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... I was referring to your original "solution" to Spencer's problem, which you posted publicly on your website as a "refutation" of a comment of my own. Your explanation of how you found that solution led directly to a positive feedback loop, which I mentioned to you at the time. That has been a couple of years now. ... [Jane Q. Public, 2014-11-27]

Once again, I explained that the equations I'm using account for an infinite series of reflections. But as MIT explained, this infinite sum converges to a finite temperature. If Jane thinks he's found a mistake in MIT's derivation, please let everyone know exactly where.

And Jane, that wasn't a couple of years ago. I refuted your Sky Dragon Slayer nonsense 3 months ago, not a couple of years ago. It probably just feels like years because you've been cussing and screaming and insisting you're right and I'm wrong for hundreds of pages. Seriously, look at the index at the top of that comment, which has links to this never ending “conversation” LINK, LINK, LINK. BACKUP 1, 2, 3, 4, 5, 6, 7, 8, 9.

But you have never acknowledged your original error. Ever moving the goalposts, ever finding new "explanations" for how your "solution" somehow didn't ACTUALLY violate conservation of energy. [Jane Q. Public, 2014-11-27]

Jane, have you ever considered the possibility that I didn't make an error, and that you simply don't understand physics as well as professional physicists do? For instance, you screwed up the very first equation because you don't know how to apply conservation of energy to a boundary around the heated source. I've tried to show you how to derive that equation, but you've repeatedly refused. Why?

Furthermore, you won't even ask a physicist you respect if electrical heating power depends on the cooler chamber wall temperature. This would be even easier than writing down a single equation. Just ask Prof. Cox (or any other mainstream physicist) and their answer might finally help you see why your Sky Dragon Slaye

... I was referring to your original "solution" to Spencer's problem, which you posted publicly on your website as a "refutation" of a comment of my own. Your explanation of how you found that solution led directly to a positive feedback loop, which I mentioned to you at the time. That has been a couple of years now. ... [Jane Q. Public, 2014-11-27]

Once again, I explained that the equations I'm using account for an infinite series of reflections. But as MIT explained, this infinite sum converges to a finite temperature. If Jane thinks he's found a mistake in MIT's derivation, please let everyone know exactly where.

And Jane, that wasn't a couple of years ago. I refuted your Sky Dragon Slayer nonsense 3 months ago, not a couple of years ago. It probably just feels like years because you've been cussing and screaming and insisting you're right and I'm wrong for hundreds of pages. Seriously, look at the index at the top of that comment, which has links to this never ending “conversation” LINK, LINK, LINK. BACKUP 1, 2, 3, 4, 5, 6, 7, 8, 9.

But you have never acknowledged your original error. Ever moving the goalposts, ever finding new "explanations" for how your "solution" somehow didn't ACTUALLY violate conservation of energy. [Jane Q. Public, 2014-11-27]

Jane, have you ever considered the possibility that I didn't make an error, and that you simply don't understand physics as well as professional physicists do? For instance, you screwed up the very first equation because you don't know how to apply conservation of energy to a boundary around the heated source. I've tried to show you how to derive that equation, but you've repeatedly refused. Why?

Furthermore, you won't even ask a physicist you respect if electrical heating power depends on the cooler chamber wall temperature. This would be even easier than writing down a single equation. Just ask Prof. Cox (or any other mainstream physicist) and their answer might finally help you see why your Sky Dragon Slaye

Oops, I actually plotted inverse density versus half-value layer thicknesses. This doesn't affect the conclusion, but here's a plot with a corrected file name.

Oops, I actually plotted inverse density versus half-value layer thicknesses. This doesn't affect the conclusion, but here's a plot with a corrected file name.

Again, metric tons per square meter = thickness * density. That means the half-value layer should be inversely proportional to the shield's density. So if metric tons per square meter are relevant to the half-value layer, the half-value layer should be inversely proportional to the shield's density.

Yes, on a first glance it should. But in fact it does not. It highly depends on the material you use.

Again, to a good approximation, those half-value layers are inversely proportional to the shield's density.

I plotted those half-value layers against the inverse densities of concrete, steel, lead, tungsten and uranium. The blue squares are for the iridium source, and the red circles are for the cobalt source. Since those points lie close to a straight line, radiation absorption is determined primarily by density. So metric tons per square meter is a good first order approximation, at least for those materials.

Jane's obligations include continuing to spread misinformation about ocean acidification even after I've repeatedly debunked him.

So I predict that Jane's answer won't include any equations that could be used to calculate the enclosed source temperature. Instead, he'll probably grace us with another lengthy, incoherent rant about "problems" in my analysis which are (as usual) too vague to be expressed in equations. In the extremely unlikely event that Jane musters up the courage and competence to actually write down an equation that could be used to calculate the enclosed source temperature, it will almost certainly violate conservation of energy.

Once again, energy is conserved, which means that if you draw a boundary around some system (like the heated plate), power going in minus power going out equals the rate at which energy inside that boundary changes. At steady-state, that rate is zero because the system doesn't change. So at steady-state, power in = power out.

I've specified the dimensions. The heated plate is a sphere with radius 6371 mm, surface area A_h, temperature T_h and emissivity epsilon_h. The enclosing plate is a 1 mm thick concentric shell with emissivity epsilon_c, an inner radius of 6378 mm, surface area A_c1 and temperature T_c1 on the inside, and A_c2 and T_c2 on the outside. The chamber walls at temperature T_c are a concentric sphere with inner radius 6386 mm, so there's a 7 mm gap on both sides of the enclosing shell. The plates and walls are oxidized aluminum, which are treated as gray bodies.

Since the enclosing shell has no edges and has nearly the same area as the heated plate, MIT's infinite plate approximation describes net heat flow (in W/m^2):

net heat flow = sigma*(T_h^4 - T_c^4)/(1/epsilon_h + 1/epsilon_c - 1) (Eq. 2)

At steady-state, net heat flow (in W/m^2) equals the electrical input. Note that MIT's Eq. 2 reduces to my Eq. 1 for blackbodies where epsilon_h = epsilon_c = 1.

The plates and chamber walls are made of oxidized aluminum with emissivity = 0.11.

Here's my Eq. 2 using Jane's variable names:

net heat flow = sigma*(T(s)^4 - T(w)^4)/(1/E(s) + 1/E(w) - 1) (Eq. 2J)

Note that it reduces to my simpler blackbody Eq. 1 if E(s) = E(w) = 1.

If you'd like me to clarify what my variable names for a particular equation would be in your terminology, just ask.

At steady-state, net heat flow out (in W/m^2) equals "electricity". The first step is to calculate that constant variable "electricity" which describes electrical power per square meter heating the sphere to 150F without an enclosing shell. I calculated 29.4 W/m^2, which is less than with the simpler blackbody plates because aluminum isn't a perfect emitter or absorber.

I calculated 29.4 W/m^2, which is less than with the simpler blackbody plates because aluminum isn't a perfect emitter or absorber.

Show your calculations where we can see them. I'm not doing this just for me, I want to show other people just how much a clown you actually are. I am not going to install Sage today just to check your math, and probably neither is anybody else who sees this. ... I have reasons for wanting it public-readable, and I will accept nothing else. [Jane Q. Public, 2014-09-02]

#Calculate constant electrical power/area heating 1st plate.
var('sigma T_c T_h electricity epsilon_h epsilon_c')
eq1 = electricity == sigma*(T_h^4 - T_c^4)/(1/epsilon_h + 1/epsilon_c - 1)
soln1 = solve(eq1.subs(T_c=255.372,T_h=338.706,sigma=5.670373E-8,epsilon_h=0.11,epsilon_c=0.11),electricity)
soln1[0].rhs().n()

ANSWER: 29.3986743761843

Can we agree on that? If not, a month ago I said we could use Wikipedia’s equation which includes areas. After I mentioned view factors, Jane agreed that the relevant view factor is 1.0 or

You USED this before to ASSUME all surfaces were at the same temperature! I quoted you saying it in a post above, and you referenced that passage just the other day. In fact this was the source of much of the misunderstanding here. ... [Jane Q. Public, 2014-09-03]

Once again, I never said that. In reality, I said that both sides of a thermal superconductor are at the same temperature. This was the source of much of the misunderstanding here, and you strongly objected to the notion of a thermal superconductor. Again, that's why I calculated the small temperature difference across an aluminum shell with finite conductivity. [Dumb Scientist]

Yes, you did say that, and anybody who wants to can read it on your website. And you wrote it BEFORE any discussion with me of "thermal superconductors". I will quote it again here:

Electric input of 509 W/m^2 is constant and the walls are held at 0F (255K). Therefore, the second plate has to radiate the same power out as the heated plate did before it was enclosed. So energy conservation at equilibrium requires that the second plate be at 150F (339K).

