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Post by icefisher on Feb 7, 2016 4:24:39 GMT
OK Andrew I will break it down in little chunks for you so you can understand what I am thinking.
1. Engineering Toolbox Net Radiation Loss Tool. the first thing you need to understand about this is its a tool, not a radiation law. Being a tool it does not need to give you more information. You can use the tool to calculate heat losses into an environment, it does not give you the heat losses by the environment. To figure those you may use additional calculations to restrict heat loss via insulation. The tool assumes no additional heat losses, which of course you can calculate them using conduction, convection, and reradiation from the backside of the receiving surface. Again Andrew its a tool, not a law. If you think it operates differently then give me a calculation of that shows where the heat comes from as the exterior walls cool to the outdoors. Demonstrate the use of the tool the way you would use it.
2. Photon Backradiation Cartoon. The key to your Planck mathematics to support this ridiculous model is that the IR blocking layer will only heat to 50% of the blackbody equilibrium value, yet we know that is not true. Observation says it will heat to the same value as the surface of the steel ball. Intuitively, and I believe its fair to claim that the only ways to slow the cooling of a surface is with insulation or reflection. I can demonstrate those. If you want to add a method I would like to see it demonstrated. A perfectly reasonable request. Now in saying that I think it would be possible to have enough CO2 to create some insulation. Perhaps Venus has enough.
3. So I am completely unconvinced that the amount of cooling from CO2 insulation value amounts to even a 1/4 degree. I would like to see calculations, calculations based upon observed effects, that would show differently.
4. Now water vapor being 25 times more prevalent in the atmosphere than CO2 might add up to something. But its really hard to say if it would result in any warming since water vapor starts by extracting a lot of heat from the surface and has a limited lifetime in the atmosphere. At any rate what ever insulating effect it had it would need reduction for that.
5. We know the surface is slowed from cooling. So does that not prove the greenhouse effect? Well no! In Spencer's backyard experiment he noted great reduction in cooling when clouds flew overhead at night. When there were no clouds temperatures plummeted at a steep slope. Now clouds are highly reflected and we already know that this can slow cooling. Of course clouds also slow warming.
6. The surface is warmer than it would be if there were no atmosphere. Well that does not prove anything. It could be for a different reason.
7. We know that warm objects radiate in all directions. Well thats definitely plausible. But it falls a bit short proof wise. But it also may not matter if what is radiated is not little finite packets of fixed energy. The wave mechanics view seems to be supported by the heat in the thermosphere. If its a vibration that heats everything lower frequency vibrations tend to dampen higher frequency vibrations rather than enhance them. The slowing of cooling calculated from a finite packet of energy transfer model needs validation by observation.
Conclusion: Due to a lack of demonstration of the principles of backradiation I remain skeptical of a radiation based greenhouse effect not verifiable via heat exchange calculations based on observation. I am unconvinced that the slowing of cooling noted in the atmosphere is a result of backradiation or a slowing of cooling by gases. Though I will note that water could be slowing cooling and I may not perceive the cooling of the oceans and lakes occurring to support it. And finally we know the big numbers in slowing of cooling and slowing of warming is due to clouds and they are not a gas.
There are a few obvious things going on: Obviously at a minimum due to insulation value so restriction of cooling is going on. That seems intuitive and I suppose you can call that a greenhouse effect. The big question is how much cooling goes on. You have a nice numeric equation to estimate it but its just numerology if one cannot demonstrate that the equation is being correctly applied.
