Click images for more details



Recent comments
Recent posts
Currently discussing

A few sites I've stumbled across recently....

Powered by Squarespace

Discussion > Numerical calculation of Planetary Black Body Temperature

TBYJ / Rhoda / Paul Dennis
Thanks, guys. You've done just what I was asking which is at least to take N&Z (and others who dispute the CO2 effect) seriously.
One more question, however. How relevant are averages? Postma (or somebody) has been arguing along the lines that the average insolation applied equally across the planet all the time can never raise the temperature sufficiently to account for the actual global mean which is why the IPCC needed to invent back radiation.
The example I heard quoted was what happens if you light a gas ring and then put the kettle on the kitchen table to boil! The average temperature will obviously never rise high enough to boil the kettle.
Some form of "point source" (eg the sun during the day) which concentrates radiation is necessary.
Does this make sense? I can see how it could but I can also detect an argument against it.

Jan 8, 2013 at 10:55 AM | Registered CommenterMike Jackson

It's the other way round Mike... 'average temp' doesn't really exist anywhere on the earth (except int he sense that a stopped clock is right twice a day - some places may accidentally be at this average), but in order to get a handle if it's rising or falling they aggregate world temperatures into a single figure which you can plot on a graph and scare people with hockey sticks.

It not being real doesn't mean it doesn't matter, though. It means that some temperatures that make up the average have gone up. Some areas may have stayed the same (e.g. UK temps over the last 40 years) but other areas must then have gone up even faster to weigh the average up.

Jan 8, 2013 at 11:14 AM | Unregistered CommenterTheBigYinJames


very briefly as I don't have much time for today. The Diviner data that I've seen has a modelled equatorial average temperature of 240K (from memory). This is less than my calculation. However, I used a constant albedo of 0.12 and constant emissivity in my calculation. I note from the Araveda 2012 paper that the albedo is actually a function of the incidence angle such that it increases when the sun is not directly overhead. They also note that there is evidence that the emissivity may also anisotropic so that could also be important. The effect of the variable albedo is to reduce the calculated mean temperature such that it will be closer to that measured estimated by the Diviner instrument. I haven't tried to incorporate this into my model but it wouldn't be difficult to do so. Perhaps TBYJ could have a look at this.

Jan 8, 2013 at 11:24 AM | Unregistered CommenterPaul Dennis

Sorry, I got the threads mixed up.

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.

(note from other thread (from RKS) rhoda and N&Z now agree almost exactly).

Jan 8, 2013 at 11:52 AM | Unregistered CommenterRoger Longstaff

Roger, I'm not done refining the model yet, so my figure isn't final. I'm not completely convinced that point density reduces as a plain sine (or cosine depending on the coordinate system) of latitude, It may well under-represent the poles, which would make the average higher than it should. I'm going to experiment with varying the number of points rather than their weighting.

Paul, I'd love to incorporate varying lunar albedo, if you have the relationship formulae to hand.

Jan 8, 2013 at 12:57 PM | Unregistered CommenterTheBigYinJames

Roger and TBYJ,

check your personal email in boxes. Hopefully you'll find something of interest.



Jan 8, 2013 at 1:11 PM | Unregistered CommenterPaul Dennis


look out for equation 1 in Varavada et al 2012. It gives the change of albedo with time from dawn to dusk.

Jan 8, 2013 at 1:13 PM | Unregistered CommenterPaul Dennis

Albedo? You're talking about the fourth root of 0.88, and it won't make much difference. I have got their 213K (exactly 213.83) by modelling the hot side as stationary and using the diviner temps for sunset and sunrise. I have to admit that to get 213 is actually a fluke, but taking estimations into account I am somewhere around there anyway. Anyhow, only the back side temps make a difference, they give 50K increase to the average just by rotating once a month.

An observation. all this is fun, but it really does not help the Earth case, because it is so different in so many ways as to be impossible to apply one to the other. Nowhere on earth gets anywhere near its SB temp, or at least nowhere livable. So although 213K is 75K short of earth's (reputed) mean, most of that shortfall may be due to day length alone. Never mind trying to calculate a local albedo formula for this planet.

Jan 8, 2013 at 1:28 PM | Registered Commenterrhoda

rhoda, I think that the point is that N&Z relate the calculation of a planet's mean temperature to their pressure dominated theory, and the question was, as the Earth and the Moon have the same constitution and insolation (but nor rotation), are the IPCC calcs for the average grey body temperatures correct? So this is simply checking the first stage of their thesis.

