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Comment: Re:No, no. Let's not go there. Please. (Score 1) 721

by dave420 (#47906823) Attached to: Why Atheists Need Captain Kirk

Sorry - the 10 commandments are good rules? The first few are about God being insecure! What the hell do they have to do with laws?

You seem to be confusing vocal atheists with the vast majority (we know they are the minority as otherwise you'd not be able to move for militant atheists). You are also not factoring in the amount of demonisation atheists suffer by various theist groups, who pounce on any atheist voicing his/her opinion and extrapolate that to every single atheist out there.

Your claims of "large movements" etc. are just your opinion - without any facts to back them up they are not particular pertinent to the discussion.

Comment: Re:illogical captain (Score 1) 721

by dave420 (#47906781) Attached to: Why Atheists Need Captain Kirk

And I'm sure even a lowly science student could outline an experiment which shows you are incorrect.

In your example, the ignorance and arrogance lies with you: You are ignorant of the fact that it's inherently possible to disprove your claims, and your arrogance is stopping you from realising your own ignorance, automatically assuming you are correct.

You are a blind man claiming he can see colours.

Comment: Re: illogical captain (Score 1) 721

by dave420 (#47906771) Attached to: Why Atheists Need Captain Kirk

You are confusing people's characteristics and why they do things. Those people also wore trousers, but I doubt you're claiming they killed for trousers.

Stalin, Mao, and Pot were the religion. This has been covered thousands of times already, so you not realising this speaks very poorly of how well-though-out your stance is.

Comment: Re:When the cat's absent, the mice rejoice (Score 4, Insightful) 210

Well, I'd be with you if the government was poking around on the users' computers, but they weren't. The users were hosting the files on a public peer-to-peer network where you essentially advertise to the world you've downloaded the file and are making it available to the world. Since both those acts are illegal, you don't really have an expectation of privacy once you've told *everyone* you've done it. While the broadcasting of the file's availability doesn't prove you have criminal intent, it's certainly probable cause for further investigation.

These guys got off on a narrow technicality. Of course technicalities do matter; a government that isn't restrained by laws is inherently despotic. The agents simply misunderstood the law; they weren't violating anyone's privacy.

Comment: Re:Crude? (Score 2) 79

by hey! (#47904781) Attached to: Original 11' <em>Star Trek Enterprise</em> Model Being Restored Again

Compare that to some of the ST:TNG props that I've seen that look fine on screen, but when examined closely look like someone gave a 5-year old a couple of shots of vodka and turned them loose with a paintbrush.

There's a certain wonder to that too.

I had the same reaction when I saw the ST:TNG props in person. You wouldn't buy a toy that looked that cheesy. The wonder of it is that the prop makers knew this piece of crap would look great onscreen. That's professional skill at work. Amateurs lavish loving care on stuff and overbuild them. Pros make them good enough, and put the extra effort into stuff that matters more.

Comment: Re: Great one more fail (Score 1) 450

by hey! (#47904749) Attached to: High School Student Builds Gun That Unlocks With Your Fingerprint

These kinds of responses are conditioned on certain assumptions that may not hold for all users.

For example, let's assume that you have no need whatsoever to prevent other users from using your gun. Then any complication you add to the firearm will necessarily make it less suitable, no matter how reliable that addition is. An example of someone on this end of the spectrum might be a big game hunter who carries a backup handgun.

On the other hand suppose you have need of a firearm, but there is so much concern that someone else might use it without authorization that you reasonably decide to do without. In that opposite situation you might well tolerate quite a high failure rate in such a device because it makes it possible to carry a gun. An example of someone on this end of the spectrum might be a prison guard -- prison guards do not carry handguns because of precisely this concern.

This isn't rocket science. It's all subject to a straightforward probabilistic analysis *of a particular scenario*. People who say that guns *always* must have a such a device are only considering one set of scenarios. People who say that guns must *never* have such a device are only considering a different set of scenarios. It's entirely possible that for such a device there are some where it is useful and others where it is not.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 138

by khayman80 (#47898793) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

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. [Jane Q. Public, 2014-09-13]

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!

Can we agree that the required electrical heating power would be zero if the chamber walls were also at 150F?

... I held the power constant, just as Spencer stipulated. ... [Jane Q. Public, 2014-09-13]

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.

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.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 138

by khayman80 (#47898707) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

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. [Jane Q. Public, 2014-09-13]

No. Draw a boundary between the source (T1=150F) and the chamber walls (T4=0F) before the hollow sphere is added. Power in = power out. Variable "electricity_initial" flows in at whatever rate is needed to keep T1=150F. Net heat transfer flows out from source to chamber walls. Power in = power out:

electricity_initial = p(14) = (e)(s) * ( T1^4 - T4^4 )

So are you disputing that power in = power out through a boundary where nothing inside that boundary is changing with time? Or are you disputing that the radiation from the chamber walls passes through a boundary drawn just inside them?

And again, if you keep ignoring that "power in" half of the equation that all Sky Dragon Slayers miss, you'll have to keep wondering why your "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.

