiation are the emissivity and the temperature of the object. Period. The material it’s made of makes absolutely no difference. I first learned this when I was studying blacksmithing. My book pointed out that you can tell the temperature of steel by the color. Deep red is coolest, then cherry red, then yellow, then orange, then white.

    The book also pointed out what seemed odd to me at the time. This was that it doesn’t matter in the slightest what the metal is. Blood red is about 1200°F (650°C) whether you are heating steel or copper or tin or whatever.

    Thermal radiation at lower temperatures is just the same. It’s a function of the temperature and the emissivity, not of the thermal mass or the type of material or the amount of heat contained in the material or any other variable.

    So when you say:

    Wouldn’t we conclude from this that it’s very important to know what substance is doing the radiating?

    No, it’s no more important than whether steel or tin is doing the radiating, because the result is the same. My desk’s radiation only depends on temperature and emissivity, not on it being wood or how heavy it is or its thermal conductivity or mass.

    Finally, you say:

    Further, might we not conclude that whatever radiates from air to land is irrelevant with regard to the surface temperature?

    Absolutely not. Radiation is real and measurable, and it contains energy. That energy is transferred to whatever absorbs the radiation, regardless of the temperature of the absorbing object.

    Remember that this is a flow of energy, not a flow of heat. Heat is the net energy flow. So if a hot object is absorbing 400 W/m of radiation, say radiation from my desk, and that same hot object is radiating at 600 W/m2, there will be a net flow of radiant energy from the object to my desk.

    Here’s the thing. If there is radiation from objectA to objectB, that means that objectB is in sight of objectA. But that also means that you can see objectA from objectB, so objectA must be absorbing radiation from objectB. It is the NET of these two energy flows that is the flow of heat.

    Note that the radiation doesn’t care about the temperature of the object it hits. Radiation from a cooler object is absorbed by a hotter object. People think this violates the 2nd Law. But if you remember that they both radiate to each other, there must be a larger radiant energy flow from the hotter object to the cooler object.

    As a result, although there is a radiant energy flow from cool to hot, the heat only flows from hot to cold, and there is no 2nd law violation.

    People also ask, if the global average background radiation is 340 W/m2, why can’t I boil water with it? The problem is, you’ll never be able to get anything warmer than the point at which it is radiating 340 W/m2. At that point it is radiating away the 340 W/m2 it is absorbing, so it will warm no further … and 340 W/m2 is down near freezing.

    People ask, how about getting mechanical work out of the radiation? Couldn’t it run a heat engine? Sure, but to do that, you need to have a colder place to reject the waste heat. Heat engines don’t run on temperature. They run on temperature differential. Where will you reject the heat if the hot end of your engine is around freezing?

    People ask, how about something like a photoelectric cell? The problem is, the radiation doesn’t have enough energy in it to knock an electron out of its easy chair … so the photoelectric effect just doesn’t occur with thermal IR in the temperature range we’re talking about in any known materials.

    Anyhow, I tend to ramble, but there’s a lot of misinformation out there on this subject.

    My best to you,

    w.

  385. Willis Eschenbach says:

    Greg House says:
    December 21, 2012 at 4:46 pm

    Willis Eschenbach says, December 21, 2012 at 9:52 am: “As to “experimental proof of back radiation”, say what? I’ve been posting graphs of actual measurements of “back radiation”[...]
    ====================================================

    Willis, this is unfortunately another example of a “straw man argument”. I never said or asked anything about an “experimental proof of back radiation”.

    It was about apparent absence of experimental proof of “WARMING/(slowing down cooling) via back radiation”. I hope you will understand the difference.

    Greg, your suspicion is misplaced. It is not a straw man argument, I don’t deal in straw. Instead it is what goes by the technical name of a “misunderstanding”.

    I took the interpretation I did because I truly couldn’t conceive that your argument was actually that

    a) back radiation exists, but

    b) it doesn’t transfer energy to the surface when it is absorbed.

    I figured if you didn’t think DLR transfers energy to the surface, you must believe it doesn’t exist. Because if it exists, as it seems you are saying, I don’t understand where you think the energy goes when the radiation is absorbed by the surface.

    If you want experimental proof that the surface end up warmer with DLR than without, because DLR transfers energy to the surface, consider that the oceans are liquid. I ascribe that to the ~ 340 W/m2 of DLR that is absorbed by the ocean … if you have another explanation for where the 340 W/m2 necessary to prevent the oceans from freezing is coming from, bring it out.

    All the best,

    w.

  386. Greg House says:

    Willis Eschenbach says, December 21, 2012 at 9:52 am: Actually, Greg, all Ryan has shown is that to get the area of a rectangle, the correct formula is A times B, and guess what? If you use any other formula, you get a wrong answer … whoa, shocking news, huh? He has conclusively proven is that if you use averages incorrectly, you will get an incorrect answer …
    That is the weakest argument against using averages that I’ve ever heard, by an order of magnitude, which is why I forbore commenting earlier, from a sense of mercy.

    ==========================================================

    Mercy? Well, I see mercy actually on the Ryan’s side. But OK, if you wish, let us get a little bit deeper into that thing about averages.

    One of the problems of that bogus calculation of “Earth without atmosphere” temperature is the extraction of average temperature from average radiation. This is a blunder, unacceptable even on junior high school level. The point is, that in reality you can have different average temperatures produced by the same average radiation. I guess, you do not understand that, and neither do many other “climate scientists”, so let us come back to the Ryan’s example to illustrate that point.

    Let us say, the “average side” of two squares is 5. Is it possible to extract from that the “average area” of those two squares? No problem for a climate scientist, I guess, his calculation would be 5×5=25. Note this 25.

    It is apparently possible that the sides of our two squares are 1 and 9 (the average is still the same 5). Now look at the areas, the are 1×1=1 and 9×9=81. The average of 1 and 81 is (1+81)/2=41. Note this 41.

    So, the correct average area for our two squares is 41, the result “climate science” would get is 25. The difference is enormous.

    Now, you can also take 2 and 8 as the sides of our two squares (the average is still the same 5), the the average area will be (2×2 + 8×8)/2=34. Still different from 25 and different from 41 as well. So, you can see, from the same “average side of a square” you can produce different “average areas”.

    I am just curious, is basic math knowledge not required from climate scientists? Anyway, I hope we have identified the problem with extracting average from average in that bogus calculation of “Earth without atmosphere” temperature.

  387. Greg House says:

    Willis Eschenbach says, December 21, 2012 at 5:30 pm: It is not a straw man argument, I don’t deal in straw. Instead it is what goes by the technical name of a “misunderstanding”.
    I took the interpretation I did because I truly couldn’t conceive that your argument was actually that
    a) back radiation exists, but
    b) it doesn’t transfer energy to the surface when it is absorbed.
    [...]If you want experimental proof that the surface end up warmer with DLR than without … consider … I ascribe that to … if you have another explanation …

    ==========================================================

    And yet we have another example of a straw misunderstanding. I did not say that either. Again, I said: “It was about apparent absence of experimental proof of “WARMING/(slowing down cooling) via back radiation”. I hope you can understand the difference now.

    What you ascribe to what and how you explain it is not an experimental proof of “WARMING/(slowing down cooling) via back radiation, I hope we can agree on that.

    We do not want to twist, squeeze and otherwise torture the notion of “experimental”, do we?

  388. jae says:

    Willis sez (finally):

    “jae, you might try this. Go to the setup page. Run the various locations for 15 July again, and this time, instead of just getting the graph for downwelling longwave radiation, get the air temperature as well (right column).”

    Well, W, been there, done that already, many times. I don’t think you are getting my point. First of all, the temperature variations are WAY wider than you state, and, more importantly, we are comparing areas which differ wlith respect to GHGs by 2-3 times, WITHOUT ANY SIGNIFICANT CHANGE IN DOWNWARD RADIATION!. Just where is there some indication of a GHE effect in this data? I’m right back at the Phoenix/Atlanta comparison–If GHGs dominate, then Atlanta should be MUCH hotter than Phoenix. But it is much cooler. WHY?

    I got a conundrum, too, W!

  389. jae says:

    Willis: BTW, my conundrum is bigger and better than your conundrum! :-)

  390. Willis Eschenbach says:

    Greg House says:
    December 21, 2012 at 5:55 pm

    Willis Eschenbach says, December 21, 2012 at 9:52 am:

    Actually, Greg, all Ryan has shown is that to get the area of a rectangle, the correct formula is A times B, and guess what? If you use any other formula, you get a wrong answer … whoa, shocking news, huh? He has conclusively proven is that if you use averages incorrectly, you will get an incorrect answer …
    That is the weakest argument against using averages that I’ve ever heard, by an order of magnitude, which is why I forbore commenting earlier, from a sense of mercy.

    ==========================================================

    Mercy? Well, I see mercy actually on the Ryan’s side. But OK, if you wish, let us get a little bit deeper into that thing about averages.

    Thanks, Greg, but I don’t wish that in the slightest. Ryan made an incorrect and frankly ridiculous argument against averages, which was that if you used averages incorrectly, you got wrong answers. Seriously, that was his whole argument …

    I refrained from commenting on it, I didn’t want to bust his bubble. You requested that I comment on it. I did so, as you quote above. I notice that you have refrained from commenting on the substance of my objections. Instead of even acknowledging that the insights you recommended to me about averages had turned out to be substance-free, suddenly you wish to investigate averages.

    What in any of that makes you think that I wish to “get a little bit deeper into that thing about averages”? I have no such desire, in fact, quite the opposite—I actively want to stay out of that discussion. It is difficult enough making progress on the topics already in hand, thanks.

    w.

  391. Willis Eschenbach says:

    jae says:
    December 21, 2012 at 6:36 pm

    Willis sez (finally): …

    Aw, go soak your head with your damn “(finally)”. I have a life, I have a wife, and a daughter visiting for Christmas, I have a day job, I have other interests. I spent today taking care of my blind 85-year-old father in law. So you can stuff your “(finally)” in that part of your pelvic anatomy where the TSI is zero.

    We now return you to your regularly scheduled science programming …

    w.

  392. Greg House says:

    Willis Eschenbach says, December 21, 2012 at 9:52 am: “Rather than focus on the words of one person at NASA, lets develop the TSI from first principles. The formula to calculate TSI is simple.
    TSI = L / (4 Pi A2) where L is the luminosity of the sun, 3.84e+26 Watts, and A is the unit distance from the sun to the earth, 1.50e+11 metres.
    … we get TSI = 3.84e+26 W / (4 * 3.1416 * (1.5e+11)2) which simplifies to 1,364 W/m2. Note that this is perpendicular to the sun, as it is measured on a sun-centered sphere whose radius is the distance from the earth to the sun.”

