All we need to solve the problem of the Comcasts and the Time-warners of the world is to expose them to competition.

If that were true, we wouldn't need common carrier regulation for shipping companies. That's where common carrier started (hence the term "carrier"). It was put in place to keep carriage networks, which are naturally limited in the efficient number of competitors, from exploiting their natural n-opolies by making preferred carrier deals with incumbent manufacturers.

In the case of wired data carriage networks, once there is one set of cables in the ground, the cost of putting each subseqent set in the ground faces an barrier-to-entry that rises more quickly than the natural barriers on, for example, retail stores. In the case of wireless, the limits on frequency band interference do the same thing.

Practical reality does not match the idealistic theories we wish were true, whether those be socialism, anarchy, or anything in between. Give up the -isms and consider observable reality. Learn from history, not religion-peddling pundits.

Since the earliest days of USENET and IRC Chat, the geek has a flawless record of making one-on-one communication over the Internet as painful a process as possible for the non-technical user.

Don't be facetious. One-on-one communication could be much more painful. In the specific case of secure (ie: end-to-end encrypted) communication, Tox is approaching the theoretical limit of simplicity. Key exchange has a mathematically bound minimum complexity in order to be secure. The reason Skype is not secure is precisely because it is easier to use than Tox.

Or, slightly differently: Tox is an example of geeks making one-to-one comm as easy as it possibly can be, for the given requirements.

And how do you exchange key? Do they plan a web of trust à la GPG?

That was one of my first questions. The answer is; however you want. They provide an "easy" (hence vulnerable) method for doing so, but you can check the public key hash against your securely transferred value before approving a key if you want.

Or, slightly differently; this is not a key exchange system, just a comm system you can use once you have authenticated a key to your level of security requirement.

There needs to be a way to disable the cameras for a short period of time. I don't think we need to see police officers using the restroom. Then there are times when officers have private conversations that are not work related. Do you really think it is valid to have anyone monitored every second from start of shift to end of shift? Would you work under those conditions?

It is a good question; how about this: The officer can click the "this is private" button any time they want. That segment of the video is still recorded, but is not included in routine reviews. If there is reasonable cause, IA can look at the protected video. If an officer is putting too much time in private mode, their superior or IA can ask what's going on. If an IA officer is abusing their privilege to look at private mode video, they get canned (pursuant to an IIAA investigation, presumably).

I'm not saying we should solve simpler problems before moving on to more complex problems. [Jane Q. Public, 2014-09-01]

Okay, then we disagree. It's always helpful to solve simpler problems before moving on to more complex problems. The simpler problem is easier to learn, and often serves as a sanity check on the more complex problem.

That got a minor mention later in his article, is not included in his diagrams, and is NOT the problem I originally presented to you. As I have said many times before, AFTER you refute Latour's calculations regarding Spencer's original challenge, which did not have the passive body enclosing the heat source, I would be happy to move on to the other issue... with no additional stipulations or additions to the problem Spencer describes. But you haven't gotten there yet. Cart before the horse, with a straw-man riding the cart. [Jane Q. Public, 2014-09-01]

Again, Latour's calculations allowed for K = 1: "K is the fraction of radiation from the first bar absorbed by the second colder bar, 0 < K <=1."

The only way K = 1 is if the cold plate completely encloses the first heated source. Otherwise, radiation from the side of the source opposite the cold plate couldn't possibly be absorbed by the cold plate, which would force K < 1. So once again, the fact that Dr. Latour included the possibility that K = 1 means that his claim applies to all geometries.

If not, why doesn't he deal with edge effects? The only ways to eliminate edge effects are if the plates are infinite, or if the cold plate completely encloses the heated source.

Why don't you just shut up and do it? Why have you been so mightily struggling, like a fish on a hook, to avoid it? [Jane Q. Public, 2014-09-01]

Again, I don't have enough time to program a finite element model to account for the fact that a non-fully-enclosing plate would cause plate temperatures to vary across their surfaces. But even if I did, the first thing I'd do after debugging it would be to check the finite element solution in a case where a simple analytic solution can be obtained. Namely, a fully-enclosing passive plate, where the plate temperatures are simple numbers.

By the way, since you keep insisting that only a particular geometry could refute Dr. Latour's treatment, could you please show where he specified the dimensions of the plates? Or where Dr. Spencer did? Otherwise, even if I had enough time to do so, how could I possibly program this complicated finite element model with the specific geometry that would finally convince you the Slayers are wrong?

