Emissivity of the ocean

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Emissivity of the ocean08-03-2017 03:52
Leitwolf ★☆☆☆☆ (117)	Hallo Everyone Ever since I stumbled over this one it gave me a lot o headache. https://scienceofdoom.com/2010/12/27/emissivity-of-the-ocean/ I tried to figure out what the total hemispheric emissivity of water would theoretically be, based on the named measured data and graphs. Now without going into detail on how I did it (sophisticated minds should be able to guess it), the result is about 0.84. That figure of course would put emissivity definitely below absorptivity (~0.93 - 0.92), meaning that the ocean surface would tend to be warmer than a perfect black body, which again is quite typical for polished or smooth surfaces. Then of course, oceans would be the main driver of the "greenhouse" effect. So... I would love to have some educated opinions on this issue.
29-03-2017 21:32
Leitwolf ★☆☆☆☆ (117)	So .. here is the story. Absorptivity (A) of water looks like this (ok, this is actually emissivity, so think it inverted): Emissivity (E), this time with regard to infrared, looks extremely similar: Now comes geometry onto the scene. As we want to know total hemispheric A and E, we must consider that solar radiation is coming in from one direction only. Those low angles, where reflectivity runs high (and A runs low) make up for only a small fraction of total sunlight received, as these approach the horizon. With emissions into a hemisphere it is a different scenario, as these low angles have a much larger weight. The little drawing may illustrate the problem. Next, to determine both functions, I reproduced the functions on A and E synthetically. It may not be perfect, but it comes pretty close and I will consider the slight deviations seperately. First I want to isolate the geometric momentum. The geometry itself looks like this (the chart illustrates the relative weight of every single gradient) Now we combine geometric weighting with the named synthetic function. This eventually gives us 0.934 for hemispheric A, and 0.835 for hemispheric E. Which consequently leads us to a minor Heureka moment: (0.934 / 0.835)^0.25 * 280 = 288°K. In other words: the radial properties of water would make (the water-) planet earth just as warm as it actually is. Ok, in all fairness I should add the specific A function is slightly lower than the specific E function, which would reduce the outcome by roughly 1°K. If on the other side we took the cool skin effect into account (due to constant vaporization), that again increase temperature. Then there are many other factors to take into account, most of all the additional heating(!!!) by clouds. Yet, this explains the temperature of earth, without any greenhouse gases, and of course proves there is no such thing a GHE. Edited on 29-03-2017 21:36
29-03-2017 22:20
Surface Detail ★★★★☆ (1673)	Where did the 280 figure come from?
29-03-2017 23:43
Leitwolf ★☆☆☆☆ (117)	Great when pictures simply disappear! Ok .. Emissivity again Correction: "Absorptivity (A) of water looks like this (ok, this is actually emissivity, so think it inverted):" It must be reflectivity instead of emissivity. "Where did the 280 figure come from?" Obviously it is the temperature of a perfect black body on earths position. You should know that. Edited on 29-03-2017 23:49
30-03-2017 00:10
Surface Detail ★★★★☆ (1673)	Leitwolf wrote: Obviously it is the temperature of a perfect black body on earths position. You should know that. Well, yes, I did suspect that (though 279K is the figure I've usually seen quoted), but it's as well to be sure. I don't have time to check though your work and assumptions at the moment, but I'll be back soon!

30-03-2017 00:42
Leitwolf ★☆☆☆☆ (117)	Ok .. 279 maybe the commonly stated figure, but 280°K is more accurate. Reason: Earths distance to the sun is usually put to 150 mio km for calculating this figure.. But rather the semi-major axis is 149.6 mio km. And, as sun does not shine from its center, the surface shining onto us is even closer, like 149 mio km on average. It is just a small difference, but it is worth almost exactly 1°K.