    ===========================================================

    Really? I was just about to believe you, Willis, but then one expression you used struck me: “it is measured”. Then I thought, why would Willis talk about “measured TSI” after he presented a pure calculation of that TSI? This is a contradiction. And contradictions, you know, are signs of, let us say, errors.

    So I asked myself, how do they know how much the luminosity of the Sun (L) is? Yeah, this is easy: they calculate it based on TSI! Like that: L=(4 Pi A2)TSI . (http://en.wikipedia.org/wiki/Solar_luminosity)

    That is nice. The luminosity of the Sun is derived from TSI, and TSI is derived from the luminosity of the Sun!

    Congratulations, Willis, this is an outstanding example of a logical fallacy called “circular reasoning” (http://en.wikipedia.org/wiki/Circular_reasoning). Thank you.

    So, here we are again. You provided the NASA link, not me, and according to it there is no scientific reason for your “disk” interpretation, it is hemisphere, and your cutting solar power in half is still an error.

  393. Greg House says:

    Willis Eschenbach says, December 21, 2012 at 7:36 pm: What in any of that makes you think that I wish to “get a little bit deeper into that thing about averages”?
    ==========================================================

    Willis, my expression “But OK, if you wish, let us get a little bit deeper into that thing about averages” was rhetorical and was connected to your failure, as I saw it, to understand the point of Ryan’s about the false extracting average from average concerning radiation and temperature.

    Even if you do not care, the readers still might be interested.

  394. Willis Eschenbach says:

    jae says:
    December 21, 2012 at 6:36 pm

    “jae, you might try this. Go to the setup page. Run the various locations for 15 July again, and this time, instead of just getting the graph for downwelling longwave radiation, get the air temperature as well (right column).”

    Well, W, been there, done that already, many times. I don’t think you are getting my point. First of all, the temperature variations are WAY wider than you state, …

    To recap the bidding, Jae is objecting to this statement of mine about the temperature variations:

    You’ll see that the air temperature for all of the locations is roughly similar, they run from about 20°C to about 30°C, some a bit higher, some a bit lower.

    So, I did what most folks somehow never get around to doing. I downloaded the data. Here’s the results:

    As you can see, the low temperatures are clustered around 20°C and the high temperatures around 30°C, as I said.

    … and, more importantly, we are comparing areas which differ wlith respect to GHGs by 2-3 times, WITHOUT ANY SIGNIFICANT CHANGE IN DOWNWARD RADIATION!. Just where is there some indication of a GHE effect in this data? I’m right back at the Phoenix/Atlanta comparison–If GHGs dominate, then Atlanta should be MUCH hotter than Phoenix. But it is much cooler. WHY?

    I see significant changes in the radiation. The minimums vary from 336 to 396, a difference of 60 W/m2 and the maximums vary from 417 to 472, a difference of 55 W/m2. A doubling of CO2 is slated to cause 3.7 W/m2 change in forcing, this has a difference of 50 or 60 W/m2, and you think there should be even more variation …

    This is why I asked you what kind of signal you were expecting.

    All the best,

    w.

  395. Willis Eschenbach says:

    Greg House says:
    December 21, 2012 at 8:17 pm

    Willis Eschenbach says, December 21, 2012 at 9:52 am:

    “Rather than focus on the words of one person at NASA, lets develop the TSI from first principles. The formula to calculate TSI is simple.
    TSI = L / (4 Pi A2) where L is the luminosity of the sun, 3.84e+26 Watts, and A is the unit distance from the sun to the earth, 1.50e+11 metres.
    … we get TSI = 3.84e+26 W / (4 * 3.1416 * (1.5e+11)2) which simplifies to 1,364 W/m2. Note that this is perpendicular to the sun, as it is measured on a sun-centered sphere whose radius is the distance from the earth to the sun.”

    ===========================================================

    Really? I was just about to believe you, Willis, but then one expression you used struck me: “it is measured”. Then I thought, why would Willis talk about “measured TSI” after he presented a pure calculation of that TSI? This is a contradiction. And contradictions, you know, are signs of, let us say, errors.

    Oh, man, now you’re going to pick on one word? Mea maxima culpa, replace “measured” with “calculated”. Can you focus in the issues now?

    So I asked myself, how do they know how much the luminosity of the Sun (L) is? Yeah, this is easy: they calculate it based on TSI! Like that: L=(4 Pi A2)TSI . (http://en.wikipedia.org/wiki/Solar_luminosity)

    That is nice. The luminosity of the Sun is derived from TSI, and TSI is derived from the luminosity of the Sun!

    Congratulations, Willis, this is an outstanding example of a logical fallacy called “circular reasoning” (http://en.wikipedia.org/wiki/Circular_reasoning). Thank you.

    So, here we are again. You provided the NASA link, not me, and according to it there is no scientific reason for your “disk” interpretation, it is hemisphere, and your cutting solar power in half is still an error.

    I give up, Greg. I’m done. I cry uncle, I surrender. I’m going to leave you in sole possession of the field. You are determined to believe what you want to believe, in spite of the fact that the various college textbooks and online resources say the same thing—that the TSI is ~ 1368 W/m2, and that averaged over the surface, that’s 342 W/m2. You could google it and get a raft of citations, but instead you insist on hanging on to your twisted interpretation of a single NASA web page. I’ve tried a host of ways to explain it to you, as have others. You remain unmoved, not the slightest change in your position. And unfortunately, you seem to interpret that immobility as a virtue, rather than seeing it for what it is, a refusal to learn …

    You are free to believe anything you want, Greg. Just don’t expect me to pay any further attention to your ideas and claims. I’ve tried that, didn’t work, you simply blew me and my ideas off entirely. I’m returning the favor.

    Sadly,

    w.

  396. Willis Eschenbach says:

    Greg House says:
    December 21, 2012 at 8:28 pm

    Willis Eschenbach says, December 21, 2012 at 7:36 pm:

    What in any of that makes you think that I wish to “get a little bit deeper into that thing about averages”?

    ==========================================================

    Willis, my expression “But OK, if you wish, let us get a little bit deeper into that thing about averages” was rhetorical and was connected to your failure, as I saw it, to understand the point of Ryan’s about the false extracting average from average concerning radiation and temperature.

    Speculation about what I might wish is never rhetorical. Instead, it was a (likely subconscious) attempt to shift responsibility for the change in subject (to averages) onto my shoulders. It was a subtle rhetorical flourish, to convince the reader that you are being generous and complying with my wishes in changing the subject.

    In fact, that wasn’t my desire at all. YOU wanted to change the subject, not me, I had no wish to do so. Sorry, but it was your idea, and trying to pawn it off on me won’t wash.

    Even if you do not care, the readers still might be interested.

    Yeah, I notice they’re beating down your door to discuss it …

    w.

    PS—Ryan’s claim was that the use of averages was suspect because you can make mistakes with them, viz:

    Ryan says:
    December 18, 2012 at 4:57 pm

    Hmmm, Take the average of the lengths of a rectangle 1cm x 9cm and square the average to find the area, then you’ll realise just why playing with averages is not a good idea in non-linear systems.

    Indeed, you can make mistakes with averages, as he illustrates. So what? You can also make mistakes with addition, and with square roots. So what? Does that mean that playing with addition is not a good idea in non-linear systems? Making mistakes is a problem in any system.

    Bottom line for me is that until Ryan or you or someone demonstrate that I am actually making mistakes with the averages, we don’t have much to discuss …

  397. Mack says:

    Willis says……”someone demonstrate that I am actually making mistakes with averages,we don’t have much to discuss”
    The figure of 1360w/sq.m is a YEARLY global AVERAGE. It is the only real measurement of incoming solar incoming radiation we have .It is a bulk amount which cannot be divided. It is the net amount from a sun that flickers. It actually is irrelevant wherefrom ie from what direction this energy comes from .; regard it coming from all directions in space if you want because the sun also shines over your head at night-time when dealing with this average. A YEARLY AVERAGE. and it has to remain simply as that. Forget about Cosine of this , angle of that, Throw away your slide-rule ,distance calculations, size calculations ,even area calculation (because this has been standardised at the TOA and Earth;s surface as 1 sq.m.) and concentrate on this net bulk amount. This real and measured net bulk amount has to remain simply as that and that then attenuates down to a real and measured yearly global average of about 340w/sq m at the surface.
    There’s nothing much more in the way of discussion I can offer other than that Willis.but there is the mistake you are making with averages. You don’t need to answer if you’re still not talking to me. ;)

  398. Willis Eschenbach says:

    Mack says:
    December 22, 2012 at 4:18 am (Edit)

    Willis says……

    ”someone demonstrate that I am actually making mistakes with averages,we don’t have much to discuss”

    The figure of 1360w/sq.m is a YEARLY global AVERAGE. It is the only real measurement of incoming solar incoming radiation we have.

    Not true in the slightest. We have satellites that measure incoming solar radiation 24/7.

    It is a bulk amount which cannot be divided.

    That sentence has no meaning I can discern.

    It is the net amount from a sun that flickers. It actually is irrelevant wherefrom ie from what direction this energy comes from .; regard it coming from all directions in space if you want because the sun also shines over your head at night-time when dealing with this average. A YEARLY AVERAGE. and it has to remain simply as that. Forget about Cosine of this , angle of that, Throw away your slide-rule ,distance calculations, size calculations ,even area calculation (because this has been standardised at the TOA and Earth;s surface as 1 sq.m.) and concentrate on this net bulk amount. This real and measured net bulk amount has to remain simply as that and that then attenuates down to a real and measured yearly global average of about 340w/sq m at the surface.

    Umm … errr … not much I can say about that, because it doesn’t make sense. Cosine of this? Angle of that? Attenuates down? IIRC, I said nothing about either cosines, angles, or attenuation, so I fear I can’t comment..

    There’s nothing much more in the way of discussion I can offer other than that Willis.but there is the mistake you are making with averages. You don’t need to answer if you’re still not talking to me. ;)

    Whoa, back up there. You say “there is the mistake you are making with averages” … but where? What is the mistake? You seem to have skipped over the part where you quote my words and show me that they are wrong.