Jan 8, 2013 at 1:57 PM | Unregistered CommenterRoger Longstaff


the albedo varies from 0.62 at dawn and dusk, through to 0.12 at noon. These seem like major changes to me and should have an effect on the mean diurnal incident energy received. I've just incorporated the Varaveda model for albedo changes and calculated an equatorial mean temperature of 264K. Without the albedo change the mean equatorial temperature was calculated at 286K.

Jan 8, 2013 at 1:58 PM | Unregistered CommenterPaul Dennis

Here is a very interesting observation. Apollo 15 landed near the Hadley Rille at 26 degrees N. The temperature profile at this site has a constant subsurface temperature of 230K. Vasavada et al 1999 modelled this subsurface temperature as 250K. I'm not worried by the difference in these two estimates. They are however very much larger than the mean temperatures that some are suggesting.

My lunchtime reading was also Gong and Jin, 2012, 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

These authors show a latitudinal profile from S to N poles of the brightness temperature at 37GhZ (Tb(37)) which seems to equate with the near constant subsurface regolith temperature. This profile ranges from 275K at the equator to less than 100K at the poles. I'd hazard a guess that the area weighted mean is about 250K.

There's a lot of data out there if one looks. I really can't afford any more time now to keep on looking!!!

Jan 8, 2013 at 2:32 PM | Unregistered CommenterPaul Dennis

I've made some changes to my model, which I'll describe in a bit, but it brings my figures almost the same and N&Z, which I find a bit troublesome.

First let me explain how my numerical model works, and why I changed it. This is going to be in-depth, so apologies if you already know how I did this.

Whenever you do numerical modelling, the idea is you split the physical problem up into small parts, each of which you can apply the general equations to with their localised conditions, then you can sum and average up all the little bits to give you the overall macro effect.

In the case of this problem, it means splitting up a sphere into lots of little areas, then working out the insolation at that point and thus the local temperature. These temperatures can then be averaged to give the average temp of the whole sphere.

My first attempt simply divided the 180 degrees of the visible disc into so many parts (I think i ran with 1000 to start with, but put it up to 10,000 later). I divided the sphere top to bottom and then left to right, giving thousands of little areas to work out the local temp for.

For each area I looked at the angle it had with the sun (instead of degrees 0-180 I used radians, which measure the same range as 0 to pi, but it all means the same thing). This is worked out by using Sine of the angle - examples:

At the top of the world, the angle is 0. Sin(0) = 0, so multipled by the Solar Flux = 0 W/m^2, giving a local temperature of... 0. ( I actually add in the cosmic background of 2.7k to this)

At the bottom of the world, the angle is pi (or 180). Sin(pi) = 0, so multipled by the Solar Flux = 0 W/m^2, giving a local temperature of... 0.

At the equator, the angle is pi/2 (or 90). Sin(pi/2) = 1 so multiplied by the solar Flux less albedo = 1000 W/m^2 at the equator, giving whatever that temp is

All the angles in between will give somewhere between 0 and solar flux.

I repeat the calculation with the left-right angle, since your longitude from right on the edge to the centre has the same distributions of flux as the up down.

So the final version of the insolation = 1000 x sin(up-down angle) x sin(left-right angle), from which a local temperature at that point can be calculated.

As I worked out out temp, I added it to a running total, then divided it by the number of points I'd looked at (1000x1000, or 10,000x10,000). This appeared to give very low temperatures, which were similar to N&Z, which was where the noddy model was when I went home last night.

On the journey home, I realized that this calculation was flawed, and that the polar regions were being over-represented. Here's why.

At the equator, I was dividing up the left-right portion by 1000 points. Imagine a ring around the equator with 1000 points spaced equally around it. At the poles I was still dividing up the left-right portion by 1000 points, but the distance around that ring near the pole was much shorter - imagine a ring around the north pole, maybe only 50km or 100km long but still with that same 1000 points, so the 1000 points were much closer together.

This meant the points were all buching up near the poles, and pulling down the average temperature by giving the low polar temps more prominance within the average.

My first thought was to weight the measurements in order to counteract this effect. If instead of giving each measurement a count of 1, I gave them a count of sin(up-down-angle) this would mean that the greater numbers of points near the poles would not impact the total, since we'd only be adding a value of less than 1 to the count - the ones at the pole would hardly be contributing at all, to counteract the sheer number of them)

When I made the count variable increment by sin(angle) instead of 1, this gave the much higher readings of around 215K for earth and 228K for the moon. Which is where it stood overnight.

