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.
[Language .. Robt]
As I was quoting Mr Springer, and you allowed his language through in the first place, I saw no reason to edit the foul language. However, upon reflection, Mr Springer’s foul mouth should be no excuse for my own, even in the circumstance where I am quoting him. My apologies, and I shall endeavour to be more circumspect in the future.
If we start with earth at 255 deg K and introduce CO2 eventually water vapour will be added and increasing CO2 to today’s level will warm the earth to qbout 288 deg C. Then start to remove CO2, is the process reversible, will all water vapour eventually drain from the atmoshere or will removal of all CO2 leave a residual amount of water vapour so that the temperature does not revert to 255 deg K? In that case the real contribution of CO2 becomes clear. Has anyone worked this out?
enneagram says:
December 26, 2011 at 4:52 am
“Remember: How soon atmosphere cools down during an eclipse.”
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Or after dark in a desert.
“Sea water also containe silt, algae and other impurities which are brought to the surface by turbulence and which are good absorbers in the IR spectrum.”
Sure they do but then they warm up the water molecules around them to speed up the rate of evaporation.
And it isn’t a surface tension issue. The ocean skin is the topmost 1 to 3 mm which is always cooler than the ocean bulk below.
Let’s call it planet Earth, forget the water, the atmosphere, the animals -us- , I still do not understand how the whole systeme works according to your calculations. If you think we should not be here, please tell us, somehow, we’ll manage.
son of mulder, there was a Schmidt et al. paper (the control knob paper) that did this (removed CO2 starting with the current atmosphere) and found that the temperature dropped about 30 degrees, and not all the water vapor goes, but most does, while the albedo goes up due to the extra sea ice that covers a large area when you cool it that much, and this helps with the positive cooling feedback too.
Richard Verney and Steven Wilde, some questions for you. I think the earth conducts a yearly experiment, which, if looked closely at, can provide some underdstanding of the different effects a world with more water has verses a world with less.
This is seen on a hemispheric bases bi-annually, as the earth’s seasonal energy pulse can reveal on a hemispheric scale some of what happens with the ocean / atmosphere on a monthly daily and hourly basis. Sunlight, falling on the Earth when it’s about 3,000,000 miles closer to the sun in January, is about 7% more intense than in July. This is close to 90 W/M2 more TSI. Because the Northern Hemisphere has more land which heats easier then water most people state that the Earth’s average temperature is about 4 degrees F higher in July than January, when in fact they should be stating that the stating that the ATMOSPHERE is 4 degrees higher in July. In January this extra SW energy is being pumped into the oceans where the “residence time” within the Earth’s ocean land and atmosphere is the longest. There are also other factors, such as the Northern hemispheres winter increase in albedo exceeds the southern hemisphere’s winter albedo due to the far larger northern hemisphere land mass.
Please consider answering my three questions in the following paragraph.
So at perihelion we have a permanent loss to space of ? W/2m SWR, due to increased NH albedo, and, in the SH a temporary loss of SWR to the atmosphere, as at perihelion the SWR is falling on far more ocean, where it is absorbed into the oceans for far longer then if that SWR fell on land. Do these balance (unlikely) or is the earth gaining or losing energy during perihelion???
The TOA seasonal flux should tell us and climate models should accurately predict the observation. Does anyone know if the climate models accurately predict what happens seasonaly on earth?
Stephen Wilde says:
December 26, 2011 at 8:09 am
“On a snowball earth most of the water vapor is frozen out of the atmosphere and the ocean is effectively removed from the equation by being covered with permanent sea ice.”
How could that happen when solar shortwave continues to be received into the oceans from equator poleward ? It is that which maintains our liquid oceans and not non condensing GHGs.
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I’m not sure how it happens, Stephen, but it does. There is no essential argument over whether the earth has experienced snowball episodes in the past. This is indisputable testimony taken from the geologic column. There is some dispute over whether the freezing was complete and perennial even at the equator but there is little dispute other than that.
What exactly triggers it is probably random catastrophe. Perhaps the equivalent of nuclear winter caused by an asteroid impact or the eruption of a supervolcano in synchronicity with a solar grand minimum and the optimal orbital configuration in the Milankovitch cycle. A perfect storm of some sort I suspect.
While I can’t say exactly how a snowball earth happens to start I cannot deny that the earth has entered into such episodes in the past. Obviously such espisodes are not quite permanent in nature so that raises the question of what snaps the earth out of it. Most people who’ve thought about it reach the same conclusion about volcanic activity.
In the meantime the snowball earth gives us a perfect opportunity to see what happens when the water cycle is, for all practical purposes, shut down by freezing temperatures over most of the globe and with it nothing left to provide a greenhouse effect but non-condensing greenhouse gases. The result is evidently a snowball for many millions of years until something happens to trigger a melt.