    Simple instructions, Mack. If you want to bust me for any of my manifold sins of commission, omission, or emission, here’s the steps:

    1. Find where you think I made a mistake.

    2. Quote what I said that you think is in error, with a link to the original.

    3. Demonstrate, through math, logic, citations, authorities, and the like, just exactly why you think I’m wrong.

    A vague ramble followed by some handwaving on the order of ‘thereis the mistake you are making with averages’, without saying just where “there” might be located, just doesn’t cut it.

    Regards, thanks for the reply,

    w.

  399. jae says:

    Thanks for all the effort, Willis. I really appreciate it.

    Notice the generally inverse relationship between temperature and downwelling IR. Just like Phoenix vs. Atlanata. Why is this, I wonder?

  400. Willis Eschenbach says:

    Thanks, jae.

    Not sure what you mean by the “inverse relationship between temperature and downwelling IR”. Here’s the situation in Bondville. I show the DLR, and also the DLR calculated from the temperature (using emissivity of 0.95 and a temperature drop to altitude of 5°C).

    As you can see, the temperature and the DLR move basically in lockstep, there is no inverse relationship. When SAT goes up, so does the DLR. Not only that, but the DLR follows very closely the theoretical relationship

    W = epsilon sigma T4

    just as we would expect.

    All the best,

    w.

  401. jae says:

    Willis:

    Thanks again.

    No surprise there, but that’s not what I’m trying to get at. I am trying to compare sites with large amounts of water vapor to sites with small amounts to determine whether the GHE (downwelling radiation) from water vapor actually causes an increase in temperature–i.e., if more water vapor, all other things constant, leads to higher temperatures via the GHE. The “inverse relationship” I have noted is BETWEEN sites, not at a single site, like you plotted. Like Phoenix and Atlanta. Phoenix is much hotter, despite having much less GHE from water vapor (the inverse relationship). I think the presence of water (even in vegetation) causes a negative feedback, and I think the data clearly show that. My analyses show that dryer areas are generally hotter than humid areas, when other variables (elevation, latitude, cloudiness) are constant (or nearly so). I think that is due to the heat loss through evaporation and increased haze/clouds. Negative feedback.

    Hope I’m making sense here.

  402. Willis Eschenbach says:

    Thanks, jae.

    I’m not sure that you can pair up sites like that. For starters, I don’t see how you’d account for other factors. For example, Phoenix is underneath the descending part of the Hadley cell circulation, where despite being at the same latitude, Atlanta is not. As a result, Phoenix is bathed in dry, hot descending air.

    For that reason, it is both hotter than Atlanta, and the temperature swings are much larger. But that is not a result of the lower water vapor, at least not directly. It is a result of Phoenix getting a constant influx of hot, dry descending air from the Hadley Cell.

    So for all of those reasons, rather than look at different sites, I’d look at one site under different humidity conditions.

    I don’t see anywhere that there is a continuous record of humidity available at the SURFRAD site, which is a shame. Seems to me the way to investigate the question that interests you would be to look at one site with both DLR and RH records, to see how they co-vary.

    Not sure where you’d find such a record, but I’d be interested in your results. Problem with RH is that it is such a local phenomenon, for example when you walk out of the forest and into the field, the RH drops dramatically.

    Maybe you could use the “water content” files from the weather satellites to give you a rough look at the RH over a wide area.

    Over to you,

    w.

  403. Willis Eschenbach says:

    A few more thoughts.

    If you do want to compare sites in the SURFRAD network, you can’t compare the radiation directly. You have to first subtract out what you expect to find given the temperature. Here’s how I do it.

    Get the underlying data from the site, there’s a button on each page. I copy it and paste it into an Excel spreadsheet.

    From looking at the map, the driest site seems to be Desert Rock, Nevada. The moistest site seems to be Goodwin Creek, Mississippi.

    You can see the difference in the diurnal temperature range. In the desert, there is little moisture, so the earth can cool very quickly. During the day, there are few clouds, so the ground gets hot quickly and continues to warm throughout the afternoon. So the range is wider in the desert than in Mississippi, and the temperature is hotter. Mostly that’s from the difference in clouds, though.

    So, I get the data. Then I use Solver in Excel to solve for the best fit.

    The relationship between the ground temperature T (kelvins) and the DLR is going to be roughly expressed by:

    DLR = sigma epsilon (T-X)4

    where sigma is the S-B constant (5.67e-8), epsilon is emissivity, T is surface temperature, and X is how much cooler the effective radiation level is than the ground temperature.

    I use Solver to give me the best solution for “X”. To isolate just that variable, I set the emissivity to 0.95, a “greybody”.

    In the case I showed above, Bondville, the best fit was with the effective radiation altitude temperature being 5° cooler than the surface.

    Now, what would we expect in a very dry place? Since the atmosphere is absorbing less ULR, we would expect the atmospheric temperature to be less. And in fact, when I ran the numbers, Desert Rock, Nevada has an effective radiation temperature of not five, but about seventeen degrees cooler than the surface.

    For Goodwin Creek, using the identical conditions, the best fit is with the effective radiation temperature about the same as the surface. In other words, in that muggy atmosphere, the upwelling radiation is absorbed near the surface, and the downwelling radiation is coming from right near the surface.

    In other words, in the desert, because the air is dry we get less downwelling longwave radiation than expected, and in Mississippi where the air is wet we get more radiation than expected.

    So I would say that the water vapor is working in the expected manner in both locations. And that’s how I would probe the data to determine that.

    Regards,

    w.

  404. Willis Eschenbach says:

    jae, reflecting on the situation in Desert Rock, Nevada brings a curious thought about your Atlanta-Phoenix paradox.

    There is undoubtedly an increase in DLR from the direct effect of water vapor. However, what this overlooks is that there is also a strong negative feedback from clouds.

    If there is no water vapor in the air, as in Desert Rock, you get less DLR, it’s true.

    But you also get less clouds. No water vapor means no puffy white solar reflectors. So (in some dry locations) that the decrease in DLR (from less water vapor, a notorious GHG) is overwhelmed by the increase in solar radiation because there are no clouds.

    So the positive feedback from the water vapor is reduced, and in some cases totally overwhelmed, by the negative feedback from the clouds. Gottal love nature …

    w.

  405. Bob Koss says:

    Jae, Willis,
    The USCRN sites are slowly becoming more comprehensive regarding the variables they are tracking. For example the AL_Gadsden_19_N site has a full set of hourly data for the last couple years. As well as the usual temperature readings they also include solar radiation, relative humidity, surface temperature, several levels of sub-surface temperature and moisture content, precipitation. Few other sites are as comprehensive, but some like CA_Fallbrook_5_NE are missing only sub-surface readings. They appear to be starting a separate data set for the sub-surface readings.

    Perhaps jae can further his investigation by making use of the hourly data for only a couple years.
    http://www.ncdc.noaa.gov/crn/qcdatasets.html

  406. Greg House says:

    Willis Eschenbach says, December 21, 2012 at 10:46 pm: “You are determined to believe what you want to believe, in spite of the fact that the various college textbooks and online resources say the same thing—that the TSI is ~ 1368 W/m2, and that averaged over the surface, that’s 342 W/m2. …I’ve tried a host of ways to explain it to you … refusal to learn …”
    =======================================================

    Willis, as I said before, it was not apparently you who initially introduced this “averaging an average”, as I see it. Therefore your reference to “college textbooks” etc. does not add anything to a scientific debate.

    Regardless where else that thing is mentioned, you failed to prove scientifically that your “342 W/m2″ is correct.

    Your first attempt was a reference to a NASA web page and it turned out that it does not support your thesis.

    Your second attempt was presentation of a bogus calculation, where you derived the disputed TSI from something that has already been derived from that same disputed TSI, and I am still not sure, whether you understand or not how wrong such tricks methods are and that they are completely unacceptable in real science.

    Now you are referring to textbooks. I am sorry, but you could write a textbook yourself and include your bogus TSI calculation there, it still would be wrong. The same goes for your cutting the solar power in half by averaging an average.

  407. jae says:

    Willis:

    Thanks for all your thoughts, which I will try to digest thoroughly…

  408. jae says:

    Willis:

    The reason I have to stick with the Atlanta/Phoenix comparison is that it is the only pair of locations where one is very dry, one very wet, where there is data, and where the latitude and elevation are the same. I’m sure that the Hadley Cell effect exerts some influence, but I would expect the greenhouse effect from the extra water vapor in Atlanta to exert some positive influence also! The influence seems negative.

    I guess what amazes me is that the difference in the amount of GHGs in a humid area like atlanta, relative to a dry place like Phoenix is equivalent to adding over 66 times as much CO2 to the atmosphere in Atlanta (simply assuming that CO2 is as efficient as HOH in capturing IR–which it is not by any stretch). (Water vapor is normally 1-3% of the atmosphere; whereas OCO is only about 0.03%). So why would we see almost exactly the same amount of downwelling radiation (about 400 Wm-2) in both locations on a nice July day? I guess it’s that logarithmic relationship between concentration and radiation? If so, we certainly don’t need much water vapor in the air to obtain the max. greenhouse effect, right? And given the overlapping of radiation bands for CO2 and HOH, I don’t see how CO2 can have any effect at all.

  409. Stephen Wilde says:

    “So why would we see almost exactly the same amount of downwelling radiation (about 400 Wm-2) in both locations on a nice July day? I guess it’s that logarithmic relationship between concentration and radiation?”

    It is due to the relationship between atmospheric density and insolation.

    It is not DWIR that is being measured but the temperature of the air around the instrument and that is controlled by the balance between KE and PE at the height of the sensor which is in turn set by the slope of the lapse rate.

    Surface temperature is set by density and insolation with the slope of the ideal lapse rate set by gravity.

    GHGs can alter the slope of the decline of temperature with height but not the surface temperature because that is set by density (mass) and insolation.

    Furthermore any effect of the GHGs on the slope in one layer is offset by an equal and opposite change in the slope in another layer.

    In order to achieve the necessary adjustments there is simply a change in the global air circulation.

    Compared to the changes caused naturally by sun and oceans our CO2 is insignificant.

  410. squashscorer says:

    Willis writes “There is undoubtedly an increase in DLR from the direct effect of water vapor. ”

    The water vapor must be coming from somewhere and so there is evaporation involved. That lowers the surface temperature and reduces the amount of DLR producing ULR. Whilst there is probably a net increase over land especially, your statement is far too strong when there are negative feedbacks involved.

