On the Flat Earth Rants of Joe Postma

Reposted from Dr. Roy Spencer’s blog

June 4th, 2019 by Roy W. Spencer, Ph. D.

Willis Eschenbach and I have been defending ourselves on Facebook against Joe Postma’s claims we have “flat Earth” beliefs about the radiative energy budget of the Earth. The guy is obviously passionate, as our discussion ended with expletive-laced insults hurled my way (I suspect Willis decided the discussion wasn’t worth the effort, and withdrew before the fireworks began).

Joe advertises himself as an astrophysicist who works at the University of Calgary. I don’t know his level of education, but his claims have considerable influence on others, which is why I am addressing them here. He has numerous writings and Youtube videos on the subject of Earth’s energy budget and greenhouse effect, and the supposed errors the climate research community has made. I get emails and comments on my blog from others who invoke his claims, and so he is difficult to ignore.

Here I want to address just one of his claims (repeated by others, and the basis of his accusation I am a flat-Earther), recently described here, regarding the value of solar flux at the top of the atmosphere that is found in many simplified diagrams of the Earth’s energy budget. I will use the same two graphics used in that article, one from Harvard and one from Penn State:

Screen-Shot-2019-01-18-at-1.23.33-PM-550x354

Joe’s claim (as far as I can tell) is that that the solar flux value (often quoted to be around 342 W/m2) is unrealistic because it is for a flat Earth. But as an astrophysicist, he should recognize the division by 4 (“Fs(1-A)/4” and “S/4”) in the upper-left portion of both figures, which takes the solar constant at the distance of the Earth from the sun (about 1,370 W/m2) and spreads it over the spherical shape of the Earth. Thus, the 342 W/m2 value represents a spherical (not flat) Earth.

Just because someone then draws a diagram using a flat surface representing the Earth doesn’t mean the calculation is for a “flat Earth”.

Next in that article, Joe’s (mistaken) value for the solar constant is then used to compute the resulting Earth-Sun distance implied by us silly climate scientists who believe the solar constant is 342.5 W/m2 (rather than the true value of 1,370 W/m2). He gets twice the true, known value of the Earth-Sun distance, simply because he used a solar flux that was off by a factor of 4.

Now, I find it hard to believe an actual astrophysicist could make such an elementary error. I can ignore Joe’s profane personal insults, but he ends up influencing many people, and then I have to deal with their questions individually. Sometimes it’s better if I can just point them to a blog post, which is why I wrote this.

341 thoughts on “On the Flat Earth Rants of Joe Postma

  1. Really?

    “.. which takes the solar constant at the distance of the Earth from the sun (about 1,370 W/m2) and spreads it over the spherical shape of the Earth. Thus, the 342 W/m2 value represents a spherical (not flat) Earth ..”

    That IS the problem, the sun doesn’t shine at the whole surface of the earth at the same time .. (i.e. flat earth) at 342 W/m2 (-18C.)

    That means, it’s not real, as empirical observations obviously demonstrate ..

    • Roald, it’s an average. And as far as “it’s not real”, consider this: NO ONE has a “REAL” description of the atmosphere’s behavior, because it is far more complex. You can’t just point to the ideal gas laws for a model, because energy transfers and inversions occur, either, so THAT’s not real. And if you don’t consider precipitation, THAT’s not real.

      The only “REAL” diagram for the atmosphere’s behavior wouldn’t fit on a one page illustration, because it’s a constantly-changing dynamic system: all the emission and absorption and state change and convection behaviors are NEVER constant. So, let averages speak to the facts that the earth has multiple pathways for energy to enter and leave the system, and it’s nice to have a reasonable way to grasp the basics. Or did you start with calculus in the first grade?

      • “Roald, it’s an average. And as far as “it’s not real”, consider this: NO ONE has a “REAL” description of the atmosphere’s behavior ..”

        Exactly!! And using the average makes it incorrect (i.e. FLAT EARTH!)

          • No, but when it is empirical, obvious, logical and factual 100% wrong, it should not be used!

            Unless you have a nefarious motive for deception ..

        • The sun illuminates a disc of area pi r^2 24/7. The total surface of a sphere is 4 pi r^4. So the average insolation of the surface of the sphere is one quarter. That is high school maths not rocket science or astrophysics.

          Not only is Joe Potsmoka dumb, he is lying about his technical qualifications.

          • I am not disputing the math, that is well known, that is not the point either.

            The point is, by averaging the suns flux to the whole sphere, you are immidiately completely wrong.

            For example does that allow for the average temperature of -18C.

            Is the average temperature of the surface of the earth – 18C.?

          • Roald J. Larsen June 4, 2019 at 4:41 pm

            … Is the average temperature of the surface of the earth – 18C.?

            Nope. Because of the atmosphere, the surface of the Earth is considerably warmer. -18 is what it would be without the atmosphere. It is also the average temperature of the Moon’s surface. It’s also the average temperature you would get if you scanned the planet from outer space with an IR thermometer. It’s the effective radiation level (ERL) blackbody temperature.

            The only way the planet can get rid of heat energy to outer space is by radiation. To balance the energy it receives from the Sun, the Earth radiates with an effective temperature of -18C.

          • CommieBob,
            Actually you are incorrect. It works out to -18 if you assume an albedo of 0.3. But with no atmosphere, there would be no clouds, and the albedo would be aproximately that of the moon, around 0.12….and the temperature of the planet would be hotter than it is now. Easy to work out yourself from SB equation.

          • Except that the surface area of a sphere is given by 4 π r^2 (not 4 π r^2, as you have shown). Typo?

          • DMacKenzie June 4, 2019 at 6:23 pm

            …the albedo would be aproximately that of the moon,

            That sounds reasonable and yet, for some reason, 0.3 is “commonly assumed for the Earth in the absence of its atmosphere”. link I can’t quickly find out why that is the common assumption and, yes, common assumptions are often wrong.

          • What assumptions do you need to justify this procedure?
            1. There is no cosine effect
            2. Albedo is independent of angle of incidence
            3. *EVERY* effect of incoming energy is linear, so it doesn’t matter if you average the energy, then compute effect or compute effect at each point, then average.

            There’s probably more assumptions needed, but, to me, all three of these are clearly false.

          • Commie B

            Actually you make another mistake as well. You cite the incorrect comparison. The idea of the Earth with and without GHG’s argument is that GHG’s increase the temperature of an atmosphere compared with one that has no GHG’s, not “no atmosphere”.

            Yes it is correct that the albedo would perhaps be the same as the moon’s but that is not the point. The appropriate argument to present when claiming that GHG’s increase the near-surface air temperature is to make a comparison under two conditions: with and without GHG’s, not with an atmosphere with GHG’s and without atmosphere at all.

            An atmosphere without GHG’s would still be heated by the sun heating the surface, 1370 W/m^2 at the equator and less at “the edges”. Absent any GHG’s, the atmosphere could not cool by radiation and the near surface temperature would be much higher than the effective -18 C it is now.

            The average near surface air temperature is not necessarily related to the effective radiating temperature (which is an average as well).

            This raises yet another defect in the argument of the Calgarian astrophysicist. The fact that the Earth is a sphere means the effective radiating temperature of the atmosphere above the equator at noon is considerably different from that at noon near the North Pole. Because cooling radiation power is and t^4 function, averaging the temperature over the area/4 and average insolation/m^2 is not a valid approach.

            The 1370 W/m^2 heating at the equator produces far greater emitting power than 342 in that same area. Taking an area receiving 0 W insolation and increasing it to 342 produces a value that is not near the truth for the globe. It is like saying that the average of 10 random numbers, or to the 10 numbers multiplied by 4, is linearly related to the average of those same numbers to the 4th power. Nonsense.

            If the “cold” spots were 0 Kelvins (no heat in, all heat out) how hot must the hots spots be to reach an average of 1370 on the whole disk when the energy radiated out is a function of the 4th power of temperature?

            Remember the goal is not to calculate the average effective emission temperature, it is to predict the actual near surface temperature. In the absence of very large heat transfers within the system, hot spots will emit more than 4 times the power of the cold spots per sq m, so dividing by 4 doesn’t give anything like the correct answer.

            Having arrived imperfectly at their -18 C, they claim the difference to +15 is caused by the effect of GHG’s. Monckton demonstrates that it is actually GHG’s plus feedbacks, from which an ECS can be calculated. That value is about 1.1-1.5 per doubling of GHG concentration, far below the values mooted by the IPCC.

            This whole thing is a mess. Give it to real heat transfer wizards like Prof Adrian Bejan. HE said the calculation is so simple it wasn’t even interesting.

          • Crispin in Waterloo but really in Johor June 5, 2019 at 12:45 am

            You can make a compelling case that GHGs aren’t the main event.

            It is very important to note that despite radically different compositions, both atmospheres (Earth and Venus) have approximately the same dry lapse rate. This tells us that the primary factor affecting the temperature is the thickness of the atmosphere, not the composition. Because Venus has a much thicker atmosphere than Earth, the temperature is much higher. Venus Envy

          • If your goal is to figure out what the impact the atmosphere has on the temperature of the surface, you need to use the albedo of the earth as is, not the albedo of what it would be without an atmosphere.
            Only change one variable at a time.

          • Misprint?: Should not “the surface area of a sphere is 4 pi r^4” be rather 4 pi r^2?…

          • Crispin old buddy,
            I’m not sure I’m with you on this “Absent any GHG’s, the atmosphere could not cool by radiation….” Without greenhouse gases, the atmosphere is transparent to IR. The sun-warmed ground then radiates directly to -270 outer space. The N2 and O2 would warm by convection, but as you say, would not radiate that warmth to space. But I think the ground radiation will exceed the “missing” GHG radiation, with a net result of a cooler average surface temp. I did not crunch this on my trusty HP 35, so could be mistaken.

        • Area of a circle = pi*r^2 Let represent Earth’s cross section in a solar flux of 1370 w/m^2
          Area of sphere = 4*pi*r^2 Let represent surface area of Earth.
          Earth’s cross section relative to Sun divided by Earth’s surface area = pi*r^2 / 4*pi*r^2 = 1/4
          Thus Earth’s surface receives 1370 w/cm^2 / 4 = 242.5 w/cm^2

          Postma fails basic geometry, as well as astrophysics.

        • Then the only model complex enough to use is the one currently operating around the planet earth. Any other will have some simplification, some aberration, some distortion. Get yourself a brain of God-like brilliance and look at the whole big picture.

        • Roald, I think your point is that our very complicated climate does not operate on averages. It operates on conditions on a scale from kilometers down to picometers.

          Averaging over a range that wide range is dubious at best. Unfortunately there are no computers even partially capable to solve the equations due to their digital nature.

          • “I think your point is that our very complicated climate does not operate on averages”

            The climate doesn’t even operate globally. Yet we speak of “global climate change”. I notice suddenly that many areas are warming “faster than others” which was the basis for rejecting the Mideval Warm Period and Little Ice Age. Yet, now it’s great to use these local phenomena to prove “global” warming. All of this is based on simplification of a very, very complex phenomena, as you state.

          • Weather over time is climate, human focus on climate because it has become a source of income for too many activists and criminals.

            Weather pattern changes from time to time, place to place.

            Averages can be useful, but not when it is used to fool people

            By averaging the sun’s input to the whole globe you get a “cold” sun that can’t drive the weather.

            That is deceptive!

        • Roald, you realize that only one square meter on the whole globe receives 1370 W/m^2, the one that is at right angles to the sun’s rays. Those square metres around the first one are slightly canted away from the sun and so they receive a bit less! And the third tier of square meters around the seceond tier are canted even more. Going further away finally to the square meters of earth’s surface near the periphery of the planet, they are hardly receiving any watts because the suns rays there are close to parallel to the surface of these distant square meters.

          Therefore, from the sun’s viewpoint, the earth presents itself as a disk. Are you following me here? Now the sun, perforce, can only illuminate half the sphere so the other half is in the dark, receiving no watts per sq m. The half of sphere being illuminated, meanwhile, is being heated the most where the sun’s rays impinge vertically on the center of the disk but, since the sun’s rays impinge evermore obliquely on the earth’s surface as we move away from this centre, insolation gets weaker.

          The area of a sphere is 4pi r^2, the area illuminated at one time is half this and the diminishing strength of the sun moving out from the centre to the periphery of the planet is half again, i.e. pi r^2, the area of a circular disc. I hope that helps.

          • 1370 is the radiation received at ground level on a horizontal surface when the sun is directly overhead (as you stated) IF there is no atmosphere. 1000 watts/m^2 (on general average with a general atmosphere and general humidity at sea level.

          • You are arguing something that is beside the topic of the article.

            Nobody is disputing the math, that is not the topic, the issue how you use the math, using an average input of energy as the start point when calculating the energy budget.

            It doesn’t depict the reality logically, mathematically or empirically.

      • No, the sun doesn’t shine at the whole world at the same time.

        But, by averaging the sun flux that do come in to the whole world, you get sunshine 24/7 – and for that to happen you need the earth to be flat.

        • Roald! Please! Pick any spot on the planet, and take the average of the illumination across a cloudless day. It’s going to roughly follow a half-wave sinusoid. Radiation outward from that point will be out of phase, but will also wax and wane in a quasi-sinusoidal pattern. At the end of a 24 hour cycle, one should be able to compute roughly how much radiation was absorbed, and how much was emitted. The actual complexity is not describable on a page, and I don’t care WHAT you think properly describes the forcings involved, NOBODY provides a complete picture. That does not, however, invalidate the idea of describing the average absorption and emission from the surface. And since the “average” insolation covers both the more intense illumination with the sun perfectly “above” the spot, and near-night illumination over the spots near sunrise and sunset, that “average” is descriptive and useful even though such an “average” spot DOES NOT EXIST (i.e., absorption and emission are out of phase with each other, therefore no spot will have exactly the right absorption and emission at the same time). This does not make such descriptions either “unreal” or “not physical” or “flat earth”. It simply should be recognized as limited, but still instructive, because although it is incomplete, and not dynamic, it still shows the paths for incoming and outgoing radiation.

          Joe’s complaints are out of ignorance concerning the reality — irrefutably measurable reality — of down-welling radiation that delays the departure of insolation energy back into space, thus creating a warmer surface atmosphere than would be present if not for the radiative behavior of water vapor and the minor radiative gases. I’ve had this argument with him personally, and he has said over and over again that radiation from the atmosphere cannot be absorbed by the surface, but without evidence other than a wrongful claim of a 2nd Law violation (please tell me you don’t need a lesson on this!), and without reference to the actual measured data from stations recording such down-welling radiation.

          The real argument with the alarmist side is the sensitivity to additional CO2, but Joe is arguing that there IS NO sensitivity to CO2, hence the need to condemn descriptions that include that as a factor.

          • Down-welling isn’t delaying the departure of radiation (the reason there exists an infrared/atmospheric window that states no atoms (carbon, hydrogen, oxygen and nitrogen atoms absorb in the interval between 8 micrometres and 14 micrometres). Convection of air carrying heat to space expands and compressing (down-welling) around a cell. All molecules in the air emit radiation, instruments on the ground measure in millivolts. The instrument can identify both solar(shortwave) and infrared (longwave) radiation. Solar insulation is 340 W m2 and outgoing is 340 (150 W m2 from clouds, 150 from solar upwelling + 38 (sensible heat). This eliminates any greenhouse effect or nonsense about CO2 being a earth’s climate control knob.

          • There is no sensitivity to CO2. If there is then the specific heat of air must have changed. I have heard of no such change.

            Either the specific heat tables are correct or climate science says they are wrong.

            Why do you think Anthony’s CO2 jar experiment had the results it did?

          • Correct, another thing, if CO2 did actually do what the Rent & Grant Seeking Community of activists and criminals say it does, we wouldn’t be having this discussion, or any other discussions for that matter, we wouldn’t even be here given the geoligical levels of CO2 …

        • I’m with you, Roald – there’s more to a tilted rotating sphere than just area.

      • Michael J,

        At the link you provided, look about halfway down the page, and there you will find the exact diagram that Postma is talking about. And that’s precisely the assumption of the math there. That’s what the math means. Dividing by four cannot have any other physical meaning other than a flat-Earth disc with the surface area of the Earth sphere, or a flat-Earth disc with the radius of the Earth at twice the distance from the sun, or a magical reality where light on a hemisphere can bend around the Earth sphere to dilute itself by one fourth, creating a perpetual twilight planet, 24/7/365.

        https://youtu.be/S-ezv5AckDk

    • Also it does not take account of the spherical shape of the Earth – the angle of incidence of sunlight. It only accounts for the area: an area is not a shape.

      • Hermit.Oldyguy, the angle of incidence of sunlight IS part of the reason why you divide by 4 to go from the area of a disk to the area of a sphere with the same radius as the disk. For the sunlight impinging on the projected disk at right angles to be spread over a larger area, it HAS to be at oblique angles.

        • @Roy Spencer, sir, with all due respect, any shape can have 4 times the area of a given circle.

        • The real problem is that the Earth is not uniform, the surface absorption and reflection characteristics are different, from one place to another.

          The planet receives solar irradiance in tranches, as you say over an area of 1/4 that of a sphere, yet it radiates over the entire surface of the sphere 24/7.

          Further, the solar irradiance is often not absorbed at the surface, but perhaps at altitude in the atmosphere, the top of tree canopies (consider that in tropical rain forests less than 3% of solar irradiance reaches the surface and a large percentage of that absorbed in the canopy powers photosynthesis), over oceans it may be asorbed some 2 to 200 metres below the surface.

          The accepted energy budget cartoon is an over simplification of a very complex system.

          • RV no problems only solutions. The cartoon is a simplification of a complex system. That is basically the function of doing an average.
            Yes the surface is not uniform and for the most part (oceans) not a true surface but permeable to some depth.
            Nonetheless there is a surface of sorts that absorbs all the unreflected radiation and emits back the energy received and it can be measured to a reasonable extent. Only a very small part of that absorbed in the canopies powers photosynthesis, the majority of it (>>96.999%) is absorbed by the leaves and branches which warm up and despite your assertion of only 3% reaching the surface the canopy is the surface at that point for all practical purposes. You mean 3% reaches the ground proper but the relevance of this escapes me.
            The planet receives solar irradiance over an area of 1/2 a sphere (+) in actuality but the amount of energy is only equal to that that would reach a disc of half this area.
            The energy budget should be a reasonable approximation.
            There may be some variation due to the Crispin in Waterloo t^4 comment counteracted by the Tom in Oregon City overall argument.
            One imagines that Roy is well versed with any slippage this might entail working with spheres and differing wavelengths and the difference is too slight to be significant despite Postma’s assertions, at least I hope so.
            While a T^4 to a times 4 average might vary a lot with large differences in individual numbers one must remember the temperature variations are on a Kelvin scale and hence a lot closer to each other in size.