We would probably be in a snowball episode right now if it weren’t for solar output being a few percentage points higher than in past ice ages. Models (hate to refer to one but they aren’t all bogus) indicate that whenever continental ice sheets reach past 25 degrees latitude toward the equator the feedback from higher albedo over such a large area makes them unstoppable and results global freeze. Indisputable evidence of glaciation in the tropics well inside the 25 degree tipping point is the evidentiary basis for a complete freeze although as I said there is some controversy over whether the freeze was indeed complete and perennial.
Richard M says:
December 26, 2011 at 8:07 am
Why do you only consider radiation? The GHG-less atmosphere will receive energy by conduction. Since the atmosphere has low emissivity that heat will not be radiated away quickly as it can be with GHGs. However, once it gets warm enough it will start conducting heat back to the surface. Where is this equilibrium point? Is it 255 K and if so, why?
I’ll admit I don’t have any training in this area so maybe there is a simple explanation. I’ve never seen it though and can’t help feeling this could a positive feedback situation where the Earth would heat up.
The conductivity of air is 0.024 W/mK. That means that a cube of air with a side of length 1metre and a temperature difference of 1 degree C from top to bottom would dissipate 0.024 watts. Now the adiabatic lapse rate is 6C per kilometre so the actual temperature difference across a cube of air at the earth’s surface would be only 6 thousandths of a degree. So the actual loss by conduction is only about 120 microwatts. This is miniscule compared with radiation. Convection and evaporation are the next largest with even these being many thousand times more important than conduction. In addition you mke the statement “when it gets warm enough”. However there is no mechanism whereby the air temperature could exceed the surface temperature and so “start conducting back to the surface”
Joel Shore
Would like to fool you by saying that the Earth without greenhouse gases would still have the same albedo(i.e. 0.3)
Yet he advances no evidence to support this unphysical assumption!
This is the fiction that supports the figure of 33K as the greenhouse effect value
Using figures from an IPCC approved source such as the KT97 energy distribution diagram we can show this figure is nonsense.
The Earth without the greenhouse gases and clouds.
Examine the diagram KT97
Of the fraction 4 X 77W/m2 most would fail to be reflected by the now non existent clouds.
Of the fraction 4 X 67W/m2 absorbed by atmosphere all would pass through since there is now no absorption.
This indicates that almost all of the 1368W/m2 will reach the surface.
KT work out for us that 30W/m2 of 198W/m2 is reflected by surface giving an albedo of 0.15.
However since much more IR is now reaching the surface the figure of 0.15 is too high and a figure of 0.1 or less might be more realistic.
It appears (using the IPCC figures from KT 97) that the effect is now 16K rather than 33K
Jim D says:
December 26, 2011 at 8:36 am
“son of mulder, there was a Schmidt et al. paper (the control knob paper) that did this (removed CO2 starting with the current atmosphere) and found that the temperature dropped about 30 degrees, and not all the water vapor goes, but most does, while the albedo goes up due to the extra sea ice that covers a large area when you cool it that much, and this helps with the positive cooling feedback too.”
I have no argument with that. The earth is in an ice age right now and it evidently requires nothing more than a slight cyclical shift in orbital alignment which makes northern hemisphere winters warmer and summers cooler to kick off a process at the end of which London and New York are beneath mile of ice.
The thing of it is though that CO2’s greenhouse effect is logarithymic in nature and the first 100ppm does most of the heavy lifting thus an increase from 280ppm to 400ppm will have only a very modest effect and due to other factors at work in heat distribution the effect is felt mostly on land and mostly at higher latitudes. Given that we’re in an ice age right the f now and given that anthropogenic emission helps warm up the joint mostly where the glaciers have their footholds and not where it’s already comfortably warm all year round it’s difficult to escape the conclusion that rising atmospheric CO2 is a good thing.
This is an interesting theoretical exercise, but one which seems to have little application to anything in the real world. Without greenhouse gases, earth would of course have no oceans and no ice, and simply be a lifeless rock, much like the moon. It really wouldn’t matter what the “average” surface temperature was, as one side of the planet would be boiling hot and one side freezing cold, as there’d be no greenhouse gases or oceans to modulate that “average”. So whatever the average would be, it would swing so wildly across a 24 hour period, that the average would be quite irrelevant. Bottom line is: whatever the “average” would be without greenhouse gases, we enjoy a much warmer planet, and more importantly, one that varies less severely, because of greenhouse gases. Also, as some or apt to do, the discussion of course goes into the relative importance of each type of greenhouse gas, and everyone knows that water vapor is more potent, but that does not indicate their overall effect. Here the discussion must turn to condensing versus non-condensing gases, and the difference between water vapor and CO2. As water vapor will condense with changes in pressure and temperature, the amount varies wildly over the earth’s surface, CO2 on the other hand, remains relatively well mixed, but more importantly, it does not condense out with the temperatures found on earth. Thus, were one to suggest we remove all the CO2 from the atmosphere (in addition to killing all the plant life) we’d also see all the water vapor condense out over time as the earth continued to cool, and we’d return to a snowball earth in fairly short order.
