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.

... power in = power out. ... Using irradiance (power/m**2) simplifies the equation: electricity + sigmaT(c)**4 = sigmaT(h)**4

This is a joke, right? Trying to see if I'd catch it? Again, among other things you are substituting irradiance for power without factoring in any area. ... [Jane Q. Public, 2014-08-29]

Again, start with power in = power out through a boundary with surface area "A". Using irradiance (power/m^2) simplifies the equation because we can divide both sides by "A" to obtain irradiance in = irradiance out.

... I mentioned this to you several times, but you haven't picked up on it: just for one thing, you're claiming to be using flux but flux has an areal component which you are not accounting for. You say power in = power out, which may be true, but that total power is being transferred via emissive power, which is in W/m^2. Nowhere are you accounting for this. As I stated before: you are conflating power and emissive power, and you can't do that. Where are your areas? It might conserve energy but without areas you do not have the information required to calculate actual radiative temperature. ... [Jane Q. Public, 2014-08-29]

Again, as long as the enclosing shell is nearly the same size as the heated plate, those areas are nearly irrelevant. And because it's a simpler problem (like a tricycle) one should master it before trying to ride a bicycle with complicated view factors. I already specified my areas. Again, neglecting area ratios predicts that the heated plate warms from 150F to 235F after it's enclosed. Accounting for area ratios similar to Earth's predicts that the heated plate warms from 150F to 233.8F.

So the tricycle isn't too inaccurate compared to the bicycle, it's much easier to learn, and it provides a sanity check on the more complicated calculation. As the area ratio approaches "1.0" the bicycle should give the same answer as the simpler tricycle. And it does.

Incidentally, that tricycle is much more accurate than Jane's prediction that the heated plate remains at 150F even after it's enclosed.

... 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]

No. A spherical heated plate with a fully-enclosing shell has spherical symmetry, so the heated and enclosing plate temperatures are constant across their surfaces. That's why the equilibrium temperature solutions are just simple numbers.

However, if the passive plate doesn't fully enclose the heated plate then the heated and enclosing plate temperatures would be complicated functions of spherical coordinates theta and phi. That's a unicycle, not a tricycle.

... There are numerous sources, including physics and engineering textbooks, which contradict your analysis and conclusions. Why don't you try the engineering textbooks Latour cited, which have examples of real-world situations? After all: ultimately what we're talking about here is the real world, not a thought experiment. [Jane Q. Public, 2014-08-29]

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 without violating the first law, the second law, or the Stefan-Boltzmann law. Just like the greenhouse effect.

... Create a realistic scenario, draw yourself a diagram, and run some actual numbers on them rather than just tossing equations around without seeing how they fit together in the real world. ... [Jane Q. Public, 2014-08-29]

How ironic. I've explained how to derive equations for increasingly realistic scenarios, ran "actual numbers" and 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, while you're too busy saying things like this?

"... non-person... disingenuous and intended to mislead ... he is either lying ... dishonest ... intellectually dishonest ... intellectually dishonest ... Khayman80's intellectual dishonesty ... Pathetic. ... you've come out the loser in every case... you can't win a fucking argument. You don't know how. You don't understand logic. You've proved this many times. Get stuffed, and go away. The ONLY thing you are to me is an annoyance. I have NO respect for you either as a scientist or a person. ... cowardice ... odious person ... you look like a fool ... utterly and disgustingly transparent ... Now get lost. Your totally unjustified arrogance is irritating as hell. ... You are simply proving you don't know what you're talking about. ... Jesus, get a clue. This is just more bullshit. ... spewing bullshit ... You're making yourself look like a fool. ... Hahahahahaha!!! Jesus, you're a fool. ... a free lesson in humility... you either misunderstand, or you're lying. After 2 years of this shit, I strongly suspect it is the latter. ... Now I KNOW you're just spouting bullshit. ... if we assume you're being honest (which I do not in fact assume) ... I wouldn't mind a bit if the whole world saw your foolishness as clearly as I do. ... stream of BS... idiot ... Your assumptions are pure shit. ... I'm done babysitting you..." [Jane Q. Public]

"Jesus, you're a dumbshit. ... your adolescent, antisocial behavior ... keep making a fool of yourself. ... you're being such a dumbass ... your analysis of it is a total clusterfuck. ... you're so damned arrogant you think I'm the one being stupid. ... you were too goddamned stupid ..." [Jane Q. Public]

"... what a despicable human being you are ... an incorrigibly rude, insufferable human being ... Now I have given you your bone, doggie. GO AWAY. ... a clusterfuck pretending to be physics ... " [Jane Q. Public]

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.