You were referring to "the second plate", as opposed to the "heated plate". That corresponds to what I have been calling the "passive" or "enclosing" plate. ... [Jane Q. Public, 2014-09-03]

Once again, no. I never said that all surfaces were at the same temperature. I've already explained that the final outer temperature of the enclosing shell doesn't happen at the same time as the initial temperature of the heated plate. Initially, the heated plate is at 150F and the enclosing shell is cooler than 100F. But because power in > power out, the plates slowly warm to a new steady-state. By the time the outer temperature of the enclosing shell is ~149.6F (accounting for area differences), the heated plate is ~233.8F. This doesn't change even if we neglect area differences: the enclosing shell and the heated plate are never at the same temperature. Again, that's why I called them T_c and T_h.

So once again, I never said that all surfaces were at the same temperature.

If you don't particularly mind, could we finally take the very first step in this calculation? Please?

Yes, I mind very much. There is no point in doing any calculations at all until we rid you of the false assumptions you have been making about this experiment (as I have been trying to do). They have been leading to incorrect results, and moving on would be a waste of everybody's time. ... There is no equilibrium in this experiment, either thermal or radiative. Period. You may not assume them, or use formulas that are only appropriate for equilibrium. Get past that and move on, or stay stuck here. That's up to you. But unless and until you do, there is simply no need for me to go any further. Your refutation to this point has been demonstrated to be invalid. [Jane Q. Public, 2014-09-03]

I'm very sorry. I take full responsibility. Can we please move on?

... you can take out the epsilons, since I thought we had already agreed we don't need them. (If they represent emissivity.) ...

Those ESA absorptivities are for absorption of sunlight. Consider the first diagram here which shows that 6000K sunlight has much shorter wavelengths than the radiation from objects at the temperatures we're considering. In fact they hardly overlap. But the emissivities are for radiation emitted by much cooler objects. That's one reason why those ESA emissivities aren't equal to their absorptivities.

I repeat that these are under conditions of Earth-absorptive-surface insolation, which was what Spencer's experiment was supposed to emulate. But whatever.

But the emissivities are for radiation emitted by much cooler objects.

No, they are not. They are reported for the previously stated conditions: 1367 W/m^2 in incident radiation. But again: whatever. I already stated that this is not important enough to argue the point. I'm not conceding your point, but I'm willing to move on with gray bodies.

Once again, I never said that. In reality, I said that both sides of a thermal superconductor are at the same temperature. This was the source of much of the misunderstanding here, and you strongly objected to the notion of a thermal superconductor. Again, that's why I calculated the small temperature difference across an aluminum shell with finite conductivity.

Yes, you did say that, and anybody who wants to can read it on your website. And you wrote it BEFORE any discussion with me of "thermal superconductors". I will quote it again here:

Electric input of 509 W/m2 is constant and the walls are held at 0ÃF (255K). Therefore, the second plate has to radiate the same power out as the heated plate did before it was enclosed. So energy conservation at equilibrium requires that the second plate be at 150ÃF (339K).

You were referring to "the second plate", as opposed to the "heated plate". That corresponds to what I have been calling the "passive" or "enclosing" plate. And you further referred to a supposed thermal equilibrium that doesn't exist.

Which fantasy would you prefer we believe? A thermal superconductor that makes no sense in this context, or an equilibrium which does not exist in this context? And you don't have the excuse that you meant "steady state", because the figure you gave would only be appropriate for actual equilibrium.

But enough of old arguments. Let's move on.

I was using this definition: "When incoming solar energy is balanced by an equal flow of heat to space, Earth is in radiative equilibrium and global temperatures become relatively stable."

Great. Except that it doesn't pertain to Spencer's challenge for several reasons. First, the chamber walls in Spencer's experiment are not "empty" space, but a material body that is being actively refrigerated, while the "enclosing passive plate" is being heated on the other side. So that plate is not in radiative equilibrium with the chamber wall or with anything else for that matter. In fact that would be impossible. There are other reasons why that description does not match Spencer's challenge, but that is irrelevant for now. One is enough.

Dr. Spencer disagrees: "Eventually the second plate will also reach a state of equilibrium, where its average temperature (letâ(TM)s say 100 deg. F) stays constant with time."

It is unfortunate that Spencer plays almost as fast-and-loose with terms as you do. That is a steady state. It is NOT "equilibrium". They are different things.

If you don't particularly mind, could we finally take the very first step in this calculation? Please?

Yes, I mind very much. There is no point in doing any calculations at all until we rid you of the false assumptions you have been making about this experiment (as I have been trying to do). They h

Obviously we'll have to agree to disagree. But I thought you wanted to do some actual calculations? As you say, it's pretty damned hard to prove anything without calculating it all the way through. So why don't we take the first step?

... it's pretty damned hard to prove anything without calculating it all the way through. ... So why don't you draw a diagram, and simply perform all the calculations? ... I'll even go with your own example of the passive plate enclosing the heat source, for now. ... Then an enclosing plate is introduced, at a temperature (initially) less than that of the source. We can, if you wish, assume it is a hollow sphere, of some reasonable thickness, so the interior and exterior areas differ, and of a smaller external radius than the outside wall, so again they don't touch. Vacuum in between. And we begin our analysis. The starting point and equilibrium are both relevant points that should be calculated. ... I don't insist, but to avoid ambiguity and to make things expressible on a standard keyboard, this is how *I* would label things: S for heat source, so radiative temperature T of S would be T(s). Passive plate (or shell) P. Outside enclosure or wall W. Absorptivity A so absorptivity of P would be A(p). Emissivity E. ... [Jane Q. Public, 2014-09-02]

Here's my Eq. 2 using your variable names:

net heat flow = sigma*(T(s)^4 - T(w)^4)/(1/E(s) + 1/E(w) - 1) (Eq. 2J)

Note that it reduces to my simpler blackbody Eq. 1 if E(s) = E(w) = 1.

If you'd like me to clarify what my variable names for a particular equation would be in your terminology, just ask.

I've specified the dimensions. The heated plate is a sphere with radius 6371 mm and surface area A_h. The enclosing plate is a 1 mm thick concentric shell with an inner radius of 6378 mm, surface area A_c1 on the inside, and A_c2 on the outside. The chamber is also a concentric sphere with inner radius 6386 mm, so there's a 7 mm gap on both sides of the enclosing shell. Again, the plates and walls are oxidized aluminum.

At equilibrium, net heat flow out (in W/m^2) equals "electricity". The first step is to calculate that constant variable "electricity" which describes electrical power per square meter heating the sphere to 150F without an enclosing shell. I calculated 29.4 W/m^2, which is less than with the simpler blackbody plates because aluminum isn't a perfect emitter or absorber.

Can we agree on that? If so, we can move on to the next step, which is calculating the final outer surface temperature of the enclosing shell once it reaches equilibrium.

Electric input of 509 W/m2 is constant and the walls are held at 0F (255K). Therefore, the second plate has to radiate the same power out as the heated plate did before it was enclosed. So energy conservation at equilibrium requires that the second plate be at 150F (339K).

Utter nonsense. The temperature of the outside of your enclosing sphere is determined entirely by its absorption minus its emission, with absorptivity and emissivity factored in. If your interior heat source were emitting at (your figure) 509W/m^2, and that is being absorbed by the interior surface of your enclosing sphere (which MUST have larger radius than the source, since they can't contact), then your outside surface, being of even larger area, must therefore be colder. ... So you're INVENTING ENERGY OUT OF THIN AIR. ... [Jane Q. Public, 2014-09-02]

I've already showed you that the outer surface of an enclosing shell with an area ratio similar to Earth's warms to ~149.6F. I've explained that neglecting area ratios is a tricycle: a simple approximation that helps us learn. It's like the "frictionless pulley" or "massless rope" or "blackbody" approximations. Again, in this case 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.

