Actually NASA (or was it NOAA?) changed their tune again and are saying it was 1937.

Gotta keep up with this stuff, man.

The raw, unadjusted temperature records always have said 1937. It's the adjustments that are questionable, not the historical record.

Except that I didn't. When my cable was installed I signed a small receipt acknowledging that the tech had been there. I signed no contract.

That might have been an oversight on their part, but that doesn't matter.

Further, the KIND of contract that Comcast has customers sign is known in the legal industry as a "contract of adhesion". What that means is that it was a non-negotiable, take-it-or-leave-it "contract". The problem being that contract law assumes that every party is free to negotiate before signing.

So in many genuine, legal senses of the term, it's not a "real" contract anyway, and honest judges are required in principle to view them "with a jaundiced eye", and lean toward the customer when a dispute arises.

I'm not saying all judges are honest enough to do that, but they're supposed to.

And one more thing I would like to make very clear:

The REASON there would not be as great a power DIFFERENCE if the chamber walls were also at 150F, is that the walls would themselves be radiating more power out, so there would be less heat transfer (in that case 0).

It is NOT, as you assert, because the heat source would be using less power. That's false, by the S-B equation. Its power output remains the same because (Spencer's stipulation) the power input remains the same.

The reason my solution does not violate conservation of energy, is that the power consumption of the chamber wall is allowed to vary. THAT is where the change takes place, not at the heat source. Again, this is a stipulation of Spencer's challenge.

Once again: power out of heat source remains constant, because P = (emissivity) * (S-B constant) * T^4. There is nothing in these conditions that changes this at all. Therefore, BECAUSE the power out and power in at the heat source remain constant, so does the temperature. It's all in that one little equation.

In fact let's just face this directly, with no mincing of words:

We are not AT thermal equilibrium, so that is a ridiculous straw-man argument.

One question only: do you agree with the Stefan-Boltzmann relation: power out P = (emissivity) * (S-B constant) * T^4 ??

No more bullshit. "Yes" if you agree that equation is valid, or "No" if you deny that it is valid. Just that and no more.

I'm not asking your permission. I'm just trying to find out whether you're actually crazy or just bullshitting.

Nonsense. This is textbook heat transfer physics. We have a fixed emissivity. Therefore, according to the Stefan-Botlzmann radiation law, the ONLY remaining variable which determines radiative power out is temperature. NOTHING else. That's what the law says: (emissivity) * (S-B constant) * T^4. That's all. Nothing more. This makes it stupidly easy to calculate the radiative power out, and therefore the necessary power in.

YOU are disputing the Stefan-Boltzmann law. But it is a known physical law, and this is a textbook demonstration of it. You lose.

You showed no such thing. Your calculations contradict themselves, and your methodology contradicts itself.

EVEN IF we accepted your idea that the "electrical" power required to be input to the heat source is dependent on the temperature difference between the heat source and chamber wall (a violation of the S-B law), you still contradict yourself because your answer of a hotter heat source would still then require MORE power, because the difference is greater. But that is not allowed by the stated conditions of the experiment, and you keep glossing over that simple check of your own work which proves it wrong.

So no matter how you cut it, your answer is wrong, by your own rules.

This is a simple requirement of the Stefan-Boltzmann law. The radiative power output of a given body does not depend on other nearby bodies. It's inherent in the law itself. And this is precisely where you are getting it wrong.

I find it highly amusing that you derive your own calculations from the Stefan-Boltzmann law, then deny that it is valid. Every time you try to squirm out of this you just contradict yourself again.

I am further amused that you find it "adorable" that you've been proven wrong. Be a man for a change and admit it. Or show us your own replacement for the Stefan-Boltmann law. You don't get to have it both ways.

Besides, 99.99% is not nearly reliable enough. (And besides, this number is misleading... probably outright false.)

According to calculations I did a year or two ago, in order for a "smart firearm" to be worthwhile and actually solved the problem for which is supposed to be designed, for modern arms, it needs to have AT LEAST three 9s behind the decimal point for true positives: 99.999%, and probably actually 4.