  411. jae says:

    Stephen Wilde says;

    “It is not DWIR that is being measured but the temperature of the air around the instrument and that is controlled by the balance between KE and PE at the height of the sensor which is in turn set by the slope of the lapse rate.”

    Stephen, I have always thought this is the case, but I’m trying to find a way to demonstrate empirically that the GHE does not work as advertised.

  412. Gail Combs says:

    Willis, I went digging again.

    I did not have the terms or physics quite correct. Solids like water droplets and ice crystals do emit a continuous spectrum, gases at lower pressure and temperature emit discrete spectra.

    Why does the atmosphere radiate? Because it is heated up via convection from the surface, solar radiation and surface radiation. The atmosphere radiates according to its temperature, in accordance with Planck’s law and at wavelengths where gas molecules are able to radiate.

    There isn’t any serious theory that the atmosphere doesn’t emit radiation. If the atmosphere is above absolute zero and contains gases that can absorb and emit longwave radiation (like water vapor and CO2) then it must radiate.
    http://scienceofdoom.com/2010/07/24/the-amazing-case-of-back-radiation-part-two/

    There are some interesting graphs of discrete DLR radiation on that thread BTW. That was what I was actually looking for.

    But the take home is you are looking at a mix bag and not just the energy from “bounced” earth energy.

    Water is STILL the big factor though and the only way the IPCC can prop up CO2 as the big bad boogie man is to bundle the effect of water under the category they call CO2 radiative forcing.

  413. TimTheToolMan says:

    Gail writes “Water is STILL the big factor though and the only way the IPCC can prop up CO2 as the big bad boogie man is to bundle the effect of water under the category they call CO2 radiative forcing.”

    I’m especially sceptical about any multiplicative effect of the CO2 over the ocean where surely there is already a great deal of water vapor directly above the surface. This is kinda important DLR too!

  414. Willis Eschenbach says:

    jae says:
    December 23, 2012 at 11:21 am

    Willis:

    The reason I have to stick with the Atlanta/Phoenix comparison is that it is the only pair of locations where one is very dry, one very wet, where there is data, and where the latitude and elevation are the same. I’m sure that the Hadley Cell effect exerts some influence, but I would expect the greenhouse effect from the extra water vapor in Atlanta to exert some positive influence also! The influence seems negative.

    Thanks, jae. I have shown above that if we compare a very dry and a very wet location, the water vapor has the effect on the DLR that we would expect. More water vapor = more DLR, no surprise there.

    The problem is that there are other effects of very dry and very wet. A wet location like Atlanta supports thermally generated thunderstorms in the summer, which have both a cooling and a moderating effect on the climate. Thunderstorms cool the earth through a host of mechanisms, including direct reflection of solar energy, transfer of heat from the surface aloft, transfer of cold from the atmosphere to the surface, and others.

    Phoenix generally doesn’t have that kind of cooling, thermally generated thunderstorms in the summer. Instead, because the air is so dry that clouds are rare, it gets sun, sun, sun, and more sun.

    Now, which place is going to be warmer? A place with the cooling effect of afternoon thunderstorms in the summer, or a place with almost no clouds at all and just sun, sun, and more sun?

    I’d say the place with full-bore sun would be warmer, and Phoenix bears that out. The heating effect of having no afternoon thunderstorms overwhelms the lack of heating from no water vapor. Here’s the thing. Clouds in the daytime have a very large cooling effect, hundreds of watts per square metre. As a result, the extra energy from the sun if there are no clouds is likely an order of magnitude greater than the loss of energy from the lack of water vapor.

    That’s why I say that you will have a very difficult time trying to establish your claim using those two cities. They are in totally different climate zones. I don’t see how you plan to account for e.g. the difference in summer cloud cover.

    w.

  415. Greg House says:

    jae says, December 23, 2012 at 5:19 pm: “…but I’m trying to find a way to demonstrate empirically that the GHE does not work as advertised.”
    =======================================================

    Jae, I am just curious. You know that in the Wood’s experiment the back radiation from glass does not work “as advertised”. Why do you think that back radiation from anything else like “GHG” still might work?

    Let me give you an example. A person is accused of shooting people from a large distance with a sniper rifle. The defence proves that the person is blind. Would it be reasonable, if the prosecutor said “OK, he shot with a Kalashnikov then”?

  416. jae says:

    Willis:

    You say: “Thanks, jae. I have shown above that if we compare a very dry and a very wet location, the water vapor has the effect on the DLR that we would expect. More water vapor = more DLR, no surprise there.”

    NO, Willis!! That is my central issue, here. One goes from an average of about 8 g/m3 vapor in Phoenix to over 20 g/m3 in Atlanta. That’s from about 1% of the air to 3%. That’s the equivalent of increasing the amount of CO2 in the atmosphere in Atlanta by 66 times over the amount in Phoenix–or a doubling of CO2 of 6 times!* And CO2 is no where near as strong a GHG as water vapor, so there should be even more of an effect! It looks to me that one would expect MUCH more effects from the GHG, if it did anything! All we get is 400 W/m2 at both locations???

    I agree with Wilde that the DWR is simply a parameter of the temperature.

    * 1%/0.03% = 33 “CO2 equivalents” for each percentage change in water vapor.

  417. Willis Eschenbach says:

    jae says:
    December 24, 2012 at 8:34 am

    Willis:

    You say:

    “Thanks, jae. I have shown above that if we compare a very dry and a very wet location, the water vapor has the effect on the DLR that we would expect. More water vapor = more DLR, no surprise there.”

    NO, Willis!! That is my central issue, here. One goes from an average of about 8 g/m3 vapor in Phoenix to over 20 g/m3 in Atlanta.

    jae, you insist on comparing raw temperatures between a place with a lot of clouds and thunderstorms (Atlanta), and one without much clouds and thunderstorms (Phoenix).

    Now, you are welcome to do that, but I don’t foresee success. How on earth are you ever going to be able to untangle the temperature differences from the clouds, from the differences in water vapor? In Phoenix the radiation from the cloud changes is likely a couple of orders of magnitude greater than the loss from not getting the DLR from the water vapor. How do you plan to extract the H2O signal from the increased solar?

    Finally, I have shown above that in Table Rock, Nevada, and in Goodwin, Mississippi, the actual observations show the expected effects of the difference in water vapor. You keep failing to comment on that, I suspect because it doesn’t fit your schtick of claiming that water vapor has no effect … so how about you forget about Atlanta and Phoenix, where you don’t have any damn data, and concentrate on a place where we actually do have data?

    You were all up in arms claiming that there was no difference in the DLR between the sites. I showed that indeed, there was a large difference in the DLR between the sites, and I told you how to detect it, and I pointed out that it was exactly the difference we would expect to find if water vapor causes DLR.

    At that point, you suddenly got interested in Phoenix and Atlanta again … curious, that.

    w.

    PS—You claim that a change from 8 g/m3 to 20 g/m3 of water in the air increases the DLR by the same as 6 doublings of CO2. What is your source for those figures? All you say is that 1% is 33 times 0.03%, viz:

    * 1%/0.03% = 33 “CO2 equivalents” for each percentage change in water vapor.

    but I don’t see what one has to do with the other. I see no physical reason why your calculation should have anything to do with comparing DLR from H2O and CO2. I ask additionally because that kind of change seems high. You may be right, but it just seems high.

    Also, regarding Phoenix and Atlanta, you say that “All we get is 400 W/m2 at both locations???”. I was unaware that there are DLR measurements for Phoenix and Atlanta, what is your source for them?

  418. Willis Eschenbach says:

    jae says:
    December 23, 2012 at 5:19 pm

    Stephen Wilde says;

    “It is not DWIR that is being measured but the temperature of the air around the instrument and that is controlled by the balance between KE and PE at the height of the sensor which is in turn set by the slope of the lapse rate.”

    Stephen, I have always thought this is the case, but I’m trying to find a way to demonstrate empirically that the GHE does not work as advertised.

    So jae, your claim is that Stephen Wilde is right, that the hundreds of scientists who thought they were using a radiation measuring instrument to measure radiation were actually too dumb to notice that they were using a temperature measuring instrument to measure temperature?

    Yeah, that’s the ticket …

    I swear, every time I think folks have hit rock bottom, they always manage to surprise me. Beauty is only skin deep, but stupid goes all the way to the bone …

    w.

  419. jae says:

    Willis, you say:

    “Finally, I have shown above that in Table Rock, Nevada, and in Goodwin, Mississippi, the actual observations show the expected effects of the difference in water vapor. You keep failing to comment on that, I suspect because it doesn’t fit your schtick of claiming that water vapor has no effect … so how about you forget about Atlanta and Phoenix, where you don’t have any damn data, and concentrate on a place where we actually do have data?”

    No, you didn’t show that, at all. I think all you showed is a temperature effect! You cannot compare the locations because Table Rock is at a much higher elevation and at a different lattitude.

    You also ask:

    “S—You claim that a change from 8 g/m3 to 20 g/m3 of water in the air increases the DLR by the same as 6 doublings of CO2. What is your source for those figures? All you say is that 1% is 33 times 0.03%, viz:”

    I don’t think you understand what I’m doing here. My source is simple logic. I am assuming, for grins, that CO2 has the same radiative power as H20 (actually C02 is much weaker, so my results would be extremely conservative). If I change the H20 vapor (absolute humidity) by 2% of the atmosphere (going from 1% in a dry location to 3% in a wet location (7 g/m3 to 21 g/m3), that is equivalent to changing the CO2 levels by 2/0.03 = 66 times (remember that CO2 is only 0.03 % of the atmosphere, while water vapor varies between about 1-3%). That is more than a sixfold doubling of CO2 (2X2X2X2X2X2 = 66).

    My only question is: Why is there no observed effect for this tremendous increase in GHGs?

    And finally:

    “So jae, your claim is that Stephen Wilde is right, that the hundreds of scientists who thought they were using a radiation measuring instrument to measure radiation were actually too dumb to notice that they were using a temperature measuring instrument to measure temperature?”

    No, I think you are again misunderstanding what I meaan (and presumably Wilde, also). All I am saying is that the radiation measurement reflects the amount of radiation coming from the GHGs in the atmosphere at the “effective temperature.” Just like you say. The DIFFERENCE is that I think it goes no further than that, and the radiation has no effect on the existing temperature. It is just a property of IR-active molecules in the air. It’s not a “heating mechanism,” or “retardation-of-cooling -mechanism,” no more than “back-conduction” is a heating mechanism in a steel rod stuck in a fire at one end. Otherwise there should be a much bigger difference between wet areas and dry ones, per the above reasoning.