... your prior analysis was still wrong. It is VERY easy to show this. Presume you have an initial source at T = 150 deg. F. It has a surface area of 1 m**2. Therefore (let's just assume your figure for power output here, it doesn't really matter and it's good enough for this illustration): it's emission is 509W/m**2. Let's say the EXTERIOR of your enclosing shell has an area of 2 m**2. However, your words (though in a slightly different context): power in = power out. Since the total power (W/m**2 times X m**2) must be the same in as out, the exterior of your shell cannot have the same irradiance. The same must be true if this were just one solid sphere, rather than a hollow sphere enclosing another sphere. Solving for the Stefan-Boltsmann relation at 509W/m**2 times 1 m**2 is total number of watts. If you try to multiply the same emission rate over 2 m**2 you get a DIFFERENT answer. That's just a fact. By assuming an external temperature of 150 deg. F, you have just created tangible energy from the vacuum. Congratulations. [Jane Q. Public, 2014-09-02]

When the area ratio departs far from 1.0, the tricycle becomes very inaccurate, so one should use the more complicated bicycle. But again, the Earth's area ratio is roughly 1.0025, so in that case the tricycle isn't too inaccurate.

Once again, I've already accounted for the area ratio to obtain the more complicated and more accurate solution.

But the second plate also radiates the same power in, toward the enclosed heated plate. Just like the cold chamber walls do. Now consider conservation of energy just inside the second plate (but outside the first) at equilibrium. We can solve for the insulated heated plate's temperature using Eq. 1 by setting Tc = 150F (339K). That yields an insulated heated plate temperature of 235F (386K).

No, it doesn't! The irradiation is total for the entire hollow sphere, not for each surface. You have to divide the total irradiance b

Electric input of 509 W/m2 is constant and the walls are held at 0F (255K). Therefore, the second plate has to radiate the same power out as the heated plate did before it was enclosed. So energy conservation at equilibrium requires that the second plate be at 150F (339K).

Utter nonsense. The temperature of the outside of your enclosing sphere is determined entirely by its absorption minus its emission, with absorptivity and emissivity factored in. If your interior heat source were emitting at (your figure) 509W/m^2, and that is being absorbed by the interior surface of your enclosing sphere (which MUST have larger radius than the source, since they can't contact), then your outside surface, being of even larger area, must therefore be colder. ... So you're INVENTING ENERGY OUT OF THIN AIR. ... [Jane Q. Public, 2014-09-02]

I've already showed you that the outer surface of an enclosing shell with an area ratio similar to Earth's warms to ~149.6F. I've explained that neglecting area ratios is a tricycle: a simple approximation that helps us learn. It's like the "frictionless pulley" or "massless rope" or "blackbody" approximations. Again, in this case 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.

... your prior analysis was still wrong. It is VERY easy to show this. Presume you have an initial source at T = 150 deg. F. It has a surface area of 1 m**2. Therefore (let's just assume your figure for power output here, it doesn't really matter and it's good enough for this illustration): it's emission is 509W/m**2. Let's say the EXTERIOR of your enclosing shell has an area of 2 m**2. However, your words (though in a slightly different context): power in = power out. Since the total power (W/m**2 times X m**2) must be the same in as out, the exterior of your shell cannot have the same irradiance. The same must be true if this were just one solid sphere, rather than a hollow sphere enclosing another sphere. Solving for the Stefan-Boltsmann relation at 509W/m**2 times 1 m**2 is total number of watts. If you try to multiply the same emission rate over 2 m**2 you get a DIFFERENT answer. That's just a fact. By assuming an external temperature of 150 deg. F, you have just created tangible energy from the vacuum. Congratulations. [Jane Q. Public, 2014-09-02]

When the area ratio departs far from 1.0, the tricycle becomes very inaccurate, so one should use the more complicated bicycle. But again, the Earth's area ratio is roughly 1.0025, so in that case the tricycle isn't too inaccurate.

Once again, I've already accounted for the area ratio to obtain the more complicated and more accurate solution.

But the second plate also radiates the same power in, toward the enclosed heated plate. Just like the cold chamber walls do. Now consider conservation of energy just inside the second plate (but outside the first) at equilibrium. We can solve for the insulated heated plate's temperature using Eq. 1 by setting Tc = 150F (339K). That yields an insulated heated plate temperature of 235F (386K).

No, it doesn't! The irradiation is total for the entire hollow sphere, not for each surface. You have to divide the total irradiance b

... 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. ...

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

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 bath

... 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'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 a

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,

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,

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,

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,

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,

... since you mention power... are you sure you don't have your units confused somewhere? But oops... I told you I wouldn't give you any more hints. ... I know they [the PSI Slayers] will (quite correctly) tear your arguments to shreds, and I even know how they'll do it. ... [Jane Q. Public, 2014-08-04]

... I know where you're making at least one mistake, but I already told you that you're going to have to discover it on your own. [Jane Q. Public, 2014-08-07]

It's fascinating that you'd wrongly implied my previous calculations had units confused somewhere, but haven't pointed out the actual units confusion in the eq. 4 I posted yesterday.

I made a mistake by forgetting to divide by the 1mm thickness "x" of the enclosing shell:

electricity = k*(T_h - T_c)/x (Eq. 4)

Here's the corrected Sage worksheet; the old wrong worksheet is here. I'm sorry for any confusion this caused, and I've corrected the equation at Dumb Scientist.

The corrected temperatures with the aluminum enclosing shell are so close to those with the superconducting shell that the differences don't show up with the four significant figures I'm using. So my original thermal superconductor approximation was even more accurate than I thought.

"... 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]

Jane, instead of typing all those charming statements, have you considered that it might be quicker and easier to just write down the equation describing conservation of energy aroun

... since you mention power... are you sure you don't have your units confused somewhere? But oops... I told you I wouldn't give you any more hints. ... I know they [the PSI Slayers] will (quite correctly) tear your arguments to shreds, and I even know how they'll do it. ... [Jane Q. Public, 2014-08-04]

... I know where you're making at least one mistake, but I already told you that you're going to have to discover it on your own. [Jane Q. Public, 2014-08-07]

It's fascinating that you'd wrongly implied my previous calculations had units confused somewhere, but haven't pointed out the actual units confusion in the eq. 4 I posted yesterday.

I made a mistake by forgetting to divide by the 1mm thickness "x" of the enclosing shell:

electricity = k*(T_h - T_c)/x (Eq. 4)

Here's the corrected Sage worksheet; the old wrong worksheet is here. I'm sorry for any confusion this caused, and I've corrected the equation at Dumb Scientist.

The corrected temperatures with the aluminum enclosing shell are so close to those with the superconducting shell that the differences don't show up with the four significant figures I'm using. So my original thermal superconductor approximation was even more accurate than I thought.

"... 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]

Jane, instead of typing all those charming statements, have you considered that it might be quicker and easier to just write down the equation describing conservation of energy aroun

More importantly, can we agree that in equilibrium, power in = power out?

No. I am not aware of any "conservation of power" law. [Jane Q. Public, 2014-08-02]

Energy is conserved, which means that if you draw a boundary around some system (like the heated plate), power going in minus power going out equals the rate at which energy inside that boundary changes. At equilibrium, that rate is zero because the system doesn't change. So at equilibrium, power in = power out. Jane replied:

... I already told you I was being an ass about your "power in equals power out" thing. Trying to lecture me about conservation of energy is particularly pointless, since I need no such lesson. ... [Jane Q. Public, 2014-08-04]

Jane claims he needs no such lesson because he said:

I admit to being an ass there. Mea culpa. But it's irrelevant. As long as the power used by the source and the power used by the cooler are constant as required, any relationship between them has no bearing on the experiment. [Jane Q. Public, 2014-08-02]

No, the fundamental principle used to determine equilibrium temperatures isn't irrelevant. Anyone making that claim either needs a lesson about conservation of energy, or is deliberately spreading misinformation.

The basis of all my calculations is the very relevant principle that in equilibrium, power in = power out. I've never even mentioned the power used by the cooler of the chamber walls, so Jane either needs a lesson about conservation of energy or Jane's deliberately spreading misinformation. Which is it?

