Guest Post by Reed Coray
On Dec. 6, 2011 12:12 am Lord Monckton posted a comment on a thread entitled Monckton on sensitivity training at Durban that appeared on this blog on Dec. 5, 2011. In that comment he wrote:
“First, it is not difficult to calculate that the Earth’s characteristic-emission temperature is 255 K. That is the temperature that would obtain at the surface in the absence of any greenhouse gases in the atmosphere. Since today’s surface temperature is 288 K, the presence as opposed to absence of all the greenhouse gases causes a warming of 33 K”.
Since I’m not sure what the definition of the “Earth’s characteristic-emission temperature” is, I can’t disagree with his claim that its value is 255 K. However, I can and do disagree with his claim that 255 K is “the temperature that would obtain at the surface in the absence of any greenhouse gases in the atmosphere”.
When computing the Earth’s surface temperature difference in “the presence as opposed to the absence of all greenhouse gases”, (i) two temperatures (A and B) must be measured/estimated and (ii) the difference in those temperatures computed. The first temperature, A, is the temperature of the Earth’s surface in the presence of an atmosphere that contains both greenhouse gases and non-greenhouse gases. The second temperature, B, is the temperature of the Earth’s surface in the presence of an atmosphere that contains non-greenhouse gases only–i.e., an atmosphere that contains non-greenhouse gases but is devoid of greenhouse gases.
For temperature A almost everyone uses a “measured average” of temperatures over the surface of the Earth. Although issues may exist regarding the algorithm used to compute a “measured average” Earth surface temperature, for the purposes of this discussion I’ll ignore all such issues and accept the value of 288 K as the value of temperature A (the temperature of the Earth’s surface in the presence of an atmosphere that contains both greenhouse gases and non-greenhouse gases).
Thus, we are left with coming up with a way to measure/estimate temperature B (the temperature of the Earth’s surface in the presence of an atmosphere that contains non-greenhouse gases only). We can’t directly measure B because we can’t remove greenhouse gases from the Earth’s atmosphere. This means we must use an algorithm (a model) to estimate B. I believe the algorithm most commonly used to compute the 255 K temperature estimate of B does NOT correspond to a model of “the temperature of the Earth’s surface in the presence of an atmosphere that contains non-greenhouse gases only”. As will be evident by my description (see below) of the commonly used algorithm, if anything that algorithm is more representative of a model of “the temperature of the Earth’s surface in the presence of an atmosphere that contains both greenhouse gases and non-greenhouse gases” than it is representative of a model of “the temperature of the Earth’s surface in the presence of an atmosphere that contains non-greenhouse gases only.”
If I am correct, then the use of 255 K in the computation of the Earth surface temperature difference with and without greenhouse gases is invalid.
Although there are many algorithms that can potentially lead to a 255 K temperature estimate of B, I now present the algorithm that I believe is most commonly used, and discuss why that algorithm does NOT represent “the temperature of the Earth’s surface in the presence of an atmosphere that is devoid of greenhouse gases”. I believe the algorithm described below represents the fundamental equation of radiative transfer for the Earth/Sun system assuming (a) an Earth absorption albedo of 0.3, and (b) an Earth emissivity of 1.
(1) The “effective temperature” of the Sun [i.e., the temperature of a sun-size spherical blackbody for which the radiated electromagnetic power (a) is representative of the total solar radiated power, and (b) has a power spectral density similar to the solar power spectral density] is approximately 5,778 K.
(2) For a spherical blackbody of radius 6.96×10^8 meters (the approximate radius of the sun) at a uniform surface temperature of 5,778 K, (a) the total radiated power is approximately 3.85×10^26 Watts, and (b) the radiated power density at a distance of 1.5×10^11 meters from the center of the blackbody (the approximate distance between the center of the Sun and the center of the Earth) is approximately 1,367 Watts per square meter.