        • “you divide by 4 to go from the area of a disk to the area of a sphere with the same radius as the disk.”

          The sphere is spinning, getting light at changing angle and then no light.
          The disk is not spinning and getting light all at the same angle

          Having a static flat disk with one side the same area as the Earth is not a good model.

      • What is missing more than having Earth being a sphere is that there is no night-time. The flat earth accusation by Postma becomes valid when you realize that simply dividing solar input by 4 to account for the dark side of the planet and the angled light at the edges is inadequate, as none of the mechanisms that work during the night to cool the planet are incorporated. The models he is criticizing have no night. All of the conversion of heat in the atmosphere to IR by the radiative gases and IR loss to space is completely ignored.

        Add to this the mono-focus on radiative energy and purposeful ignorance regarding the water cycle and the huge heat engine it produces and we have a seriously flawed model.

    • Roald, you are correct when you say that the sun does not shine equally on all of Earth’s surface all the time. And it is indeed important not to use the average (~342 W/m2) when the actual should be used. But in this case, the calculation is of Earth’s energy budget, and for that it is correct to use the average. That’s because the energy budget deals explicitly with average energy in and out over a period of time.

      • Mike Jonas – “the calculation is of Earth’s energy budget, and for that it is correct to use the average”

        And it’s all an average based on surface area right?

        Imagine if we were able to drag a notched trowel over the Mojave Desert such that the surface area is doubled. We can then divide the energy received from the Sun by 2 correct? And, we can double the amount of energy emitted to space since the surface area has been doubled. So the desert would receive half the amount of energy from the Sun and emit twice as much energy to space. Now what would the temperature be?

    • The main error of the so called “energy budget” is that it is a radiation calculation for only 1 second (!) in Watt/m^2 = Joule/sm^2. Joule is the unit for energy, and in one second the radiation from the sun hits only half of the earth, and not evenly all over, but 1362 W/m^2 at equator and 0 at the poles (during equinox), so the factor should rather be 2, instead og 4. The outgoing radiation will be from the whole sphere though. The rest of the calculation will carry this error, and so the “energy budget” will be erroneous. It is simply unphysical.
      In addition to this error, the analytical US Standard Atmosphere principles and calculations supported by the empirical evidence of Nikolov & Zeller explains the global temperature, which is modulated by clouds as prof. Dr. Henrik Svensmark has proved.

      • It is also zero at the equator along the line of termination. yes, eventually that point will rotate to full incidence, but at the instant of the analysis point it is zero. The equator is a transient point. The energy absorbed by the moving surface of the sphere variations both in position and time, and that variation gets larger the closer you get to the equator.

    • Roald J. Larsen

      “That IS the problem, the sun doesn’t shine at the whole surface of the earth at the same time .. (i.e. flat earth) at 342 W/m2 (-18C.)”

      Mr Larsen, this has nothing to do with flat earth thinking.

      Our Sun hits at any moment a complete hemisphere, its surface being 2 pi R².
      But while sunlit reaches a maximum at the Equator, it is zero at the Poles.

      You therefore have to apply a latitude weighting to the sunlit, by integrating the square of the cosine of sunlit’s incidence angle with the surface. Integrating cos²(x) from 0 up to pi/2 gives exactly 0.5.

      This gives by accident exactly the same as a flat disk.

      • “This gives by accident exactly the same as a flat disk.”

        Bindidon,

        Let’s be absolutely clear about this, that is no accident. It is not even the hint of an accident, it is a pristine, top notch, validated, fundamental relationship of spherical geometry, known since the time of Archimedes.

        Consider the example of shadow theatre, the shadow shape on the screen has a clear and measurable surface area. However, it is literally impossible to say what the morphology of the corporate body is that created the shadow. This information has been lost. A corporate body creates an identical shadow to that of a flat plane silhouette of that body, held at right angles to the beam of light.

        As my father once explained to me, spherical geometry is built on the strange premise that it is possible to have a triangle with three internal angles that sum to more than 180 degrees. For example, if you start at the North Pole at the time of the equinox, with the sun on the horizon, and journey due south until you reach the equator. The sun is now overhead. Turn 90 degrees right and follow the equator west for the same distance as you have already travelled south from the pole. Then turn 90 degrees right again and return due north to the pole. On reaching the pole you will arrive back following a direction 90 degrees to your outward path. You have traversed a three-sided shape with three internal angles that sum to 270 degrees.

  2. There is a simple way to look at it.
    At any given moment half the sphere is unlit so that reduces by half the power of the solar beam when averaged across the entire spherical surface.
    Then we must address the lit half on its own.
    Due to three dimensional geometry the surface of the lit half is double the surface area of a flat section across the centre of the sphere so the solar beam is spread across that doubled surface area which means that averaged across that larger surface area the power of the solar beam must be halved again.
    For the purpose of heating a surface, the area of that surface affects the maximum average surface temperature that can be achieved from a given power input so for a spinning sphere it is necessary to calculate that average power input across the entire surface area in order to arrive at the correct temperature to be expected.
    The simple solution is the divide by 4 rule so Roy and Willis are correct in this instance.

    • Stephen,

      So with Zeller-Nikolov and the recent WUWT posting backing up that surface temperature/greenhouse effect is due to atmospheric pressure, do you think that we have reached a tipping point in the discussion of greenhouse gas theory? Perhaps the tide has turned toward science and away from belief that CO2 atmospheric concentrations have any impact on temperatures. Just hoping.

      • The atmospheric pressure alone does not cause the so called greenhouse effect, it is the raising the altitude in the atmosphere (above the surface) where the average location of where radiation to space occurs (due to absorption and then re-radiation), and then using the lapse rate (pressure effect combined with condensation of water vapor) that then determines the raise in surface temperature over what would be observed were there no optically absorbing gases or aerosols. The lapse rate exists even for atmospheres with no green house gases, but those atmospheres would not raise the surface temperature.

        • The simpler explanation is that the atmospheric pressure determines the temperature at which convection will occur and remove heat from the surface much more quickly. Therefore, that gives you a first order approximation of the surface temperature. It roughly explains why Venus is much warmer than Earth and Mars is much cooler. link

      • I think the better way to say it is the “average temperature” of the planet is due to the mass of the volatile fraction of that planet – aka the mass of the atmosphere – as well as the average solar irradiance. The mass of the atmosphere can change via degassing/dissolving/condensing.

    • I’m a little confused at the blithe use of averaging when dealing with anything other than the most conceptual principles of radiative transfer. When you start using T^4 terms, the actual fluxes start to matter a lot. The fact that the whole mess is rotating doesn’t help things either. That portion of the earth rotating from the day side across the terminator to the night side has a whole different circus going on than the surface that’s in full sunlight, or the surface rotating into the day side from the night. Taking the integral over the whole surface may get you the same place in the end, but if you want to understand the difference between ±5° of the equator and what’s going on at 50°N or 50°S, you want real numbers, not averages.

      • As I have discussed on my blog, for the slow rotation rate and lack of water on the moon, this matters a lot… you cant just take the solar constant to compute the average temperature of the moon because the temperature range is huge and the non linearity of the S-B relationship has a large effect. For Earth, however, the approximation is quite good.

        • Roy,
          It may be “quite good,” but I think that it should be considered a first-order approximation. The ‘cartoons’ are only accurate for a small area that is normal to the incoming radiation. As Rocketscientist notes, reflectivity is dependent on the angle of incidence, especially for the specular reflector we call water, whicht covers 71% of the surface! Additionally, for all but the normal-incidence situation, the slant-range through the atmosphere changes with angle and therefore the amount of scattering and absorption increases towards the limbs. It really is a matter of defining an integral for the cumulative effects dependent on angle, which are not linear. Therefore, a simple average will not give an exact answer.

        • The Earth’s surface being 70% or so water, and the reflectivity of water surfaces to light being highly related to angle of incidence, it seems that the true number for heat absorption must be even lower than the 1/4 of total incident radiation that just measures the geometry simplistically, without considering reflectivity as an addition term. Is this concept included in the radiation budget considerations? I would think that a term adjusting for average angle of incidence of incoming light from the sun would be important. Light reflected back to space off the ocean surfaces is not part of the Earth’s “heat budget”, I should think.

          Anyone?

          • kwinterkorn
            An “average” angle of incidence (about 45 deg) would not give a good answer. A solution of Fresnel’s equation for reflectance shows that there is an approximately a linear increase in reflectance from water from 0 deg incidence (about 2%) to about 45 deg and then the reflectance curve bends sharply upwards at around 60 deg, reaching a maximum of 100% at 90 deg. The last 45 deg are very non-linear.

        • Large surface area receiving very low angle of direct solar energy being reflected or diffused.

          Is that where Trenberth’s ‘missing heat’ calculation went (or should I say never arrived)?

      • Agreed, you can average the incoming flux which is uniform. You can not average T^4.

        However, if you consider that daily temperature swings are typically 10 K , maybe 20K in arid zones how far off is the error?

        Does it “matter” if you use the forth power of the mean temperature instead of mean of the forth powers ?

        ( (273+20)**4 – 273**4 ) /2.0 = 907739480.0
        (273+10)**4 = 6414247921.0
        6414247921.0 / 907739480.0 = 7.06617709411515 ooh,err.

        • It depends on where you are on the globe as to how much the temperature of the day changes. For example, here in Australia the humid areas along the coast will change by 20C or more each day and night. And the central region (everything more than 100km inland) will change by 40C or more each day and night (particularly in summer).

        • ” ooh,err”
          Indeed.
          ( (273+20)^4 + 273^4 ) /2.0 = 6462311321.0
          (273+10)**4 = 6414247921.0
          6462311321.0/6414247921.0 = 1.007493
          Answer, not much at all

          • The real issue is the OLR.. do you want to average that too? How about clear sky, cloudy sky, high humidity, low humidity?

          • … and 0.007493 * 342 w/m^2 = 2.56 w/m^2
            Hmm… so worst case is that an approximation like this could only throw results off by a similar order of magnitude to the calculated contribution of increased post-industrial atmospheric CO2 to the greenhouse effect?

    • “For the purpose of heating a surface, the area of that surface affects the maximum average surface temperature that can be achieved from a given power input so for a spinning sphere it is necessary to calculate that average power input across the entire surface area in order to arrive at the correct temperature to be expected.”

      If temperature expected is the amount watts the Earth radiates, ok. Roughly.
      But I would without the details one could only get a guess of how much a planet radiates.
      You could not know what ocean temperature is. Ocean surface or entire ocean average temperature.
      You could not know surface air temperature. Before begin to guess, you would first define what elevation the air is at, that you counting as surface air temperature.
      Earth has feature of being covered mostly with water and we have definable term called sea level.
      What do count as sea level one other worlds? You could have seas of different levels. You have most of surface covered with land. You have to have argument about what level is important relative to global climate temperatures. And also numerous factors that humans can’t even imagine.

  3. Joe Postma does have a MS/Astrophysics from Univ/Calgary, and now is a staff member — not a teaching position, thankfully: http://contacts.ucalgary.ca/info/phas/profiles/486-146153

    His thesis was observational, “The observation and analysis of the Cepheid SZ Tauri”, finding and recording a new phenomenon. That was a 1-year extension from his degree in Astronomy. Probably not much radiative Physics in his education. He is part of the “Slayers” group, thus also “Principia Scientific Intl.”, and their focus is on “anything except radiative return to the surface” — violation of 2nd Law (as if photons actually cared), or photons going “back down” to the surface “know not to carry any energy”, or “it’s all atmospheric pressure (abuse of ideal gas law)”, etc. That group is infamous for hating the creation of diagrams showing the static, average, behavior of the surface and the atmosphere, and then proceeding to produce their OWN diagrams of static, average behavior, instead. Hence the “flat earth” silliness, when simple math makes it clear that the surface area being illuminated accounts for a sphere, not a flat disk.

    Joe gets around to berating, then banning, anyone who disagrees. He’s tossed me off his “Climate of Sophistry” group more than once.

    His misunderstanding of the 2nd Law is profound, sometimes arguing that the surface cannot absorb photons — thus adding energy to the surface — if their origin was the atmosphere instead of the sun..

    • I will only focus on two of your misunderstandings.

      It’s not that radiation doesn’t also radiate downwards: “.. and their focus is on “anything except radiative return to the surface” — violation of 2nd Law (as if photons actually cared), or photons going “back down” to the surface “know not to carry any energy” ..

      But 1. It’s the result of temperature (originating from the sun), and 2. At what altitude does the “heat” suddenly turn its focus on the surface? What is the temperature difference?

      Woila! What does the 2nd Law of Thermodynamic say?

      Does the cold atmosphere add heat to an already warmer surface?

      • Radiation never “focuses” on the surface. At all attitudes the direction of radiation is random, which means some fraction will be radiated down.

        • MarkW
          Not actually true.
          Total Solar radiation at sea level on the earth is defined as Global Horiztal Irradiation (GHI), which is composed of two parts: The direct beam horizontal irradiation (DBHI) and the diffuse horizontal irradiation (DHI). You have (correctly) identified only the diffuse horizontal radiation component.
          Direct Beam Horizontal Irradiation = Direct Beam Normal Irradiation x sin (Solar Elevation Angle) in a clear sky.
          (DBNI is also known as Direct Radiation, Beam Radiation, Normal Irradiation, etc.)
          GHI = DBHI + DHI

          You are correct though. Under cloudy skies, the Direct Beam Normal Irradiation is absorbed and re-reflected inside and between the clouds and vapor and dust particles, and so only diffuse radiation reaches sea level.

        • Did you know?

          Below aprx. 1000 meters over 73% of all energy is exchanged by conduction (because the atmosphere is too dense).

          That means, majority of all cooling processes are done by convection and only above aprx. 1000m. does radiation start taking over ..

          What is the temperature difference between the surface and the altitude of 1000m.?

          That means, radiation has to overcome at least 2 different obstacles in order to re-heat the surface.

          And that is impossible!

          • The Earth’s surface being 70% or so water, and the reflectivity of water surfaces to light being highly related to angle of incidence, it seems that the true number for heat absorption must be even lower than the 1/4 of total incident radiation that just measures the geometry simplistically, without considering reflectivity as an addition term. Is this concept included in the radiation budget considerations? I would think that a term adjusting for average angle of incidence of incoming light from the sun would be important. Light reflected back to space off the ocean surfaces is not part of the Earth’s “heat budget”, I should think.

            Anyone?

          • There is a common confusion at this blog between “net” transfers of heat energy and “absolute” transfers of heat energy. Net transfer of heat must ultimately be from hotter places to cooler places, but there is still absolute transfer of heat energy from cooler to hotter….which is just part of the equation that “net’s” out all the various media of heat energy transfer in all directions. Some heat energy still irradiates downward from cold air to warm surface, even if the overall balance or net flow is in the other direction. If something increases the absolutely downward radiation from cold air to warm earth, without changing any other factor, then the warm Earth will warm a little bit more than before that increase. This change in net effect is the “greenhouse gas” effect.

            Fortunately this net effect seems to change little as CO2 rises, and the various negative feedbacks do a good job in keeping the Earth’s warmth pretty stable.

      • IF the cold atmosphere is warmer than what was there before, then yes, the surface will warm.

      • Does insulation in a cold wall add heat to a heated house? No? How then can the cold insulation in those walls make the interior warmer than without insulation?

        • That is beside the point of this post, isolation works by preventing convection (something CO2 doesn’t do).

          • Foil-backed insulation works differently than simple fiber-glass ‘batting’.

            “How does a mirror work?” is your clue here …

          • Jim the silver on a mirror is a trace element.
            everyone knows it cant have an effect. its only .001% of the mirror volume

          • Steven: The silver on the mirror is not part of the mirror and does not affect the glass it’s painted on in any way. I was unaware CO2 was not an active participant in climate, but rather a coating on the atmosphere. However, I am sure your analogy is not meant to mislead….

          • steven mosher June 5, 2019 at 12:08 am
            Jim the silver on a mirror is a trace element.
            everyone knows it cant have an effect. its only .001% of the mirror volume

            Nice sarcasm gambit mosh.

            The term I was looking for, but was unavailable at the time of the original posting, was: “Radiant barrier”. We have that option in Tejas as an underspray
            applied to the roof, a means to reduce heating of the living quarters below …

          • Roy is talking about insulation in the context of conduction not convection.

        • This analogy between a wall and the atmosphere that would insulate the surface from the “cold winter” does not stand :
          – the walls help the room stay warm because they stop convection,
          – conversely, the atmosphere is were convection takes place, with a cooling effect on the warmer surface.

          Some observations first :
          – The Keihl & Trenberth diagram (1997) shows that the atmophere radiates 165 W/m² into space and absorbs 26 W/m² from the infrared upward flux emited by the surface.
          Thus the active gases in the infrared spectrum have a net cooling effect on the atmosphere of 139 W/m².
          And this cooling effect can’t do anything other than cool down the surface.

          Back to the analogy :
          The “wall – insulation – atmosphere” analogy stands only if we assume that :
          – the heat transfer by atmospheric active gases in the infrared spectrum is the analog of the heat transfer by thermal conduction in the walls,
          – since the Earth’s surface is warmer than the atmosphere, there can’t be any cooling of the atmosphere by radiative heat transfer towards the Earth’s surface (the net radiative transfer is upward, as shows the Keihl & Trenberth diagram).

          So, this two process are the only way for the atmosphere on one hand and the wall on the other hand to lose heat (into space for the atmosphere, in the cold winter for the walls).

          But now, the analogy results are inversed :
          – as the thermal conduction in the walls contributes to cool down the room, the infrared active gases in the atmosphere contribute to cool down the Earth’s surface.

          PS : active gases in the infrared spectrum are the only way for the atmosphere to lose heat into space :
          – indeed, the atmopshere do not behave as a black or a gray body as some assume. See M. Modest 2003 – Radiative heat transfer, or Kondratyev 1969 – Radiation in the atmosphere.

          • re: “indeed, the atmopshere do not behave as a black or a gray body as some assume.”

            Some may take this as “air does not radiate”, but, this must be qualified.

            Sans (that is, without) CO2 or water (H2O) being present, yes, true, but the troposphere contains non-trivial amounts of H2O vapor, which *is* visible via satellite, hence, it is known to – wait for it – radiate.

            The foregoing has probably been mentioned, but I mention it again for completeness.