    Sorry to see you get angry; usually I’m the one that blows my stack…

  420. Willis Eschenbach says:

    jae, let me start with what you close with:

    jae says:
    December 24, 2012 at 10:28 am


    Sorry to see you get angry; usually I’m the one that blows my stack…

    jae, not sure where you get the idea that I’m angry. I do get angry at times, but you don’t have to parse my words to find it, you’ll know when it happens … I assure you that I am not angry in the slightest, it’s a lovely day here after some days of rain.

    You go on to say:

    Willis, you say:

    “Finally, I have shown above that in Table Rock, Nevada, and in Goodwin, Mississippi, the actual observations show the expected effects of the difference in water vapor. You keep failing to comment on that, I suspect because it doesn’t fit your schtick of claiming that water vapor has no effect … so how about you forget about Atlanta and Phoenix, where you don’t have any damn data, and concentrate on a place where we actually do have data?”

    No, you didn’t show that, at all. I think all you showed is a temperature effect! You cannot compare the locations because Table Rock is at a much higher elevation and at a different lattitude.

    Hmmm, I see that my writing is not clear.

    In fact, I have done what I thought was a rather clever analysis to cancel out the temperature effect. What I have done is to find, not the temperature of the effective radiation level, but the difference between that and the actual surface air temperature. In that way, the surface temperature drops out of the equation. We get to see, for each location, the effective DLR radiation temperature. This, of course, is also closely related to the effective DLR radiation altitude.

    If there is lots of water vapor in the air, we would expect the upwelling radiation to be absorbed near the surface. By the same token, this would mean that the DLR striking the ground originated near the surface. In this case, of course, the temperature of the radiating layer would be close to ground temperature (whatever that might be).

    If there is no water vapor in the air, on the other hand, the atmosphere is relatively transparent to IR. In that case we’d expect the upwelling radiation to penetrate to higher, cooler elevations. And by the same token, we’d expect the DLR to be originating in those same higher, cooler layers.

    As to whether I can compare locations at different elevations and latitudes, why not? I am doing a relative comparison (surface temperature minus effective radiation temperature).

    You also ask:

    “PS—You claim that a change from 8 g/m3 to 20 g/m3 of water in the air increases the DLR by the same as 6 doublings of CO2. What is your source for those figures? All you say is that 1% is 33 times 0.03%, viz:”

    I don’t think you understand what I’m doing here.

    Indeed, that’s why I asked.

    My source is simple logic. I am assuming, for grins, that CO2 has the same radiative power as H20 (actually C02 is much weaker, so my results would be extremely conservative). If I change the H20 vapor (absolute humidity) by 2% of the atmosphere (going from 1% in a dry location to 3% in a wet location (7 g/m3 to 21 g/m3), that is equivalent to changing the CO2 levels by 2/0.03 = 66 times (remember that CO2 is only 0.03 % of the atmosphere, while water vapor varies between about 1-3%). That is more than a sixfold doubling of CO2 (2X2X2X2X2X2 = 66).

    My only question is: Why is there no observed effect for this tremendous increase in GHGs?

    I fear that “simple logic” in a complex system may not be adequate.

    Second, I can’t follow the “simple logic” in that at all.

    A large part of the problem is, it seems like you are not talking about absolute humidity (AH), which is measured in mass per unit volume (e.g. grams per cubic metre). It is never measured as a percentage that I know of.

    Instead you seem to be kind of talking about specific humidity, which is grams of water per kilogram of dry air. That you could express as a percentage, although I have not seen it done … but the problem is, normal surface values for that are in the range of 2 to 20 grams of water per kilogram of dry air. As a percentage, that’s 0.2% to 2% with the larger values occurring in the tropics and decreasing towards the poles. But that maxes out at around 2%, and you are talking about 3%.

    Alternatively, you could be talking about relative humidity. This actually is commonly measured as a percentage (of the theoretical water capacity of the air at that temperature). But if that is the case, then going from a wet location to a dry location might take you from a few percent RH to 80% or 90% RH, not to 3%.

    Here, for example, is the average specific humidity by month for the Atlanta Airport, from this site:

    Month, Specific Humidity at Atlanta Hartsfield Jackson Airport g/kg
    Jan, 3.8 g/kg
    Feb, 4.2 g/kg
    Mar, 5.2 g/kg
    Apr, 6.6 g/kg
    May, 10 g/kg
    Jun, 13.3 g/kg
    Jul, 15.4 g/kg
    Aug, 15 g/kg
    Sep, 12.1 g/kg
    Oct, 8.2 g/kg
    Nov, 5.8 g/kg
    Dec, 4.3 g/kg

    I note that from the low to the high point it varies by about 15 / 3.8 = almost four to one …

    In any case, I can’t make sense of it. I can’t figure out the units you are using or where you are getting your info. Specific humidity in the summer in Atlanta is about twice the SH in Phoenix summer. See here.

    And finally:

    “So jae, your claim is that Stephen Wilde is right, that the hundreds of scientists who thought they were using a radiation measuring instrument to measure radiation were actually too dumb to notice that they were using a temperature measuring instrument to measure temperature?”

    No, I think you are again misunderstanding what I meaan (and presumably Wilde, also). All I am saying is that the radiation measurement reflects the amount of radiation coming from the GHGs in the atmosphere at the “effective temperature.” Just like you say. The DIFFERENCE is that I think it goes no further than that, and the radiation has no effect on the existing temperature. It is just a property of IR-active molecules in the air. It’s not a “heating mechanism,” or “retardation-of-cooling -mechanism,” no more than “back-conduction” is a heating mechanism in a steel rod stuck in a fire at one end. Otherwise there should be a much bigger difference between wet areas and dry ones, per the above reasoning.

    jae, you say that ” the radiation has no effect on the existing temperature“. From this, it is clear that you are saying that this is a special, brand-new kind of radiation that doesn’t contain any energy … because if it contained energy it would transfer that energy to whatever object absorbs the radiation, thus affecting the temperature of the object in question. So if it doesn’t affect the temperature, it cannot contain energy.

    Now, if you and Stephen think you have discovered some kind of energy-free radiation, a never-observed kind of radiation that doesn’t affect the temperature of the object that absorbs the radiation, I’m afraid you’ll have to be the ones to notify the newspapers …

    All the best,

    w.

  421. Greg House says:

    Willis Eschenbach says:, December 21, 2012 at 9:52 am: “Go back and look at the calculation of the average power. I use this mysterious thing called “mathematics”. I take the total amount of solar energy intercepted by the planet, and divide it by the ENTIRE SURFACE AREA OF THE PLANET, using the following magical incantation:
    Average energy per unit area = total energy / total area.
    Study that staggeringly complex formula for a while, and you’ll see that I have indeed given you the average sunlight over the total surface of the planet, and that I have absolutely NOT done what you say, to “consider only half of the earth to receive solar power).”

    =========================================================

    Yes, you have done that implicitly by calling the half of energy “total energy”. To demonstrate that, we can simplify a little bit and imagine that the Earth without atmosphere is a disk and does not rotate continuously but simply switches sides every 12 hours.

    Then it is only for the first 12 hours after the Sun is “turned on” for the first time that only half the total area (1 hemisphere) receives sunlight. The light side would get like 120C warm and the dark side would remain 0K. After the switch the former dark side will get the same 120C, but the other side would only cool down and that not immediately to 0K, hence it’s temperature will be higher then 0K. If the cooling is slow, both sides of the disk would have almost the same temperature.

    Your calculation “average energy per unit area = total energy / total area” is used to extract “average temperature” but does not take into consideration what I described above. As a result, “the average temperature” derived from your calculation is false (not to mention the wrong operating with averages”).

    In other words, your approach does not make any physical sense, because it has little to do with the real physical process.

  422. Greg House says:

    Willis Eschenbach says, December 24, 2012 at 3:30 pm: “jae, you say that ” the radiation has no effect on the existing temperature“. From this, it is clear that you are saying that this is a special, brand-new kind of radiation that doesn’t contain any energy … because if it contained energy it would transfer that energy to whatever object absorbs the radiation, thus affecting the temperature of the object in question. So if it doesn’t affect the temperature, it cannot contain energy. Now, if you and Stephen think you have discovered some kind of energy-free radiation, …
    =======================================================

    I am sorry, but that is an inversion of logic. This would be a correct logical procedure:

    1. you assume that back radiation warms, you have your reasons, wonderful, but

    2. the reality merciless demonstrates that it does not work, then you come to the conclusion that

    3. Your initial assumption is wrong.

    So simple is that.

  423. jae says:

    Willis:

    “A large part of the problem is, it seems like you are not talking about absolute humidity (AH), which is measured in mass per unit volume (e.g. grams per cubic metre). It is never measured as a percentage that I know of.”

    Shit! You talk about bone-deep STUPID! WOW! (and THAT KIND OF RESPONSE FROM YOU IS ANGER, BRO!) And then you LECTURE me on just what absolute humidity is??? Just WHO the hell are you??

    Read my math again and just try to understand it.

    More, later, after you’ve cooled off…..

  424. Willis Eschenbach says:

    jae says:
    December 24, 2012 at 8:33 pm

    Willis:

    “A large part of the problem is, it seems like you are not talking about absolute humidity (AH), which is measured in mass per unit volume (e.g. grams per cubic metre). It is never measured as a percentage that I know of.”

    Shit! You talk about bone-deep STUPID! WOW! (and THAT KIND OF RESPONSE FROM YOU IS ANGER, BRO!) And then you LECTURE me on just what absolute humidity is??? Just WHO the hell are you??

    Read my math again and just try to understand it.

    More, later, after you’ve cooled off…..

    jae, you keep trying to convince folks that I’m angry. Sorry, that dog won’t hunt. I’m not angry. I’m in mystery. Truly, jae, if I were angry, you’d know it, because your ears would be burning and the hair on your chest would be scorched. Since those haven’t happened, you can rest assured that I’m not angry.

    I am, however, confused about the point you are trying to make.

    For example, as I said, absolute humidity is the mass of water in a given volume of air. It is typically measured in grams per cubic metre. And in fact, it is nothing more than the density of the water vapor.