(3) If the center of a sphere of radius 6.44×10^6 meters (the approximate radius of the Earth) is placed at a distance of 1.5×10^11 meters from the center of the Sun, to a good approximation the “effective absorbing area” of that sphere for blackbody radiation from the Sun is 1.3×10^14 square meters; and hence the solar power incident on the effective absorbing area of the sphere of radius 6.44×10^6 meters is approximately 1.78×10^17 Watts (1.3×10^14 square meters x 1,367 Watts per square meter).
(4) If the sphere of radius 6.44×10^6 meters absorbs electromagnetic energy with an “effective absorption albedo” of 0.3, then the solar power absorbed by the sphere is 1.25×10^17 Watts [1.78×10^17 Watts x (1 – 0.3)].
(5) A spherical blackbody (i.e., a spherical body whose surface radiates like a surface having an emissivity of 1) of radius 6.44×10^6 meters and at a temperature 254.87 K (hereafter rounded to 255 K) will radiate energy at the approximate rate of 1.25×10^17 Watts.
(6) If independent of the direction of energy incident on a sphere, the surface temperature of the sphere at any instant in time is everywhere the same, then the sphere possesses the property of perfect-thermal-conduction. Thus, for (a) an inert (no internal thermal energy source) perfect-thermal-conduction spherical body of radius 6.44×10^6 meters and uniform surface temperature 255 K whose center is placed at a distance of 1.5×10^11 meters from the center of an active (internal thermal energy source) spherical blackbody of radius 6.96×10^8 meters and uniform surface temperature 5,778 K, and (b) the inert perfect-thermal-conduction spherical body (i) absorbs electromagnetic energy with an effective absorption albedo of 0.3, and (ii) radiates electromagnetic energy with an emissivity of 1 then the perfect-thermal-conduction inert spherical body at temperature 255 K will be in radiation rate equilibrium with the active spherical blackbody at temperature 5,778 K. If the phrase “inert perfect-thermal-conduction spherical body of radius 6.44×10^6 meters” is replaced with the word “Earth,” and the phrase ” active spherical blackbody of radius 6.96×10^8 meters and uniform surface temperature 5,778 K” is replaced with the word “Sun”, it can be concluded that: If (a) an “Earth” at temperature 255 K is placed at a distance of 1.5×10^11 meters from the “Sun” and (b) the “Earth” (i) absorbs electromagnetic energy with an effective absorption albedo of 0.3, and (ii) radiates energy with an emissivity of 1, then the “Earth” will be in radiation rate equilibrium with the “Sun.” For the above conditions, the temperature of the “Earth” in radiation rate equilibrium with the “Sun” will be 255 K.
This completes the algorithm that I believe is commonly used to arrive at an “Earth’s characteristic-emission temperature” of 255 K, and hence is used to compute the 33 K temperature difference.
Even ignoring the facts that (1) it is incorrect to use the “average surface temperature” when computing radiation energy loss from a surface, and (2) in the presence of an atmosphere, (a) the blackbody radiation formula may not apply, and (b) blackbody radiation from the surface of the Earth is not the only mechanism for Earth energy loss to space (the atmosphere even without greenhouse gases will be heated by conduction from the Earth surface and both conduction and convection will cause that thermal energy to be distributed throughout the atmosphere, and the heated atmosphere will also radiate energy to space), the problem with using the 255 K temperature computed above to determine the difference between the Earth’s temperature with and without greenhouse gases is that the effective Earth absorption albedo of 0.3 used to generate the 255 K temperature is in part (mainly?) due to clouds in the atmosphere, and atmospheric clouds are created from water vapor, which is a greenhouse gas.
Thus an effective absorption albedo of 0.3 is based on the presence of a greenhouse gas–water vapor. It is illogical to compute a difference between two temperatures both of whose values are based on the presence of greenhouse gases and then claim that temperature difference represents the temperature difference with and without greenhouse gases. Without water vapor, there won’t be any clouds as we know them. Without clouds, the effective absorption albedo of the Earth will likely not be 0.3, and hence without the greenhouse gas water vapor, the Earth’s surface temperature in the absence of greenhouse gases is likely to be something other than 255 K. Thus, the 255 K “Earth characteristic-emission temperature” as computed using the algorithm above is NOT relevant to a discussion of the Earth surface temperature difference for an atmosphere that does and an atmosphere that does not contain greenhouse gases. Only if 0.3 is the effective absorption albedo of the Earth in the presence of an atmosphere devoid of all greenhouse gases is it fair to claim the presence of greenhouse gases increases the temperature of the Earth by 33 K.
