Are you presenting? If so, during what session?

I’m giving a talk in session A41K, Climate Modeling in a Transparent World and Integrated Test Beds I on Thursday December 6, 2012 from 9:00 to 9:15 am. I’ll be speaking in 3010 (Moscone West), and my talk will be called ‘The Azimuth Project: an Open-Access Educational Resource’.

I’ll try to find you and your poster. This conference sounds like it’s going to be a madhouse, it’s so big!

]]>I’ll also be at AGU.

If you happen to be walking by NG010: Nonlinear and Scaling Processes in the Atmosphere and Ocean at all Scales, From Microscales to Climate, from 800-1200am on Thursday, consider stopping by my poster on vortex generation by breaking surface gravity waves.

Are you presenting? If so, during what session?

Nick

]]>For the AGU meeting, local info is here:

http://fallmeeting.agu.org/2012/travel-housing/

So far I’ve mainly been thinking about a few kinds of networks: Feynman diagrams in particle physics, stochastic Petri nets, Markov processes, electrical circuits and Bayesian networks (aka belief networks). These turn out to be mathematically a lot more tightly related than you might at first think, and that’s bound to be a good thing. But it’ll take a while longer before I can prove results that excite people who aren’t interested in conceptual unification for its own sake.

Box models are definitely also on my to-do list, but so far my friend Eugene Lerman is way ahead of me on those.

See you! Should I, umm, be getting a hotel and stuff? I didn’t get any information about that.

]]>Let’s talk about this at the AGU meeting.

]]>Here are some little codes of the day ( http://dl.dropbox.com/u/7915230/Luna/dia.rtf )and the night (http://dl.dropbox.com/u/7915230/Luna/noche.rtf )dynamics. They run in the demo version of “Berkeley Madonna” software.

The first one solves the equation

Where $H$ is the heat capacity, the density, and $d$ is the effective depth. Since apparently the moon is mostly silicates like sand, I choose the values for sand: 800 and 2000 . Also I thought 10 cm was a reasonable value for the effective depth. $\theta $ is the latitude (in radians), the full power of the sun (1370) and is the numbers of seconds in half of the moon cycle.

For the equator with an initial temperature of 50 degrees Kelvin we get

and for the same location but starting at 130 degrees

we get

What I find interesting is that it really doesn’t matter where at what temperature you start with, by noon you have a value very near the maximum thermodynamical equilibria for the full power of the sun. To change the final temperature we have to change of latitude , here is a graph at 45 degrees.

The night equation is simpler, it does not depend on the latitude

here is a graph starting the night at 220

and here is a graph starting the night at 150

Note that in the equation there is no 4 as I thought before, this is for two practical reasons and I would like to hear a good reason.

The first reason is that if I put here, I should put it also in the day equation. But then the maximum noon temperature would be much less than the observed one, in fact changing the value of the thermodynamical equilibrium. The other reason is that if a leave it I don’t get reasonable temperatures, in order to adjust to get reasonable temperatures I have to adjust the effective depth to something like 40 cm. Which for some reason I think is to much.

The following is nice: Let the composition of the two dynamics, a map from $[0,390 ]$ to itself. We start from some temperature and after 29.5 days we have another temperature. By elementary calculus -the mean value theorem- there is a fixed point, this fix point corresponds ti a periodic orbit. Moreover the transformation is a huge contraction, as we observed before more or less independently of the initial morning temperature we arrive near the maximum, the cooling is also contraction so there is only one fixed point, so the periodic orbit is an attractor.

There is something that still is not very well explained, the difference of the minimal temperature in each latitude is much smaller than the ones that appear in the graph of http://www.diviner.ucla.edu/science.shtml

I guess this has to do with with this effective depth not being quite right.

You are right lets do first the energy balance in a square meter in the equator, knowing the temperature.

Let (1370) be the sun power per square meter, and be the number of seconds (1274400) in half a cycle (29.5 days) . If we start in the lunar morning the effective power as function of time should be something like

So the received energy in a cycle is

.

Which is one of the few integrals I can do, . So a square meter in the equators moon receives Joules in a moon cycle.

Now the for the emitted radiation, we have to interpret the graphs of the temperature in http://www.diviner.ucla.edu/science.shtml

The graph is very flat in the minimum, a consequence of the moon cooling down very quickly. To the resolution of my eyes looks that all night has roughly the minimum value.

So I propose to a first approximation for the day and in the night.

The emitted energy is then

This integral can be done but it is easier to evaluate in a online calculator

http://www.numberempire.com/ and gives . The second term, the night radiation, is of order so doesn’t change significantly the answer.

So the emitted radiation is only around of the received. We can push the minimum night to only 7 days and adjust the function to cover 22 days, which I think is way to much, and we would get then something around . Is this equal enough to not be confused?

]]>