Most people with a basic physics education can’t remember any of the basic physics….
With regard to the calculation of Earth average temperature, the equating of outgoing IR over a sphere and incoming over a disk is simply a calculation of some temperature without any useful application. What is it we are trying to illustrate? Are we trying to find the actual earth temperature in the absence of a “greenhouse effect”?
Even on a planet that rotates as rapidly as Earth, without a “greenhouse effect” and without a means to transfer heat poleward rapidly, temperature at the subsolar point is going to be quite high relative to the poles. If one assumes the subsolar point radiates all that it receives then T=T(sun)*(R(sun)/R(sun->earth))^(1/2)*(1-A)^(1/4). If Earth’s albedo (Bond’s Albedo) is 0.3, which was used to estimate the “average” surface temperature of 255K, then T=360K in the subsolar region and everywhere else is PDC (pretty damned cold), or as BenAW suggests, that the sunlit hemisphere radiates preferentially, T would be 303K, with the other hemisphere being PDC. These are very different temperatures and distributions.
On a related note, I’ve often wondered why we use average temperature rather than the M-th root of integrated temperature to the M-th power, with 4<M<5, considering the Planck function and filtering effect of the atmosphere. I mean, average temperature really doesn't correspond to anything important, does it?
My latest response to continued vain debates like this article and the comments it has provoked:
Continuing Vain Climate Debates
Dave Springer;
The emissivity of water is dependent on temperature and can be overwhelmed by the coefficient of evaporation.>>>
In physics, particularly as it applies to undestanding climate processes, we are not as concerned about what “can” happen as we are about what does happen in general over a period of time. The following image from ERBE is instructive:
http://eos.atmos.washington.edu/cgi-bin/erbe/disp.pl?olr.ann.d
As can be seen in this image, the outgoing LW radiation from the ocean surfaces over a period of one year in general exceeds that of the land masses by a considerable amount. I will leave it to you to determine if you are wrong due to the reasons I stated, or if you are wrong due to other reasons. Data trumps theory however, and your theory appears to be incorrect.
Dave Springer;
Aside from those facts of physics which you and Monckton are evidently unaware of is the fact that water is totally opaque to long wave infrared and its emissivity is limited to a surface layer just a few microns deep. The solar heated water below that depth must first be somehow transported to the surface before it can emit any radiation. Then when it reaches the surface it must compete with the evaporation coefficient.>>>
As I explained earlier, still water is opaque to IR, turbulent water is not. As for mechanisms to promote mixing, please note that rain is pretty much a daily occurance over the open ocean and the rain is in general not only a source of turbulence, but also causes mixing because it is cooler than the ocean itself. As for the warmer water below, you are neglecting conduction. If the radiance emitted was overwhelmed by evaporation, then the ERBE data would not show such high amounts of outbound LW.
Dave Springer;
The net result is reflected in many ocean heat budget studies. Approximately 70% of ocean heat loss is evaporative, 25% emissive, and 5% conductive. These are the facts. You may deny them but you cannot change them and neither can Lord Monckton.>>>
I do not deny those facts. Your contention was that escape of LW to space is negligible from the ocean surface. I’ve demonstrated that this is incorrect.
Dave Springer;
With all due respect to your expertise as an IT salesman>>>
A rather snide remark coming from someone who spent most of his career working in the computer industry designing hardware for gaming consoles and desk top computers. BTW, you may want to consider that what I did in my career for one period of time does not apply to my entire work experience, nor is it indicative of my educational background.
Dave,
I’m not convinced that there has ever been a snowball Earth. As I understand it there has always been an ice fee area either side of the equator.
So all we have is an extreme form of ice age with the ice moving down across more of each hemisphere than is usual.
Over the 4 billion year history of the Earth I am unsurprised that such things happened from time to time. The most likely cause would be large changes in the level of solar activity for whatever reason with a very negative polar oscillation spreading polar air equatorward and cool oceans failing to resist. Your ‘perfect storm’ suggestions are plausible too.