... Since the temperature of every other object is less than that of the heat source, there is no net heat flow TO the heat source, therefore the heat source does not become hotter. This is, and has been, the whole of Latour's argument, and it is valid. It is not crazy speculation by some nitwit... [Jane Q. Public, 2014-08-02]

Again, Eq. 1 describes equilibrium temperature:

electricity + sigma*T_c^4 = sigma*T_h^4 (Eq. 1)

Eq. 1 shows that Jane and "the whole of Latour's argument" are wrong. Net heat transfer doesn't have to flow from plate to source in order to cause the heat source to be hotter. Just reducing the net heat flow from source to plate is sufficient to warm the plate, as long as electrical heating power is constant.

... you're conflating electrical power with "emissive power" or irradiance, which are different things, in different units. Sheesh. You'd at least expect a "physicist" to get that much right. So I gave that much away. And you still didn't deserve it. ... Now I have given you your bone, doggie. GO AWAY. [Jane Q. Public, 2014-08-28]

No. As I originally said: "Using irradiance (power/m^2) simplifies the equation... Sage solves Eq. 1 for a constant electric input of 509 W/m^2."

So the variable "electricity" has always been in the same units as irradiance, which made the equations simpler. The electrical power used by the heater is "electricity" times the surface area of the heated plate. I've repeatedly noted that electrical heating power is constant, which means that the variable "electricity" is also constant unless the heated plate shape-shifts to change its surface area. Just to be clear, I haven't been considering shapeshifting plates.

Again, it's fascinating that Jane keeps wrongly implying my previous calculations had units confused, but didn't point out the actual units confusion in the eq. 4 I posted.

... I'm not in the slightest confused. ... [Jane Q. Public, 2014-08-28]

That's what I told Demena.

... I still know things you don't. Why do you think I've felt free to be so glib? I've been watching you make a fool of yourself, ever since you revealed what a despicable human being you are (again, just my opinion of course, but I've had some confirmation). My advice to go do something more worthwhile was sincere. Because if you don't, after you are gone, I will quite happily reveal those things and your "legacy" won't be quite what you thought it was. That's not a threat in any way, it's just a description of the truth. ... [Jane Q. Public, 2014-08-28]

Empty bluster won't stop me from continuing to debunk your civilization-paralyzing misinformation as long as I can.

... you still have yet to share with us what this "civilization-paralyzing misinformation" is. It isn't in the links you provided above. And you're still wrong about Spencer and Latour. [Jane Q. Public, 2014-08-28]

Yes it was. And you're still spreading Dr. Latour's civilization-paralyzing Slayer misinformation:

... The plate cannot cause the heat source to be hotter because that would require NET heat transfer in the other direction. ... [Jane Q. Public, 2014-08-20]

No. Again, warming the heat source doesn't require net heat transfer from the plate to the source. At equilibrium, power in = power out. Because electrical heating power is constant, the heat source warms even if net "power out" decreases. It doesn't have to reverse direction (plate to source) in order to warm the source.

Maybe an analogy would help. Suppose water flows from a bathtub faucet at a rate of 1 liter/minute. The drain is open, letting water out at 1 liter/minute. Since water in = water out, the bathtub water level is constant.

Now partially close the drain so water only leaves at 0.5 liter/minute. Since water in > water out, the bathtub water level rises.

Raising the bathtub water level doesn't require that the drain reverse direction and start pumping water up from the drain into the bathtub. Because the faucet pours a constant 1 liter/minute into the tub, raising the water level only requires reducing the water out.

... I was only partly wrong about the NATO rounds. ... I wasn't wrong, my information was just old. ... [Jane Q. Public, 2014-08-27]

Condescendingly lecturing a veteran like this was wrong: "Bullshit, dude. Maybe where your tour was... Just plain bullshit. ... Give up, man. You are trying to argue with someone who knows what she's [she's?!?] talking about. ... Jeez, dude. Do you even read your own bullshit? ... You may know more than I do about what the military is currently doing, but I do know something about 5.56 ballistics, thank you very fucking much. ... maybe you know more about what the military is doing these days, but if that's what they're doing, they're being just plain stupid. ..."