... The problem is that there is no such thing as a thermal superconductor of this kind, and you aren't seeing that it leads to contradictions. The only way it could exist would be if it had NO thermal effect on its surroundings whatever. So it's the ultimate straw-man argument. There is no way it can be legitimately used to demonstrate anything. [Jane Q. Public, 2014-09-01]

Again, we'll have to agree to disagree about thermal superconductors. That's why I've repeatedly pointed out that I've already solved this problem with an aluminum enclosing shell, and it also warms the heated plate (aka Jane's "source") to ~233.8F.

No, they didn't, because it's a different problem, being given a theoretical treatment. You keep doing that, but I'm not buying. Two infinite plates, neither of which is heated, is not even remotely the same situation, and it's also theoretical only. They're not taking into account certain real-world factors pertaining to Spencer's experiment. Latour does. Not that they're doing anything wrong... given the context of their situation: infinite non-heated grey bodies. This is not Spencer's experiment. [Jane Q. Public, 2014-09-01]

No, it's exactly the same problem. The same infinite sum of absorption and reflection. The plates are only "infinite" to avoid having to model fringing field effects around the plate edges. And note that Dr. Latour doesn't model edge effects either, so his plates are either infinite or the passive plate completely encloses the "source". Either way, there would be no edges.

Notice that the first example MIT applies their final equation to is a thermos bottle where the inside wall is heated by hot fluid.

You did not point to a calculation he performed on Spencer's situation and prove it wrong. You took what you incorrectly called an analogous situation and called that wrong. Which has been my whole point here. You keep claiming something else represents Spencer's experiment, but you won't tackle Spencer's actual, original experiment. You have consistently refused, for over 2 years. ... You continue to refuse to actually do what you said you'd done: refute Latour's treatment of Spencer's challenge. [Jane Q. Public, 2014-09-01]

Again, Dr. Spencer's actual, original experiment included the possibility of a fully-enclosing passive plate. And so did Dr. Latour's treatment of it. If you don't agree, please show where Dr. Latour specifies the dimensions of the plates before wrongly concluding that T remains 150.

In fact, as far as I can tell nobody's specified the plate dimensions except for me. Since the argument I'm refuting never specified the plate dimensions, why would the plate dimensions matter?

... I repeat: get the experiment with the two separate plates (actively heated plate and passive plate) right first. Then you can move on to a fully-enclosing plate. You say it's simpler but in a way it's not; you're trying to ride a bicycle when you haven't even managed to ride your tricycle without falling off. ... [Jane Q. Public, 2014-08-29]

... Take Spencer's original experiment, with two separated, non-enclosing plates, and show SPECIFICALLY where Latour was wrong in his calculations. THEN, if you like, you can move on to the enclosed-source situation. ... [Jane Q. Public, 2014-09-01]

Once again, the original experiment included both scenarios: fully-enclosed and not-fully-enclosed. We can agree that one should solve simpler problems before moving on to more complex problems, but we seem to disagree about which of the scenarios in Dr. Spencer's original experiment is simpler.

Again, solving a problem without spherical symmetry means you'll have to solve for equilibrium temperatures which aren't constant across the heated and passive plates. Those equilibrium temperatures wouldn't be simple numbers. They'd be complicated functions that would vary across the plate surfaces. Contrast that with a spherically symmetric enclosing plate, where equilibrium temperatures are just simple numbers.

Are you disputing those facts, or do you really not see which of these problems is more complicated?

I don't have enough time to program a finite element model to account for the fact that a non-fully-enclosing plate would cause plate temperatures to vary across their surfaces. But even if I did, the first thing I'd do after debugging it would be to check the finite element solution in a case where a simple analytic solution can be obtained. Namely, a fully-enclosing passive plate, where the plate temperatures are simple numbers.

Censorship is the suppression of speech. For example: "You can't talk about Oranges, they are evil!"
Moderation is the regulation of speech: "You can talk about Oranges, just not here. Go over there to talk about Oranges."

A related problem is the "Free Speech Zones" outside political party rallies. They do not censor speech, but they do prevent you from speaking in some portion of the public square. To the extent that Facebook has become the public square, the cost to society of speech prohibition in that forum is the same. To the extent that "Free Speech Zones" are an infringement of free speech, and Facebook has become the public square, Facebook presents the same risks to society.

