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Discussion > An experimental demo of GHE.

Martin: It may be a Bourbaki thing. One never quite knows.

Jan 8, 2013 at 9:58 AM | Registered CommenterRichard Drake

Genuine question (wrt the rotation discussion) do we have a definition of the average surface temperature of a planet? I note that N&Z now talk about the "mean surface temperature", and I am getting confused - arithmetic, geometric...?

Jan 8, 2013 at 10:44 AM | Unregistered CommenterRoger Longstaff

I'm going to get shot for goofing off work, but this is too interesting!

So far, average surface temp of moon:

Paul: 266K
BigYin: 228K
Diviner (equatorial only): 213K
Rhoda: 198K
N&Z: 154K

Nove, Diviner chart (from: is measured from orbit and is for the equator only. I assume the mean is the temperature that has the same integrated area above it as below it? Note that temperatures would be lower at higher (or lower) latitudes, therefore the planetary average must be less than 213K.

Note also that N&Z are refining their calculation to include the specific heat and thermal conductivity of thr regolith - something that will raise their result. The Diviner data clearly show the nightime surface temperature gradually cooling, so this will give a good approximation for regoloth thermal properties (along with Apollo samples).

Paul - have you only considered the sunlit hemisphere of the Moon - what did you assume for the other hemisphere?

So far, rhoda seems to be closest. Apologies to anybody I have misrepresented here.

Definitely, a three pipe problem!

Jan 8, 2013 at 11:31 AM | Unregistered CommenterRoger Longstaff

Jan 8, 2013 at 11:31 AM | Roger Longstaff>>>

N&Z make it 197.3k not 154K so Rhoda's almost spot on:-

Long story short - the Moon is the perfect airless equivalent of Earth, which means that the thermal effect of Earth's atmosphere (i.e. the Greenhouse Effect) must be evaluated against Moon's mean surface temperature. Hence, Earth's GE = 287.6K - 197.3K = 90.3K.

- Ned "

Jan 8, 2013 at 11:38 AM | Registered CommenterRKS

"Rhoda's almost spot on"

Let's hear it for the Oxfordshire housewife! Rhoda, do tell, how the hell did you do it?

Jan 8, 2013 at 11:44 AM | Unregistered CommenterRoger Longstaff

BigYin - just to say I am not after you - just looking for answers like you. Its more than 30 years since I studied physics, which I excelled at but then lost interest in (loved relativity but I could never get my head round quantum physics - God rolling dice and all that). The 500W figure was rounded down - as I said in an earlier comment on page 17 (and apologised for as the source was Wikipedia) the IR component of solar insolence at ground level is 527W, visible light is 445W, and UV 32W, when the Sun is at it's zenith. However, I have just noticed that Wiki contradicts this on the by stating "most of the power is in the visible light portion of the spectrum". Either way, the IR component can't be insignificant. I would just like to know what the magnitude of the IR component at the TOA and at sea level, ideally zenith and average. Because if increased CO2 are reflecting more outward LW back to Earth, why isn't it also reflecting more incoming IR out to space during daytime? I am sure others have asked this question, but I have never seen a comprehensive answer. Incidentally the maximum outward radiation on a typical night seems to be about 75W/m2 and daytime maximum about 100-150W/m2 (from Colorado Uni course work data I found on the web). This is when air and surface temps are about 15C, but I am not sure now they measure the outward flux, and would like to know more.

p.s. (and you are not going to like this) the Sun is not as bright when it is lower in the sky in winter or sunrise/sunset - increased path length means more atmospheric haze, which tends to scatter the shorter wavelengths more than the IR, though this is primarily due to particulates. (I think water vapour is a factor also, especially in a climate like ours).

Jan 8, 2013 at 11:46 AM | Registered Commenterlapogus


the poster I have that details the N&Z work gives a mean temperature of 154K. As I said it is very difficult to determine how they came to this temperature from the paper at tallbloke's site.

RKS and Roger the Diviner estimate for mean surface temperature at the equator is 240K and not 213K as indicated in the plot at tallbloke's site. One needs to read the Araveda paper and the thermal modelling. The temperature at the surface through which there is no energy flux represents the mean diurnal/annual temperature. According to Vasavada 2012 this temperature is 240K (reading from their graph in Figure 7). They plot an average surface temperature of 213K which is lower than the 240K. This implies to me that the surface is not in equilibrium and that there is a net energy flux from the constant temperature layer. Perhaps this is a seasonal effect?

Jan 8, 2013 at 11:49 AM | Unregistered CommenterPaul Dennis

Paul, the Vasavade paper is behind a paywall. Can you give us any further info?

Jan 8, 2013 at 12:00 PM | Unregistered CommenterRoger Longstaff


okay. I'll see if I can get hold of a copy for you. There is also another very interesting paper based on temperature measurements made by the Chinese and their orbiting satellite. This one isn't behind a paywall.