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 138

by khayman80 (#47898663) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

... It's not that I don't agree. You might come up with the right answer for some sub-calculation. I don't know, I don't care, and I'm not even going to bother to check, much less agree. The issue is that I have already solved the problem, and arrived at the correct answer (within reasonable limits). So I don't HAVE to agree or disagree with you. I've already done it, according to the correct textbook-approved physics. AND (unlike you) I checked my work and it checks out. And unlike your answer it doesn't violate conservation of energy. ... [Jane Q. Public, 2014-09-13]

I just showed that Jane/Lonny Eachus solved the "correct answer" to a different question. Instead of holding the electrical heating power constant like Dr. Spencer did, Jane/Lonny held the source temperature constant. In that case, the electrical heating power required to keep the source at 150F drops by a factor of two after the enclosing shell is added. This shows that holding the electrical heating power constant like Dr. Spencer did is different than holding the source temperature constant like Jane/Lonny did.

... 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. ... [Jane Q. Public, 2014-09-13]

No, in your example the electrical heating power drops by a factor of two after the enclosing shell is added. And once again, your calculation of the power sufficient to heat the heat source would be exactly the same if the chamber walls were also at 150F. But the right answer there is zero, because an electric heater wouldn't be necessary. Is this really so hard to understand, or are you deliberately spreading misinformation?

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 138

by khayman80 (#47898605) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

... For a given gray body, its thermodynamic temperature is related ONLY to emissivity, radiant power output, and the S-B relation (emissivity)* (S-B constant) * T^4. PERIOD. That's physics. ... [Jane Q. Public, 2014-09-13]

And that's why what you're calculating isn't Dr. Spencer's electrical heating power, because it should be "zero" if the chamber walls are also at 150F.

... I repeat: given your OWN "draw a border around it" thermodynamic reasoning, the power input (whether it is electrical, chemical, or something else) must equal that output. That's physics. You're trying to bring in energy from elsewhere, but it isn't relevant to this calculation AT ALL; it is erroneous thinking. Power input is specified to be constant. Calculating the total power in initial conditions is, as I stated before, "dirt simple". Specified emissivity is known: 0.11. Temperature is known: 338.71K. Solving for the above we get 82.12 W/m^2. We already have ALL the information needed to calculate this, given the Stefan-Boltzmann relation (above), relating these numbers. Nothing else is required, and in fact trying to introduce other factors is ERROR. That is what the accepted science says. ... [Jane Q. Public, 2014-09-13]

If you draw a boundary around the heated source, you have to account for the 0F chamber walls because they're radiating power in through the boundary. Otherwise you're not actually calculating Dr. Spencer's electrical heating power, or you misunderstand conservation of energy.

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?"

If the power input to the heated sphere is fixed, then the power output in the form of radiant temperature is fixed: (epsilon)(sigma)T^4. It's physics! It doesn't matter how you try to squirm and twist this. You have been owned. End of story. [Jane Q. Public, 2014-09-13]

Jane, didn't it seem odd that you interpreted Dr. Spencer's challenge to mean "Assuming the source temperature is held fixed, does the source temperature change after a passive plate is added?"

How is that different than asking "Assume x = 150 forever. Will x change?"

Isn't that a silly question? Shouldn't you at least consider the possibility that you've misinterpreted "power input to the heat source"?

Comment: Jane/Lonny Eachus goes Sky Dragon Slayer (Score 1) 138

by khayman80 (#47897631) Attached to: 3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting

No. Holding constant the electrical power heating the source is very different than holding constant the source temperature. Like Jane, let's assume the source temperature is constant (rather than the electrical heating power) and use Jane's equation and notation:

... we have 4 surfaces, which I will call 1, 2, 3, 4 moving outward, so 1 is the surface of the heat source, 2 the inside of the hollow sphere, 3 the outside of the hollow sphere, and 4 the chamber wall. T3 for example would be radiative Temperature of surface 3. ... [Jane Q. Public, 2014-09-10]

Draw a boundary between the source (T1=150F) and the chamber walls (T4=0F) before the hollow sphere is added. Power in = power out. Variable "electricity_initial" flows in at whatever rate is needed to keep T1=150F. Net heat transfer flows out from source to chamber walls. Power in = power out:

electricity_initial = p(14) = (e)(s) * ( T1^4 - T4^4 ) = (e)(s) * (8908858139.78) = 55.5913 W/m^2

Now add the hollow sphere and draw a boundary between the source (T1=150F) and the inside of the hollow sphere (T2). A different "electricity_final" flows in, and heat transfer p(12) flows out.

electricity_final = p(12) = (e)(s) * ( T1^4 - T2^4 )

Now draw a boundary between the outside of the hollow sphere (T3=T2) and the chamber walls (T4=0F): "electricity_final" flows in, and heat transfer p(34) flows out. Since power in = power out:

electricity_final = p(34) = (e)(s) * ( T2^4 - T4^4 )

Combine these two equations:

T1^4 - T2^4 = T2^4 - T4^4

Solve for:

T2 = T3 = 305.47K = 90.176 deg. F.

electricity_final = 27.8 W/m^2.

So if the source temperature is held constant at 150F, adding the hollow sphere reduces the necessary electrical heating power to keep the source at 150F by a factor of two, from 55.6 to 27.8 W/m^2.

Can we agree on that?

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