This is not merely a question of how you dice the legal technicalities, it is a question of the purpose and means of free speech. Free speech is more important to our society in the long run than any other right; it is the basis of having a strong GDP upon which Facebook can build its business. If Facebook becomes destructive of the system, it is our rationally self-interested duty as a society to stop it, even if the particular existing legal terms can be parsed in a way that says it is legal.

At equilibrium, the enclosing shell radiates the same power out as the heated plate did before it was enclosed. But its area is 1.0025 times larger, so its outer temperature is 149.6F (338.5K) instead of 150.0F (338.7K).

In order for what you say to be correct, then the "enclosing shell" you refer to is not the heated plate enclosing the source. Which would mean you were talking about a completely different experiment, not even the one Spencer mentioned with the heated plate enclosing the source. [Jane Q. Public, 2014-08-31]

We might be talking past each other. What you're calling the "source" is what I've been calling the "heated plate" with temperature "T_h" in all my equations. I've called the other enclosing plate the "cold plate" with temperature "T_c". As I've repeatedly and consistently stressed, "T_c" is only identical on both sides of the enclosing cold plate if it's a thermal superconductor.

I'm sorry for any confusion this caused, but as you can tell I really am talking about the experiment Dr. Spencer mentioned. We're just using different words, and again I'm sorry for not noticing this miscommunication earlier. I take full responsibility.

... But your hypothetical thermal superconductor could not store heat like a black body and remain a superconductor. That's a contradiction. So it's a different creature, from your imagination. This is why I say: leave it out. There is no way you can try to demonstrate anything else with it, either, without leading to a contradiction. And it's not part of the original experiment anyway; it's nothing but misdirection. [Jane Q. Public, 2014-08-31]

We'll have to agree to disagree about thermal superconductors. I'm sorry for trying to simplify the problem in a way that ultimately just caused us to waste so much time. Again, I take full responsibility.

But again, I've already solved this problem with an aluminum enclosing shell, and it also warms the heated plate (aka Jane's "source") to ~233.8F.

... I'm not interested. Original experiment. Latour's treatment of it. Show where he was wrong. Period. Stop prevaricating. [Jane Q. Public, 2014-08-31]

That was Dr. Spencer's original challenge. He included the possibility of a fully-enclosing passive plate. And so did Dr. Latour's treatment of it. If you don't agree, please show where Dr. Latour specifies the dimensions of the plates before wrongly concluding that T remains 150. Also, why did Dr. Latour explicitly allow for K = 1 and k = 1, which describes a fully-enclosing blackbody passive plate?

Dr. Latour really did wrongly claim that a fully-enclosing passive plate wouldn't warm the heated plate (aka Jane's "source"). I've shown that his claim violates conservation of energy. As long as the shell is warmer than the chamber walls (which it is), the net radiative heat loss from the heated plate (aka Jane's "source") is reduced. So power in > power out, which means the heated plate either warms or energy isn't conserved. Just like how a bathtub fills up.

"Stop prevaricating"? Really? I've showed that Dr. Latour was wrong because his claim violates conservation of energy. Again, in physics that's a really big mistake.

Since you just linked to this excellent example, did you notice that MIT solved this problem at the very top and got a completely different answer than Dr. Latour?

How is that prevaricating? Did you even read MIT's solution to this problem? They show how to solve it correctly.

Again, note that MIT's final expression reduces to my Eq. 1 for blackbodies, and is consistent with these equations and Eq. 1 in Goodman 1957.

Don't you see that you threw in this whole "thermal superconductor" schtick without considering what properties a thermal superconductor must actually have? In order to superconduct, it must be the same temperature everywhere, always. The only way this would be even remotely possible were if it were a perfect radiator... [Jane Q. Public, 2014-08-30]

Superconductors are distinguished from aluminum by internal properties, not radiative surface properties. That's because conduction happens inside materials, whereas radiation is emitted and absorbed on surfaces.

... The only way this would be even remotely possible were if it were a perfect radiator, with emissivity of 1. It would also be a perfect absorber, absorptivity of 1. Regardless of wavelength. So while this might not technically be true, for all practical purposes it is: a thermal superconductor would be completely transparent to all radiation... [Jane Q. Public, 2014-08-30]

No. As I've explained, emissivity = 1 and absorptivity = 1 is the definition of a blackbody. A completely transparent material would have transmittance = 1 and absorptivity = 0. Blackbodies can't be transparent.

