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Since the emissivity for every object in our system is the same, power output is proportional to the T^4. Period. End of story. Draw your boundary around the heat source. Power in = power out (your own principle). Therefore the power in is 41886.54 Watts, which is the power initially being radiated out. SPENCER stipulated that this power is held constant. It wasn't my idea. It's a condition of the experiment. [Jane Q. Public, 2014-09-19]

No. Once again, in this experiment there is a "... constant flow of energy into the plate from the electric heater... flowing in at a constant rate... the electric heater pumps in energy at a constant rate. ..."

Jane's even stumbled across this point:

... Of course it wouldn't need a separate heat source if its environment were maintained at 150 degrees. ... [Jane Q. Public, 2014-09-15]

Of course! That's why the variable Jane's holding constant isn't the electrical power supplied to the separate heat source. If Jane can realize that there's no need for a separate heat source if its environment were maintained at 150 degrees, why can't Jane see that his equation for required electrical power doesn't reflect this obvious fact?

... you have confirmed that you have not abandoned your incorrect (and actually quite ludicrous) version of heat transfer, which violates the Stefan-Boltzmann radiation law on its very face. ... [Jane Q. Public, 2014-09-15]

... or maybe we disagree about which variable to hold constant.

Instead of holding electrical heating power constant, Jane held the source's radiative power output constant. That held source temperature constant and forced electrical heating power to change. Solving this problem using both sets of boundary conditions shows that Jane's solution forces electrical heating power to drop by a factor of two after the shell is added.

These two sets of boundary conditions are very different, just like Neumann boundary conditions are different from Dirichlet boundary conditions. Upon hearing that a disagreement might be caused by holding different variables constant, a real skeptic might consider working the problem again while holding that other variable constant. But Jane can't even admit there's a difference between holding electrical heating power constant and holding the source's radiative power output constant. Jane even insists he held electrical heating power constant, despite the evidence.

So Jane won't solve this problem with the electrical heating power constant. That's unfortunate, because it's critical:

"... critical to the whole experiment is that, like the sun heating the surface of the Earth, there is energy being continuously pumped into the system from outside. ..."

1. Holding electrical heating power constant while adding an enclosing shell is like doubling CO2 while holding solar heating power constant, then calculating how much Earth's surface warms.

2. Holding source temperature constant while adding an enclosing shell is like doubling CO2 while holding Earth's surface temperature constant, then calculating how much solar heating power would have to drop to keep Earth's surface temperature constant.

Even if Jane doesn't want to solve that first problem, he should recognize that it's different from the second problem Jane actually solved.

To see this difference, solve a problem with Neumann boundary conditions:

"In thermodynamics, where a surface has a prescribed heat flux, such as a perfect insulator (where flux is zero) or an electrical component dissipating a known power."

... then solve the same problem with Dirichlet boundary conditions:

"In thermodynamics, where a surface is held at a fixed temperature.

Dr. Spencer's thought experiment placed Neumann boundary conditions on the source and Dirichlet boundary conditions on the chamber walls. Instead, Jane placed Dirichlet boundary conditions on the chamber walls and the source.

In other words, the electrical heating power is determined by drawing a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source

Since power in = power out:

electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source

= WikiLeaks releases previously unseen copies of weaponised German surveillance malware used by intelligence agencies around the world to spy on journalists, political dissidents and others.

http://slashdot.org/firehose.p...

https://wikileaks.org/spyfiles...

"Today, 15 September 2014, WikiLeaks releases previously unseen copies of weaponised German surveillance malware used by intelligence agencies around the world to spy on journalists, political dissidents and others.

FinFisher (formerly part of the UK based Gamma Group International until late 2013) is a German company that produces and sells computer intrusion systems, software exploits and remote monitoring systems that are capable of intercepting communications and data from OS X, Windows and Linux computers as well as Android, iOS, BlackBerry, Symbian and Windows Mobile devices. FinFisher first came to public attention in December 2011 when WikiLeaks published documents detailing their products and business in the first SpyFiles release.

Since the first SpyFiles release, researchers published reports that identified the presence of FinFisher products in countries aroud the world and documented its use against journalists, activists and political dissidents.

Julian Assange, WikiLeaks Editor in Chief said: "FinFisher continues to operate brazenly from Germany selling weaponised surveillance malware to some of the most abusive regimes in the world. The Merkel government pretends to be concerned about privacy, but its actions speak otherwise. Why does the Merkel government continue to protect FinFisher? This full data release will help the technical community build tools to protect people from FinFisher including by tracking down its command and control centers."

FinFisher Relay and FinSpy Proxy are the components of the FinFisher suite responsible for collecting the data acquired from the infected victims and delivering it to their controllers. It is commonly deployed by FinFisher's customers in strategic points around the world to route the collected data through an anonymizing chain, in order to disguise the identity of its operators and the real location of the final storage, which is instead operated by the FinSpy Master."

= Story continued at URL above.

== Archives:
http://web.archive.org/web/201...
https://archive.today/XRT0p

... input power at steady-state is fixed, and a value that we already know: 41886.54 W. ... [Jane Q. Public, 2014-09-12]

Again, we disagree about what's held fixed. That value you keep calculating isn't the constant electrical power heating the source.