          • “PS : active gases in the infrared spectrum are the only way for the atmosphere to lose heat into space :”

            No it is not.
            It’r the opposite.
            They emit LWIR, yes … but they also absorb LWIR.
            The two come together.
            That combination, along with the ultimate emission from a higher, colder region in the atmosphere is what the GHE is.
            Result – a warmer atmosphere WITH GHGs

            In a pure O2 and N2 atmosphere, the atmospheric window would be as wide as a canyon.
            It would cool very rapidly via radiation, with nothing to impede terrestrial IR.

      • Woila! What does the 2nd Law of Thermodynamic say?

        As an engineer, I’ve noticed that some invoke the 2nd law into what is completely a 1st law (conservation of energy) issue & that shows a misunderstanding of the issue. 1st law applications (energy in and out) are all that are necessary to tell you what you need to know.

        • beng135

          “As an engineer, I’ve noticed that some invoke the 2nd law into what is completely a 1st law (conservation of energy) issue & that shows a misunderstanding of the issue.”

          Thanks! (Says another engineer but having worked in a very different context.)

          You should ask Anthony Watts for presenting that in a guest post.

    • His misunderstanding of the 2nd Law is profound, sometimes arguing that the surface cannot absorb photons — thus adding energy to the surface — if their origin was the atmosphere instead of the sun..

      The wavelengths are very different. What’s coming from the Sun is centered in the visible range, what’s coming from the atmosphere is long wave infrared (LWIR).

      Some folks insist that LWIR can’t warm the ocean because it can’t penetrate the surface. I don’t know if that’s true but if it were, then the photons coming from the sun and the photons coming from the atmosphere would indeed have a different effect. yes/no?

      • Evaporation only occurs at the skin, yet it is the largest mechanism of heat loss by water bodies.

          • And Scott there, bragging how his CO2 laser can boil water without admitting that it has 100,000 times the intensity of Sunlight.

          • And Scott there bragging that his CO2 can boil water, yet fails to mention it’s 100,000 times the intensity of solar IR.

      • to: commieBob June 4, 2019 at 3:14 pm
        re: “The wavelengths are very different. What’s coming from the Sun is centered in the visible range, what’s coming from the atmosphere is long wave infrared (LWIR).”

        What might be called a “common misconception”; you’re referring to what? – the Planck curve? Look instead at the absolute values of the “energy” in the LWIR region from those two sources. Don’t just look at where the curve peaks …

        (You notice, when you sit in the sun, or sit next to a fire, deep heating from LWIR occurs?)

  4. Don’t be too harsh in condemning “Flat Earthers”.
    Remember our motto:
    “We are on the level”

    • Yes, and as one “Flat-Earther” famously said ….
      “We have members all over the Globe”.

    • Robert Kernodle June 4, 2019 at 3:18 pm

      Less-intense energy from the atmosphere cannot add to more-intense energy of the surface that already has this less-intense energy.

      Sorry, Robert, but that is simply not true. Once a photon leaves an object and starts traveling, it will be absorbed by whatever it hits, no matter the temperature of the object that absorbs it.

      Curious fact. If you light a candle outside in the daytime, the sun gets warmer …

      w.

      • Right, but gases behave entirely different than solids and liquids when it comes to emissivity of and opacity to photons.
        Gases only absorb and emit photons matching the energy levels in their molecular vibrational modes. No matter how high the temperature a gas becomes in the atmosphere, it will only emit photons corresponding to these energy levels.
        We’re basically arguing against the ridiculous hypothesis that the temperature of rocky planet atmospheres is controlled by the radiative emission from the molecular vibrations of gas molecules. The alternative hypothesis is entirely logical and matches basic principles like the kinetic theory of gases and the ideal gas law.

        • That depends on the ambient temperature. Also, does the ice cube reflect the candle’s heat back to it. As always, the devil is in the details.

          Anyway, Willis is correct. Any photon striking an object can impart its energy to that object. Anyway, we have the equation which describes radiant heat transfer. link link You have to ask yourself why the temperature of the cooler object matters to how fast the warmer object loses heat.

          • No the photon will be absorbed unless the object is reflective at that frequency. You seem to sort of accept it up to a point but still want to put temperatures on thing.

            Lets cut and give you the harsh reality because you are half way, you might as well make the full transition. Photons are quantum objects, in QM there is no such thing as temperature it is a human made up statistic that has certain traits in classical physics. So a photon does not even know what temperature so how would it interact with it.

            You won’t notice the effect normally because your classical convection exchanges with the air are many many magnitudes higher. Put the objects in a vacuum and you will get very different answers.

          • LdB June 4, 2019 at 7:31 pm

            The radiative heat transfer equation is easily demonstrated in the lab.

          • Oh this should be good show me where QM interacts with heat please … dying to see this.

          • Would you like for me to break any radiative heat transfer equation you care to put up .. read the link now apply your heat transfer equation.
            https://en.wikipedia.org/wiki/Laser_cooling

            Did any of your transfer equations say that heat would produce absolute zero???????

            Take extreme care when using classical laws they are all rubbish and can be easily broken by QM.

          • commieBob June 4, 2019 at 6:59 pm
            That depends on the ambient temperature. Also, does the ice cube reflect the candle’s heat back to it. As always, the devil is in the details.

            Anyway, Willis is correct. Any photon striking an object can impart its energy to that object.

            WHAT seems to be missing here is … so-called “photons” (they are mythical creations not- existent until 1920’s) are an ELECTRO-MAGNETIC phenomenon, not some ‘particle’ that bounces around … think RADIO WAVES but of really, really HIGH frequency and therefore really, really SHORT wavelengths.

            This means, in part, they can and WILL react according to EM laws (esp Maxwell Eqs), including reflection, refraction WHEN dealing with liquids/solids comprised of metals (or anything else) with a (an electron) conduction band.

            Gases, of course, are a slightly different story, esp, with ‘polar’ molecules like H2O (and CO2, when excited). See subject IR Spectroscopy for the full story there …

          • The problem with your simplification of trying to turn light into a classical wave description, you require the particle behaviour to be able to absorb discrete packets and you require the momentum (both particle behaviours). Trying to use classical wave just leads to stupid answers. That is why light duality is always taught because you get stupid answers if you use the wrong one.

          • LdB Of course temperature has an effect on QM. For instance, the emissivity and absorption spectra change with temperature.

            And while we’re cutting to harsh realities, photons are not quantum objects. Photons are the physical matter manifest from the collapse of the wave function. The photon doesn’t even exist until that happens. This has been proven again and again with experiments since the 1970s, the latest being the delayed choice quantum eraser experiment.

          • LdB June 5, 2019 at 7:42 am
            The problem with your simplification of trying to turn light into a classical wave description, …

            I will submit, that, QM and the invention of the “photon” has made more compromises and further distorted the reality of EM radiation and energy transfer than the application of so-called “classical” laws to same.

            Furthermore, it was the mis-application of classical laws which lead to the invention of the photon.

            Going even further, multiple different so-called “double-slit” experiments can be explained on a different basis than that which is commonly claimed, for instance, the double slit experiment with electrons. The ONLY double-slit experiment with any credibility is the double-slit involving light, an EM wave, or radio wave, obviously ALSO an EM wave.

        • Actually it does. You can easily run single photon at whatever frequency you want and put the photon detector in the flame if you really want it proved. Why you aren’t familiar with the effect because the normal classical thermal transfers are massive compared to the effect and you can’t notice it. You can however easily measure it in the QM domain if you really want.

        • @ Robert Kernodle
          The more interesting fact that you wont understand is if you could put enough observers sitting around earth you could actually dim the sunlight hitting earth. The universe is a little bit stranger than you think 🙂

      • “it will be absorbed by whatever it hits, no matter the temperature of the object that absorbs it”
        And if it has higher energy, what it hits that has lower energy, it will be energized.

        • It is more complex than that and does not need to be higher power and may not heat it can cool. It needs to be of the right frequency etc as well. Remember reflection, diffraction etc are other possible outcomes.

          Laser cooling for example is a good example to look at because it is using the absorbing to cool so it is like the ultimate in weird.

      • Willis Eschenbach
        You said, “Once a photon leaves an object and starts traveling, it will be absorbed by whatever it hits, no matter the temperature of the object that absorbs it.” With the caveat that the surface is absorptive at the wavelength representing the energy of the photon. What that means is that the complex refractive index of the surface must have a non-zero extinction coefficient (imaginary component of the complex refractive index) at the wavelength of the IR photon. Generally speaking, that is the case, but the situation is even more convoluted. At small, but non-negligible extinction coefficients, a photon will penetrate deeply but eventually be absorbed. On the other hand, metals have very high extinction coefficients and they are typically highly reflective. I’d have to sit down and build a model to calculate the complex refractive index for optimal absorption. Which in turn, gets more complicated because other than water, it is rare to find any reasonably transparent materials that are homogeneous. Things like beach sand and soil have particles that experience internal reflections and diffuse the photons.

        I think that Kernodle is confusing the transmission of heat by conduction where the heat travels in the direction of the cooler object. With reasonably transparent materials, such as gases, photons can and do travel in all directions. What is important is the net transport.

      • >>
        Curious fact. If you light a candle outside in the daytime, the sun gets warmer …
        <<

        Oh Boo! Pick another example! What do you mean by “outside?” Outside the Moon? The Solar System? The Galaxy? What do you mean by light the candle and the Sun gets warmer? What if I blow out the candles on my birthday cake, when you light your candle? That alone would darken the Sun for many minutes.

        For a scientific fact to actually be a fact, you have to be able to do an experiment that can be used as evidence. It also has to be done, or is able to be done–multiple times. AND it’s considered a truth by most (although not always–it could still be wrong). You’re not going to be able to light a single candle and show the Sun getting warmer–it’s never been done!

        Bah! Try again!

        Jim

        • Nothing worse than pedants filling up space with nitpicking rubbish. By “outside”, Willis obviously meant “Walk out of your home or office into the sunshine.” And it’s a fact because the laws of physics as best known to us predict exactly what Willis said. You don’t need an experiment to have good reason to believe something is a fact. Example: “If you drove a Roman chariot off a 1,000m high cliff in the Himalayas, you would most likely die.” We all believe it; we all know it. But no one ever has driven even one chariot off such a cliff, let alone enough for a statistical verification of “most likely”. We believe it because it is the compelling answer to application of known laws of physics.

          As for your total rudeness towards Willis, I have only one thing to say: Rejected! Bah! Try again!

          • >>
            Nothing worse than pedants filling up space with nitpicking rubbish.
            <<

            I quote Willis from the Monckton of Brenchley thread:

            “Me, I never deal with what I call “Third-Party Offense”, where someone gets all offended by what someone else calls a third party.”

            I think you’re getting offended as a third party.

            So you or Willis show me the experiment that “proves” what he said. There is none. Willis’s statement is stupid. Willis repeated the comment from the Monckton of Brenchley thread. It was stupid there too. It’s not a scientific fact if you’re basing it on theory.

            I’ve studied thermodynamics, and I know what Willis is trying to say. He’s basically correct, but it was a stupid example.

            >>
            You don’t need an experiment to have good reason to believe something is a fact.
            <<

            Yes you do. Look up the definition of “scientific fact.”

            Jim

        • Okay, I’m going to take back what I said to Willis and apologize. Mr. House was correct–I was being rude. Sometimes you should think before hitting the post button.

          Jim

      • ”Curious fact. If you light a candle outside in the daytime, the sun gets warmer …”

        I will assume you mean the heat from the flame of the candle.

        Ok, but then it must work both ways correct? So that means the candle (flame) also gets warmer. So now they are both warmer than before they were brought together. So where does the extra heat come from?

        • Mike, what makes you think that there is “extra heat”? All that is happening is that both are absorbing some radiative energy that otherwise would continue into space. There’s no “extra” heat at all.

          It’s like, when you walk up to the campfire, you get warmer … but again, there’s no “extra heat”. You’re just absorbing some energy that otherwise would go on to hit something else.

          w.

      • If you light a candle outside in the daytime, the sun gets warmer

        Yeah, but it takes 8 minutes for the sun to notice, and another 8 minutes for us to notice the sun’s reaction. 🙂

    • If the sum total of photons hitting the earth’s surface is greater than it was before, than the surface of the earth will warm, regardless how “intense” the energy of the surface is.

      • Not if the rate of photons emitting from Earth are greater than the addition…

    • A sneeze in a hurricane, while trivially insignificant to the wind forces, is still a sneeze.

    • Atmospheric radiation is energy originating from the surface, which again, originate from the sun.

      I.e. it’s the surface temperature dictating the atmospheric radiation.

  5. You are addressing them here because you know he cannot respond to you here, and you get your asz handed back to you when you debate elsewhere.

    Your a sham gate-keeper of a radiative greenhouse effect that doesnt exist.

    The only greenhouse effect on this planet is the h20 effect of a tub of water in a greenhouse effect.

    ”Joe advertises himself as an astrophysicist who works at the University of Calgary. I don’t know his level of education, but his claims have considerable influence on others, which is why I am addressing them here.”

    And you would similarly get your asz handed to you here if you got into it with Alan Tomalty also.

    Have a nice day.

    • Fact check. Dr. Roy, AS HE SAID, has been disputing this with Postma on Twitter. So OBVIOUSLY, Dr. Roy is not worried about his asz in re Postma …

      w.

    • So if someone debates in two forums, they have defacto admitted to defeat in one of those forums????
      As to having his asz handed to him, that’s an opinion, and not a very good one.

      The claim that the radiative greenhouse effect doesn’t exist has been shredded every place it has been presented.

      • The problem for me is not whether Postma’s claim is credible, it is his alleged shutting down of serious questions and debate on his site. Those who care for truth want to hear all reasonable sides presented.

    • Well you are being radiated on by the gas molecules within the effective emission length from one of their 0.1-0.001 eV photons. If you stand in a desert at night you can feel how effective it is and how much it matters to the temperature of the atmosphere.

  6. Here’s my comment reposted from Dr. Roy’s blog …

    w.
    ==========

    Dr. Roy, thanks for this. I keep getting the same nonsense. You hit the nail on the head when you said:

    If I measure 24 hours of solar flux at the surface of the Earth, and do a 24 hour average of that, it is the actual average amount of sunlight that my location received in 24 hours. It does NOT imply that the sun shines at night!

    People have this bizarre idea that an average of a variable, say sunshine, somehow implies that the variable is constant and equal everywhere. But that’s not the case.

    It’s just an average, folks.

    Go outside as Dr. Roy describes. Measure the amount of sunlight every five minutes over a 24 hour period. You’ll get a number that goes from hundreds of watts per square meter (W/m2) during the day to zero at night.

    Now, average them. I think we can all agree that what you get is the average sunlight that has fallen on that spot over 24 hours. AVERAGE.

    But that average does NOT mean that the sun shines at night. It’s just an average, which is very useful for a variety of calculations.

    The same is true for a global average. Let’s repeat Dr. Roy’s thought experiment around the planet. Suppose we could get thousands and thousands of people all over the globe to take the measurements just described. Everyone everywhere measures the sunlight every five minutes for 24 hours.

    We take all of those measurements and we average them.

    I think we can all agree that what you get is the average sunlight that has fallen on the globe over 24 hours. It will be on the order of 170 W/m2 or so at the surface.

    But again, that does NOT imply that either the sun shines at night, nor that the earth is flat.

    As before, it’s just an average. It doesn’t imply anything at all about the flatness of the planet.

    There are lots of valid arguments that climate models and global energy balances have problems.

    The false claim that climate scientists and climate models deal with the world as if it were flat is not one of those valid arguments …

    Regards to all, and Dr. Roy, thanks for your blog. It’s always interesting and worthwhile.

    w.

    • “If I measure 24 hours of solar flux at the surface of the Earth, and do a 24 hour average of that, it is the actual average amount of sunlight that my location received in 24 hours. It does NOT imply that the sun shines at night!”

      But that is not what you are doing (why divide it by 4 if it is the actual measurements?)

      Then the “average” would be 1,370 W/m2 ..

      • Roald, in my thought experiment, people were measuring the incoming solar all over the world.

        But the normal measure of TSI (total solar irradiance) is NOT that. It is the amount of solar energy hitting a square meter perpendicular to the sunshine.

        The sunlight is intersected by the cross-sectional area of the planet, which is pi R squared. However, that sunlight is spread over the surface area of 4 pi R squared.

        So, to get the global 24/7 average, we have to divide the ~ 1,370 W/m2 by 4, giving us about 340 W/m2.

        w.

        w.

        • Willis

          I am not arguing the math, that is correct – math.

          But that does not resemble the earth, the reality ..

          • Roald J. Larsen June 4, 2019 at 5:34 pm

            Willis

            I am not arguing the math, that is correct – math.

            But that does not resemble the earth, the reality ..

            Thanks, Roald, but say what? E = MC2 is nowhere visible to us, it doesn’t resemble what I see outside my window but it is true nonetheless. And the average US income doesn’t resemble actual money … why should they? They are extremely useful mathematical abstraction.

            Or to take a more familiar example, the map is not the territory, it’s a flat sheet of paper that doesn’t resemble the earth at all … but it is crucial to navigation.

            w.

          • Willis. All valid points.

            The earth only receive input from the sun on one side at any given time. That is the reality. You can average it to the whole sphere and get a interesting number, then what? You can not use that number for any purpose, it’s an average.

            Weather systems and patterns are not created in reality based on an average of 342 W/m2 (-18C.)
            What’s drive the climate is the energy input from the sun of about 1,370 W/m2 (actually 1370*(1 –0.3) = 960 W/m2= 88C. at Zenith).

            Based on empirical observation, that is the reality.
            Due to pressure, gravity and mass (and water vapor, i.e. latent heat), energy is preserved in the atmosphere long enough for us not to have full winter each night regardless of season.

            These facts leaves no room for magical CO2 effect.

            Math can be useful, but to be focusing on it too much might make you forget the atmosphere is not an average, well, actually it can be, but for understanding the climate and energy input, it’s a dead end.

          • Mr Larsen

            The average over 24 hours of heat absorption is relevant to heat budget discussions, because the heat loss from the surface goes on all 24 hours.

            To do a heat budget on whatever time unit, you must pick a unit that is useful and measurable, and then use it both heat in and heat out.
            So average heat absorption/day and heat loss/day make sense to create a budget. No one is being devious at this level. Other levels maybe.

          • Heat budget discussions doesn’t depict the reality logically, mathematically or empirically.

          • What is wrong with people, it’s not the math that is the issue, it’s how you use it.

      • Then the “average” would be 1,370 W/m2 ..

        No, because the solar flux of 1,370 only applies at noon on the equator.
        The average energy falling on the earth’s surface at any instant is 1,370*𝝿r^2
        Average that over the whole sphere which has an area of 4𝝿r^2
        Then the average over the whole sphere is 1,370*𝝿r^2/4𝝿r^2 = 1,370/4

        • 1362 watts/m^2.
          But only at top of atmosphere.