... a thermal superconductor ... has no "thermal mass". So it would have absolutely no effect on anything in this experiment. For practical purposes, it would not exist. Your idea that you can get around this by placing some kind of thin lining on its interior doesn't work. It's still as though it weren't there at all... all you have left for practical purposes is the thin shell, nothing else. ... [Jane Q. Public, 2014-08-30]

I've already solved this problem with an aluminum enclosing shell rather than a thermal superconductor shell. Both shells warm the heated plate to ~233.8F.

... That's why I say: no more prevarication. No more beating about the bush. Take Spencer's original challenge, apply Latour's thermodynamic treatment of it, and show where it is wrong. Anything else constitutes failure to back up your claim that Latour is wrong and -- as you have said more than once -- some kind of nutcase. You've had more than 2 years. That is plenty. [Jane Q. Public, 2014-08-30]

Dr. Spencer's original challenge included the possibility of a fully-enclosing passive plate. And so did Dr. Latour. Note that Dr. Latour never specifies the dimensions of the plates (as Jane began to) before wrongly concluding that T remains 150. This means his incorrect conclusion must apply to all geometries, including a fully-enclosing passive plate. In fact, notice that Dr. Latour explicitly allows for K = 1 and k = 1, which describes a fully-enclosing blackbody passive plate.

So Dr. Latour wrongly claimed that a fully-enclosing passive plate wouldn't warm the heated plate. I've shown that his claim violates conservation of energy. As long as the shell is warmer than the chamber walls (which it is), the net radiative heat loss from the heated plate is reduced. So power in > power out, which means the heated plate either warms or energy isn't conserved. Just like how a bathtub fills up.

Since you just linked to this excellent example, did you notice that MIT solved this problem at the very top and got a completely different answer than Dr. Latour?

Again, note that MIT's final expression reduces to my Eq. 1 for blackbodies, and is consistent with these equations and Eq. 1 in Goodman 1957.

No, I'm not wrong. You calculated the outside temperature from the inside temperature, saying it's LOWER because of its greater area. This much is correct. THEN you try to say that with a thermal superconductor, the inner temperature would be the same as outside. Except you just calculated that outside temperature from a WARMER interior. You quite literally can't have it both ways. EITHER you're claiming a superconductor has a different temperature on both sides, or you're claiming that the inside has 2 different temperatures simultaneously. [Jane Q. Public, 2014-08-30]

Remember that the inner surface of the enclosing shell is different than the surface of the heated plate. The inner and outer surfaces of the enclosing shell are at exactly the same temperature because it's a thermal superconductor. That's what I've always been saying, despite your attempts to pretend otherwise.

The surface of the heated plate at equilibrium, however, is warmer than the inner surface of the enclosing shell. It has to be.

Here is an excellent example of this (19.3.2), which illustrates why it is a straw-man argument that is not relevant to the problem at hand. In this case the walls are warmer, not cooler, and the radiation shield is blocking the thermocouple from the radiation inward from the chamber walls, so that it can get an accurate temperature reading of the air without interference from the walls. In your case, it is the opposite: the walls are cooler than the thermocouple. But in neither case is the situation a representation of equilibrium (for example in this case, air is convecting away some of the heat of the thermocouple). The shield is absorbing and emitting radiation, too, it's just that it is isolated from the chamber walls, and so is closer to the ambient temperature of the medium being measured. This is in no way related to our experiment at all. It is in a vacuum. There is no "medium" to measure, with an ambient temperature. Not even remotely. [Jane Q. Public, 2014-08-30]

I've repeatedly linked to that excellent example. Despite your incoherent protests, it's a relevant example where a passive plate reduces radiative heat loss from a warmer source, warming it to a higher equilibrium temperature. It's a real world example which shows Jane and the Sky Dragon Slayers are wrong.

See? Same shit different day. You won't sit down and do the calculations start-to-finish, instead you do one small part, then start indulging in your hallmark game of out-of-context he-said, she-said, toss in a straw-man, then claim it's all proved. ... It's simply another illustration of the depths of hand-waving you will go to, rather than actually doing all the calculations on the actual experiment from start to finish. All you're doing is tossing in more straw-men and irrelevancies. You won't do the actual experiment. The only reasonable conclusion to be drawn here is that you won't do it because you know you're wrong. [Jane Q. Public, 2014-08-30]

Don't you see the irony here? I've repeatedly done the calculations "start-to-finish" by deriving and solving equations describing the final equilibrium temperature of the enclosed plate using increasingly realistic scenarios. I've repeatedly told you that you'd only be able to understand this thought experiment if you did the same. But you still haven't. Haven't you noticed that I'm the only one here deriving equations and doing calculations?