Remember that conservation of energy at equilibrium let us calculate the 233.8F equilibrium temperature of a heated plate enclosed by a superconducting shell. But we can also account for the finite thermal conductivity of an aluminum shell using this same relevant principle by drawing a boundary within the enclosing shell.

The same relevant principle applies: in equilibrium, power in = power out. Again, electrical power flows in. But all the other boundaries we drew were in vacuum, so heat transfer was by radiation. This time the boundary is inside aluminum, so heat transfer out is by thermal conduction.

electricity = k*(T_h - T_c) (Eq. 4)

For aluminum, thermal conductivity k = 215 W/(m*K). Sage solves this equation for an equilibrium inner shell temperature of 149.9F rather than 149.6F for a superconducting shell. This warms the enclosed plate to 234.0F rather than 233.8F for a superconducting shell.

Hopefully this exercise shows how useful it is to start with the widely applicable principle that in equilibrium, power in = power out. Hopefully it's also clear that none of these equations has anything to do with the power used by the cooler. 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.

Why does Jane wrongly claim that the fundamental principle used to determine equilibrium temperatures is "irrelevant"? Does Jane need a lesson about conservation of energy, or is he deliberately spreading misinformation?

"If you don't thi

More importantly, can we agree that in equilibrium, power in = power out?

No. I am not aware of any "conservation of power" law. [Jane Q. Public, 2014-08-02]

Energy is conserved, which means that if you draw a boundary around some system (like the heated plate), power going in minus power going out equals the rate at which energy inside that boundary changes. At equilibrium, that rate is zero because the system doesn't change. So at equilibrium, power in = power out. Jane replied:

... I already told you I was being an ass about your "power in equals power out" thing. Trying to lecture me about conservation of energy is particularly pointless, since I need no such lesson. ... [Jane Q. Public, 2014-08-04]

Jane claims he needs no such lesson because he said:

I admit to being an ass there. Mea culpa. But it's irrelevant. As long as the power used by the source and the power used by the cooler are constant as required, any relationship between them has no bearing on the experiment. [Jane Q. Public, 2014-08-02]

No, the fundamental principle used to determine equilibrium temperatures isn't irrelevant. Anyone making that claim either needs a lesson about conservation of energy, or is deliberately spreading misinformation.

The basis of all my calculations is the very relevant principle that in equilibrium, power in = power out. I've never even mentioned the power used by the cooler of the chamber walls, so Jane either needs a lesson about conservation of energy or Jane's deliberately spreading misinformation. Which is it?

Remember that conservation of energy at equilibrium let us calculate the 233.8F equilibrium temperature of a heated plate enclosed by a superconducting shell. But we can also account for the finite thermal conductivity of an aluminum shell using this same relevant principle by drawing a boundary within the enclosing shell.

The same relevant principle applies: in equilibrium, power in = power out. Again, electrical power flows in. But all the other boundaries we drew were in vacuum, so heat transfer was by radiation. This time the boundary is inside aluminum, so heat transfer out is by thermal conduction.

electricity = k*(T_h - T_c) (Eq. 4)

For aluminum, thermal conductivity k = 215 W/(m*K). Sage solves this equation for an equilibrium inner shell temperature of 149.9F rather than 149.6F for a superconducting shell. This warms the enclosed plate to 234.0F rather than 233.8F for a superconducting shell.

Hopefully this exercise shows how useful it is to start with the widely applicable principle that in equilibrium, power in = power out. Hopefully it's also clear that none of these equations has anything to do with the power used by the cooler. 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.

Why does Jane wrongly claim that the fundamental principle used to determine equilibrium temperatures is "irrelevant"? Does Jane need a lesson about conservation of energy, or is he deliberately spreading misinformation?

"If you don't thi

Global-warming proponents betray science by shutting down debate ow.ly/Av6AX [CFACT, retweeted by Lonny Eachus, 2014-08-19]

"Climate science” isn’t “settled”, at all. On the contrary, it’s very Unsettled. ow.ly/Av6AX [Lonny Eachus, 2014-08-19]

Lonny's link claims that:

"... Most discussion on the science of AGW revolves around the climatic effects of increased levels of carbon dioxide in the atmosphere. How it got there in the first place- the assumption being that increased carbon dioxide arises overwhelmingly from human activities- is often taken for granted. Yet Salby believed that he had uncovered clear evidence that this was not the case, as his trip to Europe was designed to expose. ... the IPCC declared in its fourth assessment report, in 2007: “The increase in atmospheric CO2 is known to be caused by human activities.” Salby contends that the IPCC’s claim isn’t supported by observations. ... In Salby’s view, the evidence actually suggests that the causality underlying AGW should be reversed. Rather than increased levels of CO2 in the atmosphere triggering global temperatures to rise, rising global temperatures come first- and account for the great majority of changes in net emissions of CO2... temperature appears more likely to be the cause, rather than the effect, of observed atmospheric changes. Further, Salby presents satellite observations showing that the highest levels of CO2 are present not over industrialized regions but over relatively uninhabited and nonindustrialized areas, such as the Amazon. ... Salby also contends that temperature alone can largely account for the rise in atmospheric CO2 through the earlier part of the twentieth century... University of Oslo geosciences professor Ole Humlum published a landmark 2012 paper demonstrating that changes of CO2 follow changes of temperature, implying the same cause and effect. ..."

I told Jane that humans are responsible for the change in CO2 concentration. Jane even seemed to agree, calling contrary claims "ridiculous". But today Jane/Lonny regressed again, linking to an article making these ridiculous claims even after Jane said:

"I haven't intentionally disputed this. Not for many years, anyway. I suppose I might have, 4-5 years ago, when I knew next to nothing about the subject. So who are you arguing with? ... not only arguing with yourself (since I was not present), but also (again as usual) arguing about something I didn't even say. I wasn't arguing with you about those things. So why did you try to make it appear I did? Why were you trying to give the impression I said something I did not in fact say? ... it's doubly hilarious that you're trying to argue with me about something I told you in plain English I wasn't even arguing. Only you."

But Jane/Lonny Eachus is still arguing about the fact that we're responsible for the CO2 rise by linking to that absurd rant and claiming it makes climate science "very Unsettled". The rant Jane/Lonny linked repeats Salby's ridiculous argument, Humlum's ridiculous calculus mistake, and John O'Sullivan's ridiculous misinformation about satellite observations. I've told Jane that they’re ignoring simple accounting, decreasing oxygen, calculus, the seasons, increa

Global-warming proponents betray science by shutting down debate ow.ly/Av6AX [CFACT, retweeted by Lonny Eachus, 2014-08-19]

"Climate science” isn’t “settled”, at all. On the contrary, it’s very Unsettled. ow.ly/Av6AX [Lonny Eachus, 2014-08-19]

Lonny's link claims that:

"... Most discussion on the science of AGW revolves around the climatic effects of increased levels of carbon dioxide in the atmosphere. How it got there in the first place- the assumption being that increased carbon dioxide arises overwhelmingly from human activities- is often taken for granted. Yet Salby believed that he had uncovered clear evidence that this was not the case, as his trip to Europe was designed to expose. ... the IPCC declared in its fourth assessment report, in 2007: “The increase in atmospheric CO2 is known to be caused by human activities.” Salby contends that the IPCC’s claim isn’t supported by observations. ... In Salby’s view, the evidence actually suggests that the causality underlying AGW should be reversed. Rather than increased levels of CO2 in the atmosphere triggering global temperatures to rise, rising global temperatures come first- and account for the great majority of changes in net emissions of CO2... temperature appears more likely to be the cause, rather than the effect, of observed atmospheric changes. Further, Salby presents satellite observations showing that the highest levels of CO2 are present not over industrialized regions but over relatively uninhabited and nonindustrialized areas, such as the Amazon. ... Salby also contends that temperature alone can largely account for the rise in atmospheric CO2 through the earlier part of the twentieth century... University of Oslo geosciences professor Ole Humlum published a landmark 2012 paper demonstrating that changes of CO2 follow changes of temperature, implying the same cause and effect. ..."

