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Discussion > Simulation of Temperature

I've found an hour to start work on the simulation I spoke with Roger about on the numerical calculation thread.

The idea is to put 1000, or 10000 points on a globe, insolate them with solar energy, rotate the points around the planet, allow them to heat and cool according to heat energy and IR laws.

I've managed to get one point modelled (once one is done, I can just spread copies around the globe by setting lat+long). My assumptions are for a sandy surface, Specific Heat of 0.8 kJ/kg K heating a mass 1m x 1m x 50cm deep with a mass of 50kg (sand is roughly 100kg/m^3)

As a teaser of how it's going, here is the temperature graph of a point on the equator. Green is while in daylight, red in darkness. I've discovered that starting conditions are not very important, the max temp is governed by the solar insolation, so it settles down quickly.

You can see at dawn, the sand heats up until it hits thermal equilibrium with the sun late morning. It then keeps in check with the increasing insolation over noon and afternoon. At night, it cools slowly.
The shape, if you compressed it horizontally, looks a little like the Lunar Diviner graph, don't you think?

Jan 9, 2013 at 3:31 PM | Unregistered CommenterTheBigYinJames

Once you get it working as you like, apply a correction for GHE direct to the ideal planet. No water, no wind, just a percentage of outgoing being reflected by a magic mirror. I wonder what that will do.

Jan 9, 2013 at 4:47 PM | Registered Commenterrhoda

I tried my moon model with an extra 25% insolation, similar to normally quoted all-source back-radiation.

I got 17K more hotside, I left the cold side alone as the temp falls off so quickly at dusk. My average therefore rose by 8K. Is the approach too naive?

Jan 9, 2013 at 5:02 PM | Registered Commenterrhoda

It's already been a lot trickier than I thought.... the ground emits during the day depending on its temperature, especially so in the evening, which is not surprising, but not something obvious from the simple SB calculations on the other thread.

Possibly, it seems that the heat capacity of the rock does slow down the cooling a bit, even when exposed to 2K background. So the backside is warmer longer than a simple SB model would say.

Jan 9, 2013 at 5:07 PM | Unregistered CommenterTheBigYinJames

Good grief - at it again! You're hooked - admit it.

A Thought - the Diviner plot doesn't change from one function to another (before and after equilibrium) after sunrise. I guess this is because the top layer of powder, in a vacuum, has a much lower conductivity than the stuff below, and can therefore sustain a large temperature drop, with the surface coming very rapidly to SB equilibrium temperature. Have a look at the paper that Paul sent to us - I think it gives figures for this (also based on Apollo samples).

I guess that if you drastically reduced the specific heat your curve would look more like Diviner. Good luck!

Jan 9, 2013 at 7:39 PM | Unregistered CommenterRoger Longstaff

I do boring programming all day long, nice to model something interesting :)

Jan 9, 2013 at 7:42 PM | Unregistered CommenterTheBigYinJames

Further suggestion - vary the specific heat until you get the same nightime temperature drop as Diviner (about 25K from memory).

Oh God - you've got me at again. I am now writing a requirement spec to hand to a programmer.

Jan 9, 2013 at 7:49 PM | Unregistered CommenterRoger Longstaff

Just put the spec in a file called ROGER_README.TXT

Jan 9, 2013 at 7:51 PM | Unregistered CommenterTheBigYinJames

Hi James, good to see all your sums here and on the other threads. I'd advise changing your density for sand though. One cubic metre weighs much more than a hundred kg it is more like 1500 kg, I think. Again good luck with your efforts.

Jan 9, 2013 at 8:09 PM | Registered CommenterJeremy Harvey

Good one!

Jan 9, 2013 at 8:10 PM | Unregistered CommenterRoger Longstaff

Thanks Jeremy, I did think it was a bit on the light side, but I put the dimensions into a commercial dry sand website - albeit in feet and it gave the results in lbs, so I may have made a mistake with either of those two conversions. It came up with 50.02kg for a 1m x 1m x .5m volume.

I may do some more digging.

The beauty of this is I may be able to model some of the points made out of different materials, e.g. mostly water southern hemisphere.

Jan 9, 2013 at 8:25 PM | Unregistered CommenterTheBigYinJames

One further problem is that the surface will reach thermal equilibrium much faster even than depths of 10 cm (think how much cooler sand on the beach is when you dig down). So using a layer 50 cm deep in your sums as all being in direct radiative contact with space will tend to overestimate the amount. What you would need is differential equations treating heat conduction within the top part of the surface later. Your program will start to get pretty long!

Jan 9, 2013 at 8:56 PM | Registered CommenterJeremy Harvey

Well, I could model the surface as say 10 plates of 5cm thickness each, exchanging heat with each other.

Once you model something, it's easyish to simply duplicate it 10 or 100 times, it just makes the calculations take longer.

Is this the Bishophill GCM v1.0 ?

Jan 9, 2013 at 9:14 PM | Unregistered CommenterTheBigYinJames

Starting to look like it! Are you going to have to ask the Met Office to borrow their computer soon?

Jan 9, 2013 at 9:22 PM | Registered CommenterJeremy Harvey

Just a little something from a non-engineer person:

If you look carefully at descriptions of the greenhouse effect - David Archer, for instance, there is no clear demarcation between where the quantum physics ends to where the meteorologic part begins to where the climate component begins.

Jan 9, 2013 at 10:08 PM | Registered Commentershub

Quite a way away from the first whisp of an atmosphere yet, shub :) Never mind quantum physics.