Is the only reasonable conclusion to be drawn here that you won't even attempt to solve this problem because you know you're wrong?

And I want to be clear about this: I'm not demanding anything from you. YOU are the one who proclaimed Latour wrong, therefore it is your burden to demonstrate that he actually is, by showing exactly where he is incorrect. ... The whole point: You claimed Latour was wrong. But you refuse to back up your claim by showing WHERE in his calculations he was incorrect. That's your burden and you haven't been meeting it. Until you do, you have no argument to make. You can throw all the ad-hominem and straw-man arguments and irrelevancies in that you want, but none of it proves you correct. Until you actually show where Latour made a mistake, in his actual calculations related to this experiment, you're wrong by default. [Jane Q. Public, 2014-08-30]

Once again, Dr. Latour and Jane claim that enclosing the heated plate wouldn't warm it. I've shown that this would violate conservation of energy.

In physics, violating conservation of energy is a pretty big mistake.

... you KNOW Latour was correct. And it isn't just him. TEXTBOOKS about practical applications of thermodynamics say so. ... [Jane Q. Public, 2014-08-30]

Again, I already showed you that MIT's equation reduces to my Eq. 1 for blackbodies, and is consistent with these equations and Eq. 1 in Goodman 1957. I've stressed that this thought experiment has been tested for decades in the real world. Radiation shields allow for more accurate measurements of gas temperatures using thermocouples:

"The greatest problem with measuring gas temperatures is combatting radiation loss. ... surround the probe with a radiation shield ... The thermocouple bead radiates to the shield which is much hotter than the surrounding walls. Thus the radiative loss and hence temperature error is significantly reduced. The shield itself radiates to the walls."

These radiation shields have been used since at least Daniels 1968 (PDF), and they work like Dr. Spencer's insulating plate. They slow radiative heat loss from the hotter thermocouple. If Jane and Dr. Latour's Sky Dragon Slayer misinformation is correct, why have accurate thermocouples used radiation shields since at least 1968? Isn't that an example of a "real world" situation that's ultimately what we're talking about?

But its inner temperature ISN'T 149.6F [Jane Q. Public, 2014-08-30]

After twice pretending that I'd claimed the inner temperature wasn't equal to its outer temperature of 149.6F... now you make that incorrect claim yourself? Bizarrely, I have to point out that a thermal superconductor enclosing shell will have an inner temperature equal to its outer temperature, exactly as I originally said.

This reminds me of your other similar mistake that you haven't acknowledged:

A plate near the heat source is NOT even remotely the same as closing the drain on a bathtub, because the total power out of the system (it's a closed system with heat being removed, remember?) remains constant, as you have so conveniently observed. [Jane Q. Public, 2014-08-28]

Completely backwards, as usual. I've never observed any such ridiculous nonsense. That's actually Jane's ridiculous "observation" which I've already tried to correct:

"... Hopefully it's also clear that Jane's also wrong to claim that the power used by the cooler is required to be constant. The chamber wall temperature is held constant, so the power used by the cooler temporarily decreases after the enclosing plate is added, until it reaches equilibrium."

I've repeatedly said the electrical heating power is constant, and that adding an enclosing plate temporarily reduces power out until the heated plate warms to a higher equilibrium temperature.

Over a period of MORE THAN TWO YEARS, I have repeatedly tried to engage you in a thorough analysis of this experiment. EVERY TIME, you have done (usually incorrectly) a partial analysis, then declared the subject proved. But it never was. When pressed, you resorted to the same kind of bullshit you have pulled here, with ad-hominem, not-sequiturs, and straw-men. NEVER daring to face the full problem in real detail. ... You have NEVER, ONCE, tackled the problem head-on. Always a little twist here, a little change there, let's ignore areal exposure to the ambient radiation, ad nauseum. Always weaseling sideways, never quite taking on the task of REFUTING LATOUR, even though that's what you claimed to be doing, with all your misdirection. [Jane Q. Public, 2014-08-30]

You're claiming my calculations are somehow incorrect, but if you'd really found an error it would have been much faster for you to simply lead by example and show how to do the calculations correctly. That would constitute engaging in a thorough analysis of this experiment.

