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Comment Re:Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 176

It's "stupidly easy" to calculate radiative power out and power in through what boundary? The boundary you're describing has to include the source's radiative power passing out through it, without including radiative power from the chamber walls passing in. I think that's impossible, but feel free to explain exactly where such a boundary would be drawn.

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.

You can draw the boundary right around the heat source. Electric power comes in, radiative power goes out. There is no contradiction, and no inconsistency.

Once again, I agree that "power out" through a boundary drawn around the heat source is given by the Stefan-Boltzmann law. But I've obviously failed to communicate that the power from the chamber walls has to pass in through that boundary, so you're only using half the equation to calculate the electrical heating power.

And again: 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. (Since we are discussing diffuse gray bodies here, we can consider it all reflected or scattered because there is no transmissivity.) The radiation that crosses the boundary that does not strike the smaller sphere due to view factor also just passes right back out. You are ignoring (e*s) * (Ta^4 - Tb^4). Anything other than what I described does not add up.

Once again, no. Draw a boundary around the heat source: power in = electrical heating power + radiative power in from the chamber walls

Just NO. Net heat transfer is ALL from hotter to colder, by (e*s) * (Ta^4 - Tb^4).

Let me put it another way: we can easily show how you have gotten your thermodynamics backward by referring to a question you asked earlier. 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).

This clearly illustrates your ass-backward thermodynamic thinking. The radiative power output of the heat source does not change due to the temperature of the walls. At all. The only thing that changes as the wall temperature changes is the heat transfer, which would lessen as you brought up the temperature of the walls. But that isn't because the heat source is using less power, it is because you are putting more power into raising the wall temperature. You are creating a more thermodynamically energetic environment, and that requires power.

Just like your other arguments: you invent power in out of thin air, and claim you can do that because it's "moving" in the opposite direction in which heat transfer is actually taking place.

You are giving physicists a bad name, and I repeat that I am going to show this to all the world to see.

Comment Re:Time for new terminology (Score 0) 635

GISS is precisely the dataset that has been accused of the the most egregious "adjustments".

Further, it was recently found that GISS was improperly averaging in "missing" data over a period of years, which they admitted to about 2 months ago.

It is interesting that the historical HCN data disagree quite a bit with the modern versions of the data sets.

Comment Re:This may be the way to escape from Comcast (Score 2) 418

In the end, you signed a contract and are legally bound to continue to pay for almost any type of service inturruption.

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.

Comment Re:Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 176

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.

Comment Re:Jane/Lonny Eachus goes Sky Dragon Slayer (Score 0) 176

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

It's also adorable that Jane keeps ignoring the fact that his "electrical heating input" calculation wouldn't change even if the chamber walls were also at 150F. Even Jane should be able to comprehend that a 150F source inside 150F chamber walls wouldn't need electrical heating power.

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.

Comment Re:Jane/Lonny Eachus goes Sky Dragon Slayer (Score 0) 176

You're either disputing conservation of energy, or you're not calculating the actual electrical heating power. If you're calculating the actual electrical heating power, your calculation has to account for radiation from the chamber walls because it passes in through that boundary. That's why the electrical heating power would be zero if the chamber walls were also at 150F!

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.

It's so adorable that Jane keeps insisting that Jane kept the power constant, even after I showed that Jane's calculation was only able to hold the source temperature constant after the enclosing shell was added by halving the actual electrical heating power.

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.

It's also adorable that Jane keeps ignoring the fact that his "electrical heating input" calculation wouldn't change even if the chamber walls were also at 150F. Even Jane should be able to comprehend that a 150F source inside 150F chamber walls wouldn't need electrical heating power.

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.

Comment Re:Great idea! Let's alienate Science even more! (Score 1) 937

I think you may be missing my point. Let me illustrate with a scenario:

Doctor: I think you have long cancer.
You: That's an extraordinary claim, I want proof.
Doctor: Sorry, you're not my patient. I don't have time to talk.

Do you ignore what he said because he made an extraordinary claim and wouldn't meet some particular burden of proof?

Comment Re:Great one more fail (Score 1) 600

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.

Comment Re:illogical captain (Score 5, Insightful) 937

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.

Comment Re:illogical captain (Score 5, Informative) 937

Atheists will be in for a rude awakening when they die as they will realize that their belief was incomplete. Regardless, they can be just as good, (or as bad) as theists if they practice the golden rule.

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.

Comment Re:Great idea! Let's alienate Science even more! (Score 1) 937

If you claim that the existence of god or gods is a truth, it is incumbent on you to show that it is indeed a truth. Otherwise, you're just blowing smoke. Assertions are simply opinions.

So how does that work then? If you tell me something that's true, but you can't be bothered to try persuading me of it, shall I flatly refuse to believe it? Regardless of the idea's underlying merit? I don't see how that policy is profitable.

Comment Re:Hollywood Logic (Score 1) 937

I believe it was the great Joel Hodgson and Josh Weinstein, who counselled us in such matters:

If you're wondering how he eats and breathes
And other science facts,
Just repeat to yourself "It's just a show,
I should really just relax
For Mystery Science Theater 3000."

Comment Re:Great idea! Let's alienate Science even more! (Score 1) 937

Those who understand how to correctly apply the scientfic method know that the burden of proof is on the person making the assertion (the alternative hypothesis).

I never really bought into this idea of "burden of proof". It strikes me as a rhetorical / debating tactic, rather than a part of good-faith truth-seeking.

If an assertion is true, then it's true regardless of who in a debate advances it.

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