    Now, here’s the thing. You say regarding the absolute humidity:

    If I change the H20 vapor (absolute humidity) by 2% of the atmosphere (going from 1% in a dry location to 3% in a wet location …

    Now, as I said, I don’t understand that. What does “change the H20 vapor (absolute humidity) by 2% of the atmosphere” mean? The absolute humidity is not calculated in “percent of the atmosphere”, as far as I know … but then I don’t have a clue what a percent of the atmosphere might be.

    But perhaps AH might be calculated as a percent, there’s lots that I don’t know, and I’ve been wrong before. As a result, that’s why I asked

    In response, you wish to claim that the issue is that I am STUPID! WOW!. I notice that you don’t say one word regarding where I am wrong, or what I am being STUPID! WOW! about. You wish to convince folks that the issue is my ANGER … when I’ve been having a great time, I haven’t been angered in the slightest. Like I say, it’s been far too nice a day for that kind of nonsense.

    Now, if you were to explain what you mean by discussing absolute humidity in percentages, that would be a great start. If you wanted to comment on the fact that the specific humidity in Atlanta varies about four-fold over the year, and speculate on how that might relate to your work, that would be interesting.

    Heck, if you’d just comment on the analysis I’ve done of the two sites, that would be great. But so far, all I’ve gotten from you is ugly insults and vague claims that I’m wrong and you’re right … if you don’t like my math, you need to point out exactly where I’m wrong. Just claiming I’m wrong is wildly inadequate as an attempt to falsify my work.

    Finally, where’s your analysis, jae? I’ve given you an analysis of the effect of the H2O vapor on the average emission height in the two sites. You don’t like it. But so far all you’ve done is wave your hands and utter reassuring words about Phoenix and Atlanta, and tell me what an angry jerkwagon I am … which is all well and good, but where’s the science? Where’s the substance? Where’s the results of your work? Where’s the beef?

    My best to you,

    w.

  425. jae says:

    OKAY, W-bro, here are sodme calculations, which you seem to enjoy:

    One cubic meter of air at STP contains 1000 liters. And at STP, that is 1000/22.4 moles/liter of gas = 44.6 moles air/M3. The average molecular weight of a mole of air at STP is about 28 gram/mole (about equal to the molecular weight of nitrogen). Therefore, the weight of a cubic meter of air at STP is = 28 gram/mole X 44.6 moles = 1249 grams/m3. At an absolute humidity of 8 grams of water vapor per m3 the amount of water vapor is = 8/1248 = 0.006 = approx. 1 percent (LIKE IN NEVADA, W). Top absolute humidity is about 30 grams/m2 = 30/1249 = 0.0240 = 2.4 %. (not quite 3 percent as the Internet says, but pretty close).

    For reference to HOH, CO2 is A TEENSY-TENNSY-TEENSY (OK, miniscule) part of the atmosphere, compared to water vapor, and it simply amazes me that there are people out there, including SCIENTISTS (AND YOUR???) that don’t get that FACT.

    Once again, look at my math and tell me where I’m wrong…

  426. jae says:

    W:

    “Heck, if you’d just comment on the analysis I’ve done of the two sites, that would be great. But so far, all I’ve gotten from you is ugly insults and vague claims that I’m wrong and you’re right … if you don’t like my math, you need to point out exactly where I’m wrong. Just claiming I’m wrong is wildly inadequate as an attempt to falsify my work.

    Finally, where’s your analysis, jae? I’ve given you an analysis of the effect of the H2O vapor on the average emission height in the two sites. You don’t like it. But so far all you’ve done is wave your hands and utter reassuring words about Phoenix and Atlanta, and tell me what an angry jerkwagon I am … which is all well and good, but where’s the science? Where’s the substance? Where’s the results of your work? Where’s the beef?”

    W: where are the “ugly insults from me?” They are all on your “side” as I see it, like your insult saying that “ugly is skin deep, but stupid goes clear to the bone,” or something like that. Do you remember that? if not, please read your past posts again.

    You seem to be losing it W. I gave you the numbers and science and all you do is spew your inner-dogma. You provided me with data, which I appreciate, but you now think you have answered all questions. BUT, SIR, you still have not addressed my fundamental question, because you refuse to even look at my math. I really don’t think you understand my question, yet!

    THE QUESTION REMAINS: WHY IN THE HELL DOES NOT A SIX-FOLD (AT MINIMUM) INCREASE IN GREENHOUSE GASES IN HUMID AREAS, COMPARED TO DRY AREAS NOT RESULT IN A SIGNIFICANT CHANGE IN TEMPERATURE ON A GIVEN SUMMER DAY?

    IMO, YOU HAVE NOT EVEN ATTEMPTED TO ANSWER THAT, W.

  427. Willis Eschenbach says:

    jae says:
    December 24, 2012 at 9:55 pm

    OKAY, W-bro, here are sodme calculations, which you seem to enjoy:

    One cubic meter of air at STP contains 1000 liters. And at STP, that is 1000/22.4 moles/liter of gas = 44.6 moles air/M3. The average molecular weight of a mole of air at STP is about 28 gram/mole (about equal to the molecular weight of nitrogen). Therefore, the weight of a cubic meter of air at STP is = 28 gram/mole X 44.6 moles = 1249 grams/m3. At an absolute humidity of 8 grams of water vapor per m3 the amount of water vapor is = 8/1248 = 0.006 = approx. 1 percent (LIKE IN NEVADA, W). Top absolute humidity is about 30 grams/m2 = 30/1249 = 0.0240 = 2.4 %. (not quite 3 percent as the Internet says, but pretty close).

    Thanks, jae. That makes sense, although I had never seen AH expressed in those units.

    So, back to your calculations:

    If I change the H20 vapor (absolute humidity) by 2% of the atmosphere (going from 1% in a dry location to 3% in a wet location (7 g/m3 to 21 g/m3), that is equivalent to changing the CO2 levels by 2/0.03 = 66 times (remember that CO2 is only 0.03 % of the atmosphere, while water vapor varies between about 1-3%). That is more than a sixfold doubling of CO2 (2X2X2X2X2X2 = 66).

    Well, you’ve assumed that the difference between Phoenix and Atlanta is the change from 1% to 3%. I just took a look at the figures for Atlanta. Here they are:

    Atlanta Hartsfield Jackson Airport
    Month, Specific Humidity (SH), Average Temperature, Absolute Humidity(AH)
    Jan, 3.8, 52, 0.37%
    Feb, 4.2, 57, 0.40%
    Mar, 5.2, 65, 0.49%
    Apr, 6.6, 73, 0.61%
    May, 10, 80, 0.92%
    Jun, 13.3, 86, 1.20%
    Jul, 15.4, 89, 1.38%
    Aug, 15, 88, 1.35%
    Sep, 12.1, 82, 1.10%
    Oct, 8.2, 73, 0.76%
    Nov, 5.8, 64, 0.55%
    Dec, 4.3, 54, 0.42%

    As you will notice, the maximum average absolute humidity (AH) in Atlanta in your units is 1.38% …

    This is the kind of thing that I expected when I asked for your calculations, jae. Why am I doing your work? You start out by claiming the difference in AH between Phoenix and Atlanta is the difference from 1% (Phoenix) to 3% (Atlanta). But come to find out, you haven’t even bothered to find out the AH in Atlanta, and it’s nothing like 3%, it’s less than half of that …

    For reference to HOH, CO2 is A TEENSY-TENNSY-TEENSY (OK, miniscule) part of the atmosphere, compared to water vapor, and it simply amazes me that there are people out there, including SCIENTISTS (AND YOUR???) that don’t get that FACT.

    Since most climate scientists know that, including me, I’m amazed that you think they don’t. See here for a long list of citations to climate scientists saying exactly what you just said.

    Once again, look at my math and tell me where I’m wrong…

    I’m still waiting for the math regarding your claims. And I still don’t understand your math for the effect of some given increase in water vapor. I mistrust using “simple logic” for that one.

    w.

  428. Willis Eschenbach says:

    jae, one further comment. I don’t think you can use the density of air at STP as you are doing. The problem is that the absolute humidity (AH) is the density of the water vapor at the instant of measurement. It is defined as mass per cubic metre of air using the conditions (pressure, temperature) occurring at the time of the measurement. It doesn’t use the mass of 1 cubic metre of air at STP.

    So if you truly want to use a percentage, to do it properly you have to use the mass of the water as a percentage of the actual mass of the air at prevailing conditions.

    In any case, using your method, here’s the results for Atlanta and Phoenix:

    You need to think about what difference you would expect this to make, including the effect of the clouds. I hold that the effects of clouds and thunderstorms (few clouds or thunderstorms in Phoenix, lots in Atlanta) will totally swamp the overall warming effect of the difference in water vapor.

    All the best,

    w.

  429. Willis Eschenbach says:

    Jae, you ask:

    THE QUESTION REMAINS: WHY IN THE HELL DOES NOT A SIX-FOLD (AT MINIMUM) INCREASE IN GREENHOUSE GASES IN HUMID AREAS, COMPARED TO DRY AREAS NOT RESULT IN A SIGNIFICANT CHANGE IN TEMPERATURE ON A GIVEN SUMMER DAY?

    IMO, YOU HAVE NOT EVEN ATTEMPTED TO ANSWER THAT, W.

    Well, I guess opinions vary on whether I have answered the question of why Phoenix is hotter than Atlanta despite Atlanta having more GHGs. So let’s settle it by evidence. That’s where I demonstrate that indeed I have answered your question. If I look upthread, I find that I said:

    The problem is that there are other effects of very dry and very wet. A wet location like Atlanta supports thermally generated thunderstorms in the summer, which have both a cooling and a moderating effect on the climate. Thunderstorms cool the earth through a host of mechanisms, including direct reflection of solar energy, transfer of heat from the surface aloft, transfer of cold from the atmosphere to the surface, and others.

    Phoenix generally doesn’t have that kind of cooling, thermally generated thunderstorms in the summer. Instead, because the air is so dry that clouds are rare, it gets sun, sun, sun, and more sun.

    Now, which place is going to be warmer? A place with the cooling effect of afternoon thunderstorms in the summer, or a place with almost no clouds at all and just sun, sun, and more sun?

    Clearly, I have offered you my explanation for what it is that seems to puzzle you. Your claim that I “HAVE NOT EVEN ATTEMPTED TO ANSWER THAT” is shown by the evidence to be false.

    So let me ask you again, which place will be warmer? I say that the effect of having no clouds and no thunderstorms is huge. You’re looking at around 60 W/m2 from the cooling albedo effects of increased clouds alone, and those are dwarfed by the cooling effects of the thunderstorms.

