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Comment: Re:lets pump the brakes here and analyze. (Score 4, Informative) 87

Then again the United States basically ignored Palestine when it pulled israel from its magical sky god book and transported displaced Jews into it after WWII.

Not only is this incorrect, but you topped it off with a bigoted assault on religion. "Palestine" was administered by the UK, not the USA, and American Jews voluntarily moved to it mostly after the 1948 independence. At the time of independence, it had nearly indefensible borders after suffering additional partitions beyond what the UK had promised. See "Transjordan" had already been partitioned for settlement by Palestinian Arabs. The West Bank, Gaza strip, and some other small areas were snipped off after WWII. Israel mostly lived with what they were given until they were attacked, repeatedly, by the Arabs who hated them. Regardless, even if you don't think Israel should have occupied any territory gained in 1948 or 1967, you can't blame it on the USA.

Comment: Re:they will defeat themselves (Score 1) 887

by operagost (#47927963) Attached to: ISIS Bans Math and Social Studies For Children
Seems like they would never be able to use ballistic weapons, because even though they have long used computers to calculate trajectory, you still need to supply information that requires more than total ignorance of mathematics. It helps to be able to use sanity checks so you don't kill your own troops, as well. Hey, maybe we should "accidentally" let them have some hacked weapons...

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 155

by khayman80 (#47922765) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

... 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

Right?

No. Not right. Since emissivity doesn't change the input required to heat source to achieve 150F is constant, regardless of where it comes from. But as long as the walls of the chamber are cooler than the source, NONE of the power comes from the chamber walls, because of that minus sign in the equation above. Nothing has changed in that respect, and that's what the Stefan-Boltzmann law requires. The only time that changes is if the walls are at an equal temperature, in which case heat transfer is 0 and you can begin to use "ambient" temperature as input. You are still supplying the same input power, you are just supplying it a different way. If the chamber walls were hotter than the central source, then heat transfer would be in the other direction (because the sign of the solution to the equation above changes), and only THEN are you getting net heat transfer TO the central sphere. ... [Jane Q. Public, 2014-09-15]

Note that conservation of energy through a boundary around the source leads directly to an equation describing the electrical power required to keep the source at temperature T1 inside chamber walls at temperature T4. This equation is valid for T1 > T4, T1 = T4, and T1 < T4. Jane might wonder why he can't derive a single equation which works for all these cases.

Again, warming the chamber walls is like partially closing the drain on a bathtub where water is flowing in at a constant rate. This raises the bathtub water level simply by reducing the water flow out. In exactly the same way, a source heated with constant electrical power warms when the chamber walls are warmed because that reduces the net power out.

... because T(p) < T(s), no matter now much of the radiation from P strikes S, no net amount is absorbed; it is all reflected, transmitted, or scattered according to S-B. ... [Jane Q. Public, 2014-09-04]

Are you REALLY the moron you make yourself out to be? NET radiation from a cooler surface that passes the boundary is reflected, transmitted, or scattered and passes right back out through the boundary. This is a corollary of the Stefan-Boltzmann radiation law, which states that NET heat transfer is always from hotter to cooler. ... by that same law, it just passes right back out again because the same NET amount of radiative power that crosses the boundary and intercepts the smaller sphere is either reflected, transmitted, or scattered. ... [Jane Q. Public, 2014-09-15]

... Since the chamber walls are COOLER than the heat source, radiative power from the chamber walls is not absorbed by the heat source. ... [Jane Q. Public, 2014-09-15]

Hopefully these are just more badly-worded sentences because they all require absorptivity = 0. But these gray bodies have emissivity = absorptivity = 0.11. Furthermore, the gray body equation has to reduce to the black body equation for emissivity = absorptivity = 1. In that case there are no reflections, just absorption.

Once again, a heated blackbody source is heated by constant electrical power flowing in. Blackbody cold walls at 0F (T4 = 255.4K) also radiate power in. The source at 150F (T1 = 338.7K) radiates power out. At steady-state, power in = power out:

electricity + (s)*T4^4 = (s)*T1^4 (Eq. 1J.2)

Since Jane's proposed equation is missing the "(s)*T4^4" term, it doesn't reduce to this simpler Eq. 1J.2 for blackbodies where (e) = 1. So it's wrong.

It's also ironic that Jane claims to account for reflections, because:

... Calculate initial (denoted by "i") heat transfer from heat source to chamber wall. We are doing this only to check our work later. Using the canonical heat transfer equation for gray bodies...
p(i) = (e)(s) * ( T1^4 - T4^4 ) ... [Jane Q. Public, 2014-09-10]

... You are ignoring (e*s) * (Ta^4 - Tb^4). Anything other than what I described does not add up. ... (e*s) * (Ta^4 - Tb^4) ... [Jane Q. Public, 2014-09-15]

That equation is true for blackbodies with emissivity = 1, which is why it's consistent with my equation 1.

But for gray bodies it's just an approximation because it ignores reflections. After obviously failing to explain that we need to account for reflections, I decided to agree to disagree. For two gray bodies interacting with small view factors (e.g. Earth's tiny view factor of the Sun) reflections can be safely neglected. But the chamber wall completely encloses the source, so its view factor is 1. That's why MIT's equation is more accurate here: it accounts for reflections.

Again, here's MIT's equation using Jane's new variable names:

p(i) = (s)*(T1^4 - T4^4)/(1/(e) + 1/(e) - 1) (Eq. 2J.2)

Luckily this disagreement isn't important because it just shifts the emissivity values. We can translate because plugging emissivity = 0.058 into Jane's equation yields the same net heat transfer as MIT's equation with emissivity = 0.11. Furthermore, my black and gray body calculations yielded identical enclosed steady-state temperatures, so those don't depend on emissivity.

But after using Jane's equation in pointless attempts to illustrate more fundamental problems in Jane's analysis, I wanted to stress once again that MIT's equation is more appropriate for enclosing chamber walls because it accounts for reflections.

Testing can show the presense of bugs, but not their absence. -- Dijkstra

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