I told Jane that humans are responsible for the change in CO2 concentration. Jane even seemed to agree, calling contrary claims "ridiculous". But today Jane/Lonny regressed again, linking to an article making these ridiculous claims even after Jane said:

"I haven't intentionally disputed this. Not for many years, anyway. I suppose I might have, 4-5 years ago, when I knew next to nothing about the subject. So who are you arguing with? ... not only arguing with yourself (since I was not present), but also (again as usual) arguing about something I didn't even say. I wasn't arguing with you about those things. So why did you try to make it appear I did? Why were you trying to give the impression I said something I did not in fact say? ... it's doubly hilarious that you're trying to argue with me about something I told you in plain English I wasn't even arguing. Only you."

But Jane/Lonny Eachus is still arguing about the fact that we're responsible for the CO2 rise by linking to that absurd rant and claiming it makes climate science "very Unsettled". The rant Jane/Lonny linked repeats Salby's ridiculous argument, Humlum's ridiculous calculus mistake, and John O'Sullivan's ridiculous misinformation about satellite observations. I've told Jane that they’re ignoring simple accounting, decreasing oxygen, calculus, the seasons, increa

Clicking "Try Sage Online" doesn't require downloading or installing Sage on your computer. But more importantly:

More importantly, can we agree that in equilibrium, power in = power out?

No. I am not aware of any "conservation of power" law. [Jane Q. Public]

Energy is conserved, which means that if you draw a boundary around some system (like the heated plate), power going in minus power going out equals the rate at which energy inside that boundary changes. At equilibrium, that rate is zero because the system doesn't change. So at equilibrium, power in = power out.

That's the basis of all these calculations, which is why I've repeatedly asked if we could agree on it.

Once again, can we agree that in equilibrium, power in = power out?

For the moment, I'll assume we can. If not, please explain why you don't agree that in equilibrium, power in = power out.

I'm sorry that I didn't realize earlier that we have such a fundamental disagreement. I should've been building a common understanding of equilibrium and conservation of energy rather than solving increasingly complicated thought experiments. So let's take this step by step and see if we can agree on anything.

Let's start with conservation of energy just inside the chamber walls at equilibrium: power in = power out.

A blackbody plate is heated by constant electrical power flowing in. Blackbody cold walls at 0F (T_c = 255K) also radiate power in. The heated plate at 150F (T_h = 339K) radiates power out. Using irradiance (power/m^2) simplifies the equation:

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

(Eq. 1 looks better in LaTeX, but hopefully this version is legible.)

Yes/No: can we agree that Eq. 1 is based on the Stefan-Boltzmann law and correctly describes conservation of energy just inside the chamber walls at equilibrium?

If yes, the next step is to solve Eq. 1 for the constant electrical input using a calculator or the Sage worksheet I provided.

If no, could you please write down the equation you think correctly describes conservation of energy just inside the chamber walls at equilibrium?

Earlier I made an offhand remark that enclosing the heated plate is like suddenly warming the chamber walls. This simpler scenario might be more helpful. Suppose the chamber walls are suddenly warmed from 0F to 149F. What will happen to the heated plate if the electrical power heating the plate remains constant? If you claim it would remain at 150F, think carefully about energy conservation at equilibrium. When the walls were at 0F, the plate was in equilibrium because power in = power out. But now the net power radiating out is much smaller, which means power in > power out. So what happens to the heated plate?

Instead of saying "an aluminum plate warms the inner plate" perhaps I should've said "an aluminum plate warms the enclosed heated plate." Maybe this will help distinguish between the inner surface of the enclosing plate and the enclosed heated plate. I'm sorry for any confusion this caused, and corrected it at Dumb Scientist.

Nice link. Do you really expect me to read that .sws file? How about something human-readable? [Jane Q. Public]

These open source Sage worksheets show my work for these thought experiments. Clicking "Try Sage Online" would let you upload my third worksheet, and hitting shift-enter a few times would recalculate all its answers. But in case you don't want to do that, here's a formatted copy of that worksheet and its answers:

#Calculate constant electrical power/area heating 1st plate.
var('sigma T_c T_h electricity epsilon_h epsilon_c')
eq1 = electricity == sigma*(T_h^4 - T_c^4)/(1/epsilon_h + 1/epsilon_c - 1)
soln1 = solve(eq1.subs(T_c=255.372,T_h=338.706,sigma=5.670373E-8,epsilon_h=0.11,epsilon_c=0.11),electricity)
soln1[0].rhs().n()

ANSWER = 29.3986743761843

6379^2/6371^2.n()

ANSWER = 1.00251295644620

338.706*1.00251295644620^(-.25).n()

ANSWER = 338.493545219805

#Completely surrounded by 2nd plate
soln2 = solve(eq1.subs(T_c=338.493545219805,electricity=29.3986743761843,sigma=5.670373e-8,epsilon_h=0.11,epsilon_c=0.11),T_h)
soln2[0].rhs().n()

ANSWER = 385.286813818721*I

This could also be done on a calculator, which is why I explained how to derive the equations using the principle that at equilibium, power in = power out.

... 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. Accounting for aluminum's finite conductivity would mean its inner temperature would be higher than its outer temperature. If you'd like, we could see how an aluminum plate warms the inner plate higher than the 233.8F it would be at with a superconducting plate. Just let me know, and I'll do the calculations.

But I don't think that would be helpful yet, because I didn't realize we have a fundamental disagreement:

More importantly, can we agree that in equilibrium, power in = power out?

No. I am not aware of any "conservation of power" law. [Jane Q. Public]

Energy is conserved, which means that if you draw a boundary around some system (like the heated plate), power going in minus power going out equals the rate at which energy inside that boundary changes. At equilibrium, that rate is zero because the system doesn't change. So at equilibrium, power in = power out.

That's the basis of all these calculations, which is why I've repeatedly asked if we could agree on it.

Once again, can we agree that in equilibrium, power in = power out?

For the moment, I'll assume we can. If no

Jane's still talking about plate areas, but he's definitely not concerned.

Let's see how a 0.2% larger enclosing plate affects equilibrium temperatures. The heated plate is a sphere with radius 6371 mm and surface area A_h. The enclosing plate is a 1 mm thick concentric shell with an inner radius of 6378 mm, surface area A_c1 on the inside, and A_c2 on the outside. The chamber is also a concentric sphere with inner radius 6386 mm, so there's a 7 mm gap on both sides of the enclosing shell. Again, the plates and walls are oxidized aluminum.

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):

A_h*T_h^4 = A_c2*T_c2^4 (Eq. 3)

For the moment, let's pretend the enclosing shell is a thermal superconductor, so its inner temperature is also 149.6F (338.5K). Energy conservation at equilibrium just inside the enclosing shell shows that the heated sphere will warm to an equilibrium temperature of 233.8F (385.3K)

Note that 233.8F is warmer than the heated sphere's original 150.0F equilibrium temperature.

We could keep making this thought experiment more realistic, but that wouldn't change the fact that enclosing the heated plate makes it warmer. For instance, instead of correcting the temperature manually as I did in Eq. 3, we could use Wikipedia's equation which includes areas. Or we could account for the enclosing shell's finite conductivity, but that would just make the heated plate even hotter.

Again, Dr. Latour and the Sky Dragon Slayers are wrong.

We've determined equilibrium temperatures in a simple example, so let's solve a more general example.

Jane's concerned that the enclosing plate is bigger than the heated plate. But Earth's mean radius is 6371 km, and the effective radiating level is ~7 km higher, so these surface areas are only ~0.2% different. Of course, in a thought experiment this difference can be made arbitrarily smaller. Despite Jane's protests, this doesn't change the fact that enclosing the heated plate makes it warmer.

More importantly, I treated the plates as blackbodies where absorptivity alpha = 1 and emissivity epsilon = 1. This is a reasonable approximation for plates made of carbon nanotube arrays (PDF) which have alpha = ~0.99955. But more conventional plates have alpha and epsilon considerably less than 1.

The next step is to treat the plates as graybodies where absorptivity and emissivity are independent of wavelength, so they appear gray. Kirchoff's Law states that absorptivity = emissivity for graybodies.

MIT calculates heat transfer between graybody plates using an infinite sum of emission, reflection and absorption. Using my variable names, their final expression is:

net heat flow = sigma*(T_h^4 - T_c^4)/(1/epsilon_h + 1/epsilon_c - 1) (Eq. 2)

(Again, Eq. 2 looks better in LaTeX, but hopefully this version is legible.)

At equilibrium, net heat flow equals the electrical input. Note that MIT's Eq. 2 reduces to my Eq. 1 for blackbodies where epsilon_h = epsilon_c = 1.