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Post by Andrew on Feb 7, 2016 6:46:03 GMT
OK Andrew I will break it down in little chunks for you so you can understand what I am thinking. 1. Engineering Toolbox Net Radiation Loss Tool. the first thing you need to understand about this is its a tool, not a radiation law. Being a tool it does not need to give you more information. You can use the tool to calculate heat losses into an environment, it does not give you the heat losses by the environment. To figure those you may use additional calculations to restrict heat loss via insulation. The tool assumes no additional heat losses, which of course you can calculate them using conduction, convection, and reradiation from the backside of the receiving surface. Again Andrew its a tool, not a law. If you think it operates differently then give me a calculation of that shows where the heat comes from as the exterior walls cool to the outdoors. Demonstrate the use of the tool the way you would use it. 2. Photon Backradiation Cartoon. The key to your Planck mathematics to support this ridiculous model is that the IR blocking layer will only heat to 50% of the blackbody equilibrium value, yet we know that is not true. Observation says it will heat to the same value as the surface of the steel ball. Intuitively, and I believe its fair to claim that the only ways to slow the cooling of a surface is with insulation or reflection. I can demonstrate those. If you want to add a method I would like to see it demonstrated. A perfectly reasonable request. Now in saying that I think it would be possible to have enough CO2 to create some insulation. Perhaps Venus has enough. 3. So I am completely unconvinced that the amount of cooling from CO2 insulation value amounts to even a 1/4 degree. I would like to see calculations, calculations based upon observed effects, that would show differently. 4. Now water vapor being 25 times more prevalent in the atmosphere than CO2 might add up to something. But its really hard to say if it would result in any warming since water vapor starts by extracting a lot of heat from the surface and has a limited lifetime in the atmosphere. At any rate what ever insulating effect it had it would need reduction for that. 5. We know the surface is slowed from cooling. So does that not prove the greenhouse effect? Well no! In Spencer's backyard experiment he noted great reduction in cooling when clouds flew overhead at night. When there were no clouds temperatures plummeted at a steep slope. Now clouds are highly reflected and we already know that this can slow cooling. Of course clouds also slow warming. 6. The surface is warmer than it would be if there were no atmosphere. Well that does not prove anything. It could be for a different reason. 7. We know that warm objects radiate in all directions. Well thats definitely plausible. But it falls a bit short proof wise. But it also may not matter if what is radiated is not little finite packets of fixed energy. The wave mechanics view seems to be supported by the heat in the thermosphere. If its a vibration that heats everything lower frequency vibrations tend to dampen higher frequency vibrations rather than enhance them. The slowing of cooling calculated from a finite packet of energy transfer model needs validation by observation. Conclusion: Due to a lack of demonstration of the principles of backradiation I remain skeptical of a radiation based greenhouse effect not verifiable via heat exchange calculations based on observation. I am unconvinced that the slowing of cooling noted in the atmosphere is a result of backradiation or a slowing of cooling by gases. Though I will note that water could be slowing cooling and I may not perceive the cooling of the oceans and lakes occurring to support it. And finally we know the big numbers in slowing of cooling and slowing of warming is due to clouds and they are not a gas. There are a few obvious things going on: Obviously at a minimum due to insulation value so restriction of cooling is going on. That seems intuitive and I suppose you can call that a greenhouse effect. The big question is how much cooling goes on. You have a nice numeric equation to estimate it but its just numerology if one cannot demonstrate that the equation is being correctly applied. 1. Engineering net heat loss curves are a snap shot. True at that moment of time. >>If you think it operates differently then give me a calculation of that shows where the heat comes from as the exterior walls cool to the outdoors. The results calculated at the beginning of the process are true before the walls cool. When the walls are a different temperature then a new heat loss value must be selected which is true for that new moment of time. The final results are estimated by producing a large number of calculations for a large number of different moments of time. Something a school boy can understand. ------------- You failed the audit of your text. You need to make corrections before I begin finding the other errors.