          About 1000 watts/m^2 in most temperate latitudes at sea level in the real flat earth.

    • While this is true, it begs the question; “are the model calculations based on the average value spread over the whole world simultaneously and constantly, or the daylight values hitting the lit surface for half of the day and zero for half of the day?”

      • Greg, it’s neither one. They calculate the sunlight hitting each gridcell at each timestep.

        And look up the meaning of “begs the question”, you’re using it incorrectly.

        w.

        • ok, I should have used “raises the question”. Glad to see the meaning was still understood.

    • “It’s just an average, folks.”
      More than that, it’s an average of a conserved quantity. So you can use it for accounting. If you get an average inflow in a volume of 10W over 24 hrs, then you have 240 WH, or 864 kJ. And it doesn’t matter if it came in at an even rate. The average gets it right.

      • Thanks, Nick. True, and I forgot to mention that. Since energy flow is conserved, it can be added and subtracted.

        w.

        • =========================
          Nick Stokes
          “It’s just an average, folks.”
          More than that, it’s an average of a conserved quantity. So you can use it for accounting. If you get an average inflow in a volume of 10W over 24 hrs, then you have 240 WH, or 864 kJ. And it doesn’t matter if it came in at an even rate. The average gets it right.

          Reply
          Willis Eschenbach June 4, 2019 at 8:30 pm
          Thanks, Nick. True, and I forgot to mention that. Since energy flow is conserved, it can be added and subtracted.
          ===============================

          And this “…average folks.” takes into account the average clouds, and average water vapour, the average stuff/dust as this watts/24h hits the planet’s atmosphere?

          Ponderous

          rick

          • I also wondered about that, Rick. Is the measured insolation cloudless? If not, what is the average cloud cover? Does it make a difference?

    • This makes sense. How do they then go from that average to something that takes into account the changing angle of incidence as one moves north/south? It would be similar in effect, I presume, to that of the earth rotating away from the peak flux (which would be a 180 relation between earth and sun surface normals) as the day moves on.

    • Roald J. Larsen June 4, 2019 at 2:20 pm 342 is 5C. Now the solar flux “spreads it over the spherical shape of the Earth” every square metre receives 342 watts from sunlight, which is 5C. I found taking a measurement for each latitude co-ordinate (90N-0,0-90S), E-90, W-90, E-180, E-90 (70 measurements) averaging 5C 340 W. This means that the colder part of the earth is not fully measured making the average higher than if earth was measured proportionately including the colder areas. This shows there’s no greenhouse effect at all. CO2 at 0.04% and water vapor (max 2.5%) hasn’t the sufficient quantities to reflect half the radiation downwards. Energy coming in must match energy going out. And that is exactly what happens except air mass internal energy is mixed to create this 10C energy and claimed as the greenhouse effect.

      • Great explanation by Stephen of Earth’s average temperature. It should be an eye opener for folks with open mind and critical thinking skills. The GHE folks are missing the molecular physics spectroscopy.
        The 8 to 14 micron wavelength of outgoing long wave IR radiation from the Earth’s surface lets it escape to the space. No role of imaginary radiative forcing of CO2 or WV which are not present in the 8-14 micron range.

  7. I know nothing about Potsma who seems to be an abusive egomaniac. Still the convenient math of dividing the solar “constant” 1,370 W/m2 by 4 has always troubled me when thinking about climate dynamics.

    Thinking in terms of ocean heat storage, in the tropics where the full force of solar insolation (minus any cloud effects) will most greatly heat the ocean. Due to greater evaporation, tropical surface waters become saltier and denser and sink, storing heat below the surface. In contrast away from the equator precipitation often exceeds evaporation. Any evaporation at the poles is likely negligible so polar insolation is unlikely to contribute to subsurface heat storage. In fact polar regions radiate more heat back to the troposphere than is absorbed. That excess heat is derived from heat transported from the tropics.

    Thus small changes in solar insolation will have a greater effect on heat storage in the tropics, so I suspect by dividing insolation by 4, the amount of heat stored in tropical oceans due to solar insolation will be underestimated.

    • Jim Steele

      I know nothing about Potsma who seems to be an abusive egomaniac. Still the convenient math of dividing the solar “constant” 1,370 W/m2 by 4 has always troubled me when thinking about climate dynamics.

      Ah, but look again at the very title: Potsma is using a simplistic flat earth gray-body one-size-fits-all-latitudes “global average” to try to distinguish between subtle factors affecting the climate of a rotating tilted globe in an elliptical orbit around a star. If I am designing an Indy or NASCAR racing suspension, am I ever going to turn left?

      • It’s run counterclockwise so you better be always turning left, at least until you pass someone going high, but then you still have to turn left at some point

      • Actually, at Indianapolis Speedway they travel counterclockwise, so they are always are turning left. And NASCAR tracks have S-turns that go left and right.
        But I get your point.

    • Well, that’s the problem with averages, isn’t it? If I stick my head in the oven and my feet in the fridge my average temperature is still fine. That’s why it is always important to keep track of the many assumptions one has taken on the way to the answer and how they might affect the conclusion if they turn out to be incorrect.

      And always, always stick to the evidence over the theory. Nature is never wrong, while we are mistaken most of the time.

      • Javier June 4, 2019 at 5:01 pm

        Well, that’s the problem with averages, isn’t it?

        I am so tired of this argument, which boils down to “some averages are meaningless, so all averages are useless” …

        w.

        • No, averaging intensive properties (temperature, for example) from different locations, is meaningless. Some averages are just peachy. Some are less than useless.

          • err no.

            Not meaningless where surface temps are concerned.

            Willis is sitting at one end of the pool and says its 73. i sit at the other end of the pool
            and say its 71. Predict the temperature you will record if you jump into the pool
            at 1000 random places? predict the temperature exactly between me an willis.

            A perfectly meaningful experiment. you can even do it. try it!

            Hell, I know machines that are shipped based on this principle. for example,
            for a machine with no chip temperature sensors the typical approach is to take
            the board temp at one corner and the temp at another corner and to linearly
            interpolate along the line of chips to “get” the temperature of the chips.
            works most of the time. How do I know, tested it!!!! but individual chip sensors
            work better. However if you only have a couple of board sensors that will work.
            and the average temp you get is used to control the machine and fan ect. works
            go figure this meaningless average of two temperatures tells me to increase fan speed and the machine doesnt burn up!

            In short we do this “meaningless” thing all the time. we must be stupid.

            Oh ya, and with the surface temperature we never average the stations. never average
            the absolute T. (only heller does that, wrong approach) with surface temps
            folks create a new variable: change in T. and then they interpolate change in T

            Does this work?

            yup.

            how do we know it works? because we tested it.

            take 100 stations: hide 30 of them.

            use the 70 stations to predict the temperature of the hidden stations. or predict the
            change in T for the 30 stations.
            to do this you create a ‘spatial average” that is NOT done (typically)
            by averaging the measurements. Instead you use the samples to PREDICT
            the temperature where you have no measurements. You interpolate. which
            means.. PREDICT.

            thats the task.

            when you finish that prediction you can integrate the surface.

            what you end up with is a number say its 15C the “average” temperature.

            what does that MEAN?

            it means this: go out and select 1000 places that were not measured and measure them. the
            estimate (prediction) 15C will give you the lowest error of prediction.

            Finally, you can also average color, even though it is intensive.
            Used to play a role in image recognition.

            [Snipped. Language. Mod] wow

            In short the THEORY that you “cant” average these things, or that it is meaningless
            FAILS in the practical world.

            bye bye theory

            oh is the average temperature during the MWP warmer or cooler than today?

            how about the average temperature on pluto? warmer or cooler than the earth?
            or is it meaningless to say?

            alaska is cooler than death valley? or is it meaningless to say

            There are better arguments skeptics should use

          • Mosher: Meaningless where Lottery balls are concerned, and about as useful when it come to global temps, imho.

          • Mosher,
            You said, “… the average temp you get is used to control the machine and fan ect. works.” Yes, it may be a cheap way to decide how much cooling the entire system needs, but it won’t give you a good estimate of the CPU’s temperature. The entire motherboard and surrounding components are acting as an extended, assymetrical heat sink, and the fan removes the heat. If the two heat sensors report 88 deg F and 90 deg F, the CPU won’t be 89 degrees! It will be considerably hotter because it is the primary source of heat and everything else in the case will be soaking up that heat and the fan will be exhausting it. The only way one get get a reliable interpolated temperature in that situation is to have a sensor on the heat source (CPU) and some arbitrary point across the board. It is effectively like a tap on a variable resistor where one can obtain different voltages by ‘interpolating’ across the total length of the resistor.

        • Willis, That is not what it boils down to.

          The message is: averages are often misapplied. Sometimes averages provide meaningful insights, and sometimes averages obscure the real dynamics!

          So we must always dissect what the “average” really means

    • If you treat the earth as a sphere, the surface area is 4 x pi x r^2. Do you have a better equation?

      • Euclid would approve of your formula for surface area of a sphere…

        …however, only one spot on the globe will have perpendicular solar radiation at any given time, so now you need is a formula integrating incoming radiation as a function of time-of day & latitude to account for angle of incidence. One assumes this is why an imperfect but good-enough average is used.

  8. How does the photon emitted from a molecules orbiting atom from a far colder surface thermalise and create work in a molecule of a much warmer surface then Tom.

    The Photon is simply deflected by mass thats operrating in a higher resonant state.

    No work no heat.

    • I’ve seen nonsense before, but this takes the cake.
      Photons have zero information on the energy state of the emitting atom/molecule. When an electron drops from one energy state to another, a photon is emitted. Period.

      Whether a photon is absorbed depends solely on whether an atom/molecule has an absorption band at the same energy level as the photon. That is a fact that is completely independent of the “resonant state” of the atom/molecule. And there are always absorption bands, they can never be completely filled.

      • I laughed so hard, the idea a photon knows what temperature is requires just a small rewrite of QM.

        Gary Ashe has given us a new quantum spin called temperature apparently but he is a little light on evidence 🙂

    • “Deflected”? Are you serious? Whence comes the “force field” that causes this deflection? Don’t scramble for an answer, I’ll get it for you. There isn’t any, because your description of events is ludicrous. The photon hits the surface, is absorbed and then another photon, somewhere else, gets emitted.

  9. Maybe someone should point out to Joe that the
    ‘temperature’ right now (June 4th) at 250 hPa level over:
    The Arctic Ocean = Minus 54.5˚C
    The Equatorial Pacific = Minus 39.5˚C
    The Sahara Desert = Minus 40˚ C
    Antarctica = Minus 66˚ C
    That’s about 35,000 feet.
    So on the day the earth stops tilting, and moist air stops being lighter than dry air…
    That’s the day I’ll start worrying.

  10. “The guy is obviously passionate, as our discussion ended with expletive-laced insults hurled my way”

    Profanity is a loss of control.
    Profanity starts as soon as the profane one realizes they haven’t got any chance of advancing their claims.
    Profanity is also a buttress for cowards who feel stronger when they are louder, more insulting and threatening.

    These cowards can be found anywhere.
    Especially when they’ve blustered and bluffed their way into their occupations and they are terrified they will be discovered to not have knowledge or understanding necessary.

    Publish the discussion that led to the coward’s profanity and insults. Keep it handy under a link.
    Shine the light of truth on the claims, assertions and errors.
    Let visitors to your excellent blog read the discussion and make their own mind sup.
    Those that prefer profanity laced nonsense are not likely to learn reality.

    Stay well!

  11. What difference does it make? If global warming is the problem, incoming radiation isn’t the issue.
    We know the temperature of the earth and oceans, we know that they emit LWIR between 13 and 18 microns. We know that CO2 will thermalize those wavelengths. CO2 is transparent to incoming radiation, that is why it reaches the earth’s surface.

    • CO2 only scatters those wavelengths, it does not thermalize them. In order for that to happen, the absorption of those wavelengths would need to add to the kinetic energy of the gas, which it doesn’t.

      • “In order for that to happen, the absorption of those wavelengths would need to add to the kinetic energy of the gas, which it doesn’t.”

        ????? That is exactly what the spectrometer measures, the wavelengths being absorbed, and that absorption caused bending of the CO2 molecule that translates into kinetic energy.

  12. Being a graduate of the U of C, and having taken post grad level courses in heat transfer there, I would say that Joe has not attended either heat transfer nor environmental engineering classes at that fine institution, or he would recognize that Dr. Spencer’s simple radiative model and rotating planet spreadsheets are what is intended, excellent instructive tools.

  13. The Earth’s energy budget diagram is NOT for a flat Earth. It is for a perpetual twilight Earth.

    The real Earth has day and night. And the energy budget varies, depending on the time of day.

    To average the day and night energy budgets together, makes as much sense as averaging winter and summer temperatures together, and claiming that the temperature was nice.

    The Earth warms up, and cools down, every day. From 06:00 till 18:00, the energy budget is totally different to from 18:00 till 06:00. What meaningful result is obtained by averaging the hours with sunlight, with the hours without sunlight?

    Could we please have a more realistic energy budget. I have a sophisticated spreadsheet model, which calculates the daily temperature cycle for every latitude. It is based on 5 or 6 simple calculations, repeated 86,400 times (once for each second of the day). It uses a simple energy in and energy out model, combined with the Stefan–Boltzmann law.

    • “To average the day and night energy budgets together, makes as much sense as averaging winter and summer temperatures together, and claiming that the temperature was nice.”
      Nonsense. Energy is a conserved quantity. Sustaining an average for a period gives you an accumulation. There is a big difference between having a positive or negative average income for the year. A positive income, even if irregular, is viable.

      • The Stefan–Boltzmann law states that the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of the black body’s thermodynamic temperature T.

        Averaging an energy budget, when one of the terms is raised the the fourth power, gives a large possible error.

          • The watts and the joules depend on summing (or integrating) the flux components.

            The energy budget for sitting at a constant temperature of 20.0 degrees Celsius for 24 hours, is not the same as the energy budget for sitting at a constant temperature of 25.0 degrees Celsius for 12 hours, followed by a constant temperature of 15.0 degrees Celsius for 12 hours. Because the energy radiated is proportional to the fourth power of the temperature.

          • The convection and conduction characteristics need to be also added and it isn’t even as clean as that. You have time lags and all sorts of problems.

            At best what they are doing is the back of an envelope calculation and even then you would be very careful with it.

  14. I have a lot of respect for Spencer, Anthony, tallbloke and willis
    but…

    there are far too many thought experiments
    not enough real experiments

  15. The crimatologists’ simplifications and modelling of the atmosphere, I am reminded of the spherical horse.

  16. The Radiative Greenhouse Effect Theory

    1. The atmosphere warms the earth, to wit: 288 K – 255 K = +33 C. (Wrong! It cools the earth) Just how does that work?

    2. There are magical GHGs constituting a pitiful .04% of the atmosphere that trap/absorb/re-emit/”back” radiate some kind of 100% perpetual looping “extra” energy thereby warming the atmosphere, to wit: 333 W/m^2. (Wrong! Thermodynamic nonsense.) Well, just where does this magical “extra” energy come from?

    3. The surface radiates LWIR as an ideal 1.0 emissivity BB, to wit: 289 K = 396 W/m^2 (Not Possible!)

    1 + 2 + 3 are absolute horse manure.

    0 RGHE = 0 CO2 warming = 0 man-caused climate change.

    • Nick,
      Take 2 flat surfaces at room temperature, like the walls of the room you are in. Surface A is radiating to surface B at a rate of 400 watts/ sq m. And surface B is radiating back to surface A at a rate of 400 watts/sq.m. No net heat exchange, yet the radiation exists but no use can be made of it by the laws of thermodynamics. If one plate is warmer by 10 C, say like your face, it radiates at 455 watts/sq.m. The back radiation from the wall is still 400 watts/ sq m. The heat flow, in engineering terms is 455-400=55. You know this from your engineering courses. Why do you keep insisting the so-called back radiation is “horse manure”?

      • Because “net” radiation is just that, poop. Energy flows, i,e, heat, from hot to cold -PERIOD!!!! If what you say is so there would be refrigerators w/o power cords. I haven’t seen any. You?
        Also this issue does not matter since the atmosphere cools and does not heat, there is zero net upwelling energy to “back” radiate and RGHE goes straight to the trash bin.

        The fundamental error here is assuming the atmosphere warms the earth (which it does not do) and then coming up with bunch of pseudo-thermo handwavium to explain what does not exist. The result, as expected, is crap.
        ************
        The surface is warmer than ToA for the same reason a house is warmer inside than out in winter – the thermal resistance of the insulated envelope.
        To move current through an electrical resistance takes a voltage difference.
        To move fluid through a hydraulic resistance takes a pressure difference.
        To move energy (heat) through a thermal resistance takes a temperature difference.
        Q, kJ/h = 1/R * A * (surface – ToA)
        R is a complex interaction of conduction, convection, advection (winds), latent (cond & evap) and radiation.
        **************
        Emissivity & the Heat Balance
        Emissivity is defined as the amount of radiative heat leaving a surface to the theoretical maximum or BB radiation at the surface temperature. The heat balance defines what enters and leaves a system, i.e.
        Incoming = outgoing, W/m^2 = radiative + conductive + convective + latent
        Emissivity = radiative / total W/m^2 = radiative / (radiative + conductive + convective + latent)
        In a vacuum (conductive + convective + latent) = 0 and emissivity equals 1.0.
        In open air full of molecules other transfer modes reduce radiation’s share and emissivity, e.g.:
        conduction = 15%, convection =35%, latent = 30%, radiation & emissivity = 20%
        *************
        An essential component of the greenhouse theory’s up/down/”back” GHG energy loop is the notion of “net” radiation. This concept says that two surfaces, one hotter and one colder, radiating towards each other go “boink” somewhere in between producing a “net” radiation.
        For instance, the hotter 396 W/m^2 BB LWIR upwelling from the surface meets the colder 333 W/m^2 LWIR downwelling from the troposphere for a “net” 63 W/m^2 leaving the surface and reaching ToA.
        First of all, as demonstrated in my Modest Experiment 1, because of the contiguous non-radiative participation of the atmospheric molecules the 396 W/m^2 BB upwelling LWIR is simply not possible. The 63 W/m^2 is the result of a surface energy balance, 160 – 17 – 80 = 63.
        Mod Exp 1
        https://principia-scientific.org/debunking-the-greenhouse-gas-theory-with-a-boiling-water-pot/
        The 396 W/m^2 is a theoretical “what if” S-B BB calculation for 16 C, 289 K. The difference of 333 W/m^2 (396-63) appears out of the thin air of this calculation and therefore violates conservation of energy.
        The 333 W/m^2 downwelling has an S-B BB temperature of around 3.8 C or 276.8 K while actual measured tropospheric temperatures are -40 to -60 C, 233 to 213 K, S-B BB of 167 to 116 W/m^2 or half of the mystical downwelling calculation.
        **********
        There is an actual equation frequently presented which supposedly verifies this “net.”
        Qnet = σ * A * (T1^4 – T2^4) where T1 is the hotter surface and T2 is the colder surface.
        But there is only a single area, A. How can a single area have two temperatures? It can’t. T1 = T2 and the parens = 0.
        Let’s step back in the equation and have two areas at two temperatures.
        Qnet = Q1 – Q2 = σ * (A1*T1^4 – A2*T2^4)
        In a closed system (no net energy in and no net energy out) there is one amount of energy split between the two areas. A small area at a high temperature can have exactly the same amount of energy as a large area at a lower temperature.
        Q1 – Q2 = 0 & Q1 = Q2 & A1*T1^4 = A2*T2^4
        What theory means to say is that energy will flow from the higher energy system to the lower energy system until they are at an equilibrium temperature. Temperature does not necessarily reflect energy.
        ***********
        So, let’s apply the entire S-B equations for two different areas.
        Q1 = σ * ε1 * A1 * T1^4 & Q2 = σ * ε2 * A2 * T2^4
        Q1 – Q2 = σ * ((ε1 * A1 * T1^4) – (ε2 * A2 * T2^4))
        Let’s suppose what this equation actually describes is the before and after thermal states of that single area.
        Let’s work this equation in full.
        The two states are:
        Before Q1 = σ * ε1 * A1 * T14 & After Q2 = σ * ε2 * A2 * T24
        The net difference: Q1 – Q2 = σ *(( ε1 * A1 * T14) – (ε2 * A2 * T24))
        There is only one area so A1 = A2 = A
        And for a single area radiating as a BB ε = ε1 = ε2 = 1

        As demonstrated in the modest experiment the only way an area radiates BB and ε = 1.0 is into a vacuum.