In this experiment there is a "... constant flow of energy into the plate from the electric heater... flowing in at a constant rate... the electric heater pumps in energy at a constant rate. ..."

In my interpretation, Dr. Spencer's challenge is basically: "Assuming an electric heater pumps energy at a constant rate to the source, does the source temperature change after a passive plate is added?"

You've repeatedly noted that there are no other factors involved in calculating your 82 W/m^2 (41886.54 W) value. So if it's held fixed, the source temperature is also held fixed.

So it seems like in your interpretation, Dr. Spencer's challenge is basically: "Assuming the source temperature is held fixed, does the source temperature change after a passive plate is added?"

Is that right?

... The formula for radiant power is (e * s) * area * T^4. Period. This is according to the Stefan-Boltzmann law, and no other variables are required at steady-state. The initial temperature of the heat source was 150F, or 338.71K. So we agreed that the input power to the heat source is sufficient for the equation (e * s) * (heat source area) * 338.71^4. The power input doesn't change. ... the total power output (and therefore power input) at the heat source, in initial conditions, was (we agreed on this) 82.12 W/m^2 * 510.065 m^2 = 41886.54 Watts. Power in = power out. ... [Jane Q. Public, 2014-09-11]

No. We've never agreed that the unchanging power input (my "constant electrical heating power") is "82 W/m^2". I've repeatedly failed to explain that the constant electrical heating power would only be "82 W/m^2" if the chamber walls were 0K blackbodies.

In this experiment there is a "... constant flow of energy into the plate from the electric heater... flowing in at a constant rate... the electric heater pumps in energy at a constant rate. ..."

Note that the constant rate of Dr. Spencer's electric heater would equal zero if the chamber walls were also at 150F. So any calculation of this crucial constant rate would also need to be zero in the case of chamber walls at 150F.

Since Jane's "82 W/m^2" value isn't the constant electrical heating power needed to keep the source at 150F inside 0F chamber walls, it isn't held constant. Here's where Jane actually calculated the constant electrical power heating the source inside 0F chamber walls:

... Calculate initial (denoted by "i") heat transfer from heat source to chamber wall. We are doing this only to check our work later. ... = 55.5913 [W/m^2]... [Jane Q. Public, 2014-09-10]

So Jane's source needs 55.6 W/m^2 of constant electrical heating power to stay at 150F inside 0F chamber walls. This value is held constant. After the enclosing shell is added and temperatures stabilize, conservation of energy demands that net heat transfer out equals Jane's 55.6 W/m^2. Does it?

... you should at least have tried drawing your boundary around your own goddamned heat source, both for initial conditions and your final result, to check your work. But you didn't. What you got was a universe-busting violation of conservation of energy. ... [Jane Q. Public, 2014-09-11]

No, I drew that boundary for both initial and final conditions to guarantee conservation of energy. In fact, I repeatedly suggested that you check your work by drawing a boundary between the source and the enclosing shell at your proposed steady-state temperatures, then calculating power in = power out using the original constant electrical power you calculated before the source was enclosed.

Let's do that:

Jane's constant electrical power of 55.6 W/m^2 flows into that boundary. At steady-state, power in = power out. But power out through that boundary is the

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.

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.

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.

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.

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.

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.

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.

A boundary drawn around a system that isn't changing always has power in = power out. Always. Because energy is always conserved.

... Let me rephrase what I was saying: at least theoretically, the power at the chamber wall is allowed to vary, in order to keep the temperature at 0 degrees F. But, if we draw a boundary around the system, and assume that the ONLY power in is what we put in, and the ONLY power out is what is removed, then of course it must be conserved. I was simply expressing my concern that your electricity figure may not be properly observing those boundaries. If your electricity figure is simply power in - power out... [Jane Q. Public, 2014-09-07]

Maybe this will help. It seems like Jane might think I meant power in = electrical heating power, and power out = cooling power of the chamber walls.

If so, that's not what I meant, and I'm sorry for not being more clear. I take full responsibility.

Just to be clear, power in = power flowing into the boundary in question, and power in = power flowing out of that boundary.

In my opinion, solving thermodynamics problems is mostly about choosing the most informative boundaries, then calculating steady-state solutions by setting power in = power out through that boundary.

From the start, the largest boundary I drew was "just inside the chamber walls" so the chamber walls and the cooler have always been outside all the boundaries. That means any power used by the cooler is simply being moved from some point outside the boundary to another point which is also outside the boundary. Because that power never crosses the boundary, it's irrelevant.

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

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

I've been explaining for over a month that the heated plate warms after it's enclosed. I realize you don't agree, which is why I'm trying in vain to get you to finally perform a single, solitary calculation of your own.