Because clouds reflect a significant amount of incoming solar power, without water vapor I believe the effective absorption albedo of the Earth will be less than 0.3. If true, then more of the Sun’s energy will be absorbed by an Earth whose atmosphere is devoid of greenhouse gases than by an Earth whose atmosphere contains clouds formed from the greenhouse gas water vapor. This implies a higher Earth surface temperature in the absence of water vapor than the “Earth’s characteristic-emission temperature of 255 K”.
For an effective absorption albedo of 0, the temperature of the Earth in radiation rate equilibrium with the Sun will be approximately 278.64 K (hereafter rounded to 279 K). If this value is used as the Earth temperature in the presence of an atmosphere devoid of greenhouse gases, then it can be argued that the presence of greenhouse gases introduces a warming of approximately 9 K (288 Kelvin minus 279 K).
In summation, using the simplified arguments that I believe are also used to arrive at the 33 K temperature difference (i.e., assumed perfect-thermal-conduction Earth, blackbody Earth emission, greybody Earth absorption with an effective absorption albedo between 0 and 0.3, and ignoring atmospheric radiation to space for an Earth atmosphere devoid of greenhouse gases), I conclude the presence of greenhouse gases in the Earth’s atmosphere increases the Earth’s temperature by somewhere between 9 K and 33 K. Thus, I believe the claim that the presence of atmospheric greenhouse gases increases the temperature of the Earth by 33 K is based on an argument that has little relevance to the Earth’s temperature in the presence of an atmosphere devoid of greenhouse gases; and hence at best is misleading and at worst incorrect.
Note: Upon first publication – the guest author Reed Coray was accidentally and unintentionally omitted.
This “effective temperature” nonsense is the biggest scam I have ever seen portrayed as “science”.
The calculation of the “effective temperature” relies on the ratio of the area of a disk to a sphere – one quarter. No matter how the explanationis presented it boils down to a disk irradiated by 957 W/sq m (1367 W/sq m X (1 – 0.3 )) is in radiative balance with a sphere radiating ~ 239 W/sq m at an “effective temperature of 255 K.
Here’s where the con begins. We know the surface is warmer so the difference must be the “greenhouse effect”.
I say rubbish – the Sun heats the Earth during the day and the heating is variable – not constant.
Further, it is obvious that using the geometry of the “radiative balance” has no bearing on the solar insolation – if it did then does it mean that my back being turned away from the radiator means the power of the radiator is halved because of “radiative balance” ?
I know that sounds silly but if this “radiative balance” calculation was correct then as Wikipedia state:-
“If an ideal thermally conductive blackbody was the same distance from the Sun as the Earth is, it would have a temperature of about 5.3 °C”
Well this is verifiable as there is a body in space the same distance from the Sun – the Moon. And any search of the net will reveal the Moon reaches up to 396 K or 123 degrees C during the lunar day.
Let’s verify the theory shall we :-
1367 W/sq m solar constant – divide by four to get solar insolation as required by “radiative balance” gives solar insolation of ~342 W/sq m.
Stefan-Boltzmann says 342 W/sq m yields a maximum temperature of ~278.7 K or 5.3 degrees C.
DAMN IT, all my theory goes up in smoke – the maximum temperature on the moon is actually ~5.3 degrees C.
So NASA is simply wrong when it says the Moon reaches 123 degrees C – the “radiative balance” calculations as used for Earth’s “effective temperature” prove the data is, as usual, simply wrong and theory and models are right.
If you believe this nonsense you really are gullible.
The other nonsense is the implication Nitrogen and Oxygen don’t play a part in warming and cooling of the atmosphere.