Microwave brightness temperature of cratered lunar surface and inversions of the physical temperature profile
and thickness of regolith layer - RADIO SCIENCE, VOL. 47, RS1012, doi:10.1029/2011RS004791, 2012

I'll see if I can get you a copy of this too. Skiming this second paper it looks like the average temperatures are higher than N&Z postulate but that is just a first impression!

I said I wouldn't get involved today....oh dear!

Jan 8, 2013 at 12:11 PM | Unregistered CommenterPaul Dennis

Ha, they all say that!

Jan 8, 2013 at 12:39 PM | Registered CommenterRichard Drake

Alas, I WAS spot on, but now I am tending to a lower figure. By a dodgy method I am not going to trouble you with, my average temp for the equator is 168K, lit and unlit sides, no conduction, albedo 1, emissivity 1. Which means the whole sphere ought to be lower again. This is of course the nonrotating case which I reckon makes for the coldest average. I am thinking now that the rotation rate must make a difference, even one revolution per lunar month. The temp at east and west points from the Diviner would be nice to know. You just have to flog away at this, there is no real value in that quarter of the surface area disc of sunlight versus a whole sphere of outbound radiation. That is possibly true for very fast rotation rates. Away to see if I can add a third dimension to my disc.

Jan 8, 2013 at 12:46 PM | Registered Commenterrhoda

Martin, for your cube world example, shouldn't you be averaging the radiated energy from each face and then calculating the temperature from that, rather than calculating temperatures from the radiated energy of each face and averaging the result?

This is discussed at SoD

Jan 8, 2013 at 1:09 PM | Unregistered CommenterBitBucket

Ok, I found the lunar temps to be around 120K at sunset through to 93K at dawn. Using equator only and cold side mean 108K, albedo and emissivity as for moon, I am getting 213K. Note here that the hottest point does pretty much reach its SB temp but the cold side keeps quite warm compared to the non-rotating body. Am I wrong to think that puts paid to the argument that rotation rate doesn't matter?

Jan 8, 2013 at 1:13 PM | Registered Commenterrhoda


what you are observing is a changing distribution of the energy flux throughout the day. Starting at dawn as the lunar surface receives sunlight it starts to warm. Some of the incident energy goes into heating the subsurface layers down to a depth of some 40cm. The heating continues until midday, but the flux of energy into the interior reduces as a proportion of the incident flux such that the surface approaches the grey body S-B temperature. As the moon rotates beyond noon the surface temperature cools and the stored energy in the subsurface is slowly released as the subsurface cools too. This is why at night the surface temperature is much greater than 2K. It's also why I say one should be considering the temperature at the depth which remains at a constant temperature and doesn't suffer diurnal fluctuations as a guide to the average Grey-Body temperature.

Jan 8, 2013 at 1:27 PM | Unregistered CommenterPaul Dennis

Martin, for your cube world example, shouldn't you be averaging the radiated energy from each face and then calculating the temperature from that, rather than calculating temperatures from the radiated energy of each face and averaging the result?

This is discussed at SoD
Jan 8, 2013 at 1:09 PM BitBucket

BB - I'll be interested to look at the SOD page.

To rephrase what I think you are saying, as a check I have interpreted your words right:

1. Calculate the energy radiated by each face (the power in fact - the rate at which energy is radiated)
2. Sum the power radiated for all six faces.
3. Calculate the temperature for a planet at uniform temperature that radiates the same total power.

I hope this captures what you had in mind. Does it?

When you say "shouldn't you do xyz?" the answer always depends on what you are wanting to achieve.

What I was trying to do is to calculate the average temperature over the whole surface of the planet.

That is the equivalent of covering the entire planet with a closely spaced grid of equally spaced thermometers, adding up the temperatures as measured by all the thermometers and finally dividing the resulting sum by the total number of thermometers.

My understanding is that when climate scientists talk about "global average temperature", this is the measure they are trying to estimate. This notion has always bothered me because I can't see that it actually means very much. You can't calculate anything that has physical meaning with it.

If this is the measure you want to calculate, then I think my method is correct.

If, as you suggest, you were to average the radiated energies (or, much the same thing, find the total radiated energy) and then calculate the temperature that would result in that total energy being radiated, then I think you finish with a value that actually does mean something. It's the temperature a planet at uniform temperature would need to be at to radiate the same power as the one you are studying.

Thus I think that your method gives something meaningful but something different from what I was wanting to calculate.
I imagine that you are familiar with calculations of voltages and currents. It's vaguely analogous to calculating the d.c. value of a current waveform (its average value) or calculating its r.m.s. value which tells you its heating value. But the analogy is only a vague one. In particular, the d.c. value actually is useful and has physical meaning.

Jan 8, 2013 at 5:58 PM | Registered CommenterMartin A

"mean surface temperature", and I am getting confused - arithmetic, geometric...?
Jan 8, 2013 at 10:44 AM Roger Longstaff

If someone talks about "mean" without qualification I think it is normal to assume they are talking about the arithmetic mean or, if you prefer, the temperature integrated over the whole surface and then divided by the total area.