I never said I disagree with this. Please find where I said that. On the contrary; I definitely agree that it warms. In fact it must: Spencer stipulated that it was to be inserted when it was colder than the heat source. ... [Jane Q. Public, 2014-09-04]

I've been explaining for over a month that the heated plate (aka Jane's "source") warms after it's enclosed. I've only been wasting my final days because Jane's repeatedly disagreed by supporting Dr. Latour's ridiculous Sky Dragon Slayer claim that the heated plate (aka Jane's "source") simply remains at 150F after it's enclosed:

... the heat source does not become hotter. This is, and has been, the whole of Latour's argument, and it is valid. It is not crazy speculation by some nitwit, it is straightforward application of Stefan-Boltzmann law. Q.E.D., indeed. If the above inequalities hold (and they do), Latour's conclusion is the only one that is mathematically valid. [Jane Q. Public, 2014-08-02]

... The plate cannot cause the heat source to be hotter ... [Jane Q. Public, 2014-08-20]

If Jane agrees that the heated plate (aka Jane's "source") warms after it's enclosed, then that's great news! In that case, we can all agree that the mainstream physics describing the greenhouse effect is accurate, obeys the laws of thermodynamics, and proves that the Sky Dragon Slayers are wrong.

... It is the engineering textbook answer. Claiming it is nonsense does not make it so. It was your own model that violated conservation of energy. But to see why, it's easiest to solve the general case first, then look at a specific case. I told you I had reasons to solve the general case first. ... Well, then, I guess you do admit defeat. It doesn't take much time to obtain a textbook on the subject (you were given references 2 years ago and it's not that hard to find others) ... [Jane Q. Public, 2014-09-01]

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

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

... I am disputing that given reasonable chosen dimensions it is anywhere near an intractable problem. ... [Jane Q. Public, 2014-09-01]

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

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

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

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

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

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

... It is the engineering textbook answer. Claiming it is nonsense does not make it so. It was your own model that violated conservation of energy. But to see why, it's easiest to solve the general case first, then look at a specific case. I told you I had reasons to solve the general case first. ... Well, then, I guess you do admit defeat. It doesn't take much time to obtain a textbook on the subject (you were given references 2 years ago and it's not that hard to find others) ... [Jane Q. Public, 2014-09-01]

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

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

... I am disputing that given reasonable chosen dimensions it is anywhere near an intractable problem. ... [Jane Q. Public, 2014-09-01]

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

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

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

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

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

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

... It is the engineering textbook answer. Claiming it is nonsense does not make it so. It was your own model that violated conservation of energy. But to see why, it's easiest to solve the general case first, then look at a specific case. I told you I had reasons to solve the general case first. ... Well, then, I guess you do admit defeat. It doesn't take much time to obtain a textbook on the subject (you were given references 2 years ago and it's not that hard to find others) ... [Jane Q. Public, 2014-09-01]

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

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

... I am disputing that given reasonable chosen dimensions it is anywhere near an intractable problem. ... [Jane Q. Public, 2014-09-01]

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

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

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

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

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

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

... It is the engineering textbook answer. Claiming it is nonsense does not make it so. It was your own model that violated conservation of energy. But to see why, it's easiest to solve the general case first, then look at a specific case. I told you I had reasons to solve the general case first. ... Well, then, I guess you do admit defeat. It doesn't take much time to obtain a textbook on the subject (you were given references 2 years ago and it's not that hard to find others) ... [Jane Q. Public, 2014-09-01]

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

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

... I am disputing that given reasonable chosen dimensions it is anywhere near an intractable problem. ... [Jane Q. Public, 2014-09-01]

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

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

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

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

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

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

... It is the engineering textbook answer. Claiming it is nonsense does not make it so. It was your own model that violated conservation of energy. But to see why, it's easiest to solve the general case first, then look at a specific case. I told you I had reasons to solve the general case first. ... Well, then, I guess you do admit defeat. It doesn't take much time to obtain a textbook on the subject (you were given references 2 years ago and it's not that hard to find others) ... [Jane Q. Public, 2014-09-01]

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

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

... I am disputing that given reasonable chosen dimensions it is anywhere near an intractable problem. ... [Jane Q. Public, 2014-09-01]

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

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

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

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

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

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

Maybe I should explain what I meant by saying that equilibrium temperatures for a non-enclosing plate would vary across the plate surfaces. Consider Dr. Spencer's first illustration. Presumably the heated plate at "150F" has finite conductivity, so its lack of spherical symmetry means that its corners will be cooler than the plate's side's midpoints. That's because the corners are closer to the cold chamber walls than those midpoints.

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

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

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

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

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

Maybe I should explain what I meant by saying that equilibrium temperatures for a non-enclosing plate would vary across the plate surfaces. Consider Dr. Spencer's first illustration. Presumably the heated plate at "150F" has finite conductivity, so its lack of spherical symmetry means that its corners will be cooler than the plate's side's midpoints. That's because the corners are closer to the cold chamber walls than those midpoints.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

... 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

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

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

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

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

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

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

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

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

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

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

I've repeat

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

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

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

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

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

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

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

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

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

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

I've repeat

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

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

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

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

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

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

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

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

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

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

I've repeat

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

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

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

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

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

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

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

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

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

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

... There are numerous sources, including physics and engineering textbooks, which contradict your analysis and conclusions. Why don't you try the engi

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

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

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

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

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

Again, Eq. 1 describes equilibrium temperature:

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

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

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

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

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

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

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,

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,

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,

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,