        So, Qnet = σ * 1 * A * (T14 – T24) or σ * A * (T14 – T24)
        But as in the modest experiment this would apply only in a vacuum.

        But this buggered equation has also discarded a key terms.
        What’s missing?
        Work! i.e. Q1 = Q2 + W or σ * A * T1^4 = σ * A * T2^4 + W
        (Q1 = σ * A * T1^4 & Q2 = σ * A * T2^4)

        Work is needed to move energy from the colder A2 to the warmer A1 surface. Just a basic refrigeration loop and a thermodynamic law!

        So Qnet = σ * A * (T1^4 – T2^4) assumes a single area and discards both emissivity and zeroes and discards the Work term!

        • Now explain laser cooiling … laser are hot or contain energy right so how can they bring a material down to absolute zero.

          So here is the 2007 work bringing 1gram down to 0.8degK
          http://news.mit.edu/2007/super-cool

          Been done hundreds of times since and most uni’s do it as part of undergrad physics lab

          How does that work given your theory?

          • Jim Masterson

            And that would violate the third law of thermodynamics–would it not?

            Jim

            Going to “near-zero” requires the object’s heat energy (vibration of the individual atoms or molecules) be transferred to a colder molecule: So liquid H2 can absorb heat energy from a “hotter” object. To get to “absolute zero” you have to focus energy on “indivual atoms” in opposite phase to the original vibration: Hitting each target atom at “near exactly” the right microsecond with an opposite wave of the right amount of energy at exactly 1/2 phase out of sequence.
            Extremely difficult obviously, and impossible for large objects.
            The net “system” (all of the technology to create, focus, and time that “opposite phase” wave of inbound energy) IS greater than zero itself, aand so the “system” follows the First, Second, and Third Laws of Thermodynamics. Individual atoms may, or may not, follow all three laws.

            Once vacuum is lost, or power turned off, or the timing phase is changed, the whole thing warms up. Again, following all three Laws of Thermodynamics.

        • There is an actual equation frequently presented which supposedly verifies this “net.”
          Qnet = σ * A * (T1^4 – T2^4) where T1 is the hotter surface and T2 is the colder surface.
          But there is only a single area, A. How can a single area have two temperatures?

          The equation is for a single surface which necessarily has the same area for absorption as for emission, hence the area A.

          So Qnet = σ * A * (T1^4 – T2^4) assumes a single area and discards both emissivity

          It’s for a blackbody so emissivity and absorptivity are 1.0, for a grey body the emissivity and absorptivity are equal (Kirchoff’s law) so in that case an emissivity term is added. Suggest you read a text on Radiation Heat Transfer, e.g. Hottel.

  17. Joe claims that any atmospheric emitted LWIR photos cannot be absorbed and thermalised in any molecule beneath the emitting layer operating in a higher resonant state i.e. a warmer environ.

    And he is right otherwise it would be cold making warm even warmer.
    The energy is simply redundant energy in that environ and is simply deflected.

    The same as all the energy ever emitted and still to be emitted in the universe will eventually become redundan, at 0.3 kelvin,

    A black super cold universe with no visible light, just all the energy that ever existed and will exist redundant and incapable of thermalisation in mass.

    A dead universe that still has all the energy it was created with, all redundant in that environ, all energy at an equilibrium state of 0.3 kelvin.

      • Willis E, from your link:

        Can A Cold Object Warm A Hot Object? … Short answer? Of course not,
        that would violate the Second Law of Thermodynamics … BUT it can leave the hot object warmer than it would be if the cold object weren’t there.

        Okay, I’m picturing a hot object. I place a colder object near the hot object. Now the hot object is warmer than it would be without the cold object there.

        Now quick: Picture the hot object with no cold object — it has a certain temperature. Immediately picture the hot object with the cold object near it.

        Hot object, cold object absent, temperature is the hot object temperature.

        Hot object, cold object present, temperature is hotter than the hot object.

        How did the hot object get hotter? The cold object made it so.

        Changing the gramatical structure of the concept does not change the concept of a cold object’s making a hot object warmer.

        • This idea seems to me to be expressed aboit as poorly as it could possibly be.

          Here’s how I would put it: The hot object cools at a slower rate with the second object there than it does without the second object. That is, I believe, the point Willis tries to make.

          Go.

          • Here’s how I would put it: The hot object cools at a slower rate with the second object there than it does without the second object. That is, I believe, the point Willis tries to make.

            Cooling more slowly does NOT make the hot object hotter, though. If you are talking about air, then, of course, warm air contained by a barrier of some sort from colder air, keeps an object surrounded by the barrier warmer longer, within a given higher temperature range. But the insulated body does NOT get hotter. It just stays warmer longer with respect to the cold, and eventually it would cool, without additional energy input.

            That’s not what is being talked about. Air is not being talked about. Radiation is what is being talked about. And radiation cannot act like insulation, as I’m understanding it. Electromagnetic radiation is NOT like air. Photons are NOT like air molecules. These are different physical domains — different scales of physical reality. They get confused, and this is a problem, I think.

          • Suggest you look into the use of radiation shields with thermocouples in flames. The presence of a radiation shield around the thermocouple results in the temperature measured by the th/c being higher than in the absence of the shield even though the shield isn’t as hot as the th/c.

          • I think he’d be forced to say no. He’s saying, I think, that even a second object at absolute zero would absorb and re-emit or reflect some small amount of radiated energy originally from the hot object back to the hot object and not into the vacuum of space therefore, once sgain, slowing the rate at which the hot object cools, even if by a tiny amount.

        • The problem is in QM which is the law the photons are playing with there is no such thing as hot and cold … there simply is not a single quantum spin or statistic called temperature. Temperature is a human construct from how we experience the world and you sort of have to bunch a whole lot of Quantum Statistics together to make a thing called temperature.

          Good luck trying to use a made up thing to control physical QM statistics. The QM behaviour determines the outcome not the other way around 🙂

  18. Joe Postma’s misunderstandings and errors just muddy the waters and provide a silly target for the cult of CAGW. There are much better theoretical challenges to AGW/CAGW.

    CAGW however seems to be a special theory that is still promoted as ‘science’ even though observations and basic analysis does not support it.

    There are dozens of peer reviewed observations and analysis results that show there is no CAGW and almost no measurable AGW, in addition to the pause/end of warming that we have recently observed.

    The deep paleo and recent observations do not support CAGW or AGW.

    For example.

    …This study demonstrates that changes in atmospheric CO2 concentration did not cause temperature change in the ancient climate.

    http://www.mdpi.com/2225-1154/5/4/76/pdf

    The Relationship between Atmospheric Carbon Dioxide Concentration and Global Temperature for the Last 425 Million Years

    Atmospheric CO2 concentration is correlated weakly but negatively with linearly-detrended T proxies over the last 425 million years.

    Of 68 correlation coefficients (half non-parametric) between CO2 and T proxies encompassing all known major Phanerozoic climate transitions, 77.9% are non-discernible (p > 0.05) and 60.0% of discernible correlations are negative.

    Marginal radiative forcing (D RFCO2 ), the change in forcing at the top of the troposphere associated with a unit increase in atmospheric CO2 concentration, was computed using MODTRAN. The correlation between D RFCO2 and linearly-detrended T across the Phanerozoic Eon is positive and discernible, but only 2.6% of variance in T is attributable to variance in D RFCO2 .

    Of 68 correlation coefficients (half non-parametric) between D RFCO2 and T proxies encompassing all known major Phanerozoic climate transitions, 75.0% are non-discernible and 41.2% of discernible correlations are negative. Spectral analysis, auto and cross-correlation show that proxies for T, atmospheric CO2 concentration and D RFCO2 oscillate across the Phanerozoic, and cycles of CO2 and D RFCO2 are antiphasic.

    Recent observations also do not support CAGW or AGW.

    http://wattsupwiththat.com/2012/08/30/important-paper-strongly-suggests-man-made-co2-is-not-the-driver-of-global-warming/

    http://www.tech-know-group.com/papers/Carbon_dioxide_Humlum_et_al.pdf

    The phase relation between atmospheric carbon dioxide and global temperature
    …As cause always must precede effect, this observation demonstrates that modern changes in temperatures are generally not induced by changes in atmospheric CO2. Indeed, the sequence of events is seen to be the opposite: temperature changes are taking place before the corresponding CO2 changes occur.

    As the theoretical initial temperature effect of changes in atmospheric CO2 must materialize first in the troposphere, and then subsequently at the planet surface (land and ocean), our diagrams 2–8 reveal that the common notion of globally dominant temperature controls exercised by atmospheric CO2 is in need of reassessment.

    Empirical observations indicate that changes in temperature generally are driving changes in atmospheric CO2, and not the other way around….

  19. In general, the “stylized flat earth insulated gray body simplification” average model is actually correct:

    The earth does receive an “yearly average” TOA radiation dose twice a year, both times right around the equinox, when all latitude do actually receive their yearly average 12 hours per day of sunlight.

    On both days, using more carefully calibrated radiation models for every hour of the day at every every latitude on earth (for latitude-appropriate “average” atmospheric conditions of haze, humidity, and clouds!) it is possible to determine that latitudes at 41 North (spring) and 39 south (fall) do indeed receive their yearly hourly average radiation of 12 x 342 watt-hrs in a single 24 hour period!

    So, on the two days of the year receiving an “average” radiation at TOA, the earth does receive a 24 hour average of 342 watts/m^2 at two specific average latitudes – one of them being very near Washington DC.

    (Note that this is calculated using the actual measured TSI “average” of 1362.3 watts/m^2, NOT the incorrect 1367, 1370, or 1376 watts/m^2 often quoted for Trenberth’s flat earth average figures.)

    • Flux is not a dose — it is a rate of flow, isn’t it?

      Amount of sunlight is not amount of daylight. What actually would we mean when we say “amount of sunlight”? — amount of energy? — and is this what we are specifically talking about here?

      Things seem to be getting confused.

  20. A flat Earth would obviously receive the entire 1,370 W/m^2 because, well, it’s a flat surface facing the sun and not spherical.

    For the real spherical Earth, the 1370 W/m^2 should be divided by 4 because we want the average energy entering the entire Earth climate system.

    The impinging solar irradiance does not just enter the sunlit side and then stay there. The energy gets propagated around and into the entire climate all around the sphere of Earth. Hence division by 4 is the correct approach.

    One thing to note is that the terrestrial climate is never in irradiance equilibrium. One side of Earth is always brighter and warmer than the other. The 342 W/m^2 average is therefore a time average as well as a space average.

    • Thanks Pat,***
      -“A flat Earth would obviously receive the entire 1,370 W/m^2 because, well, it’s a flat surface facing the sun and not spherical. ”
      This is where half the Postma error comes in .
      The Postma concept of the flat earth is actually the whole of the earth surface which therefore has to have half of the surface not getting the sun all the time [at night].
      So the solar input can only go to one half of his “flat earth” at any one time.
      That gets rid of one of the factors of 2.
      -“For the real spherical Earth, the 1370 W/m^2 should be divided by 4 because we want the average energy entering the entire Earth climate system.”
      So we divide by 2 because half of the earth, flat or spherical is always radiating heat away at night.
      Then we divide by 2 again because the actual surface area of half a sphere is twice that of the half flat earth area being irradiated earth.
      Gets rid of the other factor of 2
      Or
      The surface area of a sphere is exactly four times the area of a circle with the same radius. You can see this in the area formula, since the area of a circle is πr2 and the surface area of a sphere is 4πr2
      N.B. Only half the sphere or half the flat earth is getting solar irradiation at anyone time.
      *** Together we may have put up an easy explanation for Roy to use.

  21. All of you are forgetting about Holder’s inequality.
    https://springerplus.springeropen.com/articles/10.1186/2193-1801-3-723
    https://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality

    “The conclusion from the above discussion is that a proper calculation of the mean physical temperature of an airless celestial body (Tna) requires an explicit integration of the SB law over the planet surface. This means first taking the 4th root of the absorbed shortwave flux at every point on the planet and then averaging the resulting temperature field across the entire surface rather than calculating a single temperature from the globally averaged absorbed solar flux.”

    • That it (Holder’s Inequality) is ritten in formal words by a formal person does NOT make it (Holder’s Inquality) right, it only allows it to be writely quoted..

  22. Yes, Pat F, I agree — a flat Earth at the Earth’s distance from the sun would indeed receive the entire 1,370 W/m^2.

    So, if you keep the distance the correct distance, then dividing that 1,370 watts by four means that the flat Earth would not only be flat, but also quite a bit larger than the intercept disc used to calculate that 1,370 watts. The divide-by-four move makes the surface of the flat disc larger, as well, and so it is doubly wrong.

    Postma’s version of the error keeps the flat Earth the same size as the intercept disc and then doubles the distance from the sun. Either way you choose to look at it, it is a flat-Earth distortion of reality, as I see it.
    You either have an absurdly large flat Earth at the standard Earth-sun distance, or you have an intercept disc of proper size to calculate the hemispheric sun flux, twice the standard Earth-sun distance.

    The impinging solar irradiance is what? — 1,370 W/m^2 — THAT energy is what gets propagated around into the entire climate all around the sphere of Earth. The divided-by-four energy is NOT what gets propagated — it’s already “propagated” by its very representation of being one fourth of what actually propagates.

    • The impinging solar irradiance is what? — 1,370 W/m^2 — THAT energy is what gets propagated around into the entire climate all around the sphere of Earth.

      No, because the impingement angle departs from 90 degrees away from the equator.

      At the poles the angle has reached zero degrees, and there’s no solar irradiance at all, except that which follows from Earth’s axial tilt.

  23. Please look at the actual empirical data. Satellite lower troposphere temperature data for the past 40 years is freely available from Dr Roy Spencer’s UAH data site. Atmospheric CO2 concentration data is available from hundreds of sites across the Globe. Mathematical analysis of some of this data has shown that the temperature is independent of the CO2 concentration, that is, there is no such thing as a “climate sensitivity”. It is a meaningless metric.
    Further, the data analysis shows that climate determines the rate of generation of atmospheric CO2. The Fourier spectra of both the temperature and the rate of change of CO2 are essentially identical, showing periodicities that are most likely related to the interaction of the bodies within the Solar System. There is nothing to show an anthropogenic effect.
    For detail see https://www.climateauditor.com
    It is not valid to take averages, the integral of the function of a variable is not the same as that function applied to the integral of the variable.

    • Another input from actual empirical data: the UAH satellite lower troposphere temperature for the past 40 years shows a global minimum rate of temperature increase at the South Polar zone of 0.01 deg. C per decade and a global maximum rate of increase at the North Polar zone of 0.25 deg. C per decade. Obviously the asymmetric rate of warming of the Earth shows that the Sun’s irradiance is causing a completely different temperature response from South to North. Surely this negates the idea of using the solar flux averaged across the whole spherical surface of the globe in any calculation of energy budget.
      It cannot have anything to do with the atmospheric CO2 concentration as that has increased almost uniformly across the globe.
      The minimal increase for the South Polar zone is reasonable as the flattening of the globe at the South Pole implies that there will be very little change in its temperature regardless of the intensity of the irradiance because the angle of incidence of the Sun’s rays is almost horizontal to the surface. So why the maximum at the North Pole zone instead of at the Tropics zone where the angle of incidence is roughly vertical ?

  24. The so-called one-dimensional radiative convection model is a toy model, that has none physical assumptions.

    Toy models are simplified models to give a rough estimate if they are conceptual correct. In this case the objective was to come up with a sciency base for CAGW, not to solve a scientific problem.

    As noted below and as we have found by observations, a doubling of atmospheric CO2 would cause 1C warming, however the 1C warming that does occur is high in the atmosphere and due to the decrease in the lapse rate the surface warming is only 0.1 to 0.2C.

    The one-dimensional radiative convection toy models cheats by assuming there is no increase in convection transferred heat due to the addition of the CO2 which is required to get any measurable warming, i.e. The 1.2C for a doubling of atmospheric CO2.

    One dimensional radiative convective equilibrium model (1DRCM) studies

    1. Failure of the fixed lapse rate assumption of 6.5K/km
    The modern anthropogenic global warming (AGW) theory began from the one dimensional radiative convective equilibrium model (1DRCM) studies with the fixed absolute and relative humidity utilizing the fixed lapse rate assumption of 6.5K/km (FLRA) for 1xCO2 and 2xCO2 [Manabe & Strickler, 1964; Manabe & Wetherald, 1967; Hansen et al., 1981].

    William: Comments:
    1. Lapse rate is the change in temperature with elevation as one goes higher in the atmosphere. As the CO2 molecules transfer heat by contact to other gases in the atmosphere and all gases in the atmosphere transfer any excess heat by convection to maintain a linear lapse rate (Temperature change in degrees Celsius/Kelvin per kilometer increase in elevation.)