... So sure, I've made some small errors. And admitted them when I did. But that is only a minority of links above, which you are apparently trying to claim are all "nonsense". Like the beta decay: after some initial confusion I asked how the oscillations take place, and someone answered. I admitted that I was wrong. ... [Jane Q. Public, 2014-08-27]

No, after delt0r answered, you insisted he must not have understood your point. After I repeated delt0r's point, you claimed that you had got yourself sorted out already and accused me of butting in and insulting you.

You've repeated this pattern ad nauseum. After your neutrino rant, you repeatedly claimed that I missed where you admitted you were wrong and asked me "why didn't you bother to repeat the part...?" when I actually had repeated that part and responded to it.

In fact, the more I read of these old streams, the more I've found where I was actually correct. (Like the one on bicycle stability for instance.) I have a copy of that paper right here and it says I was correct. ... [Jane Q. Public, 2014-08-27]

It's more likely that your Sauron-class Morton's demon told you that it says you were correct. Just like you've insisted you were still correct about punctuation despite never providing sentences with the plurals of i, a, and u.

... YOUR problem is that you claim these things are nonsense, but you haven't disproved a single one of them. Why not? ... in a lot of it I wasn't wrong at all, you just think I was. ... [Jane Q. Public, 2014-08-27]

Because you're galloping faster than any Gish Gallop I've ever seen, and because despite your protests you seldom accept refutations for longer than about 5 minutes anyway.

... One last thing, to anybody else who has bothered to wade through all his bullshit: ask yourselves why he's keeping a record of ALL the comments I made on Slashdot over a period of years that he thinks were wrong. Do YOU do that to people? No, you don't, do you? That's because YOU are probably a normal human being, who doesn't stalk or obsess over strangers. [Jane Q. Public, 2014-08-27]

I probably don't have more than about a month to live, so I'm obsessing over my legacy. The misinformation you're spreading seems like the biggest current threat to humanity, so I'll spend my final days debunking you.

... Your attempts to shame me haven't been coming off too well, you know. [Jane Q. Public, 2014-08-27]

One disturbing possibility is that you can't experience shame, which is why I'm trying to figure out why you're shamelessly posing as a woman. Maybe the way you were raised could help answer this question.

... I was seriously concerned that my dad might start thinking I was gay or something. :0) [Lonny Eachus, 2009-11-01]

I was sure by then my father must have been convinced I was gay or something. [Lonny Eachus, 2011-03-07]

... seriously thinking: "Oh, shit. My father probably thinks I'm gay or something now." [Lonny Eachus, 2011-12-23]

Well, you would have to know too that my father was a pretty serious bigot and gay-basher, both. It's how HE was raised. [Lonny Eachus, 2011-12-23]

I meant what I said to Demena. I dismissed the possibility that you're transgendered after you claimed that was quite literally not your problem. But if your gay-bashing bigot father left you confused about your gender then I'll apologize, retract my accusations, and support you as you experiment with your gender identity.

Releasing this burden might even let you stop spreading civilization-paralyzing misinformation. Jane/Lonny Eachus would have fewer stains on his legacy, and civilization would be less paralyzed. Win-win.

My point is that you've been spreading nonsense like a firehose for years, and each time your Sauron-class Morton's demon convinces you that you're right and the other person isn't very good at refutation. This doesn't just apply to your nonsense about climate change, dark matter, neutrino oscillation, the Casimir effect and Maxwell's equations, creationists, Obama birthers and 9/11 Truthers.

It also applies to your nonsense about conservation of energy, beta decay, quantum computing, nuclear isomers, Cherenkov radiation, virtual particles, infinities, string theory, cold fusion, R o s s i ' s E - C a t L E N R h o a x, peltier coolers, GPS, bicycle stability, control theory, hyperbolic trajectories, relativistic slingshots, replicators, the Kessel run, x-rays, gene therapy, Dr. Bur z y n s k i, ferret superflu, fluo ride, ethanol, petaflops, correlation/causation, failure probabilities, slavery, h o m o p h o b i a, the transgendered, punctuation, space flight, thrust, specific impulse, fly-by-wire, the FAA, airspace, inflation and the gold standard, capitalism, bitcoins, atom bombs, AK-74's, NATO rounds, firearm laws, state laws, Shock and Awe, naval bases, paranoia, Layzej's link, etc.