    Next, annually the average absolute humidity (AH) in Atlanta (0.79%) is only 80% higher than the annual average AH in Phoenix (0.45%). That’s 1.8-fold, not six-fold as you assert. The difference peaks in May and June, when the AH has risen in Atlanta but not in Phoenix. Those are the only two months where Atlanta has over twice the AH of Phoenix … and the difference is never anywhere near six-fold.

    Finally, jae, someone writing in capital letters is taken on the web as shouting. You are shouting about a “SIX-FOLD (AT MINIMUM) INCREASE IN GREENHOUSE GASES IN HUMID AREAS”. The data for the two cities you selected doesn’t show anything like that. Far from six-fold being a minimum, your cities average only a bit less than two-fold … that doesn’t help your credibility.

    That’s why I say do the numbers first. It should be you telling me about the true ratio of Atlanta and Phoenix, it’s your line of investigation, not mine. It is an interesting one, to be sure, but at the moment it seems like I’m the one doing it …

    w.

  430. jae says:

    Willis:

    I guess I’ll have to give up, because you seem to bent on not even trying to understand my point. I am talking about a “doubling” with respect to CO2 EQUIVALENTS, not a doubling of actual amounts of water vapor. The climate scientists are saying that a doubling of CO2 can cause anywhere from 1.2-4 degrees C. But that “doubling” represents a change of ONLY about 0.03 percent of the atmosphere (0.03 now to 0.06 for a doubling). I’m talking about a difference in greenhouse gas that makes the CO2 doubling very minor.

    You are right that I shouldn’t use STP for the calculations, due to air density differences. What I really should have done is calculate the volumetric percent so that the mass of the air doesn’t matter. I also used too high a value for abs. humidity in Atlanta, as you show. But the point is still the same. Let me try this again:

    7 grams of water/m3 (approx. abs. humidity in Phoenix in July) is 7/18 = 0.39 moles; 0.39 moles water vapor/44.6 moles air/m3 = 0.87% by volume
    18 grams water/m3 (approx. abs. humidity in Atlanta in July) is (18/7)(0.87) = 2.23 percent by volume.

    So 7 grams of water (abs. humidity of 7 g/m3 or 0.87% by volume) represents 0.87/0.03 = 29 “CO2 equivalents” and 18 grams (abs. humidity of 18 g/m3) = 2.23/0.03 = 74 “CO2 equivalents,” giving a difference of 45 equivalents. So, by this calculation, the increase in GHE in July in Atlanta, relative to Phoenix should be equivalent to increasing the amount of CO2 by 45 times, which would be over 5 “doublings.” And HOH is a much better GHG than CO2, so the true difference would be far greater.

    I just would expect to see a much larger GHE in Atlanta, that’s all…Unless the greenhouse effect of water vapor becomes saturated at a fairly low percentage of air volume…

  431. Willis Eschenbach says:

    jae, thanks for the post. I think that the problem is that you are dividing the amount of H2O in the atmosphere by the amount of CO2 in the atmosphere to find the relative amounts of “CO2 equivalents”. I see no theoretical justification for the procedure. I don’t see any reason to think that dividing one by the other would give you “CO2 equivalents”.

    In any case, I’m not sure what your point is. It is well known that the effect of H20 is an order of magnitude larger than that of CO2. You say “I just would expect to see a much larger GHE in Atlanta, that’s all…”. If GHE ruled the roost, you’d be right. But it doesn’t, there are other stronger factors at play.

    You keep ignoring the other effects of water vapor, so I will continue to point them out. If there are thunderstorms and clouds, the peak summertime temperatures don’t go anywhere near as high. This is a difference (clouds and thunderstorms vs clear skies) that is on the order of hundreds of watts per square metre. I have no idea why you’d think that a few watts from the difference in H2O downwelling radiation would be more important than fifty or a hundred watts from having clear skies instead of clouds.

    w.

  432. jae says:

    Willis:

    Sigh, sigh, sigh, AND SIGH, again! willis, YOU STILL DON’T GET MY POINT!

    PLEASE PAY ATTENTION!

    You say: “I think that the problem is that you are dividing the amount of H2O in the atmosphere by the amount of CO2 in the atmosphere to find the relative amounts of “CO2 equivalents”. I see no theoretical justification for the procedure. I don’t see any reason to think that dividing one by the other would give you “CO2 equivalents”.

    WHAT? READ IT AGAIN, WILLIS! AND AGAIN… I AM NOT SAYING ANYTHING LIKE THAT!

    Then you say:

    “In any case, I’m not sure what your point is. It is well known that the effect of H20 is an order of magnitude larger than that of CO2. You say “I just would expect to see a much larger GHE in Atlanta, that’s all…”. If GHE ruled the roost, you’d be right. But it doesn’t, there are other stronger factors at play.”

    Willis, Willis, Willis, you STILL don’t get my point!!!! You are one of my heros in this hillarious academic bullshit game of “climate science”. BUT LOOK, W., if a change in OCO of 0.03% can affect temperature IN ANY SIGNIFICANT WAY (the “intellegencia says 1-4C or so), then a change in water vapor of many times that should affect temperature much more. To quote you ” It is well known that the effect of H20 is an order of magnitude larger than that of CO2.” WILLIS, IT FOLLOWS THEN THAT YOU SHOULD SEE A BIGGER GREENHOUSE EFFECT IN ATLANTA THAN PHOENIX. BUT I DO NOT SEE IT! I just don’t know how to say it any clearer. I don’t know what else to say..

    SHIT, YOU MUST BE A DEMOCRAT!!

  433. Willis Eschenbach says:

    jae says:
    December 25, 2012 at 7:42 pm

    Willis:

    Sigh, sigh, sigh, AND SIGH, again! willis, YOU STILL DON’T GET MY POINT!

    PLEASE PAY ATTENTION!

    WHAT? READ IT AGAIN, WILLIS! AND AGAIN… I AM NOT SAYING ANYTHING LIKE THAT!

    Willis, Willis, Willis, you STILL don’t get my point!!!!

    WILLIS, IT FOLLOWS THEN THAT YOU SHOULD SEE A BIGGER GREENHOUSE EFFECT IN ATLANTA THAN PHOENIX. BUT I DO NOT SEE IT!

    SHIT, YOU MUST BE A DEMOCRAT!!

    And you must be a fool … Jae, that’s it. I’m through. That’s over the top. I have done what I could to assist you, and you want to talk to me as though I were a child.

    Sorry, but you’ve used up all of your second chances with me. I’m done with your nasty insinuations, your innuendoes, and your SHOUTING IN CAPITAL LETTERS. Go away, I’m done here. You can only insult me so many times, then the door slams shut.

    It’s too bad, because you had interesting ideas. But YELLING AND SCREAMING doesn’t move me, jae, it just makes you look like a jerkwagon.

    So. please go away. Don’t go away mad … just go away. It’s not any fun any more. Your refusal to either explain yourself or pay any attention to anything except your own big mouth has finally reached the level of my gag reflex. I can’t take any more of your childish BS, jae, it’s gotten to the point where it turns my stomach.

    It’s over, my friend, over and done. When your name comes up in the future, I’ll just point and laugh. You want to post on my threads? Prepare for not getting any answers from me. You’ve had your fun, acting like you actually were interested in other peoples’ ideas … then refusing to respond when they presented ideas … then yelling and whining like a six-year-old when people refused to respect your lack of details and your absence of facts and your deficiency of explanations … but the game is over.

    I’m done with you, my friend. Congratulations, your vote has been officially cancelled by me. Oh, you are free to post here, but I won’t answer, I’ve got better things to do.

    Curiously, jae, I do wish you well. I just don’t have time to faff around with your lack of cooperation.

    w.

  434. I’ve been following the conversation with one simple-minded thought in mind…the widely reported 33C of average surface temperature increase due to “greenhouse gases”. The simple-minded part is trying to uncover the mysterious mechanics of this average increase. I see the expected modulating mechanisms where peaks (high and low) get constrained, but I don’t see the necessary asymmetry (quick heating and slowed cooling via atmospheric constituents) or exothermic reactions which would do anything to the average. If you add a CO2 molecule that wasn’t otherwise there, surely it’s obvious the cooling rate increases…both via convection and via providing a radiation “portal” when CO2 collides with N2 or O2 molecules.

  435. Gail Combs says:

    jae says:
    December 24, 2012 at 10:28 am

    No, I think you are again misunderstanding what I meaan (and presumably Wilde, also). All I am saying is that the radiation measurement reflects the amount of radiation coming from the GHGs in the atmosphere at the “effective temperature.” Just like you say. The DIFFERENCE is that I think it goes no further than that, and the radiation has no effect on the existing temperature. It is just a property of IR-active molecules in the air. It’s not a “heating mechanism,” or “retardation-of-cooling -mechanism,” no more than “back-conduction” is a heating mechanism in a steel rod stuck in a fire at one end. Otherwise there should be a much bigger difference between wet areas and dry ones, per the above reasoning.
    >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

    Except there is a big difference between wet and dry areas.

    If you are talking DLR, this paper records actual measurements.

    Parameterization of atmospheric long-wave emissivity in a mountainous
    site for all sky conditions
    J. Herrero and M. J. Polo
    Received: 14 February 2012 – Accepted: 11 March 2012 – Published: 21 March 2012

    ABSTRACT
    Long-wave radiation is an important component of the energy balance of the Earth’s surface. The downward component, emitted by the clouds and aerosols in the atmosphere, is rarely measured, and is still not well understood…..

    After the Stefan-Boltzmann Law for the radiation emission of any body at a temperature
    L↓
    [down]
    T (K), downward long-wave radiation L (W m ^-2) coming from the near-surface layer of
    the atmosphere may be written as:

    L↓ [down] = Ea S Ta

    where Ea is the apparent emissivity of the sky (Unsworth and Monteith, 1975), S (W m^-2 K^-4 ) is the Stefan-Boltzmann constant, and Ta (K) is the air temperature near
    the surface (typically 2 m).

