## Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 530

... It is the engineering textbook answer. Claiming it is nonsense does not make it so. It was your own model that violated conservation of energy. But to see why, it's easiest to solve the general case first, then look at a specific case. I told you I had reasons to solve the general case first.

... Well, then, I guess you do admit defeat. It doesn't take much time to obtain a textbook on the subject (you were given references 2 years ago and it's not that hard to find others) ... [Jane Q. Public, 2014-09-01]

No, the PSI Sky Dragon Slayers told you it's the engineering textbook answer. I showed you MIT's final expression which reduces to my Eq. 1 for blackbodies, and is consistent with these equations and Eq. 1 in Goodman 1957. Physicists and engineers have been using thermodynamics for decades in the real world that contradicts Dr. Latour's Slayer nonsense.

That's why Jane, Dr. Latour and the rest of the Slayers disagree with the American Institute of Physics, the American Physical Society, the Australian Institute of Physics, and the European Physical Society.

... I am disputing that given reasonable chosen dimensions it is anywhere near an intractable problem.

... [Jane Q. Public, 2014-09-01]

I never said the problem is intractable. Just that it's more complicated than the spherically symmetric problem. Again, do you dispute that equilibrium temperatures for a non-enclosing plate would vary across the plate surfaces rather than being simple numbers like with a spherically symmetric fully enclosing plate?

Maybe I should explain that. Consider Dr. Spencer's first illustration. Presumably the heated plate at "150F" has finite conductivity, so its lack of spherical symmetry means that its corners will be cooler than the plate's side's midpoints. That's because the corners are closer to the cold chamber walls than those midpoints.

An integral over the heated plate's surface might average to "150F" but (unlike a spherically symmetric plate) it can't have that temperature everywhere as long as it has finite conductivity. But at least the single heated plate has bilateral symmetry; the left and right hand side midpoints have the same temperature.

Adding a cool plate removes even that bilateral symmetry. The left hand side's midpoint warms the least because it's still radiating to the 0F chamber walls. The right hand side's midpoint warms the most because it's now radiating to the (initially) 100F cold plate.

Since enclosing a spherically symmetric plate warms it from 150F to ~233.8F for area ratios similar to Earth's, the right hand side's midpoint won't warm past ~233.8F. But it has to warm to conserve energy because at equilibrium power in = power out.

I can't be more specific without programming a finite element model. But Dr. Latour never even allowed for the heated plate's temperature to be different on each side. As long as we're only considering materials with finite conductivity, this would only be possible for a spherically symmetric enclosing plate.

... Derived equations are available which give approximations with reasonable precision. Or you can assume particular dimensions of the general case which simplify the math. I said that was a bullshit excuse, I meant it when I said it, and I still mean it.

... [Jane Q. Public, 2014-09-01]

Dr. Latour's answer wasn't "reasonably precise". He claimed that the heated plate wouldn't warm **at all** when the cold plate was added, even if it completely enclosed the heated plate such that K = 1. This is a specific prediction of "0.0000...F" warming. Since energy conservation means that adding a cold plate **has to** warm the heated plate, he's only off by a factor of infinity.