Jane, you just quoted me saying that "its outer temperature is 149.6F ... let's pretend the enclosing shell is a thermal superconductor, so its inner temperature is also 149.6F"

Don't you see how my quote shows you were wrong to twice pretend that I'd claimed otherwise?

Spencer's INITIAL description of his thought experiment. As I have told you several time. This first, then more if you want to get into it. I will not discuss this with you in the other order, AS I HAVE TOLD YOU. Because until you get that right, you're not going to get the other one right. If you continue to argue the other case first, then we are done, and I will write you off as hopeless. ... No "enclosing shell". Two parallel plates. The original thought experiment is two parallel plates (we can make them of equal dimensions just to simplify, but it's not necessary). I repeat: we briefly discussed "even if it were enclosing" but that's a complication of the original, and we'll solve the original first. [Jane Q. Public, 2014-08-30]

Once again, solving a problem without spherical symmetry means you'll have to solve for equilibrium temperatures which aren't constant across the heated and passive plates. Those equilibrium temperatures wouldn't be simple numbers. They'd be complicated functions that would vary across the plate surfaces. Contrast that with a spherically symmetric enclosing plate, where equilibrium temperatures are just simple numbers.

Are you disputing those facts, or do you really not see which of these problems is more complicated?

... Also, I don't think we're assuming black bodies. The best we can realistically do is grey bodies that absorb in all the relevant frequencies under discussion. ... [Jane Q. Public, 2014-08-30]

I already solved the problem for graybodies, and showed that the graybody equation reduces to the blackbody equation. That's why it's useful to solve the simpler blackbody problem first, to provide a sanity check on the more complicated solution.

...Anything is better than your "thermal superconductors" that you then claim are different temperatures on different sides. Do you remember that is the second time you tried to pull that? I bet not. [Jane Q. Public, 2014-08-30]

I've never claimed that, but this is the second time you've tried to pretend I have. Once again:

... its outer temperature is 149.6F ... pretend the enclosing shell is a thermal superconductor, so its inner temperature is also 149.6F ... [Dumb Scientist]

So, first you postulate a thermal superconductor, and then assert that it has a far higher temperature on one side than on the other? What a magical world you must live in. [Jane Q. Public]

No, I said both sides of a thermal superconductor enclosing shell are at 149.6F.

Spencer's INITIAL description of his thought experiment. As I have told you several time. This first, then more if you want to get into it. I will not discuss this with you in the other order, AS I HAVE TOLD YOU. Because until you get that right, you're not going to get the other one right. If you continue to argue the other case first, then we are done, and I will write you off as hopeless. ... No "enclosing shell". Two parallel plates. The original thought experiment is two parallel plates (we can make them of equal dimensions just to simplify, but it's not necessary). I repeat: we briefly discussed "even if it were enclosing" but that's a complication of the original, and we'll solve the original first. [Jane Q. Public, 2014-08-30]

Once again, solving a problem without spherical symmetry means you'll have to solve for equilibrium temperatures which aren't constant across the heated and passive plates. Those equilibrium temperatures wouldn't be simple numbers. They'd be complicated functions that would vary across the plate surfaces. Contrast that with a spherically symmetric enclosing plate, where equilibrium temperatures are just simple numbers.

Are you disputing those facts, or do you really not see which of these problems is more complicated?

... Also, I don't think we're assuming black bodies. The best we can realistically do is grey bodies that absorb in all the relevant frequencies under discussion. ... [Jane Q. Public, 2014-08-30]

I already solved the problem for graybodies, and showed that the graybody equation reduces to the blackbody equation. That's why it's useful to solve the simpler blackbody problem first, to provide a sanity check on the more complicated solution.

...Anything is better than your "thermal superconductors" that you then claim are different temperatures on different sides. Do you remember that is the second time you tried to pull that? I bet not. [Jane Q. Public, 2014-08-30]

I've never claimed that, but this is the second time you've tried to pretend I have. Once again:

... its outer temperature is 149.6F ... pretend the enclosing shell is a thermal superconductor, so its inner temperature is also 149.6F ... [Dumb Scientist]

So, first you postulate a thermal superconductor, and then assert that it has a far higher temperature on one side than on the other? What a magical world you must live in. [Jane Q. Public]

No, I said both sides of a thermal superconductor enclosing shell are at 149.6F.