    2. It is given in degrees kelvin which is confusing as it is a change so Celsius could have been used as a change in temperature per kilometer change in elevation is the same for the Kelvin and Celsius temperature systems as the only difference in the two temperature systems is selection of what temperature zero is to be.

    ..In the 1DRCM studies, the most basic assumption is the fixed lapse rate of 6.5K/km for 1xCO2 and 2xCO2.

    There is no guarantee, however, for the same lapse rate maintained in the perturbed atmosphere with 2xCO2 [Chylek & Kiehl, 1981; Sinha, 1995].

    Therefore, the lapse rate for 2xCO2 is a parameter requiring a sensitivity analysis as shown in Fig.1. In the figure, line B shows the FLRA giving a uniform warming for the troposphere and the surface.

    Since the CS (FAH) greatly changes with a minute variation of the lapse rate for 2xCO2, the computed results of the 1DRCM studies in Table 1 are theoretically meaningless along with the failure of the FLRA.

    In physical reality, the surface climate sensitivity is 0.1~0.2K from the energy budget of the earth and the surface radiative forcing of 1.1W.m2 for 2xCO2. Since there is no positive feedback from water vapor and ice albedo at the surface, the zero feedback climate sensitivity CS (FAH) is also 0.1~0.2K.

    A 1K warming occurs in responding to the radiative forcing of 3.7W/m2 for 2xCO2 at the effective radiation height of 5km. This gives the slightly reduced lapse rate of 6.3K/km from 6.5K/km as shown in Fig.2.

    In the physical reality with a bold line in Fig.2, the surface temperature increases as much as 0.1~0.2K with the slightly decreased lapse rate of 6.3K/km from 6.5K/km.

    Since the CS (FAH) is negligible small at the surface, there is no water vapor and ice albedo feedback which are large positive feedbacks in the 3DGCMs studies of the IPCC.

    …. (c) More than 100 parameters are utilized in the 3DGCMs (William: Three dimensional General Circulation Models, silly toy models) giving the canonical climate sensitivity of 3K claimed by the IPCC with the tuning of them.

    The followings are supporting data for the Kimoto lapse rate theory above.
    (A) Kiehl & Ramanathan (1982) shows the following radiative forcing for 2xCO2.
    Radiative forcing at the tropopause: 3.7W/m2.
    Radiative forcing at the surface: 0.55~1.56W/m2 (averaged 1.1W/m2).

    This denies the FLRA giving the uniform warming throughout the troposphere in the 1DRCM and the 3DGCMs studies.

    (B) Newell & Dopplick (1979) obtained a climate sensitivity of 0.24K considering the
    evaporation cooling from the surface of the ocean.

    (C) Ramanathan (1981) shows the surface temperature increase of 0.17K with the
    direct heating of 1.2W/m2 for 2xCO2 at the surface.

    Transcript of a portion of Weart’s interview with Hansen.

    Weart:
    This was a radiative convective model, so where’s the convective part come in. Again, are you using somebody else’s…

    Hansen: That’s trivial. You just put in…

    Weart: … a lapse rate…

    Hansen: Yes. So it’s a fudge. That’s why you have to have a 3-D model to do it properly. In the 1-D model, it’s just a fudge, and you can choose different lapse rates and you get somewhat different answers

    William: Different answers that invalidate CAGW, the 3-D models have more than 100 parameters to play with so any answer is possible. The 1-D model is so simple so it possible to see the fudging/shenanigans which is fixing the lapse rate.

    • W Astley ”
      As noted below and as we have found by observations, a doubling of atmospheric CO2 would cause 1C warming, however the 1C warming that does occur is high in the atmosphere and due to the decrease in the lapse rate the surface warming is only 0.1 to 0.2C.”
      No.
      It occurs in global mean near-surface air temperature.
      Not just up in the clouds.

      The equilibrium climate sensitivity (ECS) refers to the equilibrium change in global mean near-surface air temperature that would result from a sustained doubling of the atmospheric equivalent CO.
      the ECS is likely between 1.5°C and 4.5°C

      • Angech,

        CAGW is a dead theory.

        As noted in my above comments CO2 does not correlate with temperature changes in the ancient climate or to the recent temperature changes.

        The IPCC predicted tropical tropospheric hot spot did not occur.

        Water vapor in the upper atmosphere has decreased which increases the amount of radiation that is emitted to space. This is opposite to what the IPCC predicted and is exactly what one would if the CAGW theory is completely incorrect.

        There has been almost no tropical region warming, most of the warming has been high latitude warming which has no stopped, which is the end of warming, calling it a pause in warming is only an attempt to keep CAGW alive.

        Repeating the cult of CAGW talking points, emphatically does not change the fact that observations do not support.

        CAGW is contingent on humans causing the rise in atmospheric CO2. If humans did not cause the rise in atmospheric CO2 there is no CAGW. Observations and analysis results in peer review papers show human CO2 emissions caused less than 5% of the recent rise in atmospheric CO2.

        Science is not a fight, it is solving puzzles.

        • William Astley June 5, CAGW is a theory. Certainly not dead while adherents believe.
          The IPCC predicted tropical tropospheric hot spot did not occur.
          “Water vapor in the upper atmosphere has decreased which increases the amount of radiation that is emitted to space.”
          Energy in should equal energy out, doubt that there is any true “increase” in energy out and certainly does not depend on the nature of the atmosphere.
          “Science is not a fight, it is solving puzzles.”
          GHG theory seems reasonable ie temperature should go up with an increase in CO2 or water vapour. Many feedbacks of unknown magnitude, lots of discrepancy at the moment. I think it is very hard to admit to most science but leave one section out because of its putative conclusions. You have to solve the puzzle and explain why this little bit of science should be wrong when all the rest ties in.
          Or accept it is right and then look for the complex reasons why the theory is not working. Not the science but the feedbacks.
          Thanks for trying to work things out and putting ideas out there.

  25. “… hard to believe an actual astrophysicist could make such an elementary error.”

    If they’re indoctrinated into the climate religion, another application of the geometric partitioning of energy often denied is how radiant energy absorbed by the atmosphere is redistributed. To satisfy the nominal sensitivity, about 3.4 times more energy must be returned to the surface than was absorbed to result in the next W/m^2 of TOA imbalance (forcing or ERF) in order to offset the incremental emissions consequential to an 0.8C temperature increase. Meanwhile, the geometry tell us about half of the radiant energy absorbed by the atmosphere ends up emitted into space and the other half returns back to the surface, or more than 6 times less than required for the nominal sensitivity. This 50/50 split is often denied owing to confusion regarding non radiant energy like latent heat, not realizing that whatever effect non radiant heat entering the atmosphere, plus its return to the surface has, is already manifested by the average steady state temperature and its consequential net radiant emissions. If non radiant energy is included, then certainly a larger fraction of the total energy entering the atmosphere is returned to the surface, but that’s independent of the requirement for a 50/50 split of the net radiant flux leaving the surface.

    The end result of everything that acts on solar energy (forcing) can be distilled down to a single value, which is 620 mw per W/m^2 of surface emissions above and beyond the 1 W/m^2 of emissions per W/m^2 of forcing characteristic of an ideal black body. The nominal sensitivity requires a value of 3.4 W/m^2 per W/m^2 above and beyond what an ideal BB can do which is obviously impossible since Joules are Joules and the Earth can’t tell the difference between the next W/m^2 and the average W/m^2.

  26. Maxwell was praised by Einstein for making his own discoveries possible. With the thermodynamic equations etc.. His huge body of work also included his Theory of Heat and studies of all phases and states of matter. Plus many experiments of the highest quality.
    There he showed that gases act as one, energetically, in atmospheres. The Gas Laws/Poisson Effect.. Exit GHG before it was out of diapers.
    Quantum effects among others disallow any old radiation effect (or kinetic one) from heating warmer objects. Or we would have an “IR catastrophe” as well as a UV one. Colder lacks the force to affect warmer, simple as that. Cognomens of silliness or rant do not belong here. Or else honest debate does not. Much depends on that. I do my own research to, so rely not just on authority.

    Early winter, NZ, 35deg Sth. Grass T in sun, 20C; shade 16-17C Vertically, blue sky, minus 2C
    Summer, over 40C normal. With fogging, the violent uplift is obvious. Add latent Heat transfer, unseen, and no need for agw if not for marxism.

    • A minimum atmospheric temperature, or tropopause, occurs at a pressure of around 0.1 bar in the atmospheres of Earth, Titan, Jupiter, Saturn, Uranus and Neptune, despite great differences in atmospheric composition, gravity, internal heat and sunlight. In all of these bodies, the tropopause separates a stratosphere with a temperature profile that is controlled by the absorption of short-wave solar radiation, from a region below characterized by convection, weather and clouds.
      https://www.nature.com/articles/ngeo2020?WT.feed_name=subjects_giant-planets&foxtrotcallback=true

      • “characteristic of many thick atmospheres”
        “In all of these bodies, the tropopause separates a stratosphere with a temperature profile that is controlled by the absorption of short-wave solar radiation, from a region below characterized by convection, weather and clouds5,6. However, it is not obvious why the tropopause occurs at the specific pressure near 0.1 bar. Here we use a simple, physically based model7 to demonstrate that, at atmospheric pressures lower than 0.1 bar, transparency to thermal radiation allows short-wave heating to dominate, creating a stratosphere. At higher pressures, atmospheres become opaque to thermal radiation, causing temperatures to increase with depth and convection to ensue. ”
        Seems to be specific to atmospheres with lower level IR absorbing properties, One imagines a fictitious world with no water or CO2, Sulphur Dioxide, dust etc but lots of O2 and Nitrogen would not have a stratosphere as such?

        • We reason that a tropopause at a pressure of approximately 0.1 bar is characteristic of many thick atmospheres, including exoplanets and exomoons in our galaxy and beyond. Judicious use of this rule could help constrain the atmospheric structure, and thus the surface environments and habitability, of exoplanets.

        • Now, as a rough cross-check, we enter the Venus altitude-versus-atmospheric pressure graph at 1000 millibars (the Earth’s average sea level atmospheric pressure) and go up to intersect the altitude-pressure profile line, and across to the left axis where we find the corresponding altitude of 49.5 kilometers (31 miles). This altitude is only three kilometers (or six percent) different than we found from the temperature graph.
          So, in spite of the surface temperature of Venus being on the order of 864 degrees Fahrenheit, there is a region in the Venusian atmosphere which approximates that of Earth with respect to temperature and pressure.
          https://web.archive.org/web/20080205025041/http://www.datasync.com/~rsf1/vel/1918vpt.htm
          The difference is only due to the difference in the distance between the Earth and Venus from the Sun.

    • “Quantum effects among others disallow any old radiation effect (or kinetic one) from heating warmer objects.”

      This is incorrect. A photon absorbed by matter (for example the Earth’s surface) increases the stored energy (temperature) of that matter independent of anything else. Photons absorbed by the surface whether from the Sun, the atmosphere or deep space, are indistinguishable from each other and all warm the surface (i.e. replace incremental surface emissions) regardless of respective temperatures.

      Think of it this way. Technically, outer space is heating the planet and outer space is very cold.

  27. The one quarter rule only applies to planets without atmospheres.
    As soon as one introduces an atmosphere then the surface temperature will rise higher due to conduction and convection but that was not the point of the initial post.

    • Stephen Wilde – “The one quarter rule applies to all planets ”

      I made this response above, I’m curious how you would answer:

      Imagine if we were able to drag a notched trowel over the Mojave Desert such that the surface area is doubled. We can then divide the energy received by the Mojave Desert from the Sun by 2 correct? And, we can double the amount of energy emitted to space since the surface area has been doubled. So the desert would receive half the amount of energy from the Sun and emit twice as much energy to space.

      • Irregularities across a two dimensional object are self cancelling so you can’t use that analogy for a three dimensional object.
        At all times one side of the furrow would receive more insolation but the other side would receive less so net effect zero.

        • Stephen Wilde “Irregularities across a two dimensional object are self cancelling so you can’t use that analogy for a three dimensional object.”

          What is the two dimensional object you’re referencing? Your calculations require the surface of the Earth to be Flat? That makes Joe Postma correct.

          “At all times one side of the furrow would receive more insolation but the other side would receive less so net effect zero.” ??? “At all times” – Which side receives more at night?

          When the sun is directly overhead, each side of furrow would receive an equal amount, and since the surface area has been doubled we divide by 2 to match your theory right? So, noontime would be cooler than a non-notched surface.

          The emission rate stays the same right? And since the surface area has been doubled, the total emissions also double?

  28. Average hourly solar radiation in watts per meter sq. from USCRN data.
    Mauna Loa 19.5353 -155.5761 3407.4 meters
    June 8th 2017
    0000 0
    0100 0
    0200 0
    0300 0
    0400 0
    0500 0
    0600 5
    0700 143
    0800 370
    0900 583
    1000 784
    1100 932
    1200 1011
    1300 1050
    1400 1001
    1500 902
    1600 745
    1700 550
    1800 298
    1900 81
    2000 0
    2100 0
    2200 0
    2300 0
    24hr average 352

    Just figured you folks might find actual data helpful to the discussion.

    • Bob Koss Average hourly solar radiation in watts per meter sq. 24hr average 352
      Thanks.
      Both sides can use it, I guess.
      Fits in with the removal of the albedo effect.
      8 hours above 352, 16 hours below of which 10 hours have no direct/indirect solar radiation.

    • Something I don’t see in these diagrams is conductive heat transfer between the surface and the atmosphere (To to T1). Given the observed weather effects of thermals and other mass convection, and that approx 90% of the atmosphere cannot be heated directly by radiation, I would expect to see some comment on it since it has a major effect on the where energy radiates from (planetary surface or atmospheric GHG). It’s also hard to imagine that all (or even a major part) of atmospheric temperature increase is driven by ghg radiation given the frequency and implications of inversion layer formation.

      There was a recent article here about UHI and it’s downstream detection that noted that it was indetectable at windspeeds above 7 m/s i believe (working from memory here so don’t hold me to the number). That is a solid example of the conductive-convective process, but it seemed as if the article was trying to state that UHI didn’t matter once wind became part of the picture. That is an erroneous assumption though because the heat still has to go somewhere it is just “diluted” into a greater mass of air so the total temperature rise of any single particle is lower. That is the entire working model of fluid to fluid heat exchangers.

      • sorry mods, this was not meant to be a reply to this coment. it was a general comment to the thread, please move it if possible.

    • Was this an average for clear days? What about cloudy days? Is it never cloudy over Mauna Loa?

  29. A black body absorbs and radiates at the same time. It doesn’t absorb during the day and cool at night. An orb that has a surface of black bodies will drop to zero quickly at night and rise quickly to the equilibrium temperature for the insolation at that spot. Its not so much about the flat Earth but pretending that the averages are like the exact uniform surface of a disc for absorbing and a uniform sphere for radiating.

    I’ll find my simple model and put it up again soon. Its for 8 uniform surfaces of a planet acting like black bodies. One with a 300 K spread of temperatures like at the equator of the Moon and one with 30K spread like the sea surface of the Earth. Tweaked a little ( only needed to change one surface of the ‘Earth’ from 273K to 275K) so that the mean of T^4 was equal for both. With mean T being close to the values for the Moon at the equator and Earth’s sea surface. Just the storing and spreading of heat makes a 30 K difference to mean temperatures, all else being equal. Its why you can’t have this ‘flat Earth’ analysis anywhere in the modelling or influencing the modelling ie implying an equilibrium with the oceans and the upper atmosphere.

  30. Many thanks to Ronald J. Larsen for his patience and clarity on this and to Jim Steele for having the courage to back him up.

  31. Averages are useless when tying to find a specific value, like the “human fingerprint in climate change”. If memory serves, all climate models initialise the earth as a flat, idealised, black-body radiator, which it isn’t, correct me if I am wrong?

  32. People, stay out of the weeds. We’ve won. Find something else to do, and do it as well as you have done this!

  33. Anthony,

    I think this must be related to the paper supposedly giving the “correct” value of a the temperature of the moon. This paper assumes no atmosphere and that the moon rotates. In reality, the moon has many different temperatures because there is no atmosphere to redistribute and equalise the temperature and as a result you’ve got to be pretty dumb to believe there’s only one interpretation of the “temperature” of the moon.

    However, this became important because of the group who tried to argue that pressure alone was the cause of the increased surface temperature. Because to get their “fit” to their curve-matching they had to find a value for the moon’s temperature that was different from normal. So, they started arguing that the “correct” temperature for the moon was something that made their curve match look better.

    However, it hardly helps your own case to use an energy flow diagram that leaves out 50% of the energy leaving the surface through convective currents. Nor does it help when you confuse heat flow with work flow (IR is both work and heat and you’re going to get thermodynamic bullshit unless you are very careful in your treatment of the subject).

  34. Ironic that accusations about believing in a Flat Earth comes from someone who doesn’t believe in Euclid.

  35. OK, then what accounts for the fact that for the majority of Earth history, the temp was much higher than now?
    Glacials/interglacials are another question, as are the fluctuations over the Holocene.
    I am not seeing any explanation for how the temp can fluctuate over time in any these arguments.
    Has the solar constant been constant over long spans of time (besides for the whole faint young Sun thing)?
    Thousands, millions, tens of millions of years?
    Anyone? Anyone? Bueller?

    • It is about total mass of atmosphere on sea level.
      It looks that both Earth and Venus started with similar level of CO2 atmosphere around 100bars.
      During Eons this CO2 was sequestrated by life to carbonates on Earth, Venus stayed same.
      Temperature is directly proportional to mass of atmosphere above ground.
      This higher atmospheric pressure allowed megafauna to fly, created coal deposits and compensated for lower sun output.

  36. It is probably worth forgetting about the flat Earth, it is much more useful to fully understand this video.
    http://www.youtube.com/watch?v=HeCqcKYj9Oc.
    Real climate science is out there, it is just ignored by many. If you ignore the foolish “greenhouse effect” nonsense, it is fairly easy to prove that there is no abnormal warming. Stop pandering to the opposition, take the argument to your own scientific understanding.

  37. This is the old 4 weak suns argument. Dividing the power radiated by the sun by 4 and expecting the result to make any sense.

    We have a pig turning on a spit, cooking gloriously on a fire below. We divide the power radiated by the fire by 4, create 4 equal power sources of P/4 and place them around the pig (having removed the original source), stop the pig rotating and wait for the pig to cook.

    How long will it take?

    Dividing, subtracting, adding fluxes and equating them to temperatures is meaningless.

    It like saying the average person has one testicle.