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Post by icefisher on Feb 7, 2016 8:05:45 GMT
OK Andrew I will break it down in little chunks for you so you can understand what I am thinking. 1. Engineering Toolbox Net Radiation Loss Tool. the first thing you need to understand about this is its a tool, not a radiation law. Being a tool it does not need to give you more information. You can use the tool to calculate heat losses into an environment, it does not give you the heat losses by the environment. To figure those you may use additional calculations to restrict heat loss via insulation. The tool assumes no additional heat losses, which of course you can calculate them using conduction, convection, and reradiation from the backside of the receiving surface. Again Andrew its a tool, not a law. If you think it operates differently then give me a calculation of that shows where the heat comes from as the exterior walls cool to the outdoors. Demonstrate the use of the tool the way you would use it. 2. Photon Backradiation Cartoon. The key to your Planck mathematics to support this ridiculous model is that the IR blocking layer will only heat to 50% of the blackbody equilibrium value, yet we know that is not true. Observation says it will heat to the same value as the surface of the steel ball. Intuitively, and I believe its fair to claim that the only ways to slow the cooling of a surface is with insulation or reflection. I can demonstrate those. If you want to add a method I would like to see it demonstrated. A perfectly reasonable request. Now in saying that I think it would be possible to have enough CO2 to create some insulation. Perhaps Venus has enough. 3. So I am completely unconvinced that the amount of cooling from CO2 insulation value amounts to even a 1/4 degree. I would like to see calculations, calculations based upon observed effects, that would show differently. 4. Now water vapor being 25 times more prevalent in the atmosphere than CO2 might add up to something. But its really hard to say if it would result in any warming since water vapor starts by extracting a lot of heat from the surface and has a limited lifetime in the atmosphere. At any rate what ever insulating effect it had it would need reduction for that. 5. We know the surface is slowed from cooling. So does that not prove the greenhouse effect? Well no! In Spencer's backyard experiment he noted great reduction in cooling when clouds flew overhead at night. When there were no clouds temperatures plummeted at a steep slope. Now clouds are highly reflected and we already know that this can slow cooling. Of course clouds also slow warming. 6. The surface is warmer than it would be if there were no atmosphere. Well that does not prove anything. It could be for a different reason. 7. We know that warm objects radiate in all directions. Well thats definitely plausible. But it falls a bit short proof wise. But it also may not matter if what is radiated is not little finite packets of fixed energy. The wave mechanics view seems to be supported by the heat in the thermosphere. If its a vibration that heats everything lower frequency vibrations tend to dampen higher frequency vibrations rather than enhance them. The slowing of cooling calculated from a finite packet of energy transfer model needs validation by observation. Conclusion: Due to a lack of demonstration of the principles of backradiation I remain skeptical of a radiation based greenhouse effect not verifiable via heat exchange calculations based on observation. I am unconvinced that the slowing of cooling noted in the atmosphere is a result of backradiation or a slowing of cooling by gases. Though I will note that water could be slowing cooling and I may not perceive the cooling of the oceans and lakes occurring to support it. And finally we know the big numbers in slowing of cooling and slowing of warming is due to clouds and they are not a gas. There are a few obvious things going on: Obviously at a minimum due to insulation value so restriction of cooling is going on. That seems intuitive and I suppose you can call that a greenhouse effect. The big question is how much cooling goes on. You have a nice numeric equation to estimate it but its just numerology if one cannot demonstrate that the equation is being correctly applied. 1. Engineering net heat loss curves are a snap shot. True at that moment of time. >>If you think it operates differently then give me a calculation of that shows where the heat comes from as the exterior walls cool to the outdoors. The results calculated at the beginning of the process are true before the walls cool. When the walls are a different temperature then a new heat loss value must be selected which is true for that new moment of time. The final results are estimated by producing a large number of calculations for a large number of different moments of time. Something a school boy can understand. ------------- Hmmmm, so you have to do a calculation for each photon absorbed? Then another for each photon emitted? If not what do you base the number of calculations upon?
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Post by Andrew on Feb 7, 2016 8:10:24 GMT
1. Engineering net heat loss curves are a snap shot. True at that moment of time. >>If you think it operates differently then give me a calculation of that shows where the heat comes from as the exterior walls cool to the outdoors. The results calculated at the beginning of the process are true before the walls cool. When the walls are a different temperature then a new heat loss value must be selected which is true for that new moment of time. The final results are estimated by producing a large number of calculations for a large number of different moments of time. Something a school boy can understand. ------------- Hmmmm, so you have to do a calculation for each photon absorbed? Then another for each photon emitted? If not what do you base the number of calculations upon? We are talking about childishly simple estimation methods for our purposes. >>>>>If you think it operates differently then give me a calculation of that shows where the heat comes from as the exterior walls cool to the outdoors. If you are talking about a wall that is colder then it should be obvious you now need to have another heat loss calculation.