And that's assuming the stats are correct. What does that 99.99% represent? True positives? What is its rate at rejecting true negatives? After all, that's the entire purpose it was designed for.

Further yet: how long does the battery last? What is its success rate with a dead battery? Current battery tech is not capable of delivering 99.99% reliability because batteries go bad even on the shelf.

Well, I didn't interpret the article the same way you did. I thought the article was saying that you can be logical and still feel wonder. It wasn't saying that science-oriented people need to be religious, but rather that religious people should stop seeing them as somehow inhuman and unfeeling without a belief in their God.

Why would they be in for a "rude" awakening, when one would think that any awakening at all should be a pleasant surprise?

Further, as Sam Harris argues quite well, one need not be a theist to have moral values. Science + secular society are perfectly capable of agreeing upon ethical and moral rules, without resorting to theism.

No, you don't. You just have to give enough people cheap energy that they're no way to put the genie back in the bottle.

Once you do that, the Gateses and Kochs of the world can be safely ignored. They won't have any power anymore.

"Any dick and his Cessna" is not involved in commercial interstate transportation.

I am disputing nothing of the sort. As I have explained many times now, you are not drawing your lines properly.

You keep making the same bullshit assertions, after I have proved them false. Why do you do this?

You're just going to look that much more foolish later.

PROOF that you're bullshitting everybody:

NO!!! I did not. I held the power constant, just as Spencer stipulated.

For a gray body, which you stipulated, radiant power out = (emissivity) * (S-B constant) * T^4. This is the Stefan-Boltzmann relation between radiant temperature of a gray body and its power output.

T is known: 150F or 338.71 K.

Solving for radiant power out we get 82.12 Watts/m^2. Times khayman80's stipulated area (510.065 m^2) = 41886.54 Watts.

It is this POWER that remains constant according to Spencer. Khayman80 himself asserted that "power in = power out". Therefore POWER IN = POWER OUT = 41886.54 Watts.

But because of the equation I showed above, which is a physical law, after the hollow sphere is inserted (which is COLDER than the heat source), nothing at the power source has changed. Emissivity is still the same. Power input is still 41886.54 Watts = radiant power output of 41886.54 Watts. Which (by the equation above) yields the same temperature.

I didn't assume the same temperature, I calculated it using known physical law.

ANYTHING ELSE is a direct violation of the Stefan-Boltzmann law.

NO!!!

I have told you 5 or 6 or maybe more times now, this is a VIOLATION of the very straightforward Stefan-Boltzmann law.

How it applies in this situation is quite straightforward, and not at all as complex as you are making it out to be.

Radiant power output of a gray body is calculated using ONLY the variables: emissivity and temperature. THAT IS ALL. There is no other variable dealing with incident radiation, or anything else. When the system is at radiant steady-state, power out (and therefore power in) are easily calculated, and I have calculated them.

Further, Spencer's "electrical" input power was to the heat source, not to the whole system.

YOUR OWN PRINCIPLE: power in = power out. Now you're trying to contradict yourself and say it meant something else.

It's just bullshit. You're squirming like a fish on a hook. You just don't seem to realize you have already been flayed, filleted, and fried in batter.

You're owned, man.

I do think it's cute, however, how you tried to use Spencer's statement as proof of itself.

Have I reminded you lately that your grasp of logic seems a bit off?

It doesn't seem odd at all, because established science shows that his assertion that the temperature changes is wrong.

Considering that he is wrong, why should I find it odd that he said a wrong thing.

SIMPLE CALCULATION, which I have already shown several times: power "sufficient" to heat the heat source under initial conditions to 150F: 41886.54 Watts.

Power input at the source remains constant. Spencer's stipulation. Therefore by the S-B relation, once everything comes up to radiative steady-state the input power and output power of the heat source are constant. There is no inconsistency here.

Further, because ALL the other surfaces are cooler than the heat source, ALL the net heat transfer is outward, because T(a)^4 - T(b)^4 is a positive number.

This is established science, and it doesn't depend on the incorrect opinions of either Spencer or yourself.