Jan 9, 2013 at 10:10 PM | Unregistered CommenterTheBigYinJames

After a quick skim of the Vasavada paper that Paul kindly sent to us it seems that the Diviner eqatorial peak temperature matches the SB equilibrium temperature (for closest Sun - Moon distance) to within a few degrees. It also gives derived figures for albedo, emissivity and surface conductance.

It seems to me that you could verify your model by first reproducing the Diviner equatorial curve, to within a few degrees, using fourth root temperature calculations at each point. You could then carry out the integration for the whole surface for equidistant points and then calculate the mean.

What value or use is the mean temperature (as distinct from the temperature of the mean insolation)? Well, perhaps nothing, but the IPCC seem to think it is important for their physical explanation of the GHE and N&Z use it in their ATE (atmospheric thermal enhancement) thesis. Of course, they both could be wrong but one thing is for sure - the analytical calculation and the numerical calculation of mean temperature MUST give the same result.

Good luck!

Jan 10, 2013 at 11:33 AM | Unregistered CommenterRoger Longstaff

The last post I made was not particularly directed at the model you are working on,, but as just an aside.

If you notice paleo reconstructions, it is clear that temperature curves have a multidecadal sinusoidal character. Of course, in order to get your model to spin multiple decade (as opposed to one day), you would have to spin the globe very, very fast. But if you do that, what would/could you do to get a long-scale sinusoidal pattern to show up?

Jan 10, 2013 at 12:11 PM | Registered Commentershub

Well, I wouldn't have to spin the model 'fast' to get a decadal response, just run it for a longer time. Each 'second' of the model only takes a fraction of a second to calculate remember - the day daily cycles graph i showed you earlier took less than second to run, thats a lot of modelled seconds per 'second'

Admittedly with only one point on the equator, once I scale it up, itll start to run more slowly, but there's no reason why a well populated model should take any more than a minute per modelled 'day', so running it decadally would only mean leaving it overnight.

Jan 10, 2013 at 12:17 PM | Unregistered CommenterTheBigYinJames

BigYin, I'd be very interested to see the effect of different spin rates. To see whether my speculation (is that higher or lower than a conjecture, either way the lower one) that the standard method works with very high spin only is true.

Shub, we have no reason to think that paleo will show any of the figures gained by the other three methods. Maybe Paul has a view on that, it is his field.

Roger, I really don't think the two methods you mention must give the same result. In fact they do not and they will not. We need to match to the observations and discard the method that doesn't work. Which is the standard disk/sphere method, in my opinion as an Oxfordshire housewife

Jan 10, 2013 at 12:41 PM | Registered Commenterrhoda


without changing anything else, I made the globe spain faster, and here are the results:

The daytime peak is the same for 1x 2x and 3x rotation rates, because it reaches maximum insolation quickly enough, but the night time cooling does not manage to get down so far, raising the average temperature.

Obviously at this early stage, not sure if this is actually realistic, but interesting.

Jan 10, 2013 at 1:04 PM | Unregistered CommenterTheBigYinJames

Thanks. This pinpoints where the standard method breaks down. With a slow rotation, the cold side pretty much stops radiating. No use to divide the disc of insolation by the area of the whole sphere when half of the sphere isn't radiating at all. (Or nearly. The cold side in the moon example is radiating 2watts/sq m. Two watts.). When you are messing with a fourth power law, you have to be very careful with what you call an average. The standard method works in that the total outbound radiation matches the inbound, but because half the sphere is doing ALL the work, the arithmetic averages obtained by instrumentation just do not match. Maybe GISS et al ought to raise thermometer temps to the fourth power before doing the averaging? Or better to calculate S-B radiation from the thermometer data then using statistical methods to get a whole earth figure to be matched directly to insolation. That would give a number to compare. That number would be an equivalent to the notorious 33K.

And now I am wondering whether the invalidity of the widely-claimed 33K figure has ever been challenged before. Surely someone must have looked at the question. Because to me, it's not a sensible conclusion at all. (Still a genuine question, not confrontational, not posing as having stumbled across the blazing truth that will destroy CAGW, just wondering if I've got the wrong end of the stick entirely)

Jan 10, 2013 at 1:32 PM | Registered Commenterrhoda

BigYin, is this sandy earth you are simulating? Because I notice that your average temp is not 33K short of measured temp. OK, I know how many things are missing, but the same things are missing from the original 33K story. So I have to ask, what difference does the sand make? Do changes to the surface specific heat or depth make much difference. I'm guessing not, but I'd like to see how much. Sometime in your copious free time, just a request to be to in line with all the others that you are bound to get.

Can you, by any chance, put a front end on the program so we can change the parameters ourselves? Rate, surface characteristics, insolation, you know what I mean. When you're ready guv'nor. No rush.

Jan 10, 2013 at 2:18 PM | Registered Commenterrhoda


I think your 2 W/m^2 is way out - the Apollo and Diviner data show more like 20 mW/m^2. This means that surface temps come up to SB temps very quickly, and the nightside cools very slowly. As the temperature field on the sunlit side is always constant, slowing the rotation would allow the nightside to cool more, therefore lowering the mean temperature of the planet, averaged over the whole surface.

Have I got this right?

Jan 10, 2013 at 2:34 PM | Unregistered CommenterRoger Longstaff

rhoda, I slowed the spin down to 100th of the speed.

I also corrected an albedo problem (I wasn't using it! - all temps in previous graphics are too high!)

Here's what the graph looks like now.

And the average is at 185K, i.e. approaching the non-rotating models.

Jan 10, 2013 at 2:45 PM | Unregistered CommenterTheBigYinJames