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Post by icefisher on Feb 7, 2016 8:42:37 GMT
Hmmmm, so you have to do a calculation for each photon absorbed? Then another for each photon emitted? If not what do you base the number of calculations upon? We are talking about childishly simple estimation methods for our purposes. If you are talking about a wall that is colder then it should be obvious you now need to have another heat loss calculation. actually you don't need another calculation. Your thought is right, but watts/m2 heat loss gives you the flow of energy in time 100watts per square meter equals 100 joules per second per square meter. sure you can do a calculation every microsecond using a millionth of a joule per mirosecond in place of a microwatt but you come up with the same answer.
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Post by Andrew on Feb 7, 2016 9:10:24 GMT
We are talking about childishly simple estimation methods for our purposes. If you are talking about a wall that is colder then it should be obvious you now need to have another heat loss calculation. actually you don't need another calculation. Your thought is right, but watts/m2 heat loss gives you the flow of energy in time 100watts per square meter equals 100 joules per second per square meter. sure you can do a calculation every microsecond using a millionth of a joule per mirosecond in place of a microwatt but you come up with the same answer. Unless there are steady state conditions where the temperatures are constant you must repeat the calculations for temperature changes
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Post by icefisher on Feb 7, 2016 9:53:20 GMT
actually you don't need another calculation. Your thought is right, but watts/m2 heat loss gives you the flow of energy in time 100watts per square meter equals 100 joules per second per square meter. sure you can do a calculation every microsecond using a millionth of a joule per mirosecond in place of a microwatt but you come up with the same answer. Unless there are steady state conditions where the temperatures are constant you must repeat the calculations for temperature changes You got the thought right now all you have to realize is thats what I am doing. The Stefan Boltzmann constant is a constant of proportionality between w/m2 or j/s/m2 and temperature. You have to fill the heat loss or temperature will drop. You can calculate the temperature of the surface by that flow knowing nothing else than the formula. It does not significantly matter if you do it one calculation or a trillion calculations. The answer will be the same. There is no compounding interest. It doesn't matter what you use for a temperature step down increment. And of course if you do put some insulation behind the irradiated surface, the flow of energy out the back of the surface will slow. I am not sure what rate you think it would slow by but presumably it would have to become zero for all 100 watts from a heated surface emitting 100 watts to be radiated back at the heated surface, and since there would be no emission out the back surface it would have to be absolute zero the way you are splitting the incoming radiation. That would be perfect insulation
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Post by Andrew on Feb 7, 2016 12:34:57 GMT
Unless there are steady state conditions where the temperatures are constant you must repeat the calculations for temperature changes You have to fill the heat loss or temperature will drop. What on Earth are you talking about? >>give me a calculation of that shows where the heat comes from as the exterior walls cool to the outdoors. Demonstrate the use of the tool the way you would use it. What the are you describing in that sentence? ??
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Post by icefisher on Feb 7, 2016 14:30:52 GMT
Simple Andrew. You gave us the one molecule, virtually non-insulated model with the radiance emittance of the two cool surfaces a long time ago.
If you don't like my conclusions and do not think they are representative of your view of the world, you describe the calculations with insulation. . . .say a moderate amount and a lot. Show us how the radiance emittance of the various surfaces will change as you add insulation.
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Post by Andrew on Feb 7, 2016 20:51:13 GMT
Simple Andrew. You gave us the one molecule, virtually non-insulated model with the radiance emittance of the two cool surfaces a long time ago. If you don't like my conclusions and do not think they are representative of your view of the world, you describe the calculations with insulation. . . .say a moderate amount and a lot. Show us how the radiance emittance of the various surfaces will change as you add insulation. You avoided the questions. You fail the audit
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Post by icefisher on Feb 7, 2016 21:51:41 GMT
Simple Andrew. You gave us the one molecule, virtually non-insulated model with the radiance emittance of the two cool surfaces a long time ago. If you don't like my conclusions and do not think they are representative of your view of the world, you describe the calculations with insulation. . . .say a moderate amount and a lot. Show us how the radiance emittance of the various surfaces will change as you add insulation. You avoided the questions. You fail the audit I suppose you knowing even less about auditing that you do about science can be forgiven. But the briefest definition of an audit is "obtaining evidence that when all weighed provides substantial proof the veracity of an assertion." Your assertion is "there is a purely radiation based greenhouse effect", my assertion is "I am highly skeptical that there is any substantial evidence of a purely radiation based greenhouse effect". Thus far zero evidence has been presented that supports the notion that a radiation greenhouse effect exists. I laid out 7 separate issues in support of skepticism. Your response is I am ignoring laws of physics but you have as yet to name one. I don't need any evidence of my skepticism, its self evident. It does not claim for example that you are skeptical. Further you have not even attempted to refute even one of the seven points. You demur on even showing your calculations regarding the application of the Engineering Toolbox Net Radiation Loss chart.