    • Are you ignoring the fact that much of the heat comes from hot air rising?
      What is the speed at which the pig is turning?
      Are you ignoring the end effects?
      When you say “ready”? Do you mean all above a specified temperature, that it’s nice and roasted outside and the OUTSIDE is ready to eat? Or what is your definition?
      What about cooking sauces?
      Will you be basting the pig?
      But most importantly, does the pig come with an atmosphere that redistributes the heat?

      • re: “Are you ignoring the fact that much of the heat comes from hot air rising?”

        When did you last spend time in front of a roaring fire? We disposed of some old lumber in late 2017 and I can testify to the large amount of radiant (IR) heat coming off a “burn barrel” (let alone the fire itself) – it is ASTOUNDING the amount of thermal energy burning carbon-based “wood” can produce … but that’s not the point.

        I think you wildly over-estimating the hot-air thing; open air pit =/= forced convection oven.

      • OK, replace the fire with an overhead grill, Mike. Replace the pig with an inanimate object, if you like. My point is still the same. 4 weak suns will not be as effective as one hot sun in raising the temperature of the object.

        As Roald says: Sophistry.

    • leitmotif: And yet, the sunlight being intercepted by the Earth (before albedo effects) does get spread over the full sphere of the Earth. That is basic geometry. Area is important. Watts **per square meter**. Averages are useful metrics, and just because the average temperature where you live is seldom ever actually measured at any give instant, doesn’t mean the average temperature isn’t a useful metric. The average temperature in Miami is very different from that in Fairbanks, and no amount of nitpicking about the average temperature value almost never actually being measured at either location will change that fact.

      • Define average temperature: (max + min)/2? Something more relevant? More useful?

        Average number in a telephone directory. Useful metric?

        • leitmotif

          “Define average temperature: (max + min)/2? Something more relevant? More useful?”

          Did you ever process ANY temperature series providing for TAVG, TMIN, TMAX?
          Do you REALLY think that TAVG records all are the same as (TMIN+TMAX)/2 ?

          Did you ever compare TAVG with (TMIN+TMAX)/2 where both are available?

          My guess: the answer is ‘no’ for all three questions.

          • Bindidon.

            My point is …. whatever?

            What does an average temperature mean?

            Do you use the method you like to support your argument?

            Average temperatures are totally meaningless.

  38. A lot of people get hung up on whether the 2nd law is violated. When a body emits a photon in a random direction it has no knowledge of where it is going and whether the destination is warmer or colder. An object radiates depending on its temperature. That is all there is to it.

    At the quantum level you have to think about probabilities. A cold body will likely receive more photons from the adjacent hot body than the hot body will receive from the adjacent cold body.

    Strictly speaking, they both receive heat from each other, but the net effect is that the heat travels from the hotter body to the colder one. The 2nd law is not violated.

  39. Fundamental error 1:
    Greenhouse theory says the atmosphere warms the earth. Since this is incorrect an entire field of bogus physics, i.e. up/down/”back” and BB LWIR, cold to hot, perpetual 100 % efficient, etc. “climate science” must be fabricated to explain how this non-existent process works.

    Fundamental error 2:
    Dividing ISR by 4 to average/spread it evenly over the entire spherical ToA 24/7 is really dumb.
    As far as the sun cares ISR sees the earth as a flat disc where ASR = (1-α) * ISR (1,368 W/m^2)
    To model the perpendicular energy distribution over the hemispherical lit side apply the following equation: ASR = ISR *(1-α) * Cos Latitude. Any engineer siting solar panels can explain why.

    Reality:
    The atmosphere cools the earth and Q = U A (Surface T – ToA T) explains why the surface is warmer than ToA. No GHG hocus-pocus physics needed.

  40. It is weird how there are some people denying the GHE and come up with totally unreasonable arguments. Some would argue with the law of thermodynamics which would not allow a cold(er) atmosphere to heat a warm(er) surface, but they do obviously not understand what the GHE is supposed to be and possibly take Al Gore as a reference. Then we had this (perfect) gas equation showing that if you have pressure, density and so on you can calculate the temperature, which for some reason should falsify the GHE theory, while indeed being just tautology. And we have this “flat earth” issue..

    On the other side we have a lot of critcal climatologists (or people of private interest) which take the GHE for granted. They oppose global warming, or parts of it, but are not at all critical over the GHE itself.

    This is so weird from my perspective, because I can tell the GHE theory is totally wrong, it is just not for the reasons named above. And it will not require some stupid, fancy approach to come to this result, but simply careful and precise analysis. The GHE theory is not constituted by one mistake only, but quite a number of which and they all go one direction. Sorting them out will marginalize the GHE due to GHGs to about 5K or so.

    I am just about putting it all together in a nice little essay that should end the global warming non sense once and for all (it is just about time I guess). Btw. if you want to contribute you will be welcome..

    In the meanwhile you can have a look at this paper which deals with some of the issues with the “GHE” (there are some flaws in it and it is really just a part of the bigger story)..

    https://de.scribd.com/document/369953233/The-Net-Effect-of-Clouds-on-the-Radiation-Balance-of-Earth-2

    • Leitwolf,

      The atmosphere does not warm the earth, it cools it.

      That fact alone is sufficient to condemn RGHE theory to the historical dust bin of failed theories.

      What more is needed?

  41. Well, this was an interesting read and the comments were interesting, but without “Passion” (no foul language). I cannot comment on the the whole argument of averages, but to me you use an average in simple models because the computations are complex for “real” numbers. As I understand it, the Earth is a complex chaotic model that has it’s secrets and surprises. One cannot model that. As models grow (computers becoming faster allowing more variables), we get some answers. Still, we are talking a chaotic system. Consider that the surface of the Earth is not totally flat (but flatter than a billiard ball!).

    Keep conversing and improving!

    By the way, foul language is not “Passion”, it is a loss of reasoning.

  42. “Now, I find it hard to believe an actual astrophysicist could make such an elementary error.”
    Not fair.
    It is a bit like that 3 doors problem.
    Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?
    Many readers of vos Savant’s column refused to believe switching is beneficial despite her explanation. After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them claiming vos Savant was wrong (Tierney 1991).
    The surface area of a sphere is exactly four times the area of a circle with the same radius. You can see this in the area formula, since the area of a circle is πr2 and the surface area of a sphere is 4πr2
    N.B. Only half the sphere or half the flat earth is getting solar irradiation at anyone time.
    *** Together we may have put up an easy explanation for Roy to use.
    I still have trouble with the doors!!!

    • I still have trouble with the doors!!!

      Angech i think her explanation was the best. With the initial pick, the odds are 1 in 3. Those odds don’t go up just because a goat has been revealed in one of the other doors. (door number one still has a 1 in 3 chance of success) Switching, therefore, will give you a 2 in 3 chance of success. i’d elaborate, but my battery is dying (☹️). Maybe sometime over at Dr Curry’s…

    • I still have trouble with the doors!!!

      Angech, i think her explanation was the best. With the initial pick, the odds are 1 in 3. Those odds don’t go up just because a goat has been revealed in one of the other doors. (door number one still has a 1 in 3 chance of success) Switching, therefore, will give you a 2 in 3 chance of success. i’d elaborate, but my battery is dying (☹️). Maybe sometime over at Dr Curry’s…

    • I still have trouble with the doors!!!

      Angech, i think her explanation was the best. The odds of success are 1 in 3. Just because a goat is revealed in another door doesn’t mean that those odds go up. (door number one still has a 1 in 3 chance of success) Therefore, switching doors will give you a 2 in 3 chance of success. Maybe i’ll elaborate sometime over at Dr Curry’s*…

      *i’m having trouble posting here and am ticked off by it, so i wanna keep this one short (😖)

    • >>
      I still have trouble with the doors!!!
      <<

      The three-door version seems to cause people problems.

      Try it this way:
      There are a million doors. One door has a car behind it, and all the other doors have a goat behind them. You pick a door. The host then opens 999,998 doors showing only goats (which he can always do whether you picked the car or not). So there are two doors left–the one you picked and one other door. Your original pick had a 1 in a million chance of being right.

      Essentially, the host has shown you where the car is. Your odds don’t change. Staying with the original door gives you a 1 in a million chance to win. Switching gives you a 999,999 in a million chance to win–the odds don’t change to fifty-fifty because there are now two doors.

      You can do the same thing with fewer doors, say 1,000, then 100, and 10. In each case you should see that it improves your odds to switch. Finally, with three doors, it’s the same problem. Your first pick is 1 out of three. Switching gives you a 2 out of three chance to win–not fifty-fifty as many think.

      Jim

    • I still have trouble with the doors!!!

      I show you Ace, Jack, Jack; I mix them up and you pick one to hold (no looking). I look at the other cards, show you a Jack, tear it up & through it away. I then let you chose from the remaining two cards try to get the single Ace. Do you keep your original or switch.

      Then I do it a little different; I show you Ace, Jack, Jack. I mix them up and you pick one to hold (no looking). I don’t look at the two cards this time, but I remind you that at least one of the remaining two cards is a Jack (which you already should know unless you are an Ocasio voter), and I put the remaining two cards on the table. I tell you that I could look & tear up the remaining Jack in the two card stack for appearance sake, but I won’t do it this time.

      Now you can pick from the two cards on the table or keep the one you have. Do you pick the two or keep the single?

      • Do you pick the two or keep the single?
        You can pick any of the three, as odds without extra information is still only 1 in 3.
        It is only when you actually tear up the jack, from the table, that you have extra information that changes the odds and makes it wise to switch.
        Change the terms.
        Show the card in your hand is a jack.
        Do you switch or stay with the jack?
        Obviously the card is on the table . Still only 50/50 but you would switch.
        If you were then shown a jack on the table you would know the other was the car.
        So 2 out of three times the car is on the table .
        1 in 3 the car is in your hand.
        So when you remove one jack from the table the car is still 2 out of 3 times likely to be on the table compared to 1 in 3 in your hand.
        Ie switch to table as double the odds when you remove one of the unknowns.

    • You had a 2/3rd chance you were wrong.
      But if you know the game host is going to show you the goat.
      Therefore, if wait for host to show you the goat, then you have a 1/2 chance of being wrong
      if you pick again.
      Or by discarding your first pick, you gain a 1/3rd chance of winning.
      Though it depend if the host will always show the door with the goat and always
      give you chance to make another choice.
      If that the case, one can ask, do want 2/3rd chance you are wrong or 1/2 chance you are
      wrong.
      And if don’t change our choice you will always have 2/3rd chance to lose.
      So look at as being your first choice didn’t matter, and always going to wait to get
      the 1/2 chance of winning.
      Or game host is going to have two goats or goat and a car. 50% chance of either.
      Picking the one discarded gives 50% of winning, which is better 33% chance of winning.
      Or host always has 2/3rd chance to get the car, you always have 1/3rd chance to get car,
      by trading with host you get his odds.
      I guess last is best way to say it.

  43. Here’s essentially the same question applied to the ocean:

    Is this the daily average depth of solar penetration into the ocean or is it the daily peak solar input?

    Did the 99.99% absorbed solar energy depth as shown result exclusively from daytime clear sky peak solar irradiance or is it from the 24-hr average that includes zero nighttime solar penetration?

    Obviously, for all practical purposes, the sun’s ocean penetration depth is driven by daytime sunshine, when solar insolation is at its highest peak, not from the daily average insolation value that includes nighttime darkness.

    Also, the daily tropical evaporation cycle responds to daytime sunshine, instantaneous solar insolation, not the daily average value.

    So it comes down to what you’re trying to figure out whether to use peak or average insolation.

    • Bob Weber
      A lot of people with technical backgrounds read and comment on this blog. From the insistence on using averages, it would seem that most of them have forgotten whatever calculus they may have learned as undergraduates!

  44. Here is my video response:

    I will do another follow-up today I think to address one of the central comments made in the OP here, which will help clarify things.

    [Thank you for the courtesy of your reply here. .mod]

    • Since 1 W/m2 = 3.6 KJ/hr/m2, you can do a “real” energy balance in either one, and easily convert one to the other.

      You can also do a balance in either watts or in KJ/hr.

      Your claim is as foolish as saying that a real energy balance is only conducted in metric units and not imperial units … sorry, Nick, but you can use either unit. In climate science, such balances are generally done in W/m2, but please, feel free to use KJ/hr/m2 or KJ/hr if you want to stand out …

      w.

    • OK, let’s do that. Jupiter receives many more kJ/h from the sun than does Earth. So, will Jupiter’s atmosphere be warmer than Earth’s atmosphere?

  45. Does anyone know a reasonable value for the inherent radiative heat loss (in W/m^2 for the surface of the sphere) due to the fact that the earth is exothermic (i.e it generates and radiates heat from its interior)?

    I know it’s small, but I’m curious.

  46. I got about halfway through this thread and gave up on it. I don’t profess to understand all the ramifications of the discussion, but to look at an average, call it unrealistic, and dismiss it, is not very persuasive to me. I’m reminded of the saying attributed to statistician George Box: “all models are wrong, some are useful.” For sure, an average is going to be “unrealistic.” That does not matter. Is it a useful approximation? That should be the issue.

  47. Joe is correct this is a first order thermal system. Input To the system is a variable half wave rectified Sine wave. Steady state Inputs/average values cannot be used to analyze the system maximum temerture is proportional to the Peak value of the input.
    Average temperate is controlled by the system time constant

  48. This is a follow-up to the previous video where I respond to Roy Spencer’s accusation that I am incorrectly and needlessly “ranting” about the fraud of flat Earth theory. I have some better graphics and a little bit more detail to discuss.

    • Joseph
      I’m not a great fan of using averages to characterize variable systems. I think that using calculus to account for the changes that occur with the curvature of the Earth is a more rigorous approach than using some average.

      However, it appears to me that your second diagram should be showing a gross input of 675 W/m^2. That is 1370/2 because the surface area of a hemisphere is 2 pi r^2. Therefore, the average input spread over the entire hemisphere, per unit area, is half as great as would be received per unit area of a flat disk of the same diameter, albeit the total amount of energy received is approximately the same (ignoring secondary effects such as scattering and forward reflection.)

      In the case of Mercury, which is tidally locked to the sun and only one side is presented, it is clear that is what is happening. So, we are confronted, perhaps, with a definitional problem. The radiant energy received by the sun-facing side of Mercury, per unit area, is one-half the energy that would be received per unit area of a flat disk facing the sun. However, the total energy received is equivalent to that received by a flat disk at the same distance. That is the simple case. Now, in the case of Earth, because it is rotating, the entire surface gets illuminated, but with a 50% ‘duty cycle.’ Taking into account time, We should again divide by 2 to get the average amount of radiant energy, per unit area, per rotation. That is mathematically equivalent to dividing TOA insolation by 4 to get the average per unit area of the full sphere.

      As I stated in my opening paragraph, I’m not a fan of trying to do everything with averages. I would like to see a more rigorous, calculus-based approach to dealing with the rotation of an astronomical body so that it could be refined subsequently with the addition of land masses and water, and the angular influences on specular reflections, and dealing with the clouds. As I stated above, I think that the ‘cartoons’ Roy provided are a good first-order approximation to convey the processes for those who are mathematically challenged. However, I think that simplifying the processes to a ‘flat-land’ approach as you suggest, is bound to introduce errors and limit the flexibility of dealing with the real Earth.

      • Clyde Spencer June 5, 2019 at 9:11 pm
        Joseph
        I’m not a great fan of using averages to characterize variable systems. I think that using calculus to account for the changes that occur with the curvature of the Earth is a more rigorous approach than using some average.

        However, it appears to me that your second diagram should be showing a gross input of 675 W/m^2. That is 1370/2 because the surface area of a hemisphere is 2 pi r^2. Therefore, the average input spread over the entire hemisphere, per unit area, is half as great as would be received per unit area of a flat disk of the same diameter, albeit the total amount of energy received is approximately the same (ignoring secondary effects such as scattering and forward reflection.)

        The TSI is the flux perpendicular to the Earth therefore the Earth intercepts the flux passing through a plan of area 𝛑r^2. You’d get the same value if you use and perform the double integral over latitude and longitude.

        • Phil
          I think you missed two points. 1) I was remarking about how surface reflectivity, atmospheric absorption, and atmospheric scattering changes with the angle of incidence; 2) I was commenting about energy per unit area of the surface.

  49. This is a follow-up to the previous video where I respond to Roy Spencer’s accusation that I am incorrectly and needlessly “ranting” about the fraud of flat Earth theory. I have some better graphics and a little bit more detail to discuss.

    [Delete the duplicate entry? .mod]

    • What Joe appears to fail to understand is that the division by four occurs in a correct application of the first law of thermodynamics: that is the sum of energy flux in over time is balanced by the sum of energy flux out. Average energy flux in is 1370*𝛑r^2, average energy flux out occurs over 4𝛑r^2 therefore the energy balance yields energy flux out per unit area= (1370*𝛑r^2)/4𝛑r^2, i.e. 1370/4.
      It is nothing to do with a flat earth.
      Of course if you present a diagram of a unit area on the earth and its energy budget, let’s say over one square meter it will necessarily appear flat. In a model of the earth the equations would have to include the effect of curvature (and they do), the methods of calculus involve consideration of a small area which can be approximated as flat, curvature effects are accounted for by the cos terms.

  50. The earth receives solar insolation over pi r ^2, in 2d.
    It radiates heat away over 4 pi r ^2, in 3d.
    So the earth’s radiative balance is a question akin to Poincare’s Conjecture – how to reconcile 2d with 3d.
    And probably about as complicated.
    Postma is as likely to be right as Nikolov and Zeller, or Miskolczi, or anyone else.

  51. At his blog, Joe is digging himself deeper now showing his abysmal ignorance of these topics. He states:

    “Think of cooking a turkey if it helps. Or invert the example: can you use 4-times the energy (with its equivalent transform to temperature via the Stefan-Boltzmann Law) but applied in ¼ of the time to get the same end result?”

    All real ovens, whether gas or electric, use cyclic on/off application of power. Let’s say the oven requires 500 watts to maintain the cooking temperature of the turkey. The oven will turn on full power for some period of time, then fully off for another period. If this oven has a full power rating of 2000 watts, it will turn fully on for time X, then fully off for time 3X. So it is applying 4 times the power for 1/4 of the time, repeatedly!

    The temperature in the oven will not be as constant as if it applied a steady 500 watts, but in reality it will be so close (typically varying 1 or 2 degrees) that applying the “average power” approximation gets you very close.

    The man does not even understand how his oven works!

    This supposed astrophysicist DOES NOT UNDERSTAND HOW HIS OVEN WORKS!!!

