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Discussion > Simulation of Temperature

Nice smooth lines, do you mind posting your algorithm, doesn't have to be code, just the main calculations, to see where it differs from mine?

Jan 11, 2013 at 9:29 AM | Unregistered CommenterTheBigYinJames

BB - it looks good (more like Diviner / Apollo). Is it just for the equator?

I think you need to adjust heat loss on the darkside to give approx 25K over 14 (Earth) days of lunar darkness - probably with a multi-layer model.

But kudos to all you guys who are actually doing the coding!

Jan 11, 2013 at 10:44 AM | Unregistered CommenterRoger Longstaff

I noticed he's using 200 Kj/Kg K, whereas I'm using 0.8 (the specific heat of dry sand)

When I put the heat capacity up to 200, I get a zigzag up in the 300s.

My question for BB (and vicariously, SoD, I suppose) is that if the specific heat can control (i.e. slow) the heat loss when it's cooling, why doesn't it control (i.e. slow) the heat gain when it's warming?

The reason mine has a knee is that for a certain number of joules (W/m^2 over 1m^2 for 1 second = joules) you get x degrees K of a temp change for a certain mass with a certain specific heat capacity, same for cooling, only a negative change.)

I notice the SoD calculations don't seem to involve mass at all (from my cursory look at the page) which is a primary term in my temperature calculation: changeinTemp = kjoules / (MassKg * SpecificHeat);
Specific heat includes a mass term in it, so not sure how SoD arrived at its graph without considering it.

Jan 11, 2013 at 11:01 AM | Unregistered CommenterTheBigYinJames

Aha!
Just for fun, I set the mass of the sq meter of sand to 10kg, and voila, I get a shape very similar to BB, with no knees.

http://i47.tinypic.com/2ijkc2h.png

Jan 11, 2013 at 11:27 AM | Unregistered CommenterTheBigYinJames

BigYin, BB,

Vasavada models the regolith in two layers, with an increase in bulk density below 2cm from 1300 to 1800 kg/m^3, and in temperature dependent thermal conductivity from 0.0011 to 0.0094 W/m/K at 200K. The geothermal heat flux is taken to be 0.016 W/m^2 (all from page 6 of is paper).

I think that if you could find a way to model that then you could reproduce the Diviner curve almost exactly.

Jan 11, 2013 at 11:47 AM | Unregistered CommenterRoger Longstaff

For reference, my code is here, for a while. Note that it is unstable when the heat capacity is too low. It might be complete rubbish.

Jan 11, 2013 at 12:42 PM | Unregistered CommenterBitBucket

BB, Sorry to reveal my ignorance, but what language is that?

Jan 11, 2013 at 8:37 PM | Unregistered CommenterRoger Longstaff

Its in C. I'm out of date, I know.

Jan 11, 2013 at 9:02 PM | Unregistered CommenterBitBucket

Just FYI, here is a simplified representation of the surface temperatures of the Apophis asteroid which ESA's Hirshel telescope took a look at when it went past this week: Apophis pass by January 2013

Not sure how fast it is spinning, or it's distance from the Sun, but surface temps ranged from a maximum of 350K to 150k at the poles. It's albedo was accurately calculated to be 0.23. The good news is that they don't think there is any chance of an impact when it returns in April 2029 (Friday the 13th).

Jan 12, 2013 at 6:36 AM | Registered Commenterlapogus

So, what's the temperature of Earth supposed to be? There is (or maybe is no longer) an idea that the temperature of Earth is supposed to be 255K. And that as it is 'really' 288K that the difference of 33K is made up by global warming. I think these simulations combined with Diviner results blow that apart. I'm not espousing N&Z, as I understand it they claim rotation rate is immaterial, and we have blown that apart too.

Have we refuted even the concept of a right temperature given only insolation?

Jan 12, 2013 at 10:40 AM | Registered Commenterrhoda

Well, I think we've demonstrated that integrating insolation over a non-rotating sphere gives the incorrect 'average' temperature. This is because, as you have pointed out, the dark side is not really emitting anything much but is maintaining a flat non zero temperature - so to average insolation over a 2K darkside is misleading, because it never happens, even for slow spin worlds like the moon, the physics of the rock means it holds onto temperature better than a black body.

I don't think the simulation proves absolute figures yet, because it's not matching even lunar figures yet, but give me a bit more fiddle time (I've been doing some more work on the model, but I haven't released it to the website yet)

Jan 12, 2013 at 10:46 AM | Unregistered CommenterTheBigYinJames

BigYin,

As and when you are satisfied with the equatorial plot for the Moon (presumably a close approximation to Diviner) there remains the numerical integration. I visualise this as calculating the fourth root temperatures for equal areas over the entire surface, and then taking the mean. Is this correct?

If so, could you use a Lambert projection for this integration?

http://en.wikipedia.org/wiki/Lambert_azimuthal_equal-area_projection

It seems there might be something available in Matlab, or perhaps in another language - what do you use?

Jan 12, 2013 at 11:26 AM | Unregistered CommenterRoger Longstaff

I agree, the "physics of the rock" means that the mean temperature must be a function of the rotation rate. I expect that is why N&Z have revised their figure to 197K - are they still saying that rotation doesn't matter?