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Post by Andrew on Feb 7, 2016 23:19:16 GMT
You avoided the questions. You fail the audit I suppose you knowing even less about auditing that you do about science can be forgiven. But the briefest definition of an audit is "obtaining evidence that when all weighed provides substantial proof the veracity of an assertion." Your assertion is "there is a purely radiation based greenhouse effect", my assertion is "I am highly skeptical that there is any substantial evidence of a purely radiation based greenhouse effect". Thus far zero evidence has been presented that supports the notion that a radiation greenhouse effect exists. I laid out 7 separate issues in support of skepticism. Your response is I am ignoring laws of physics but you have as yet to name one. I don't need any evidence of my skepticism, its self evident. It does not claim for example that you are skeptical. Further you have not even attempted to refute even one of the seven points. You demur on even showing your calculations regarding the application of the Engineering Toolbox Net Radiation Loss chart. I began auditing issue number 1, where you totally rejecting Stefans law while saying you are not. At this point in time I am waiting for you to answer my questions.>>a calculation of that shows where the heat comes from as the exterior walls cool to the outdoors. 1. What do you mean by 'where the heat comes from'? 2. What do you mean by 'as the exterior walls cool to the outdoors'. Are the walls getting colder as I had first believed or is something different happening? >>You have to fill the heat loss What do you mean by 'you have to fill the heat loss'?
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Post by icefisher on Feb 7, 2016 23:43:53 GMT
I began auditing issue number 1, where you totally rejecting Stefans law while saying you are not. 1. Engineering Toolbox Net Radiation Loss Tool. the first thing you need to understand about this is its a tool, not a radiation law. Being a tool it does not need to give you more information. You can use the tool to calculate heat losses into an environment, it does not give you the heat losses by the environment. To figure those you may use additional calculations to restrict heat loss via insulation. The tool assumes no additional heat losses, which of course you can calculate them using conduction, convection, and reradiation from the backside of the receiving surface. Again Andrew its a tool, not a law. If you think it operates differently then give me a calculation of that shows where the heat comes from as the exterior walls cool to the outdoors. Demonstrate the use of the tool the way you would use it.In no example here or in the Woods thread where I detail the failure of the Woods project to produce even one degree out of 60 expected degrees do I calculate anything at cross purposes to Stefan's Law. Its not possible unless I made a mistake. If I made a mistake you should point at the precise calculation where the mistake is or you are. . . .uh. . . .lying. At this point in time I am waiting for you to answer my questions.>>a calculation of that shows where the heat comes from as the exterior walls cool to the outdoors. 1. What do you mean by 'where the heat comes from'? 2. What do you mean by 'as the exterior walls cool to the outdoors'. Are the walls getting colder as I had first believed or is something different happening? >>You have to fill the heat loss What do you mean by 'you have to fill the heat loss'? I don't need to answer those questions Andrew. The request if for you to provide an example of how you would use the calculation in say a real world environmenht besides one that you have already done so in the case of a single molecule thick surface. If you are confused by what I mean a real world environment. Say a radiant heat source inside of a room with insulated walls and an exterior temperature. If you don't understand that request I can't help you.
