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Post by Andrew on Jan 22, 2016 13:32:57 GMT
you don't understand that a brick is not a gas? you are trolling Andrew! Don't you have anything better to do? Another audit fail Why are you mentioning insulation and surface area majorly impacting or invalidating the results of spencers radiation thought experiment in a vacuum when the warming result can be verified only with reference to the engineers net radiation heat loss curves where insulation and surface area can play no role in invalidating the results? ? How have you created that result for the vacuum??? how is it relevant to the bricks???
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Post by icefisher on Jan 22, 2016 17:56:14 GMT
you don't understand that a brick is not a gas? you are trolling Andrew! Don't you have anything better to do? Another audit fail Why are you mentioning insulation and surface area majorly impacting or invalidating the results of spencers radiation thought experiment in a vacuum when the warming result can be verified only with reference to the engineers net radiation heat loss curves where insulation and surface area can play no role in invalidating the results? ? How have you created that result for the vacuum??? how is it relevant to the bricks??? Simply Andrew Radiation is a weak force. Your model shows that a molecule cooling to outerspace can only be heated by radiation to 50% of the rate radiation shines on it absorbing all radiation and losing half out to space. . . .reducing the cooling rate of the radiator by 50%. Fine I can accept that as a possibility. It seems that with a gas the rate of diffusion is also the same .5. Engineering Toolbox Heat Flux for free convection. That would explain why Woods Greenhouse experiment failed to produce any more heat in the IR blocking greenhouse than the IR transparent greenhouse. And either results in an identical amount of warming for the greenhouse. the obvious question is why? The IR transparent greenhouse creates as much warming as the IR opaque one. Seems nothing else matters Andrew. If you can't address this issue your theory (or if you prefer the inculcated popular theory) does not stand up.
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Post by Andrew on Jan 22, 2016 20:06:15 GMT
And so yet another audit fail passes by Why are you mentioning insulation and surface area majorly impacting or invalidating the results of spencers radiation thought experiment in a vacuum when the warming result can be verified only with reference to the engineers net radiation heat loss curves where insulation and surface area can play no role in invalidating the results? ? How have you created that result for the vacuum??? how is it relevant to the bricks??? >> Your model shows that a molecule cooling to outerspace can only be heated by radiation to 50% of the rate radiation shines on it absorbing all radiation and losing half out to space. . . .reducing the cooling rate of the radiator by 50%. I have no idea what that is supposed to mean or describe
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Post by icefisher on Jan 22, 2016 20:15:33 GMT
Indeed another failure by the auditee. The police eat that up as a sign of guilt when the interviewee evades the question
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Post by Andrew on Jan 22, 2016 20:21:27 GMT
Indeed another failure by the auditee. The police eat that up as a sign of guilt when the interviewee evades the question No evasion by me. I never changed any goal posts and you constantly claimed I did. That makes you a lying cheating son of a pregnant dog. If you do have something intelligent to say on this topic i am still waiting to here about it. Why are you mentioning insulation and surface area majorly impacting or invalidating the results of spencers radiation thought experiment in a vacuum when the warming result can be verified only with reference to the engineers net radiation heat loss curves where insulation and surface area can play no role in invalidating the results? ? How have you created that result for the vacuum??? how is it relevant to the bricks??? >> Your model shows that a molecule cooling to outerspace can only be heated by radiation to 50% of the rate radiation shines on it absorbing all radiation and losing half out to space. . . .reducing the cooling rate of the radiator by 50%. I have no idea what that is supposed to mean or describe. If you can write it in proper English it might help
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Post by icefisher on Jan 22, 2016 20:28:34 GMT
If you take a steel ball with no source of heat inside of a hollow steel sphere and heat the outside of the steel sphere until it is a uniform equalized temperature of 15c how hot is the steel ball?
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Post by Andrew on Jan 22, 2016 20:31:44 GMT
If you take a steel ball with no source of heat inside of a hollow steel sphere and heat the outside of the steel sphere until it is a uniform equalized temperature of 15c how hot is the steel ball? I am supposing the answer you want will be 15C but no doubt you can surprise me.
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Post by icefisher on Jan 22, 2016 23:32:03 GMT
If you take a steel ball with no source of heat inside of a hollow steel sphere and heat the outside of the steel sphere until it is a uniform equalized temperature of 15c how hot is the steel ball? How big is this hollow steel sphere? 12 inches or 12 feet? Nice catch Code. Distance does make a difference. But lets say the diameter of the steel ball is the same as earth at 12,742km and the inside diameter of the sphere is the diameter of the earth plus the atmosphere up to the stratopause or about 12,850km An answer within a 1/2 degree should be good enough.