If anyone is stupid enough to believe that 99% of the atmosphere doesn’t become heated and radiate IR in accordance with the temperature then they have either been conned or lack the cognitive ability to recognize such an impossibility.
Nitrogen and Oxygen may not absorb much IR but they MUST become heated and MUST radiate.
The amount of IR from water vapour is at best a trace amount and the amount of IR from CO2 is tiny – unless CO2 possess some hidden magical power that science has not detected yet.
Here (pdf download) is a NASA archive document, titled, “Earth Albedo and Emitted Radiation — NASA SP-8067” showing how the 255 K effective radiant temperature of Earth was derived — by experimental measurements using satellites. The document is dated from 1971, which means it was written — and the data were collected — prior to the politicization of NASA. Which means we can safely assume those who wrote it are trustworthy, and the data are not fudged. This is a very sad commentary about the times we live in now, but we’ve all been burned too many times to ignore that truth.
In any case, after a long analysis they report the average TOA emitted radiant power is 237(+/-)7 W/m^2. This is an empirically measured value, not calculated from a model. The word “average” is very important because, as the document points out, the locally radiated power varies greatly.
Here’s what the document says: “As a first approximation, suitable for estimating the effects of Earth emission and albedo on spacecraft subsystems and elements having relatively long thermal time constants, the Earth may be treated as a uniform and diffuse (Lambertian) emitter and reflector. The mean annual values to be used are:
Emission: 237(+/-)7 Watts m^-2 (0.34(+/-)0.01 cal cm^-2 min^-1)
Albedo: 0.30(+/-)0.02
These values were derived fron analyses of data acquired by spacecraft (ref. 38). It should be recognized that they were derived from albedo values that range from about 0.10 to 0.80 and from long wave radiation values ranging from about 105 to greater than 350 Watts m^-2 (0.15 to 0.50 cal cm^-2 min^-1) over limited geographical regions. (my bold)”
This 237 W/m^2 translates directly to an average Stefan-Boltzmann blackbody TOA temperature of 254 K. A distant spacecraft looking back at Earth radiating as a thermal point-source would measure that temperature.
The greenhouse warming is just the difference between the bottom-of-the-atmosphere (BOA) surface temperature and the TOA radiant temperature. As Earth is at overall energy equilibrium — energy in = energy out — and outgoing energy is by radiation into space, then Earth radiates as though it is at equilibrium with an energy source equivalent to a radiant temperature of 254 K. So, the surface is warmer than would be expected by the observed thermal equilibrium, and that is the GH difference. Note: observed, not deduced.
GH average temperature just a kind of convenient fiction to illustrate how the water-rich atmosphere impacts the surface, making Earth habitable down here for us all. Nevertheless, the atmosphere does make the BOA average ~33 K warmer than the TOA average.
Reed Coray said in part:
“(the atmosphere even without greenhouse gases will be heated by conduction from the Earth surface and both conduction and convection will cause that thermal energy to be distributed throughout the atmosphere, and the heated atmosphere will also radiate energy to space)”
Non-“greenhouse” gases do not significantly radiate thermal energy to space. A planet with an atmosphere free of GHGs, such as an atmosphere made of neon and argon, has thermal radiation to space only from its surface.
However, I do see some of the points on theoretical temperature of Earth’s surface if it had an atmosphere free of GHGs and its current albedo, as in 255 K not being the “likely correct average”. For example, 255 K would not be the average, but the “root mean 4th”, or 4th root of global-average 4th power of surface temperature. The average would be less. This means existing GHGs are boosting the global average surface temperature by more than the mentioned 32 K.
Dave Springer says:
December 26, 2011 at 6:48 am
“The ocean has an albedo of effectively 0 and it covers 70% of the surface.”
True but in a very limited set of circumstances that are not part of the real world: Yes, at very low degrees incidence angle with no waves, hardly the case all over the Earth at any one time. Water is mostly a mirror. Albedo equations are not proper for water, Fresnel equations are. Nevertheless, water is considered to have a very low albedo, I believe that to be wrong.
Best,
J.
