Welp, they sure split that baby.

(No seriously. Remember, the point of Solomon's judgement was to use a decision that's bad for both sides to determine who the real winner should be in the end. Same here. I'm betting we'll see Boeing whine, delay, and run over budget while SpaceX gets down and builds some rockets, but either way, in a few years we'll see who the manned spacecraft baby really belongs to.)

I'm sick of this bullshit belief that "liberal arts" refers to non-STEM majors in the humanities and social sciences, and is college in "easy mode". Quick history lesson: it's called "liberal" arts because from Roman times through the Renaissance, they were the skills that made one worthy of being a free person, as opposed to the manual skills appropriate for a slave. They included both artistic subjects like grammar, rhetoric, and logic, and scientific "arts" like astronomy and math. Of course meaning changed over the years, but today liberal arts colleges try to create well-rounded generalist thinkers, jacks of all trades and masters of at least one.

I've got a BA in physics from one of the top liberal arts colleges in the nation. You might think that's a joke, but my PhD advisor at MIT didn't. I'm now a tenured professor in physics, and my college buddies do stuff like dark matter research at Livermore, software development for Google and Microsoft, etc.

Enough bragging and tech namedropping, the point is that a liberal arts education can get you an excellent technical education. Unfortunately, too many major universities offer a "liberal arts" program which *is* college easymode, intended for folks who go to college for the social scene. But getting a liberal arts at these places is like buying organic local produce from Walmart: sure, they have it, they've got everything, but it's so contrary to the philosophy of the place that you're right to be skeptical.

"Is there any place for degrees in the humanities and social sciences in tech?" Now that's a reasonable question, to which I think the answer is obviously "yes", and my friends the Latin major computer programmer and the religion major tech writer would agree. But if you think "liberal arts" can't provide a top-notch education in STEM subjects, you're not qualified to read a resume.

What you're describing is a random number generator with a key to initialize it. Some of the good ones might be good enough (or might not). Anything you can keep in your head is going to be crap and fairly easily breakable. Either way, you're still better off to just exchange regular secret keys at your meeting, which can be concealed in a variety of ways. Even real one time pads can be fairly easily concealed - a "blank" USB key, for example.

... 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.

You didn't say so, but I'm assuming you're encrypting your message using the book page as a one time pad, then obscuring it using steganography. If someone sufficiently motivated were after your criminals, they could break that. Steganography isn't much protection when someone knows there might be hidden messages. And your one time pad, while one time, isn't random. Book pages have quite a bit of structure.

Any structure in a one time pad makes it vulnerable. To the point where people have gone to great lengths to construct them using the best random numbers obtainable, from devices ranging from antennae monitoring the ionosphere to quantum devices.

One time pads have been, and probably are, used extensively. You send a bunch of random data to someone via some secure method, which is usually very slow (like hopping on an airplane with a DVD full of random numbers on your person). You can then exchange messages securely using a convenient and fast channel, such as e-mail. See the utility there?

You can interpret it that way. That's not the only way to interpret it.

Yup, and the solution available to any rational being is the same: since by hypothesis the two choices are indistinguishable, flip a coin to create a new situation in which one of them has a trivial weight on its side.

Starving to death (or letting everyone die) is obviously inferior to this to any rational being (which the donkey and the robot are both presumed to be) and adding randomness is a perfectly general solution to the problem.

Buridan's donkey is not in fact an example of a rational being, but rather a passive, uncreative being, who must for some unspecified reason decide without acting on the situation, as if it was living in some bizarrely unrealistic world like Plato's Cave, where it could only know the world via shadows on the wall which it cannot act on in any way.

Why anyone thinks thought-experiments about such limited beings, which are completely unlike humans in their inability to act on the world to change their situation, is beyond me.

Ah, got you. Still needs appreciable power, but being a continuous load, that's not a major issue. The water makers on board are RO too, feeding and washing a couple of hundred (very) sweaty bodies. But for big fresh water requirements (hundreds of cu. m. ) we bring in non-potable water on one of the flotilla boats.

Every one of the links you have posted comes from a mainstream Aussie media outlet, when those stories stop appearing you have a real corruption problem. The NSW ICAC judicial inquiry has forced the resignation of at least a dozen MP's who took illegal donations from property developers and is still going strong. Now think about real oppression (say) Mugabe or Saddam, they tow the line or risk summary execution.

Internet snooping by cops is a double edged sword, sure it can be used as a tool of oppression (if the political climate is ripe) but it has also been used to solve some high profile murder/rape cases. In the Jill Megan (sic?) case the cops didn't spy on anyone, they simply explained the situation to the banks (on the weekend) and the banks voluntarily gave them what they needed to track the bastard down.

Disclaimer: I have a female cousin who has served in the Victorian police for over 3 decades. I'm not claiming all Aussie cops are saints, but certainly the vast majority have their heart in the right place and are doing a tough job as best they can.

... 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... [Jane Q. Public, 2014-09-15]

But if the chamber walls are also at 150F, they're not cooler than the source and the input required to heat the source to 150F is zero.

... do you still maintain that after the enclosing passive sphere is inserted, the central heat source raises in temperature to approximately 241 degrees F? You haven't said anything about that in a while, so I'm just checking. [Jane Q. Public, 2014-09-15]

Once again, if the electrical heating power is held constant, the heat source has to warm. Once agin, Jane's heat source keeps the source temperature constant by halving its electrical heating power. Jane/Lonny Eachus might ask himself why his required electrical heating power goes down by a factor of two after the enclosing shell is added.

You're right: if you change what it says by deleting some of the words, then it says something different.

In the next sentence, it says in particular what you're not allowed to use or run, including proxy services.

Use or run: It's not merely that you're not allowed to run proxy services: you're not allowed to use them, either.

If that's stupid-- well, how about that.

As I said: the interpretation of this text could be ambiguous. You could do the lawyer thing and claim to interpret it the way you say. But the clear straightforward text is: proxies are listed on the list of things you are specifically not allowed to use or run.

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

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?

You asked me if I believed the power usage of the heat source would be the same if the walls were also at 150F. The answer is YES, and here is why: You are proposing to bring the whole system up to a level of higher thermodynamic energy, rather than just the heat source. And you are somehow proposing that it doesn't take more energy to do that. But of course it does. The power required to bring the heat source up to 150F remains the same, because the Stefan-Boltzmann law says it has to be. But NOW, you are ALSO bringing the walls up to that higher temperature, and THAT would require even more power (because of the slightly larger surface area). [Jane Q. Public, 2014-09-15]

Again, that's completely ridiculous. I've explained why the power used to set the chamber wall temperature is irrelevant. Any power used 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.

For example, you could simply place the vacuum chamber somewhere with an ambient temperature of 150F. That would require zero power, but once again it doesn't matter even if the vacuum chamber were on Pluto. Because that power never crosses the boundary.

Either way, as long as the chamber walls are held at 150F, the heat source would need absolutely no electrical heating power to remain at 150F. Zero. Period.

You asked me if I believed the power usage of the heat source would be the same if the walls were also at 150F. The answer is YES... [Jane Q. Public, 2014-09-15]

Here's our disagreement. Conservation of energy demands that a heat source at 150F requires no electrical heating power inside 150F vacuum chamber walls.