Today, as I hinted, I wasn't so happy about this sin(angle) weighting factor, since I wasn't sure it was actually doing what I hoped - just because the sin function reduces from 1 to 0 from equator to pole, doesn't mean it actually reduces at the correct rate to counteract the point density. It seemed 'right' but I was suspicious of it because I hadn't proved it from first principles. It was possible, as I mentioned, that the profile was unjustly penalising the poles this time, so the average was being pulled up by hotter equatorial temperatures, which is just as bad.

So I had the idea that instead of weighting each point differently, I'd made sure there were just fewer points as you got nearer the pole. So the top to bottom would still be 1000, or 10,000, but the left to right would be a fraction of that depending on the up-down angle. So it would be 1000 at the equator, but as you moved up, less and less, until right at the pole you only needed 1 point.

To make sure I understood the exact profile I had to make sure the number of points on each ring varied proportionately with the length of that ring. It's simple to work out the circumference of the
ring at any point ont the up down angle - it's 2 pi X, where X = R sin(theta). R being the radius of the earth.

The length of the ring at the equator angle pi/2 (or 90) is 2 pi R sin (pi/2) - sin pi/2 is 1, so L1 = 2 pi R.
The length of the ring at an any other angle = 2 pi R sin (angle)

So the ratio of circumferences = 2 pi R sin (angle) / 2 pi R = sin(angle)

So for every angle up and down, I just had to multiply the 1000 (or 10,000) by sin(angle) to get the number of points which should exist on that line to keep them spaced equally and the numbers of them to retain their density.

So I made the changes to the code:

      static void CalcTempPolarWeightingNumberRatio(double TotalSolar, double Albedo, double Emissivity, string name)
         double allT = 0;
         double count = 0;
         for (double lat = 0f; lat <= Math.PI; lat = lat + (Math.PI / steps))
            int innercount = Convert.ToInt32(steps * Math.Sin(lat)) + 1;
            for (double lon = 0f; lon <= Math.PI; lon = lon + (Math.PI / innercount))
               //lit side
               double T = backgroundTemp + Math.Pow(((1 - Albedo) * TotalSolar * Math.Sin(lon) * Math.Sin(lat)) / (Emissivity * SBconstant), 0.25);
               allT += T;
               count += 1;
               // dark side
               allT += backgroundTemp;
               count += 1;
         Console.WriteLine("{0} : {1:f1} K", name, allT / count);

As you can see, every point gets an equal count rating of 1, but the inner loop, the one which does the left right sweep, only does a fraction of the step count according to sin(angle). I had to add 1 to this because right at the pole, sin(angle) was giving 0, and you can't divide an arc into 0 points!

I stepped through this and stopped it at a few points, and it indeed was giving fewer points the closer to the poles, and it seemed to be in propertion to the circumference of that circle of latitude.

The results were surprising, to say the least:

Average Grey Body Temperatures
Earth : 148.7 K
Moon : 157.3 K

With the N&Z number coming out for the moon at 154K, this has me worried. Coincidence? have I made a grievous mistake? Perhaps Paul Dennis can see what I've done.

Jan 8, 2013 at 2:52 PM | Unregistered CommenterTheBigYinJames

I get a hotside average of 320K, cold side 108K (diviner) average whole equator 213.83. I'll put in that albedo, what's is the formula? Or I can apply the same 8% reduction 264 / 286 and get 295 hotside and 202K average.

Jan 8, 2013 at 2:55 PM | Registered Commenterrhoda


what you have done, and I think N&Z possibly, is to consider the spehre doesn't rotate. i.e. there is no incident energy on the far side. So you have a vast area of the sphere (2*Pi*r^2) which is at 2.7K, and an area on the near side which is much hotter.

The moon and Earth do spin so the energy is averaged over the whole surface. The result is you get a far higher average temperature. The reason for this is because T scales as the 1/4 power of the radiated energy. If you spread the energy over a much wider surface area then the temperature reduces, but is effectively spread out over a wider area thus ensuring a higher average temperature. In your model (ignoring the CBR temperature) you are suggesting that all the energy is irradiated from the near side and none from the far side.

The fact that the moon and Earth have surfaces that are thermally conductive ensures that the incident energy is more evenly averaged across the surface. Energy is stored in the sub-surface and then released at night. For the moon which has a long diurnal cycle the depth of heat penetration is on the order of 50cm!

Jan 8, 2013 at 3:12 PM | Unregistered CommenterPaul Dennis

Yes, my sphere doesn't rotate, but I'm thinking specifically of the N&Z criticism that even with this simple model, the IPCC figure of 33K is not right because if you integrate over the speherical surface you get a much larger figure. Was the IPCC model they are critical of not simply a non-rotating grey body too?