Extending what BB suggested, it would make sense to talk about the "fourth root of the mean of the fourth power" of the temperature, but I doubt that anyone actually estimates that measure.

Jan 8, 2013 at 6:06 PM | Registered CommenterMartin A

Thanks Martin.

BTW we could use your advice on the numerical integration thread.

Jan 8, 2013 at 7:56 PM | Unregistered CommenterRoger Longstaff

Martin, I would never claim anything to be "my method". I was just reporting what I understood from SoD. Your 5.58PM sums that up well. I agree that it seems a more valuable meaningful (not that my opinion is meaningful). BTW a similar discussion is now active in Numerical calculation of Planetary Black Body Temperature. I thought you were not around so I commented there too.

Jan 8, 2013 at 8:00 PM | Unregistered CommenterBitBucket

...that was supposed to say, "I agree that it seems a more valuable or meaningful method"

Jan 8, 2013 at 8:31 PM | Unregistered CommenterBitBucket

From by own, and Perhaps Mike Jackson's and Roger Lonstaff's, point of view I think the aim of bringing the work of N&Z into the discussion on GHE has proved it's worth.

The fact that N&Z collaborated closely with members of the NASA JPL team responsible for commissioning the Diviner instrument, who I imagine know the best way to interpret the measured results, when reaching their calculated Lunar temperature of 197.3K [NOT the previous figure of 154K from their earlier work when fewer data were available, and which seems to have caused some confusion in the thread discussions] is enough for me to accept their figures till peer review determines otherwise.

We now seem to accept that the 'accepted' 33K attributed to GHE is nonsense, and therefore the best guessed contribution of CO2 [and water vapour] based on this 33K nonsense is also wrong. So what is the REAL cause of the atmospheric greenhouse effect ,NOT radiative as radiation alone cannot explain such a large rise in temperature.

If one wishes to dismiss the contribution of Ideal Gas Laws, As proposed by N&Z and others and which was accepted for planetary warming prior the the 'radiative' GHE fad of the 70's] then a serious reworking of radiative GHE is need urgently.

Rhoda has set up the thread to explore experimental ways to measure [radiative?] GHE without much response so far, but at least an acceptance of the much larger ACTUAL contribution of GHE is another stepping stone on the way to defining a more realistic theory of climate.

Jan 9, 2013 at 12:30 PM | Registered CommenterRKS

And I don't accept "33K is nonsense" on this thread either :)

Jan 9, 2013 at 12:32 PM | Unregistered CommenterTheBigYinJames

If one looks through all the relevant Tallbloke threads as they evolve, Ned Nikolov alters his figure of 154K for mean Lunar temperature to 197K as improved data from Lunar Diviner become available. In fact I quote earlier from an email from him that the temperature, arrived in conjunction with NASA JPL scientists responsible for the Diviner instrument [I presume they have looked at the correct way to interpret the data], is 197.3K. To use the earlier figure could be confusing.

Jan 9, 2013 at 12:42 PM | Registered CommenterRKS

And I don't accept "33K is nonsense" on this thread either :)

Jan 9, 2013 at 12:32 PM | TheBigYinJames>>>>>

You don't accept that your own calculations for Lunar mean temperature don't apply to the bare, atmosphere less, Earth which is known to be made of the same material as the Moon and to share the same insolation. This is the 'bare Earth' temperature 'estimated' and used by the IPCC as the basis for the 33K contribution of GHE.

Jan 9, 2013 at 12:50 PM | Registered CommenterRKS

No, my own calculations, integrating the diurnal insolation over the sphere in the way N&Z recommend, showed that the GHE was around 50K for earth. That's close enough for cash, for me. 33K is the SB grey-body temperature for an airless earth. I can only get my model to approximate N&Z if I assume the earth is not rotating and that half of the surface is at 2K cosmic background.

Point of note:

Earth and Moon don't share the same insolation, earth has a larger albedo due to clouds (not part of the GHE, so it was wrong for N&Z to remove them, work out the difference, then attribute the whole larger effect to GHE alone)

Jan 9, 2013 at 12:58 PM | Unregistered CommenterTheBigYinJames

Earth and Moon don't share the same insolation, earth has a larger albedo due to clouds (not part of the GHE, so it was wrong for N&Z to remove them, work out the difference, then attribute the whole larger effect to GHE alone)

Jan 9, 2013 at 12:58 PM | TheBigYinJames>>>>

To allow for GHE you must first calculate the mean temperature of Earth without ANY atmosphere, which includes everything to do with atmosphere such as clouds and oceans which, of course, cannot exist without atmospheric pressure.

We are then left with identical insolation as the Moon, prior to the addition of atmosphere, and by which the mean temperature of an airless Earth, prior to the effects of GHE, must be determined.

You cannot determine the conditions of an airless Earth using the same albedo etc as if it had an atmosphere and oceans.

Jan 9, 2013 at 1:19 PM | Registered CommenterRKS