Neo, I like your comments. A question – how much of the thermal gradient/depth readings can be ascribed to compression?
gas laws-
http://physics.about.com/od/glossary/g/temperature.htm
kinetic energy transfer.
http://en.wikipedia.org/wiki/Avogadro's_law
same pressure = same temperature
not radiation physics
To ALL:
It is amazing to see how much confusion can be created out of a simple problem. Here are a few points for all of you to consider since no one has addressed the issue in a physically correct way including Reed Coray:
1) When assessing the strength of the GH effect, one needs to compare two temperatures – (a) current mean temperature of Earth (Ts) with atmosphere, and (b) the estimated ‘gray-body’ temperature of Earth (Tgb) WITHOUT atmosphere. Discussions involving a hypothetical atmosphere with no ‘greenhouse’ gases is physically meaningless, because the GH effect measured as a surface thermal enhancement, i.e. GHE = Ts/Tgb may not be due to heat-absorbing gases at all …
2) Calculating Earth’s mean gray-body temperature without atmosphere requites an explicit spherical integration of the temperature field. In other words, temperature at each point on Earth’s sphere needs be calculated first by taking the 4th root of the absorbed solar radiation at that point, and then the resulting temperature field must be averaged. This approach is different from the conventional method attempting to estimate the mean planetary temperature from the spherically AVERAGE absorbed solar radiation using the S-B law. Due to the non-linear relationship between temperature and radiative flux, and a highly non-uniform distribution of temperatures on an airless planet, one CANNOT obtain the true (arithmetic) mean temperature from an AVERAGE absorbed radiation using S-B equation! This is explained by the Hoelder’s inequality between non-linear integrals in math. According to this inequality, the effective emission temperature calculated from the spherically average radiation flux will ALWAYS be significantly HIGHER than the true mean temperature of the planet. That’s is why effective emission temperatures are not physically compatible (and should not be equated) with palatable mean temperatures!
3) When one computes the MEAN surface temperature of an airless Earth through EXPLICIT integration of the temperature field resulting from a differential solar heating of the sphere and assuming an average surface albedo of 0.12, one arrives at 154.3K. Since the actual near-surface temperature of Earth is currently 287.6K, the true ‘Greenhouse Effect’ turns out to be 133.3K. In terms of the proposed thermal enhancement ratio above, this means GHE = 287.6/154.3 = 1.86. This raises the question: Can a handful of trace gases, which amount to less than 0.5% of atmospheric mass trap enough radiant heat to boost the planet surface temperature some 86% above that of a moon-like airless gray body? According to classical thermodynamics, this not possible. Hence, the atmospheric GH effect is not caused by heat-absorbing gases, but is likely due to an entirely different mechanism … The nature of this mechanism is explained in a special posting to occur soon on this website … Stay tuned!
Ian S says:
December 26, 2011 at 5:13 am
“A thought experiment. If the earth had an albedo of 0 and you could put whatever imaginary layer you wanted around it (with physical properties, i.e. you can’t use one-way glass that doesn’t exist in the real world). What is the maximum temperature?
Could you get the earth to have a higher temperature than a theoretical perfect black-body? If so, does that seem logical?”
The answer is yes. There is such a thing as a “hot mirror” – an object that is largely transparent to wavelengths shorter than 700 nm or so, and largely reflecting of wavelengths longer than that, up to maybe 2,000 nm. This is an example of a “dichroic filter”. It is not fundamentally physically impossible to create spherical ones whose infrared-specific reflection extends through “low temperature thermal IR” out to 20,000 or 50,000 nm.
Put a black rock inside one of these “oversimplified” (but not fundamentally physically impossible) dichroic puppies with a vacuum in between, put it in sunlight, and then the black rock will get hotter until its visible outgoing radiation is in balance with its visible incoming radiation. At that point, the black rock will warm up to theoretically 1,476 degrees K, and appear orange in daylight and a brightish, possibly-slightly-whitish, and quite yellowish shade of orange in someone’s living room or bedroom. The rock may be lava at that point.