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Post by icefisher on Jan 23, 2016 2:54:09 GMT
If you take a steel ball with no source of heat inside of a hollow steel sphere and heat the outside of the steel sphere until it is a uniform equalized temperature of 15c how hot is the steel ball? I am supposing the answer you want will be 15C but no doubt you can surprise me. Andrew I want you to use the Engineering Toolbox Chart and anything else you use to calculate this stuff and give an answer.
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Post by nonentropic on Jan 23, 2016 3:33:49 GMT
OK this is so simple its silly since the center is hollow and thus a vacuum it has no temperature as expressed as a molecular velocity.
and you have already told us the sphere is 15C. QED.
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Post by Andrew on Jan 23, 2016 4:03:16 GMT
OK this is so simple its silly since the center is hollow and thus a vacuum it has no temperature as expressed as a molecular velocity. and you have already told us the sphere is 15C. QED. Nonentropic, He will be getting confused by the different circumferances of these two objects where for the smallest object all of the radiation from it can only leave that small object and only be absorbed by the larger object whereas for the larger object far more radiations are absorbed by itself than reach the smaller object. I am supposing if we get a circumferance calculator and radiation net transfer heating curves we will be able to demonstrate the known laws of science and mathematics can survive the question. Thinking about it now I think I already did this challenge many years ago for Icefisher and I do not recall using a circumferance calculator. The answer will be 15C anyway. I checked the earlier conversation. Icefisher specified the areas of each sphere as simple ratios to make it simpler to calculate.
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Post by icefisher on Jan 23, 2016 4:07:13 GMT
OK this is so simple its silly since the center is hollow and thus a vacuum it has no temperature as expressed as a molecular velocity. and you have already told us the sphere is 15C. QED. Well we did put a smaller steel ball (solid) suspended in the middle of the otherwise hollow sphere just to be clear.
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Post by Andrew on Jan 23, 2016 4:32:43 GMT
OK this is so simple its silly since the center is hollow and thus a vacuum it has no temperature as expressed as a molecular velocity. and you have already told us the sphere is 15C. QED. Well we did put a smaller steel ball (solid) suspended in the middle of the otherwise hollow sphere just to be clear. Is it suspended by wires? No matter how it is constructed the answer is going to be 15C. If you want to disprove the maths and science of the last 150 years you will need to show your working rather than getting others to show one more time just how daft you can be.
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Post by icefisher on Jan 23, 2016 4:40:30 GMT
OK this is so simple its silly since the center is hollow and thus a vacuum it has no temperature as expressed as a molecular velocity. and you have already told us the sphere is 15C. QED. Nonentropic, He will be getting confused by the different circumferances of these two objects where for the smallest object all of the radiation from it can only leave that small object and only be absorbed by the larger object whereas for the larger object far more radiations are absorbed by itself than reach the smaller object. I am supposing if we get a circumferance calculator and radiation net transfer heating curves we will be able to demonstrate the known laws of science and mathematics can survive the question. Thinking about it now I think I already did this challenge many years ago for Icefisher and I do not recall using a circumferance calculator. The answer will be 15C anyway. I checked the earlier conversation. Icefisher specified the areas of each sphere as simple ratios to make it simpler to calculate. forget the differences in diameter. i asked for an answer to the closest half degree which rules out any difference to the answer. So it appears you have given the correct answer: 15C. I probably should have simply said make the sphere hot enough so the steel ball surface was 15C. So the next thing we are going to do is float some glass beads in the space between the steel ball and the inside surface of the sphere.Uh lets just call it a gas why not! So what temperature would the gas be when its all warmed as much as its going to warm, without any lapse rate?
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Post by Andrew on Jan 23, 2016 6:04:51 GMT
For an almost maximum size internal sphere that is almost the same size as the outer sphere then while all radiations from the small sphere can only reach the larger sphere a small amount of radiations from the larger sphere will reach the larger sphere and miss the small sphere. Therefore as a good approximation for almost the same size spheres, almost the same amount of radiation will pass back and forth between the objects.
However, for a very small sphere, a very large number of emissions from the larger sphere are absorbed into the the larger sphere and only a much smaller number actually arrive at the surface of the small sphere.
The ratio of the different sphere sizes is only 1.0085. We know that some of the radiations from the larger sphere do not reach the smaller sphere and will only reach the larger sphere. As the ratio increases towards infinity then more and more of the radiations of the large sphere will only reach the larger sphere.
So if you did the geometry, while it is true as the small sphere gets smaller and fewer emissions leave the smaller sphere, you will find the exact same number of emissions leave the larger sphere to be absorbed on the smaller sphere, while for any size difference it is always the case that all emissions from the smaller sphere always reach the larger sphere.
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