So the water cycle never did stop despite vast areas of ice cover and in due course the water cycle regained its power once more.
to springer, wilde and the author – in this vast psychodrama that was never about science, you guys have found some. and to think the entire circus could have been avoided if you’d had a say about it…lol
i think you have the correct view on how the planet is heated and refrigerated.
arthur clarke once wondered why it was called ‘earth’ when it was mostly water…
i like the meme ‘hot water bottle effect’. it might be supplemented with the ‘sauna effect’.
the greenhouse meme is precisely as useful as the notion of phlogiston – it is a useful indicator of how well a person’s mind is navigating the catastrophist diorama.
“we’d also see all the water vapor condense out over time as the earth continued to cool, and we’d return to a snowball earth in fairly short order.”
How could that ever happen with solar shortwave still entering the equatorial oceans and raising vast quantities of water vapour with or without the aid of non condensing GHGs?
Dave Springer notes: “Sunlight penetrates the ocean to some 100 meters where it is absorbed along the way by impurities.”
Why can you see the mid-ocean ridges thousands of feet below the surface as viewed from space?
“I like the meme ‘hot water bottle effect’”.
Yes, it is starting to enter the lexicon.
Original iteration here:
http://climaterealists.com/index.php?id=1487&linkbox=true&position=5
June 25th 2008.
Dave Springer says:
December 26, 2011 at 7:45 am
“What you get is a snowball earth.”
Only in a flat world with 1/4 of solar radiation.
I would be very interested to see calculation that leads to snowball earth in a rotating sphere-world with 1365 W/m2 not a 1/4 of it, including day & night. This would not let water freeze in the tropics – no way, sorry.
With ocean circulation this expands north and south so snowball earth needs additional conditions to happen, cannot be caused by missing carbon dioxide from the atmosphere.
R. Gates says:
December 26, 2011 at 9:03 am
“one side of the planet would be boiling hot and one side freezing cold, as there’d be no greenhouse gases or oceans to modulate that “average””
The earth is a sphere that is spinning no way to have one side boiling hot and one side freezing cold – what you say does not make sense R. Gates, sorry.
Bomber_the_Cat says:
December 26, 2011 at 8:35 am
David, what you should be aware of is that NO2 is not nitrogen, it is Nitrous Oxide, which is a greenhouse gas. The graphic for Oxygen is actually for Oxygen and Ozone shown on the same plot, for some reason. The little blip in the Long Wave Infrared (LWIR) is caused exclusively by the Ozone; Oxygen does not absorb nor emit significantly in the LWIR, as the graphic shows. Neither does Nitrogen, which isn’t shown at all in the graphic.
I feel a spasm of chemistry pedantry coming on. So:
Nitrous oxide is N2O, laughing gas, as used in whipped cream aerosols, and not yet banned, even though it’s a greenhouse gas. NO2 is nitrogen dioxide.
(NO, on the other hand, is Nitric oxide, responsible for the vasodilation effect of such useful drugs as trinitroglycerine and sildenafil)
Sorry. Back to the discussion.
Harry Dale Huffman says:
December 26, 2011 at 9:16 am
Harry, the problem I have with your work is that you base the whole thing on the pressure related (adiabatic) lapse rate which is of course the same for any planet such that all other things being equal the ‘temperature’ (I prefer energy content) of the planet will be related directly to atmospheric pressure at the surface and distance from the sun.
The trouble is that the actual (so called ‘environmental’) lapse rate is very different from planet to planet so Venus CANNOT be directly compared to Earth.
On Earth we have a fully functioning and highly effective water cycle which Venus does not have so the outturn on each planet is NOT comparable.
Earth has a powerful Hot Water Bottle Effect and a very weak Greenhouse Effect. Venus has a very powerful Greenhouse Effect and NO Hot Water Bottle Efffect.
Chalk and cheese come to mind.
The albedo of the earth without clouds and ice is 0.07. This gives an effective no-atmosphere temperature of 273K or ~0°C. However, have a non-GHG atmosphere with aerosols and you will have lapse rate cooling. In that case the present GHG warming is ~9K.
So, IPCC climate science overestimates CO2 climate sensitivity by at least a factor of 33/9 = 3.66. But there’s also the -0.7W/m^2 ‘cloud albedo effect’ cooling in AR4. Being a figment of Sagan’s incorrect aerosol physics, it doesn’t exist and could be warming. So reduce CO2 climate sensitivity by a further 44%.
The maximum CO2 climate sensitivity is 0.45. in reality, it’s probably much smaller because the IR physics is wrong as well, not taking account of self-absorption at IR band saturation.
The other thing wrong with the IPCC farrago is no ‘back radiation’, it’s really Prevost Exchange Energy so can’t do thermodynamic work. All in all, this is probably the biggest scientific fraud in History: time it was ended.
@ur momisugly wilde:
good job. now if you only had willis’ awesome jimmy.buffetness… nah- you’ll do without… lol