Jan 8, 2013 at 3:21 PM | Unregistered CommenterTheBigYinJames

Embarrassing error, emissivity on the wrong side. Now looking at 302K hotside 205K average, with crude albedo correction of 8%. And by a very dubious method applied it to a sphere and got average 192K, not corrected for albedo variation, in which case 182K. Final answer? I doubt it.

Jan 8, 2013 at 3:28 PM | Registered Commenterrhoda


Just for interest, what happens if you force all of the dark side to 100K (a rough average from Diviner)? This should (very crudely) approximate for the thermal conductance that Paul mentions. What would the new average be? (I'm guessing it may come out at about 197K, like N&Z).

I am still worried about the definition of "average" or "mean". I like your method of weighting the integration that you explained (if I understood it correctly). What does anybody think it means on the Diviner graph (equatorial mean = 213K)?

Jan 8, 2013 at 3:29 PM | Unregistered CommenterRoger Longstaff

The IPCC model is that of a rotating sphere. The incident solar energy is given by pi*r^2*(1 - alpha)*S
where S is 1366 W/m^-2.

This is the solar energy averaged on a diurnal cycle over the whole sphere ad so it is implicit that the sphere is rotating.

Jan 8, 2013 at 3:30 PM | Unregistered CommenterPaul Dennis


I think they have simply averaged all the temperature measurements on the equatorial mean. i.e. it is an arithmetic mean, or the average. I don't think, however, that this represents the mean, say averaged over the whole year. My reason for saying this is that the temperature profiles shown in the 2012 paper (figure 7 from memory) show that the mean at depth is 240K. Their average temperature profile implies there is still some excess of heat in the system, perhaps caused by summer insolation. The moon does have very weak seasons.

If you force the dark side to 100K, then you have to make compensations on the near side. TBYJ's calculations don't allow for any heat storage.

Jan 8, 2013 at 3:35 PM | Unregistered CommenterPaul Dennis


the energy for that 100K retained heat has to come out of the daylight budget, so it isn't very fair to introduce this into the code without subtracting it from the daytime flux, but for curiosity sake, I ran it anyway with the darkside at 100K, and it comes out at 205K for the moon.

Jan 8, 2013 at 3:36 PM | Unregistered CommenterTheBigYinJames

Ah, well I think we've almost identified the differences between N&Z and the IPCC.

My figures now match N&Z because they don't assume rotation, and neither do I.

Now I will model the average diurnal insolation, and see where I get to.

Jan 8, 2013 at 3:38 PM | Unregistered CommenterTheBigYinJames

An intriguing idea:

Did N&Z mistake that 33K as the result of a simplistic averaging of solar flux over a disk, whereas the 33K comes from a full spherical integration of insolation averaged over a diurnal cycle..,, which just happens to give you exactly the same answer?

Jan 8, 2013 at 3:52 PM | Unregistered CommenterTheBigYinJames

BigYin, I think you've cracked it! I vote you win the prize (what was the prize BTW?). Was that the final version of your code that you gave above, if not, can you share it please (including the 100K forcing for the dark side).

Paul, I completely agree with your comments, but intuition tells me that with the top 2 cm of powder being a good insulator (from the Varavada paper?) the effect on temperature during insolation would be minimal. I may be wrong about this, need to check.

Jan 8, 2013 at 3:53 PM | Unregistered CommenterRoger Longstaff

"which just happens to give you exactly the same answer?"

Sorry, I have lost the plot here. Wasn't N&Z's point that given 197K calculated for the Moon, Earth's GHE was more like 90K than the IPCC figure of 33K ?

Jan 8, 2013 at 3:58 PM | Unregistered CommenterRoger Longstaff


I haven't written the code yet, so I might be way off, but something along the lines of...

What N&Z thought the IPCC had done:

Solar Flux - albedo, model as a flat disc - come out with a simplistic average temperature = 33K GHE

What the IPCC actually did

Solar flux - albedo, integrated spherically (according to N&Z) but averaged over one diurnal cycle, now hot side is cooler but cold side is warmer, thus higher average = 33K GHE

And N&Z sought to prove them wrong by doing:

Solar flux - albedo, integrated spherically, but assuming non rotation (as I did) = 150K GHE

As I said, i've not done the code yet for the case of smearing the insolation out over 24 hours, but if it comes out anywhere near 255K for the earth, then it might be an explanation. We could then ask them. but let me do it first.

Jan 8, 2013 at 4:05 PM | Unregistered CommenterTheBigYinJames