Suppose the plates and chamber walls are made of oxidized aluminum with emissivity = 0.11. In this case, Sage solves Eq. 2 for a constant electric input of 29.6 W/m^2, which is lower than before because aluminum doesn't radiate as well as a blackbody.

Using Eq. 2 and the same reasoning as before, fully enclosing the heated plate warms it to the same equilibrium temperature of 235F (386K). Fully exposing the plate to the cosmic microwave background radiation cools it to 13F (263K), which is lower than before because the CMBR is a blackbody and aluminum chamber walls aren't.

So even for graybody plates, MIT's mainstream physics refutes Dr. Latour's nonsensical claim that the enclosed heated plate remains at 150F. They also use this equation to explain how thermos bottles insulate drinks, and describe the same radiation shields used since at least

We've determined equilibrium temperatures in a simple example, so let's solve a more general example.

Jane's concerned that the enclosing plate is bigger than the heated plate. But Earth's mean radius is 6371 km, and the effective radiating level is ~7 km higher, so these surface areas are only ~0.2% different. Of course, in a thought experiment this difference can be made arbitrarily smaller. Despite Jane's protests, this doesn't change the fact that enclosing the heated plate makes it warmer.

More importantly, I treated the plates as blackbodies where absorptivity alpha = 1 and emissivity epsilon = 1. This is a reasonable approximation for plates made of carbon nanotube arrays (PDF) which have alpha = ~0.99955. But more conventional plates have alpha and epsilon considerably less than 1.

The next step is to treat the plates as graybodies where absorptivity and emissivity are independent of wavelength, so they appear gray. Kirchoff's Law states that absorptivity = emissivity for graybodies.

MIT calculates heat transfer between graybody plates using an infinite sum of emission, reflection and absorption. Using my variable names, their final expression is:

net heat flow = sigma*(T_h^4 - T_c^4)/(1/epsilon_h + 1/epsilon_c - 1) (Eq. 2)

(Again, Eq. 2 looks better in LaTeX, but hopefully this version is legible.)

At equilibrium, net heat flow equals the electrical input. Note that MIT's Eq. 2 reduces to my Eq. 1 for blackbodies where epsilon_h = epsilon_c = 1.

Suppose the plates and chamber walls are made of oxidized aluminum with emissivity = 0.11. In this case, Sage solves Eq. 2 for a constant electric input of 29.6 W/m^2, which is lower than before because aluminum doesn't radiate as well as a blackbody.

Using Eq. 2 and the same reasoning as before, fully enclosing the heated plate warms it to the same equilibrium temperature of 235F (386K). Fully exposing the plate to the cosmic microwave background radiation cools it to 13F (263K), which is lower than before because the CMBR is a blackbody and aluminum chamber walls aren't.

So even for graybody plates, MIT's mainstream physics refutes Dr. Latour's nonsensical claim that the enclosed heated plate remains at 150F. They also use this equation to explain how thermos bottles insulate drinks, and describe the same radiation shields used since at least

We've determined equilibrium temperatures in a simple example, so let's solve a more general example.

Jane's concerned that the enclosing plate is bigger than the heated plate. But Earth's mean radius is 6371 km, and the effective radiating level is ~7 km higher, so these surface areas are only ~0.2% different. Of course, in a thought experiment this difference can be made arbitrarily smaller. Despite Jane's protests, this doesn't change the fact that enclosing the heated plate makes it warmer.

More importantly, I treated the plates as blackbodies where absorptivity alpha = 1 and emissivity epsilon = 1. This is a reasonable approximation for plates made of carbon nanotube arrays (PDF) which have alpha = ~0.99955. But more conventional plates have alpha and epsilon considerably less than 1.

The next step is to treat the plates as graybodies where absorptivity and emissivity are independent of wavelength, so they appear gray. Kirchoff's Law states that absorptivity = emissivity for graybodies.

MIT calculates heat transfer between graybody plates using an infinite sum of emission, reflection and absorption. Using my variable names, their final expression is:

net heat flow = sigma*(T_h^4 - T_c^4)/(1/epsilon_h + 1/epsilon_c - 1) (Eq. 2)

(Again, Eq. 2 looks better in LaTeX, but hopefully this version is legible.)

At equilibrium, net heat flow equals the electrical input. Note that MIT's Eq. 2 reduces to my Eq. 1 for blackbodies where epsilon_h = epsilon_c = 1.

Suppose the plates and chamber walls are made of oxidized aluminum with emissivity = 0.11. In this case, Sage solves Eq. 2 for a constant electric input of 29.6 W/m^2, which is lower than before because aluminum doesn't radiate as well as a blackbody.

Using Eq. 2 and the same reasoning as before, fully enclosing the heated plate warms it to the same equilibrium temperature of 235F (386K). Fully exposing the plate to the cosmic microwave background radiation cools it to 13F (263K), which is lower than before because the CMBR is a blackbody and aluminum chamber walls aren't.

So even for graybody plates, MIT's mainstream physics refutes Dr. Latour's nonsensical claim that the enclosed heated plate remains at 150F. They also use this equation to explain how thermos bottles insulate drinks, and describe the same radiation shields used since at least

Again, he's completely wrong. The hotter bar absorbs cold back-radiation, and T does not remain 150F. That's why I refuted Dr. Latour by showing that a completely enclosed heated plate reaches an equilibrium temperature of 235F (386K), which is less than the infinite temperature he claimed.

Hahahahahaha!!! Jesus, you're a fool. THAT ISN'T WHAT HE CLAIMED. Quite the contrary. He claimed that a completely enclosed plate DOES NOT reach infinite temperature, which of course agrees with observations. [Jane Q. Public, 2014-08-01]

Again, Dr. Latour claimed that mainstream physics, which includes absorption of cold back-radiation, "would constitute creation of energy, a violation of the first law of thermodynamics." You've even repeated his claim:

... the temperature would go up until it outshone the rest of the universe, or it would cool down to zero. ... [Jane Q. Public, 2013-12-24]

"I told you what I think happens to that 10 Joules. By the first law of thermodynamics it doesn't just disappear so what happens to it?"

Yes, I know you told me but that doesn't happen. It would be a violation of the First Law of Thermodynamics. ... The system would never reach equilibrium, but would continue warming to infinity (if such a thing as infinite temperature existed). It would soon destroy itself from all this extra energy that is coming from nowhere. ... [Jane Q. Public, 2014-02-17]

That's why I've repeatedly told you that:

"Dr. Latour was wrong to claim that mainstream physics predicts the heated plate warms infinitely."

"I refuted Dr. Latour's claim that mainstream physics predicts infinite warming..."

If you're retracting your claim that absorbing cold back-radiation (i.e. mainstream physics) would violate the first law and "continue warming to infinity" then that's great news!

Here's one way you are wrong. In any realistic system, the enclosing plate would be of larger dimensions than the internal source, however slightly. So while the total re-radiated energy might be the same, it is spread over a larger area, so the energy density (and therefore temperature) would be lower. How did you allow a layman to catch you in such an elementary error? Not that I had any obligation to do so. Your argument is with him, not me. Just consider it a free lesson in humility. [Jane Q. Public, 2014-08-01]

The key phrase is "however slightly" because that difference can be made arbitrarily small. Since the only objection you've raised is arbitrarily small, does that mean you now see that Dr. Latour is wrong to claim that the heated plate will stay at 150F after the second plate is added, because he wrongly claims that absorbing cold back-radiation would violate the first law?

If not, maybe it would help if we kept checking my calculations step by step. For the simplest case of blackbody plates with arbitrarily similar areas, this equation represents conservation of energy at equilibrium:

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

His actual argument is that "k is the fraction of re-radiation from the second bar absorbed by the first hotter bar... k must be identically zero, so no cold back-radiation is absorbed and T remains 150. Quod Erat Demonstrandum, QED."

Again, he's completely wrong. The hotter bar absorbs cold back-radiation, and T does not remain 150F. That's why I refuted Dr. Latour by showing that a completely enclosed heated plate reaches an equilibrium temperature of 235F (386K), which is less than the infinite temperature he claimed.

Maybe it would help if we checked my calculations step by step. Start with conservation of energy just inside the chamber walls at equilibrium: power in = power out.