    The downward long-wave radiation measured for 5 consecutive years at RP Station, converted to Ea according to Eq. (1), is shown on Fig. 2a and summarized in the probability density function (pdf) in Fig. 3. The lower values of Ea belong to clear sky situations, and in the pdf they smoothly fit a Gaussian with a mean value of 0.68 and a standard deviation of 0.0565. During very clear days, with a low temperature and relative humidity, it exhibits values ranging from 0.5 to 0.6. In the pdf, 0.77 sets the limit between clear sky and partly covered situations; higher values of Ea denote the presence of clouds in the atmosphere. A seasonal pattern is easily observed in Fig. 2a, where the lowest emissivity values from clear skies are reached during winter. This emphasizes the importance of long-wave balance for cooling the soil and snow under high mountain clear skies. These measurements are similar to those found by Frigerio (2004) in Argentina, at 2300 m a.s.l., with night values of atmospheric emissivity under 0.7 with clear skies. Figure 2b represents daily variation of εa , that is, the difference between maximum and minimum daily value. It exhibits a marked seasonality, where wider daily variations of Ea in winter are in accordance with wider variations in temperature and relative humidity.….

    Figure 4a shows the relationship between the measured values of Ea , Ta , and relative
    humidity, Wa , for all sky conditions
    . In RP Station records, Wa represents the effect of
    the presence of water in the atmosphere better than the water vapor pressure. That
    relationship is especially strong for clear and completely covered skies, as shown by
    the low magnitudes of the standard deviation (std) in Fig. 4b for the values of Ea under 0.7 and over 0.9, respectively.
    Partly covered skies appear as a transition zone
    between these two boundary situations. There are some differences in these relationships between daytime and night-time values, but they were not found significant for
    these particular data…..

    Do not forget the emissivity for water droplets and ice crystals is 0.98.

    Some real world data:
    For May 2012, Barcelos, Brazil (Lat: 1 South)
    Temp: monthly min 20° C
    monthly max 33° C
    monthly average 26° C
    Average humidity 90%
    Temperature range was 13° C

    For May 2012, Adrar, Algeria (Lat: 27 North)
    Temp: monthly min 9° C
    monthly max 44° C
    , monthly average 30° C
    Average humidity around 0%
    Temperature range was 35° C

    For Tampico, Mexico, (Lat: 22.3° N) – At an Airport on Atlantic ocean
    May humidity was avg 70%
    min 24.1 ° C
    mean 28.8 ° C
    max 33.6 ° C
    Temperature range was 9.5 ° C
    Varied from mostly cloudy to scattered clouds with afternoons clear. 3 days with a bit of drizzle (0.01 to 0.2 inches of rain) one day T-storm

    Another city similar in latitude and closer in altitude to Adar Algeria.
    Laredo, TX Lat 27.5° N Altitude is 126 m or 413 ft above sea level (Airport)
    22.2° C
    28.8° C
    35.6° C
    humidity 62%
    Range is 12.4° C

    Take a good hard look at those pieces of real world data and ask yourself what it is telling you.

    #1. The solar eclipse data tells you the earth & air temperature response (in low humidity) to a change in solar energy is FAST!

    #2 The effect of the addition of water vapor (~ 4% globally) is not to raise the temperature but to even the temperature out. The monthly high is ~10C lower and the monthly low is ~10C higher when the GHG H2O is added to the atmosphere in this example. The average temperature is about 4C lower in Brazil despite the fact that Algeria is further north above the tropic of Cancer, For Tampico Mexico and Laredo texas the average is 1C lower that Algeria with the humidity @ 60 to 70% (dipping towards the 50% mid day.) It is read at an airport. Some of the difference is from the effect of clouds/albedo but the dramatic effect on the temperature extremes is from the humidity.

    I took a rough look at the data from Brazil. Twelve days were sunny. I had to toss the data for two days because it was bogus. (see link) The average humidity was 80% for those ten days. The high was 32 with a range of 1.7C and the low was 22.7C with a range of 2.8C. Given the small range in values over the month the data is probably a pretty good estimate for the effects of humidity only. You still get the day-night variation of ~ 10C with a high humidity vs a day-night variation of 35C with very low humidity and the average temp is STILL going to be lower when the humidity is high.

    ALTITUDE:
    Tampico Mexico ~ Elevation: 15 metres (49 feet)
    Barcelos, Brazil elevation ~ 30 meters (100 ft)
    Adrar, Algeria ~ Elevation: 280 metres (920 feet) a drop in temperature of ~ 4C due to altitude: http://www.engineeringtoolbox.com/air-altitude-temperature-d_461.html

    One would expect a drop in temperature of ~ 4C due to altitude for Adrar, Algeria.

    This data would indicate GHGs have two effects. One is to even out the temperature and the second is to act as a “coolant” at least if the GHG is H2O.

    The latent heat of evaporation could be why the average is 4C lower when in Brazil vs Algeria. As one of the commenters here at WUWT mentioned using temperature without humidity to estimate the global heat content is bad physics.

    Data Brazil:
    http://classic.wunderground.com/history/station/82113/2012/5/22/MonthlyHistory.html
    http://www.climate-charts.com/Locations/b/BZ82113.php
    Map: http://www.worldatlas.com/webimage/countrys/samerica/brsa.gif

    Data Algeria:
    http://www.wunderground.com/history/airport/DAUA/2012/5/20/MonthlyHistory.html
    http://www.climate-charts.com/Locations/a/AL60620.php
    Map: http://www.worldatlas.com/webimage/countrys/africa/dzafrica.gif

    Data for Mexico:
    http://classic.wunderground.com/history/airport/MMTM/2012/5/19/DailyHistory.html
    http://www.tampico.climatemps.com/
    http://www.zonu.com/fullsize1-en/2009-09-17-5311/Satellite-image-of-Tampico.html

    Photos Adrar, Algeria and Barcelos, Brazil
    http://algeriaspace.blogspot.com/2007/07/photos-satellite-ville-adrar-algerie.html
    http://www.maplandia.com/brazil/airports/barcelos-airport/

  436. Willis Eschenbach says:

    kencoffman (@kencoffman) says:
    December 26, 2012 at 6:11 am

    I’ve been following the conversation with one simple-minded thought in mind…the widely reported 33C of average surface temperature increase due to “greenhouse gases”. The simple-minded part is trying to uncover the mysterious mechanics of this average increase. I see the expected modulating mechanisms where peaks (high and low) get constrained, but I don’t see the necessary asymmetry (quick heating and slowed cooling via atmospheric constituents) or exothermic reactions which would do anything to the average. If you add a CO2 molecule that wasn’t otherwise there, surely it’s obvious the cooling rate increases…both via convection and via providing a radiation “portal” when CO2 collides with N2 or O2 molecules.

    Ken, for the mechanisms see my two posts, The Steel Greenhouse and People Living in Glass Planets.

    All the best,

    w.

  437. Willis Eschenbach says:

    Gail Combs says:
    December 26, 2012 at 6:21 am

    … Take a good hard look at those pieces of real world data and ask yourself what it is telling you.

    Thanks, Gail. What the real world data tell me is that it is naive to try to tease out a heating signal from water vapor by just grabbing a bunch of cities and looking at their temperatures.

    The part that you and jae keep sailing right on past is that you have to, you must, you need to, you are required to account for the confounding variables. You can’t just say, as jae does, that Phoenix is warmer than Atlanta despite having less water, so that proves the water vapor doesn’t heat anything. That’s grade school understanding of attribution.

    The confounding variables that you need to account for include variables unrelated to water, such as altitude or soil type or distance from the ocean, along with much more difficult confounding variables such as clouds and thunderstorms and evaporative cooling and ground cover that are a function of water itself. These latter ones are harder to account for, since they vary in proportion with water vapor.

    So no, you cannot just grab a handful of cities in Brazil, Mexico, and Algeria, look at the temperatures and the relative humidity, and declare the game over.

    I have shown above one way that actually works that shows the difference in the DLR, which I described above. This was to calculate how high the effective radiative level was above the ground (or alternately how cool the radiative level was) in moist and dry areas. This calculation shows that the water vapor indeed does increase the DLR …

    But of course, to you and jae this finding is anathema, so you just ignore it.

    Another way to reduce, if not get rid of, the confounding variables is to look at different times in the same location. Look at the data for Goodwin Creek, for example, month by month. See if the DLR varies in phase with the water vapor.

    But please, Gail, please, don’t do what jae does and claim that he has discovered some new kind of DLR radiation that doesn’t contain energy. DLR contains energy, we’ve measured it many many times. We know just how much energy the DLR contains, and we know that energy can’t be created or destroyed … which means that when DLR is absorbed an object, that object ends up warmer than it would be in the absence of DLR. In other words, jae’s claim that DLR that contains no energy is just his desperate attempt to salvage his incorrect understanding.

    w.

  438. jae says:

    Willis:

    So you can comment to Gail and disparage me and my thoughts on your tirade, and I cannot argue back? You are, indeed a Democrat! Pure Obama-style-authoritarian, fella….

    Do you have ANY concept of fairness??

    But I’m relieved, in a way, that I am now an outcast here, because, SIR, you are NOT a listener, but some kind of big-headed “boss.” YOU STILL DON’T GET MY POINT, AFTER ALL THIS TALK. Gail doesn’t either.

    And using CAPS is nowhere near as nasty, angry, hot-headed as your post was. Hypocrite!

    I’m again weary of this conversation, and I give up here, AGAIN!

    Good bye.

  439. Willis Eschenbach says:

    jae says:
    December 26, 2012 at 6:02 pm

    Willis:

    So you can comment to Gail and disparage me and my thoughts on your tirade, and I cannot argue back? You are, indeed a Democrat! Pure Obama-style-authoritarian, fella….

    I don’t have the power to stop you from arguing back, jae, or to stop you from arguing front for that matter. Nor would I stop you if I could. Nor did I say that you couldn’t argue back.

    So you are arguing against something that, as far as I know, I never said.

    Next time, quote what I said so we can all be clear about just what is the latest thing that you are upset with me about. It changes so often, I can’t keep track.

    Of course, you include the requisite number of insults …

    … Do you have ANY concept of fairness??

    … SIR, you are NOT a listener, but some kind of big-headed “boss.”

    … YOU STILL DON’T GET MY POINT, AFTER ALL THIS TALK. Gail doesn’t either.

    I’m sure it’s Gail’s fault for not understanding your brilliance, jae. I’ll speak sharply to her and tell her to shape up …

    w.

  440. Mack says:

    Kencoffman says…
    “If you add a CO2 molecule that wasn’t otherwise there, surely it’s obvious the cooling rate increases”……….. Yes Ken, you’re right , gases don’t add energy, but disperse it. All gases dissipate heat.

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