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Post by Andrew on Feb 7, 2016 23:57:25 GMT
I began auditing issue number 1, where you totally rejecting Stefans law while saying you are not. 1. Engineering Toolbox Net Radiation Loss Tool. the first thing you need to understand about this is its a tool, not a radiation law. Being a tool it does not need to give you more information. You can use the tool to calculate heat losses into an environment, it does not give you the heat losses by the environment. To figure those you may use additional calculations to restrict heat loss via insulation. The tool assumes no additional heat losses, which of course you can calculate them using conduction, convection, and reradiation from the backside of the receiving surface. Again Andrew its a tool, not a law. If you think it operates differently then give me a calculation of that shows where the heat comes from as the exterior walls cool to the outdoors. Demonstrate the use of the tool the way you would use it.In no example here or in the Woods thread where I detail the failure of the Woods project to produce even one degree out of 60 expected degrees do I calculate anything at cross purposes to Stefan's Law. Its not possible unless I made a mistake. If I made a mistake you should point at the precise calculation where the mistake is or you are. . . .uh. . . .lying. At this point in time I am waiting for you to answer my questions.>>a calculation of that shows where the heat comes from as the exterior walls cool to the outdoors. 1. What do you mean by 'where the heat comes from'? 2. What do you mean by 'as the exterior walls cool to the outdoors'. Are the walls getting colder as I had first believed or is something different happening? >>You have to fill the heat loss What do you mean by 'you have to fill the heat loss'? I don't need to answer those questions Andrew. The request if for you to provide an example of how you would use the calculation in say a real world environmenht besides one that you have already done so in the case of a single molecule thick surface. If you are confused by what I mean a real world environment. Say a radiant heat source inside of a room with insulated walls and an exterior temperature. If you don't understand that request I can't help you. I asked you to explain what you are talking about. You responded. I then began auditing your response, issue 1. I then asked you to explain what you are talking about. At this point in time I am waiting for you to answer my questions.>>a calculation of that shows where the heat comes from as the exterior walls cool to the outdoors. 1. What do you mean by 'where the heat comes from'? 2. What do you mean by 'as the exterior walls cool to the outdoors'. Are the walls getting colder as I had first believed or is something different happening? >>You have to fill the heat loss What do you mean by 'you have to fill the heat loss'?
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Post by icefisher on Feb 8, 2016 2:28:37 GMT
1. Engineering Toolbox Net Radiation Loss Tool. the first thing you need to understand about this is its a tool, not a radiation law. Being a tool it does not need to give you more information. You can use the tool to calculate heat losses into an environment, it does not give you the heat losses by the environment. To figure those you may use additional calculations to restrict heat loss via insulation. The tool assumes no additional heat losses, which of course you can calculate them using conduction, convection, and reradiation from the backside of the receiving surface. Again Andrew its a tool, not a law. If you think it operates differently then give me a calculation of that shows where the heat comes from as the exterior walls cool to the outdoors. Demonstrate the use of the tool the way you would use it.In no example here or in the Woods thread where I detail the failure of the Woods project to produce even one degree out of 60 expected degrees do I calculate anything at cross purposes to Stefan's Law. Its not possible unless I made a mistake. If I made a mistake you should point at the precise calculation where the mistake is or you are. . . .uh. . . .lying. I don't need to answer those questions Andrew. The request if for you to provide an example of how you would use the calculation in say a real world environmenht besides one that you have already done so in the case of a single molecule thick surface. If you are confused by what I mean a real world environment. Say a radiant heat source inside of a room with insulated walls and an exterior temperature. If you don't understand that request I can't help you. I asked you to explain what you are talking about. You responded. I then began auditing your response, issue 1. I then asked you to explain what you are talking about. At this point in time I am waiting for you to answer my questions.>>a calculation of that shows where the heat comes from as the exterior walls cool to the outdoors. 1. What do you mean by 'where the heat comes from'? 2. What do you mean by 'as the exterior walls cool to the outdoors'. Are the walls getting colder as I had first believed or is something different happening? >>You have to fill the heat loss What do you mean by 'you have to fill the heat loss'? Well if you don't know what you are talking about when in support of a greenhouse effect, can't explain it, can't demonstrate, nor even calculate in more ways that one. . . .uh its clear you have no clue. Of course its always up to the skeptics to prove that the pseudo-scientists are wrong. . . .can't have it any other way or heck you might be a real scientist. NOT!
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