As for greenhouse gases: For an oversimplified case, consider a blackbody planet surrounded by an “oversimplified” sheet of GHGs. Incoming radiation comes in, through the GHGs, unimpeded The outgoing radiation from the planet is absorbed by the GHG layer. Half of the GHG-absorbed radiation gets re-radiated outwards, and half gets re-radiated inwards towards the planet. So, the planet radiates twice the radiation that its sun delivers to it in order to achieve
thermal equilibrium. Its GHG atmosphere layer, if oversimplifyable to being a perfect full-performance GHG one of unifrorm temperature and not running into convection complicating things, would have the temperature that achieves outgoing radiation equal to incoming radiation to the planet.
9 to 33C? More like 9 to 16 C. Latent cooling alone drops the surface average flux from approximately 393 to about 313 Wm-2 (273K) minus 240 Wm-2 TOA OLR is 73 Wm-2 combined for greenhouse radiant and conductive/convective. The greenhouse is only the amount over 255C. We live on a water world and Trenberth in a fantasy world.
Pat Frank says that by experimental evidence and calculation the Earth with an albedo of 0.3
results in;
” Earth radiates as though it is at equilibrium with an energy source equivalent to a radiant temperature of 254 K.”
Nobody disputes these points.
What is in dispute is the claim that if the so called greenhouse gases did not emit/absorb in the IR the Earths albedo would stay fixed at 0.3.
It should be fairly obvious that much more radiation(particularly in the IR) would reach the surface.
The surface has an albedo of 0.15 for the existing solar radiation.
With a large increase of IR radiation the surface albedo is bound to fall much further.
It is an abuse of any kind of logic to use an albedo of 0.3 for a calculation of a mythical greenhouse effect of 33K.
Any theory to be taken seriously must be internally consistent.
By using IPCC accepted values the claimed 33K turns out to be nearer 16K.
The greenhouse theory as presented in popular expositions is internally inconsistent.
Leonard Weinstein says in part:
December 26, 2011 at 6:32 am
“Look up any ocean heat budget study in the literature and you will discover that the ocean loses most of its heat by evaporation not radiation”.
This may be the case on a planet with evaporative cooling and precipitating-cloud-forming convection, and these items involve a major greenhouse gas (water vapor).
I thought the discussion was for comparison of Earth to a most-Earth-like-possible planet with an atmosphere free of GHGs. Even though such a GHG-free planet would be one without bodies of water, as in a dry, deserted rock!
wayne says:
December 26, 2011 at 11:10 am
R. Gates says:
December 26, 2011 at 10:27 am
I guess then that the moon, which is also a spinning sphere, would not have a boiling side and freezing side?
>>>
The moon… spinning huh, like Venus you must mean, right? R. Gates , that sure is a cheap shot to Lars P. Or, I guess using your logic, that even with a greenhouse gas-less Earth spinning at 1 RPM, it would have the sun lit side boiling hot and the dark side freezing cold.
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Why would it be any different? It would not. Just as we get large diurnal temperature swings in the desert areas of earth because there is less water vapor in the atmosphere above the desert, if you took away all greenhouse gases planet wide you would get a diurnal temperature swing that would be even more extreme. It might not be as dramatic as that of the moon (-230C at night to 120C during the day) because of some heat coming from earth’s interior, but it would still be extreme enough to make Earth a very inhospitable place.
Diatomic and mono-atomic molecules have very very low emission in the thermal (IR) bands. Only at higher pressure and temperature does interactions between molecules create the dipoles allowing such thermal emission. But thermodynamics will not be denied! What happens is that such gases (if warmer than their surroundings) will cool by expansion.
http://www.heliosat3.de/e-learning/remote-sensing/Lec7.pdf
To Donald L. Klipstein
I hope I can interest you further to consider this comparison.
A black body is often approximated by a spherical cavity with a small hole in it. The energy gets in at one angle, but bounces around ‘trapped’ at many other angles with a small percentage chance of finding the hole again.
