It was a typo/brain fart. I though it was the American Amateur Radio, I remebered it was the ARRL right after I hit the post button.

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.

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.

... 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?

... 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"?

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?

Except that if you hurt yourself you are not breaking the law.
There are lots of safety rules. You have to obey the regulations.

I am not just talking about Amature radio but FCC licensing. That includes TV, Radio, Public service bands, HAM, and even wifi. The power regulations and frequency regulations are mostly in place to keep you from stomping on everyone else. AARL has a special relationship with the FCC that allows them to do testing and have a large voice in setting the rules for HAMs. But HAMs are a small section of licensed users.
Frankly AARL works on the notion that someone that makes the effort to study and pass the tests will not do anything hyper stupid like highjacking a TV stations band. Frankly much like the EAA the AARL seems to do a good job with keeping the dangerous idiots out.
But what I said stands. The licensing is about interference. If you hurt yourself you will not get a visit from FCC but if you interfere with someone else you could get a visit from the FCC

... input power at steady-state is fixed, and a value that we already know: 41886.54 W. ... [Jane Q. Public, 2014-09-12]

Again, we disagree about what's held fixed. That value you keep calculating isn't the constant electrical power heating the source.

In this experiment there is a "... constant flow of energy into the plate from the electric heater... flowing in at a constant rate... the electric heater pumps in energy at a constant rate. ..."

In my interpretation, Dr. Spencer's challenge is basically: "Assuming an electric heater pumps energy at a constant rate to the source, does the source temperature change after a passive plate is added?"

You've repeatedly noted that there are no other factors involved in calculating your 82 W/m^2 (41886.54 W) value. So if it's held fixed, the source temperature is also held fixed.

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?"

Is that right?

The key would be how many places had mag strip first. Brazil is moving faster now than 10 years ago.

or smoothwall or moonwall.

That is what the person I was replying to claimed. You can replace baseload with Wind with natural gas peaking plants but I think Nuclear is a better long term source.

Microsoft has a shot if they do not destroy Nokia. I flat out think that Nokia makes the best looking phone. LG, Samsung, and Apple all use great displays, Android has the best feature set, Nokia the best camera,Motorola has the Motovoice feature that I really want, and Apple has great stability and battery life.
Windows Phone 8.1 actually looks pretty good now.
For me the perfect phone would be a Nokia with the screen from the G3 running Android with Motovoice.
Others would have a different prefect phone.

25 million people in Australia.
313 million people in the US.
It takes a while longer to roll out big changes like that. I have had a chip and pin card for over a year but not every location can take it.

Funny but I have actually written an IOS app and it is in the app store. I have a mac and I love it. This rev of the iphone has bigger screens... Like Android phones. NFC like Android phones. As to the rest of the statement... You do know that the Apple app store was the original home of the fart app...

IOS is fine for people that like it. I find it too restrictive and dull for my tastes but it is good OS and the phones are good hardware. They are just not worth the worship that heaped on them.

Why is Google better than Apple?
Simple, Compare an IOS phone with all the Google apps removed to an Android phone without any Apple apps.

NFC is cool and boy did people think it was neat when I paid for something at the 7-11 with my Galaxy Nexus a few years ago. You are right that it might finally take off in the US now that Apple is doing it. They are even following the standards so they can use the existing NFC machines at Walgreens, 7-11, and McDonald's. It is really not innovative but an example of the clout that Apple has with the carriers.