The plate is heated by constant electrical power flowing in. The cold walls at 0F (T_c = 255K) also radiate power in. The heated plate at 150F (T_h = 339K) radiates power out. Using irradiance (power/m^2) simplifies the equation:

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

(Eq. 1 looks better in LaTeX, but hopefully this version is legible.)

Yes/No: can we agree that Eq. 1 is based on the Stefan-Boltzmann law and correctly describes conservation of energy just inside the chamber walls at equilibrium?

If yes, the next step is to solve Eq. 1 for the constant electrical input using a calculator or the Sage worksheet I provided.

If no, could you please write down the equation you think correctly describes conservation of energy just inside the chamber walls at equilibrium?

Maybe it would help if we checked my calculations step by step. Start with conservation of energy just inside the chamber walls at equilibrium: power in = power out.

The plate is heated by constant electrical power flowing in. The cold walls at 0F (T_c = 255K) also radiate power in. The heated plate at 150F (T_h = 339K) radiates power out. Using irradiance (power/m^2) simplifies the equation:

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

(Eq. 1 looks better in LaTeX, but hopefully this version is legible.)

Yes/No: can we agree that Eq. 1 is based on the Stefan-Boltzmann law and correctly describes conservation of energy just inside the chamber walls at equilibrium?

If yes, the next step is to solve Eq. 1 for the constant electrical input using a calculator or the Sage worksheet I provided.

If no, could you please write down the equation you think correctly describes conservation of energy just inside the chamber walls at equilibrium?

Once again, if Dr. Latour understood the second law refers to net heat, he'd agree that adding a cold plate makes the heated plate lose heat slower. That's okay because net heat still flows from hot to cold, i.e. more heat moves from hot to cold than vice versa.

Again, he must have forgotten this nebulous correction which you still haven't linked. I linked to an archive of his blog post that I made yesterday, but here's another archive I just made showing that his blog post is still live today and still contains nonsense like this: "k is the fraction of re-radiation from the second bar absorbed by the first hotter bar... k must be identically zero, so no cold back-radiation is absorbed and T remains 150. Quod Erat Demonstrandum, QED."

He's completely wrong. The hotter bar absorbs cold back-radiation, and T does not remain 150F. That's why I refuted Dr. Latour by showing that a completely enclosed heated plate reaches an equilibrium temperature of 235F (386K), which is less than the infinite temperature he claimed.

Apparently unlike you, sir, I have a basic understanding of math and physics. Please explain to us all where the Stefan-Boltzmann radiation law is in error. I am sure we would all love to know. [Jane Q. Public, 2012-11-20]

... just what part of the S-B law do you find controversial? [Jane Q. Public, 2014-07-29]

Again, the greenhouse effect is based on the Stefan-Boltzmann law. As I've explained: greenhouse gases re-emit some of [the upwelling long-wave IR], and it bounces around the troposphere until it gets to a height known as the "effective radiating level". Above this height (roughly 7km), there aren’t enough greenhouse gases to keep "most" of the IR from escaping to space altogether. This effective radiating level controls the outflow of heat from the Earth. Stefan-Boltzmann tells us that power radiated is proportional to temperature^4, and temperature decreases with height in the troposphere. Adding greenhouse gases raises the height of this effective radiating level, where it is cooler, which therefore decreases the outflow of heat from the Earth. This is the greenhouse effect, and it isn’t saturated because the effective radiating level can just keep getting higher (e.g. Venus).

Andrew Dessler also explains how the greenhouse effect depends on the Stefan-Boltzmann law. He even explains that an isothermal atmosphere wouldn't have a greenhouse effect: the Slayers' holy grail! Ironically, the greenhouse effect disappears if the upper troposphere isn't colder than the surface. The cold upper troposphere isn't a problem for the greenhouse effect. It's a fundamental requirement, along with the Stefan-Boltzmann law.

Once again, if Dr. Latour understood the second law refers to net heat, he'd agree that adding a cold plate makes the heated plate lose heat slower. That's okay because net heat still flows from hot to cold, i.e. more heat moves from hot to cold than vice versa.

Again, he must have forgotten this nebulous correction which you still haven't linked. I linked to an archive of his blog post that I made yesterday, but here's another archive I just made showing that his blog post is still live today and still contains nonsense like this: "k is the fraction of re-radiation from the second bar absorbed by the first hotter bar... k must be identically zero, so no cold back-radiation is absorbed and T remains 150. Quod Erat Demonstrandum, QED."

He's completely wrong. The hotter bar absorbs cold back-radiation, and T does not remain 150F. That's why I refuted Dr. Latour by showing that a completely enclosed heated plate reaches an equilibrium temperature of 235F (386K), which is less than the infinite temperature he claimed.

Apparently unlike you, sir, I have a basic understanding of math and physics. Please explain to us all where the Stefan-Boltzmann radiation law is in error. I am sure we would all love to know. [Jane Q. Public, 2012-11-20]

... just what part of the S-B law do you find controversial? [Jane Q. Public, 2014-07-29]

Again, the greenhouse effect is based on the Stefan-Boltzmann law. As I've explained: greenhouse gases re-emit some of [the upwelling long-wave IR], and it bounces around the troposphere until it gets to a height known as the "effective radiating level". Above this height (roughly 7km), there aren’t enough greenhouse gases to keep "most" of the IR from escaping to space altogether. This effective radiating level controls the outflow of heat from the Earth. Stefan-Boltzmann tells us that power radiated is proportional to temperature^4, and temperature decreases with height in the troposphere. Adding greenhouse gases raises the height of this effective radiating level, where it is cooler, which therefore decreases the outflow of heat from the Earth. This is the greenhouse effect, and it isn’t saturated because the effective radiating level can just keep getting higher (e.g. Venus).

Andrew Dessler also explains how the greenhouse effect depends on the Stefan-Boltzmann law. He even explains that an isothermal atmosphere wouldn't have a greenhouse effect: the Slayers' holy grail! Ironically, the greenhouse effect disappears if the upper troposphere isn't colder than the surface. The cold upper troposphere isn't a problem for the greenhouse effect. It's a fundamental requirement, along with the Stefan-Boltzmann law.

As a Hydra, I scoff at mortal concepts like flattery. I also seem to have grown a sixth head judging by Jane's claim that I quoted myself complimenting myself. If Jane's referring to these compliments then my sixth Hydra head also has a real name which is different than mine. We Hydras are powerful and tricksy, and certainly not paranoid delusions. Nope.

... Explain to us what Venus vs. Mercury have to do with Pierre Latour's thermodynamic argument in regard to greenhouse warming? ... [Jane Q. Public]

Again, if the Slayers are right, why is Venus hotter than Mercury? Instead of regurgitating bad arguments you find in 30 seconds and which you don't even read carefully, please read carefully before regurgitating even more misinformation for me to debunk.

... you have failed for 2 years to refute Latour. ... You know you can't refute Latour...

I refuted Dr. Latour's claim that mainstream physics predicts infinite warming, and explained how the greenhouse effect is based on the Stefan Boltzmann law and requires a cold upper troposphere. Again, a real skeptic would be checking my calculation that a completely enclosed heated plate would reach an equilibrium temperature of 235F (386K).

... the part I was referencing was the part about Venus. ... I knew next to nothing about the subject. ... [Jane Q. Public]

Do you see how crackpot websites which make "ridiculous" claims that you might have made when you "knew next to nothing about the subject" might not be the best source of science education?

... I just did you a favor and looked up something you asked for on Google. His arguments are not my own and I did not even read them carefully. I merely looked them up for you because you seemed to wanted to argue about yet another straw-man that had next to nothing to do with anything I had said. ... [Jane Q. Public]

Venus vs. Mercury has everything to do with the Slayer nonsense you're spreading. You're just regurgitating even more misinformation that I have to debunk. That's the exact opposite of a favor! It's the same absurd behavior I've repeatedly asked you to stop.

Again, thanks for finally being honest. You’re not interested in valid science, just something you can use to argue, even if it doesn’t hold up under scrutiny. You’ve used this "principle of superficiality" to spread civilization-paralyzing misinformation which seems plausible at first glance to non-scientists, but doesn’t hold up under scrutiny. In fact, I said as much last year:

"... each contrarian is more effective at superficial "science communication" than the average scientist. ... Once you get a contrarian started, a stream of regurgitated-but-superficially-plausible nonsense spews forth. Just consider Jane Q. Public. ..."