    • Ed B,

      I don’t think you understand that your description is not an equivalent comparison, and it amounts to a diversion and an opportunity to berate somebody attempting to explain a basic flaw.

      Now explain to me how a blower works.

      • RK:

        JP objects vociferously to the use of average power input for rough calculations and conceptual illustrations, when the power input is really time varying. He calls this the “flat earth fallacy”. In the earth’s case, the average (over time and area) power input is 1/4 of the peak.

        It was JP himself who brought up the turkey-in-the-oven analogy, arguing that it would be crazy to think that you could roast a turkey this way, with time-varying power inputs.

        But, in fact, all ovens do indeed work this way, with time varying power input. The thermal power source for an oven is either full-on or full-off. If the time-averaged power requirements are less than the full-on power, the oven goes through on/off power-input cycles.

        This is DIRECTLY analogous to the day/night cycle for a location on the earth, and it is common to use the average power value as a first approximation. JP cannot realize this, no matter how many times it is pointed out to him (and it has been explained to him multiple times).

        Because of the time-varying input power, the temperature is not constant, and will vary somewhat each cycle. But that does not mean that the average is not a decent first-cut approximation.

        • Ed B, I carefully considered your reply:

          It was JP himself who brought up the turkey-in-the-oven analogy, arguing that it would be crazy to think that you could roast a turkey this way, with time-varying power inputs

          But, in fact, all ovens do indeed work this way, with time varying power input.

          Varying the oven’s power input relies on oven insulation to maintain a steady temperature. We are talking about the air molecules in the oven being contained and restrained from convecting to cool. How is this, in any way, correctly related to air in Earth’s atmsophere being heated by sunlight?

          The thermal power source for an oven is either full-on or full-off. If the time-averaged power requirements are less than the full-on power, the oven goes through on/off power-input cycles.

          While the oven is full on, the turkey is cooking at full power. When the oven is full off, the oven insulation still maintains temperature for the off time, still cooking the turkey at the maintained temperature. The Earth does NOT have this insulation, and the sun is NOT blinking on and off to maintain a steady temperature. Are you really saying that the Earth’s atmosphere is like an oven, and the sun is like a blinking-on-and-off oven? On the other hand, I would agree that there are lots of turkeys living on Earth, trying to cook up greenhouse mythology.

          This is DIRECTLY analogous to the day/night cycle for a location on the earth, and it is common to use the average power value as a first approximation. JP cannot realize this, no matter how many times it is pointed out to him (and it has been explained to him multiple times).

          No, it is not directly analogous. Plainly.

          Because of the time-varying input power, the temperature is not constant, and will vary somewhat each cycle. But that does not mean that the average is not a decent first-cut approximation.

          In the oven you so well describe, the temperature IS constant, with the aid of insulation designed to accomodate the power blinking on and off accordingly. How is this like the sun-Earth-atmosphere relationship again? I think you are going to great pains to deform the analogy incorrectly, slightly contradicting yourself in the process.

          • RK:

            I see you have never taken an actual thermodynamics or heat transfer course, and you, like JP, don’t actually understand how your oven works, so you don’t realize the direct analogs.

            In both cases, we have a very hot object (the sun, the heating element) transferring power to to a relatively warm object (the earth, the oven interior including the turkey) that exists in a colder ambient (deep space, the kitchen).

            Both sources transfer power in a cyclic manner to the object. In the earth’s case, it is because a given location on earth rotates away from the sun each night. In the oven’s case, it is because the thermostat turns power to the element on and off.

            In the earth’s case, the cycle is 24 hours long. In the oven’s case, the cycle is typically a few minutes long.

            Both objects (earth and oven) have a certain thermal capacitance responding to the cyclic application of power. Both objects have a certain thermal resistance to losing thermal energy to ambient (space, kitchen).

            You are just plain WRONG when you state that the temperature of the oven “IS constant” through the cycle because of the insulation. The insulation is NOT perfect (the thermal resistance is NOT infinite), so the temperature drops whenever power is not being applied.

            You do understand that when you turn off your oven for good, it cools down to the kitchen’s ambient temperature, don’t you? That also means that its temperature does drop WHENEVER the power is not applied, including the OFF part of the cycle.

            I design control systems with cyclic power application for a living, so I am keenly aware of the nature of the dynamic response to this type of power application. In a typical oven, the temperature drops about 5 degrees during the OFF cycle before the thermostat turns the power full-on again. This band is called the hysteresis of the system. Your house’s thermostat has about a 1 degree hysteresis.

            Similarly, a location on the earth, as it rotates away from the sun, will start to cool down because it has no more solar power input but it is still transferring energy to space. Because there is thermal resistance to this transfer, the cooling is gradual, and the bigger the active thermal capacitance, the less the temperature drop due to the energy loss to space. This continues until dawn, when the next diurnal cycle begins, and the temperature starts to rise again. In many places, the diurnal temperature variation is about 10 degrees.

            So yes,the two systems are directly analogous, and if you or Joe actually understood basic thermodynamics or heat transfer, you could grasp this.

            P.S. I am not EdB. That is someone else.

          • Ed Bo

            I see you have never taken an actual thermodynamics or heat transfer course, and you, like JP, don’t actually understand how your oven works, so you don’t realize the direct analogs.

            I realize that your analogy is not what you think it is.

            In both cases, we have a very hot object (the sun, the heating element) transferring power to to a relatively warm object (the earth, the oven interior including the turkey) that exists in a colder ambient (deep space, the kitchen).

            The oven heating element transfers power, intermittently, in real-time doses, and these real-time doses have real-time effects. The simple, radiative greenhouse heuristic, on the other hand, starts with the diluted sunlight as the power input, and attributes to this diluted power the effects of the full sun in locations where its power does not exist in any full doses ever. With this diluted input power, as illustrated, there is never a place anywhere that the full power applies in real time to have real (i.e., cooking) power.

            Both sources transfer power in a cyclic manner to the object. In the earth’s case, it is because a given location on earth rotates away from the sun each night. In the oven’s case, it is because the thermostat turns power to the element on and off.

            The power is full on, however, in those moments, and this is NOT what the radiative greenhouse heuristic shows — it shows a one-fourth-diluted sun power impinging on the entire planet all at once, instantaneously. There is no accounting there for cycling, as you put it. The math does NOT mean cycling, the way you put it.

            In the earth’s case, the cycle is 24 hours long. In the oven’s case, the cycle is typically a few minutes long.

            Still you speak of cylces of full on and full off power, in the real-time moments during which they occurs. Again, this is NOT what the radiative greenhouse heuristic shows — it shows a given power, one-fourth of the sun power, impinging on the ENTIRE planet, all the time, each second, each day. There is no accounting for the fact that the power in those cycles is fully on — it clearly is one fourth fully on ALL THE TIME, EVERYWHERE.

            Both objects (earth and oven) have a certain thermal capacitance responding to the cyclic application of power. Both objects have a certain thermal resistance to losing thermal energy to ambient (space, kitchen).

            Yes, but this thermal capacitance is reacting to the full power in the moments that it occurs. There are NOT any moments of full power in the radiative greenhouse heuristic. That heuristic shows the planetary output resulting from the full-time impingement of a quarter-power sun in those moments of cycling — NO full-power sun. EVER ! So, it’s NOT your oven, ED — its somebody else’s oven trying to cook that turkey at one fourth power that is one-fourth power ALL THE TIME. The on time is one-fourth power, and the off-time is one-fourth power. That’s what the diagram shows. That’s what the math there means.

            You are just plain WRONG when you state that the temperature of the oven “IS constant” through the cycle because of the insulation. The insulation is NOT perfect (the thermal resistance is NOT infinite), so the temperature drops whenever power is not being applied.

            When an oven says 350 F., most people trust that this is roughly correct, or else the turkey would not cook properly. Now, technically, you might be right that the temperature drops somewhat from this exact figure, but this is an insignificant technicality that really does not bear on the main flaw of your analogy. The fact is that the turkey cooks because of cycling of full-power-on moments with full-power-off moments.

            You do understand that when you turn off your oven for good, it cools down to the kitchen’s ambient temperature, don’t you? That also means that its temperature does drop WHENEVER the power is not applied, including the OFF part of the cycle.

            You know perfectly well that I understand the oven’s cooling to kitchen ambient temperature. Asking me is an obvious swipe at my intelligence, which is, alas, an obvious swipe at your own for thinking that I cannot detect this. (^_^)

            During cook time, the oven temperature does not drop appreciably anywhere near ambient temperature — it remains at a level that enables the turkey to keep cooking properly, just in time for the next power surge that prevents it from cooling too much to continue cooking properly.

            I design control systems with cyclic power application for a living, so I am keenly aware of the nature of the dynamic response to this type of power application. In a typical oven, the temperature drops about 5 degrees during the OFF cycle before the thermostat turns the power full-on again. This band is called the hysteresis of the system. Your house’s thermostat has about a 1 degree hysteresis.

            So, my understanding is correct, it seems.

            Now you seem to by trying to dazzle me with your expertise on a detail, which again is rather trivial in comparison to the main point we are talking about. Citation of your expertise on these fairly irrelevant details and your use of the associated technical lingo does not deter me from my conclusion.

            Similarly, a location on the earth, as it rotates away from the sun, will start to cool down because it has no more solar power input but it is still transferring energy to space.

            Again, yes, but the power that it DID have was the full power of the sun in that moment when it did have power, NOT one-fourth power.

            Because there is thermal resistance to this transfer, the cooling is gradual, and the bigger the active thermal capacitance, the less the temperature drop due to the energy loss to space. This continues until dawn, when the next diurnal cycle begins, and the temperature starts to rise again. In many places, the diurnal temperature variation is about 10 degrees.

            Yes, but again this would be a reaction to the full power applied at the beginning of the cycle. You continue to speak as though you are talking about full power applied in the moments of the cycle you speak of. You are NOT talking about full power, however, if you are using the radiative greenhouse heuristic as your illustration. That’s NOT what this diagram represents at all.

            So yes,the two systems are directly analogous, and if you or Joe actually understood basic thermodynamics or heat transfer, you could grasp this.

            What I understand is that you are using a false analogy that assumes something that is not so.

            P.S. I am not EdB. That is someone else.

            I know your handle — it is Ed Bo — I knew it, when I typed “Ed B” — I abbreviate last names, in order to be less formal and less direct in addressing people sometimes, because it seems a bit less finger-pointery, and keeps a bit of anonymity in the exchange for viewers who are not familiar with the players in the discussion. Feel free to address me as “Robert K” or “RK”, if you want.

          • Interesting that Ed Bo says that JP doesn’t know how an Oven works and yet it is he that does not know how a Gas Cooker Oven works.
            It is on full power until the thermostat temperature setting is reached, it does not “turn off”, it only turns down sufficiently to maintain that temperature.
            It would normally only go back to Full On if the door is opened and substaintial heat is lost or a higher temperature is chosen.

          • RK:

            I’m afraid you’re still missing the whole point, which is why I have begun to question your basic competence.

            You obviously consider the average-power model to be a reasonable approximation for an oven with on-off control. And I’m glad you are no longer disputing me when I claim that the temperature falls during the off cycle. (When you earlier flat out argued against me and claimed “in the oven you so well describe, the temperature IS constant”, you were arguing that the temperature did not go down when the input power was off. I’m sorry, but you deserved the mockery that followed.)

            Now, at least you acknowledge the existence of a temperature drop, saying “During cook time, the oven temperature does not drop appreciably anywhere near ambient temperature.” Similarly, during the night half of the earth’s diurnal cycle, the earth temperature does not drop appreciably anywhere near ambient temperature. In the case of the earth, “ambient temperature” is 3K (-270C).

            By the way, on its website, the GE appliance division says that the typical variation for its ovens over a cycle is between +30F and -30F about the set point, for a total variation of 60F!

            You still have not given any real argument why you consider the average-power model to be a decent approximation for the oven with cyclic power input, but not for a location on the earth that has cyclic power input. That diagram is only a simple conceptual illustration using the average-power model.

            There is NOTHING about the science of the “greenhouse effect” that is in any way “based” on this simplified model, just as there is nothing about the science of an oven that is in anyway based on the simplified average-power model for the oven. Any serious calculations do split the earth into small sections at small time intervals, respecting the location varying and time varying inputs for each section and time step.

            JP’s continued inability to realize that the diagrams he mocks are just simplified conceptual illustrations, rather than the analytical basis for the effect, just goes to underline his complete confusion in these matters.

          • I cannot comprehend how stupid some people can be. The oven switches on and off to maintain a fixed TEMPERATURE. You cook at a fixed temperature in an oven and the control system of the regulates this by modulating the power inpout to maintain the temperature which is monitored by a thrmocouple. Stick a live turkey in an oven and cook at 280f for 6 hours and the turkey will die and get cooked. Put the live turkey in an oven at a lower temperture for a much longer period and the turkey gets warmer but lives quite happily. The total energy input is exactly the same. Very hot temperature for a few hours, then nothing, versus a moderate temperature over a long time. The outcomes are very different!
            As has been pointed out, the world and physical processes occur in real time…not some arbitrary average. I think it has also been shown in numerous places that chemical reactions have an activation energy. You have to supply a certain amount of energ to indce a reaction to occur. The example of the burning paper and the heater is a perfect example. To burn, paper must be raised to a certain temperature. If you do not raise the paper to that temperature it will never burn. So although (in the heater case) the the energy supplied over a period of 24hrs is the same for both examples, the physical reality of what can be done with that total enery is very different. In one case the paper burns, in the other it never can. This is what you normally learn at school at the age of 14.

          • Similarly, during the night half of the earth’s diurnal cycle, the earth temperature does not drop appreciably anywhere near ambient temperature. In the case of the earth, “ambient temperature” is 3K (-270C).

            Yes, and in the case of the moon with a much longer diurnal cycle the temperature drops much closer to the baseline. For a very small diurnal cycle one can make a reasonable approximation by using averages, Earth is near to that case.

    • This is typical diversionary nonsense.

      You simply cannot achieve the 180+°C required to cook anything in an oven by using “average” energy flux.

      As a simple blackbody calculation a body 180°C emits ~2390 W/m2 therefore by the law of energy in = out requires similar input to reach 180°C. You can cook something in an hour for a total energy input of 2390 Watt hours.

      But you cannot cook something by adding 10 hours worth of 239 W/m2 for the same energy input of 2390 Watt hours.

      It doesn’t add up so why not simply admit it ?

      • Rosco:

        Like Joe and Robert, you have absolutely no idea how your oven works! Don’t you realize that your kitchen oven, whether gas or electric, applies power in a cyclic on/off manner? And that these ovens are perfectly capable of cooking things at 180C? This is a simple empirical everyday FACT!

        Let’s say your well insulated oven (not able to radiate directly to the kitchen), when at 180C inside, loses thermal power to the kitchen ambient (~25C) at a rate of 500 watts. If the full-on power of the oven is 2000 watts, the thermostat will turn on the power for 1/4 of the cycle, and off for 3/4 of the cycle, producing an average power input of 500 watts.

        During this cycle, the temperature inside the oven will rise several degrees (say, to 183C) while the power is on, and drop several degrees (say,to 178C) when the power is off. Typically this cycle is a few minutes long. And the turkey cooks just fine. And nobody thinks the extra cost and complexity of allowing the oven to apply partial power continuously is worth it!

        • whether gas or electric, applies power in a cyclic on/off manner?

          Apparently you don’t understand how they work. A gas oven absolutely does NOT cycle itself on/off .. I used to be a chef, I have used many gas stoves and ovens. You are a blooming idiot !!!

          • No Squidly:

            It is absolutely standard for thermostatically controlled gas ovens to cycle on and off. Every gas oven I have ever used does this. Websites are full of people’s questions as to why their ovens do this. They have answers such as “Your oven is constantly cycling on and off automatically to maintain a temperature range close to what you have set at the oven thermostat.”

            Home furnace systems work this way as well, and so do hot water systems.

            My point stands

          • It sounds like GE ovens are electric.
            Gas Ovens never “Switch Off”, they only lower their flame to maintain the temperature.
            I have cooked with gas forthe last 60 years and not one of them ever switched off and switched back to full on unless the door was opened.

          • AC:

            Read the very first sentence on the web page Phil links to. It says:

            “Gas and electric ovens will cycle on and off throughout the cycle to maintain the set temperature.”

            Notice the first word! Are you claiming GE doesn’t know how its own ovens work?

            Automatically controlled variable gas valves are expensive and finicky compared to simple on/off valves. Ditto for the electronics needed to control them. They are very difficult to justify in a cost-competitive market.

            That is why every gas oven, gas home furnace, or gas water heater I have ever used employs on-off control.

          • Well, I have had 5 Gas Cookers in the UK and not one of them Turned itself off, it turned itself down just low enough to maintain the set temperature.
            The only cooker I know of that turns itself off is one that does so on stupid safety grounds when you open the door.
            So my 5 gas cookers do not work the same as yours or GEs and neither did Squidly’s when he was cheffing.

  52. We spend enormous effort debating the how a trace gas in the atmosphere retards the passage of IR radiation to space thereby reducing the cooling of the planet, but we tend to ignore a much larger trapping of heat by the oceans.

    Most of the incoming solar energy strikes the planet in the Tropics, about 40% of the earth’s surface area. This area is 36% land mass, 64% water. The greater portion of the radiation penetrates the oceans where it is effectively trapped as heat. Consider how this heat can dissipate. Water is opaque to IR, convection is suppressed because the water nearer the surface is already warmer, sideways conduction is possible. The main distribution is by currents but the heat in the oceans can remain there for very long periods. The oceans control our climate.

  53. On any given day on earth some surface area at + 40°C and some at -40°C(rough min/max). 273+/-40 is approximately a +/- of 15%. 1.15^4 ~ 1.75, 0.85^4 ~ 0.5 that is a 3.5 to 1 non-linear variation in W/m^2 . 275/273 ~ 1.007. 1.007^4 ~ 1.03% change for 2°C warming. Whatever co2 warming there may be is totally lost in the errors of using linear averages on non-linear equations.

  54. Climate change needs what physics needs: A Theory of Everything. What I have read here and elsewhere makes it abundantly clear that there are many schools on climate change theory and none actually agree. There are arguments all the time over what the theory does and does not say. That’s okay in theoretical physics—no one is attempting to create world government and shut down capitalism with string theory. It’s very, very bad in climate change. Until someone can explain global warming in a way everyone agrees upon, we don’t need action. Trying to fix an undefined entity is just plain foolish.

  55. So, if this energy from the sun is now hitting a disk then the bright side of the disk ‘should’ equate to the average hot temperature of the Earth (average Tmax?), so the dark side of the disk should be cold (average Tmin?).
    If not why so?

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