Jan 12, 2013 at 11:37 AM | Unregistered CommenterRoger Longstaff

Roger,

I won't have to worry about integrating, because this is a simulation, and as you said I'll have 1000 or more (I'll make it a variable to play with) points spread over the surface I can just take an arithmetic mean of the current temperatures of all of them - each point will have its own 'curve' and be at different positions on the day night cycle, so the arithmetic mean of them all will be the actual arithmetic average temperature.

As long as my points are distributed equally (a problem I finally solved on the Numerical Computation thread, so I'm not concerned) then this should all work out OK and give us an 'average' temperature curve to go along with the one for a point on the equator.

(I'm going to make the latitude of the point you want to see a variabel too, so you can look at polar curves etc - or perhaps just show curves at the equator, 30, 60 and polar all at once - along with average)

As for them revising their numbers...perhaps N&Z are reading these threads ;) hi Ned!

Jan 12, 2013 at 11:50 AM | Unregistered CommenterTheBigYinJames

Thinking about it... average temperature won't be a curve, it'll be a flat line... ho hum.

Jan 12, 2013 at 12:00 PM | Unregistered CommenterTheBigYinJames

By "numerical integration" I meant exactly what you said - multiple calculations, sum then average. Good that you already know how to do this - you're way ahead of me.

We now seem to be in a right mess! If N&Z say rotation doesn't matter it seems they are wrong. Also, if the IPCC say that the mean temperature is 255K they are clearly wrong (if the mean Diviner equatorial temperature is 213K then the mean for the whole surface must clearly be less than this, as the equator receives the most insolation).

So if both the IPCC and N&Z are wrong what does it mean? Probably the subject for a new thread beacuse this one is too important to be diverted.

It would be good if the other modellers (BB and Paul Dennis) would add their comments here - and maybe even Ned?

Jan 12, 2013 at 12:08 PM | Unregistered CommenterRoger Longstaff

The sim actually doesn't say what a day is, earth or lunar. I would like to see it as generic as possible, with parms for the rock numbers too. That is very hard to get right, or at least I speculate that in vacuum on sand the top layer will be reaching full S-B in minutes, albeit a very thin layer. Not 10cm, but 3 or 4 mm.

If N&Z are claiming that rotation makes no difference, they must be using a black body. No heat retention at all. But that makes the idea of 'effective temperature' a nonsense and the method of calculation (disk insolation and sphere radiation) is a nonsense too. The effective temperature can only work where there is no such thing as temperature in terms of mass x temp = heat, because no mass is ever heated. I'm mulling the idea of a post on this to take the results out of this simulation thread.


Seems while I was trying to comment, forgetting to login, ducking the captcha, cutting and pasting, Roger has made a similar suggestion. Shall we leave the sim work here and take discussion of the results to a new thread?

Jan 12, 2013 at 12:13 PM | Registered Commenterrhoda

Could one of you summarize what has been done for the benefit of readers, like a small abstract?

Jan 12, 2013 at 2:25 PM | Registered Commentershub

shub, I would defer to BigYin (cos it's his thread), but I think that he, rhoda and I are in agreement.

My summary (to date) is essentially contained in my post of 12.08pm.

Jan 12, 2013 at 2:53 PM | Unregistered CommenterRoger Longstaff

shub,

we're still 'in progress' really, but early runs of the simulator bear out the 'intuitive' answer that rotation and heat capacity make a big difference to average temperature.

Over the next few days I hope to get the simulator to approximate the Lunar diviner readings, but we're still a long way away from saying the model is modelling reality.

Jan 12, 2013 at 3:01 PM | Unregistered CommenterTheBigYinJames

For interest, Diviner results for all lunar latitudes can be seen here:

http://www.diviner.ucla.edu/science.shtml

The mean temperature would seem to be somewhere between 150K and 200K

Jan 12, 2013 at 4:45 PM | Unregistered CommenterRoger Longstaff

Nice data about the moon on that diviner site. The stuff about the earth is dodgy though. It certainly cannot be used for direct comparison. The all-time high is deprecated now, the all-time low is at 11,000 feet, which though true isn't good for anything.

I'm going to compare the lunar temps with my own static all-moon model to see if I got the latitude method right. I am only modelling the lit side and using diviner temps for the average. Another final number coming soon.

Jan 12, 2013 at 6:07 PM | Registered Commenterrhoda

Jan 11, 2013 at 12:42 PM BitBucket

"const double pi = 3.1428;"

Probably won't make any real difference but hard-wired into my brain cells is pi = 3.141592654.

Jan 12, 2013 at 6:17 PM | Unregistered Commentersplitpin

I agree rhoda - only the diviner graphs are of any use on that site.

If your static model gets the latitude data right please tell us how you do it. If you then take the average for the whole surface using about 80K for the unlit hemisphere then you should get fairly close to the right answer IMO.

Jan 12, 2013 at 6:19 PM | Unregistered CommenterRoger Longstaff

I think there's a mistake in that graph. It doesn't match my temps...yeah, OK, I know. Two things. I am using 1366 insolation, 0.12 albedo, 0.95 emissivity. I can't get to 390K with this. Plainly I need more insolation. Second, I can't match the latitude lines to the graph, specifically I can only see nine lines but there ought to be ten. Just me, or an omission?

Jan 12, 2013 at 7:11 PM | Registered Commenterrhoda