I believe that band-pass, or high-pass filters are just another type of ‘hole’. The energy gets in at one frequency and is converted (through absorption and emission) to different frequencies with a small percentage chance of hitting the escape frequency again.
I like to think of things in terms of energy densities. It helps get rid of the confusion of temperature which is material dependent and adds complexity.
Whether we are talking about angles or frequencies, the end result remains the same (I believe) and you can not create an average energy density inside the spherical cavity that is higher than the energy density outside the cavity. [Of course if our energy source is coherent (i.e. radiation in a single direction, or at a single frequency) then we can be focus the energy to obtain extremely high energy densities, however the average energy density will remain the same.] Average energy density conservation is basically just a different way of looking at second law of thermodynamics and how you can not transfer heat from a cold object to a hot object and special filters of any kind we can imagine are not a form of Maxwell’s demon.
R. Gates says:
December 26, 2011 at 10:27 am
______
I guess then that the moon, which is also a spinning sphere, would not have a boiling side and freezing side? But it does. The essential point is that the oceans and greenhouse atmosphere prevent a strong diurnal temperature variation. The oceans provide the primary heat “sink” of the planet and the greenhouse atmosphere provides a transport mechanism for the heat.
————–
Dear R. Gates, yes the oceans are the primary “sink” of the planet as you can also see from my other comments in this thread – we agree on this point, quelle surprise, this is exactly the main point that is forgotten in greenhouse. But please don’t try to defend greenhouse theory based on the oceans, as it was already said many times the downwelling infrared does not even reach under the first millimetre of water. Greenhouse does not add to the warming of the oceans. And the 24 hours versus 28 days to make one rotation is a fundamental difference that allows for only 12 hours cooling or 14 days. And of course the oceans with their heat capacity, the way how they warm and how they lose heat play the main role – not greenhouse R. Gates.
Does anyone have access to this paper? http://www.agu.org/pubs/crossref/2011/2010JD015343.shtml
A precis of results would be appreciated..
Dave Springer says on December 26, 2011 at 7:24 am:
“Land surfaces radiate. Water evaporates. Write that down. Land has an albedo of 0.15. Water has an albedo of 0.00. Write that down too.”
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What you are saying there seems, at first glance, to be correct but unfortunately it ignores one very important fact, which is that the soil anywhere on this planet of ours does have a fairly high ‘water-content’. Many of the Earth’s desert-soils, or sands, have very low water contents – which is the – or at leas one, very important reason for rapid cooling after “Sun-down”
I come from “Way up North” (60° 57′ 06″ N) and in that district, when I was a youngster, the “Winter Ground Frost” used to reach down to a depth of – roughly 2 + meters (6′ 6″) and I am convinced it is the water-, and not the soil molecules that freeze solid . If the snow was gone by mid-April, the ground frost would also all be gone by mid. to late May. —
And therein lays another “We cannot mention that point” that “climate scientists” treasure, which, of course shows the importance of “Conduction” and by that I mean heat/energy transferred from the top of the surface – downward, or inwards, as well as from the surface to the Atmosphere. And it must mainly happen by evaporation from anywhere where water is present
I see that some people think N2 and O2 are incapable of emitting radiation. If that is a fact, wouldn’t that make them into very powerful GHGs?
This “what if” analysis of removing all green house gasses but keeping the earth’s albedo unchanged sure promotes discussion. It kind of reminds me of a Saturday Night Live skit they did years ago where they would ask a question such as “What would have happened if Genghis Khan had a machine gun” and had a panel discuss it. It seems that the writers for that program understood the ridiculous nature of such proposals better than climate scientists!
Myrrh says:
December 26, 2011 at 5:50 pm
Will this do?
http://www.gewex.org/BSRN/BSRN-11_presentations/Tues-wacker.pdf
@ur momisugly Ned Nikolov:
December 26, 2011 at 3:36 pm
Absolutely fascinating poster & contents Dr. Nikolov. I wholly agree with the path you and your colleagues are taking, without a doubt. Congratulations and looking forward to your paper.