...I was not present... [Jane Q. Public]

Actually, you did respond. Repeatedly. Sure you weren't present?

... the part I was referencing was the part about Venus. ... I knew next to nothing about the subject. ... [Jane Q. Public]

Do you see how crackpot websites which make "ridiculous" claims that you might have made when you "knew next to nothing about the subject" might not be the best source of science education?

... I just did you a favor and looked up something you asked for on Google. His arguments are not my own and I did not even read them carefully. I merely looked them up for you because you seemed to wanted to argue about yet another straw-man that had next to nothing to do with anything I had said. ... [Jane Q. Public]

Venus vs. Mercury has everything to do with the Slayer nonsense you're spreading. You're just regurgitating even more misinformation that I have to debunk. That's the exact opposite of a favor! It's the same absurd behavior I've repeatedly asked you to stop.

Again, thanks for finally being honest. You’re not interested in valid science, just something you can use to argue, even if it doesn’t hold up under scrutiny. You’ve used this "principle of superficiality" to spread civilization-paralyzing misinformation which seems plausible at first glance to non-scientists, but doesn’t hold up under scrutiny. In fact, I said as much last year:

"... each contrarian is more effective at superficial "science communication" than the average scientist. ... Once you get a contrarian started, a stream of regurgitated-but-superficially-plausible nonsense spews forth. Just consider Jane Q. Public. ..."

...I was not present... [Jane Q. Public]

Actually, you did respond. Repeatedly. Sure you weren't present?

... the part I was referencing was the part about Venus. ... I knew next to nothing about the subject. ... [Jane Q. Public]

Do you see how crackpot websites which make "ridiculous" claims that you might have made when you "knew next to nothing about the subject" might not be the best source of science education?

... I just did you a favor and looked up something you asked for on Google. His arguments are not my own and I did not even read them carefully. I merely looked them up for you because you seemed to wanted to argue about yet another straw-man that had next to nothing to do with anything I had said. ... [Jane Q. Public]

Venus vs. Mercury has everything to do with the Slayer nonsense you're spreading. You're just regurgitating even more misinformation that I have to debunk. That's the exact opposite of a favor! It's the same absurd behavior I've repeatedly asked you to stop.

Again, thanks for finally being honest. You’re not interested in valid science, just something you can use to argue, even if it doesn’t hold up under scrutiny. You’ve used this "principle of superficiality" to spread civilization-paralyzing misinformation which seems plausible at first glance to non-scientists, but doesn’t hold up under scrutiny. In fact, I said as much last year:

"... each contrarian is more effective at superficial "science communication" than the average scientist. ... Once you get a contrarian started, a stream of regurgitated-but-superficially-plausible nonsense spews forth. Just consider Jane Q. Public. ..."

...I was not present... [Jane Q. Public]

Actually, you did respond. Repeatedly. Sure you weren't present?

... the part I was referencing was the part about Venus. ... I knew next to nothing about the subject. ... [Jane Q. Public]

Do you see how crackpot websites which make "ridiculous" claims that you might have made when you "knew next to nothing about the subject" might not be the best source of science education?

... I just did you a favor and looked up something you asked for on Google. His arguments are not my own and I did not even read them carefully. I merely looked them up for you because you seemed to wanted to argue about yet another straw-man that had next to nothing to do with anything I had said. ... [Jane Q. Public]

Venus vs. Mercury has everything to do with the Slayer nonsense you're spreading. You're just regurgitating even more misinformation that I have to debunk. That's the exact opposite of a favor! It's the same absurd behavior I've repeatedly asked you to stop.

Again, thanks for finally being honest. You’re not interested in valid science, just something you can use to argue, even if it doesn’t hold up under scrutiny. You’ve used this "principle of superficiality" to spread civilization-paralyzing misinformation which seems plausible at first glance to non-scientists, but doesn’t hold up under scrutiny. In fact, I said as much last year:

"... each contrarian is more effective at superficial "science communication" than the average scientist. ... Once you get a contrarian started, a stream of regurgitated-but-superficially-plausible nonsense spews forth. Just consider Jane Q. Public. ..."

...I was not present... [Jane Q. Public]

Actually, you did respond. Repeatedly. Sure you weren't present?

... the part I was referencing was the part about Venus. ... I knew next to nothing about the subject. ... [Jane Q. Public]

Do you see how crackpot websites which make "ridiculous" claims that you might have made when you "knew next to nothing about the subject" might not be the best source of science education?

... I just did you a favor and looked up something you asked for on Google. His arguments are not my own and I did not even read them carefully. I merely looked them up for you because you seemed to wanted to argue about yet another straw-man that had next to nothing to do with anything I had said. ... [Jane Q. Public]

Venus vs. Mercury has everything to do with the Slayer nonsense you're spreading. You're just regurgitating even more misinformation that I have to debunk. That's the exact opposite of a favor! It's the same absurd behavior I've repeatedly asked you to stop.

Again, thanks for finally being honest. You’re not interested in valid science, just something you can use to argue, even if it doesn’t hold up under scrutiny. You’ve used this "principle of superficiality" to spread civilization-paralyzing misinformation which seems plausible at first glance to non-scientists, but doesn’t hold up under scrutiny. In fact, I said as much last year:

"... each contrarian is more effective at superficial "science communication" than the average scientist. ... Once you get a contrarian started, a stream of regurgitated-but-superficially-plausible nonsense spews forth. Just consider Jane Q. Public. ..."

...I was not present... [Jane Q. Public]

Actually, you did respond. Repeatedly. Sure you weren't present?

... the part I was referencing was the part about Venus. ... I knew next to nothing about the subject. ... [Jane Q. Public]

Do you see how crackpot websites which make "ridiculous" claims that you might have made when you "knew next to nothing about the subject" might not be the best source of science education?

... I just did you a favor and looked up something you asked for on Google. His arguments are not my own and I did not even read them carefully. I merely looked them up for you because you seemed to wanted to argue about yet another straw-man that had next to nothing to do with anything I had said. ... [Jane Q. Public]

Venus vs. Mercury has everything to do with the Slayer nonsense you're spreading. You're just regurgitating even more misinformation that I have to debunk. That's the exact opposite of a favor! It's the same absurd behavior I've repeatedly asked you to stop.

Again, thanks for finally being honest. You’re not interested in valid science, just something you can use to argue, even if it doesn’t hold up under scrutiny. You’ve used this "principle of superficiality" to spread civilization-paralyzing misinformation which seems plausible at first glance to non-scientists, but doesn’t hold up under scrutiny. In fact, I said as much last year:

"... each contrarian is more effective at superficial "science communication" than the average scientist. ... Once you get a contrarian started, a stream of regurgitated-but-superficially-plausible nonsense spews forth. Just consider Jane Q. Public. ..."

...I was not present... [Jane Q. Public]

Actually, you did respond. Repeatedly. Sure you weren't present?

... the part I was referencing was the part about Venus. ... I knew next to nothing about the subject. ... [Jane Q. Public]

Do you see how crackpot websites which make "ridiculous" claims that you might have made when you "knew next to nothing about the subject" might not be the best source of science education?

... I just did you a favor and looked up something you asked for on Google. His arguments are not my own and I did not even read them carefully. I merely looked them up for you because you seemed to wanted to argue about yet another straw-man that had next to nothing to do with anything I had said. ... [Jane Q. Public]

Venus vs. Mercury has everything to do with the Slayer nonsense you're spreading. You're just regurgitating even more misinformation that I have to debunk. That's the exact opposite of a favor! It's the same absurd behavior I've repeatedly asked you to stop.

Again, thanks for finally being honest. You’re not interested in valid science, just something you can use to argue, even if it doesn’t hold up under scrutiny. You’ve used this "principle of superficiality" to spread civilization-paralyzing misinformation which seems plausible at first glance to non-scientists, but doesn’t hold up under scrutiny. In fact, I said as much last year:

"... each contrarian is more effective at superficial "science communication" than the average scientist. ... Once you get a contrarian started, a stream of regurgitated-but-superficially-plausible nonsense spews forth. Just consider Jane Q. Public. ..."

...I was not present... [Jane Q. Public]

Actually, you did respond. Repeatedly. Sure you weren't present?