There is one aspect I found was not mentioned within that poster and text that may, only possibly, help account for some of the remaining small variances between each of the bodies and their individual atmosphere chemical composition. I will drop an e-mail via Dr. Zeller if you would be willing to consider a bit of input even though it may have already been considered. — Wayne Jackson
Michael Hammer said:
“Doubling water vapour at a first approximation doubles cloud mass (either denser clouds or more of them both of which increase albedo)”
As I see it, increasing global temperature increases density of clouds. So, albedo of many of them may increase from 99.9% to 99.95% or something-like-that should global warming achieve a 10-11 degree C/K temperature rise.
Meanwhile, I see vertical convection with clouds getting more efficient at circulating heat as the oceans warm from AGW. While, need to circulate heat decreases because the Arctic warms more than the tropics when the globe warms. So, I expect warming of the globe by increase of “greenhouse gases” would result in a slight *decrease* of cloud cover.
However, I suspect IPCC and its favored scientists do over-estimating of this effect, and neglect to consider that a warmed atmosphere with more-efficient more-compact clouds has its global relative humidity decreased – partially negating the “water vapor positive feedback”.
As a result, I like to consider “real-world” combined water vapor and cloud albedo feedbacks
equal to the IPCC-favored value or the MODTRAN-derivable figure for water-vapor positive
feedback. As in, for oversimplification, use some published or available figure for the water vapor feedback, next consider cloud albedo feedback to be zero, Plug this into GCMs (global circulation models), and I hope honest scientists see what I like to see – a “less-partial snowball” Earth has global climate sensitivity decreasing as CO2 increases, towards “regulating” global
average surface temperature of 23 C (9 degree rise) if CO2 increases by a factor of 20 and
Antarctica repeats a previous latitude warmer than its current one that gets it largely covered by 2 miles of ice.
>>
Dave Springer says:
December 26, 2011 at 10:29 am
There are a zillion class M stars at all stages of their life cycles for us to look at and relatively speaking a class-M star like ours is before it becomes a red giant in its cups is one of the most constant dependable things in the universe.
<<
There may be a “zillion” class M stars, but the Sun is not one of them. I’m assuming you mean spectral class. The Sun is a main sequence, spectral class G2, dwarf star (some say G2.5). That would give it a somewhat yellow color. M class stars are red.
Jim
Hmmm…
Is Dr Nikolov’s paper above agreement with Harry Huffman’s claims/
http://www.wcrp-climate.org/conference2011/posters/C7/C7_Nikolov_M15A.pdf
http://theendofthemystery.blogspot.com/2010/11/venus-no-greenhouse-effect.html
@ur momisugly Ned Nikolov:
December 26, 2011 at 3:36 pm
This approach is different from the conventional method attempting to estimate the mean planetary temperature from the spherically AVERAGE absorbed solar radiation using the S-B law. Due to the non-linear relationship between temperature and radiative flux, and a highly non-uniform distribution of temperatures on an airless planet, one CANNOT obtain the true (arithmetic) mean temperature from an AVERAGE absorbed radiation using S-B equation!
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Yes Ned, a fellow commentator here, I believe ‘anna v’, was pressing that very point over two years ago of the impossibility to the non-linear distribution when jumping from temperatures to radiative power and vice versa, and I have never forgotten that very aspect, knowing all “consensus” climate science calculations were off by at least this factor. But you have the deep math training that is one of my weaknesses. I would have been immediately be jumping into numeric integration.
I now see your point, using your equation #2 explicit integration:
T.gb = 2/5 • √√( (1362.0 + 13.25e-05) • (1 – 0.12) / (0.955 * ‹kSB›) )
= 154.28 K
TrueGHE = 287.6 K – T.gb
= 133.32 K
ends at your starting conclusions in this aspect, I see it now. Thanks to you both to swim against the collective current, everyone does seem to know simple Stefan-Boltzmann application is not correct in this arena, but you both are the first to understandably frame it to near completeness.
I hope others here will read and follow your work (http://www.wcrp-climate.org/conference2011/posters/C7/C7_Nikolov_M15A.pdf), it is a fresh breath.