30-03-2017 01:58
Surface Detail ★★★★☆ (1673)	Hmm, I think I can see what you're getting at (but if would help a lot if you labelled you diagrams and graphs!). If I understand correctly, you're considering points on the surface of a water-covered Earth and establishing the difference between the effective emissivities and absorptivities, taking into account the direction of incidence of radiation. However, I suspect you may have oversimplified the geometric aspects. You derive a figure for the hemispherical emissivity of water of 0.835. The direction of radiation emitted from a point is, of course, independent of incoming radiation and so will be same for all points of the Earth's surface. Why, then, does you figure for emissivity differ from the book figure for the hemispherical emissivity of water of 0.96, listed under Emissivities of common surfaces?
30-03-2017 03:27
Leitwolf ★☆☆☆☆ (117)	Well .. we would need to read the book. But there is good reason why do not rely on such data. There is plenty of non sense data out there. Like putting emissivity to 1, when calculating the GHE, or claiming the cloud albedo effect was 44 - 50W/m2 (while it is about 2/3 of a 106W/m2 for the total albedo), or elevating clear sky albedo for earth 0.15 and beyong. All that is non sense, but apparently claimed to support the theory of the so called GHE. However, if it done correctly, the result is as above.
30-03-2017 10:17
Surface Detail ★★★★☆ (1673)	Leitwolf wrote: Ok .. 279 maybe the commonly stated figure, but 280°K is more accurate. Reason: Earths distance to the sun is usually put to 150 mio km for calculating this figure.. But rather the semi-major axis is 149.6 mio km. And, as sun does not shine from its center, the surface shining onto us is even closer, like 149 mio km on average. It is just a small difference, but it is worth almost exactly 1°K. Actually, no, you're wrong about this. It is indeed the semi-major axis that is important, not the distance from the sun's surface. Why? Because the Earth receives a fraction of the sun's total output equal to the ratios of the areas of a disk of the Earth's radius and a sphere with a radius equal to the semi-major axis of the Earth's orbit (rather than the Earth-sun distance). So 280K is more accurate. Edit: See Effective Temperature of the Earth for the calculation. Edited on 30-03-2017 10:19
30-03-2017 10:37
Surface Detail ★★★★☆ (1673)	Leitwolf wrote: Well .. we would need to read the book. But there is good reason why do not rely on such data. There is plenty of non sense data out there. Like putting emissivity to 1, when calculating the GHE, or claiming the cloud albedo effect was 44 - 50W/m2 (while it is about 2/3 of a 106W/m2 for the total albedo), or elevating clear sky albedo for earth 0.15 and beyong. All that is non sense, but apparently claimed to support the theory of the so called GHE. However, if it done correctly, the result is as above. You can read some of the book here: Thermal Radiative Transfer and Properties My guess (you've not shown it, so it remains a guess) is that your calculation of hemispherical emissivity from the emissivity as a function of polar angle is wrong. How did you work it out? Did you integrate the function over the surface of surface of a hemisphere to find the total emissivity as shown in the book? Did you use an analytical or a numerical approach? (Note to ITN: this is an example of a problem that does actually require calculus).
30-03-2017 19:34
Into the Night★★★★★ (21588)	Surface Detail wrote: Leitwolf wrote: Well .. we would need to read the book. But there is good reason why do not rely on such data. There is plenty of non sense data out there. Like putting emissivity to 1, when calculating the GHE, or claiming the cloud albedo effect was 44 - 50W/m2 (while it is about 2/3 of a 106W/m2 for the total albedo), or elevating clear sky albedo for earth 0.15 and beyong. All that is non sense, but apparently claimed to support the theory of the so called GHE. However, if it done correctly, the result is as above. You can read some of the book here: Thermal Radiative Transfer and Properties My guess (you've not shown it, so it remains a guess) is that your calculation of hemispherical emissivity from the emissivity as a function of polar angle is wrong. How did you work it out? Did you integrate the function over the surface of surface of a hemisphere to find the total emissivity as shown in the book? Did you use an analytical or a numerical approach? (Note to ITN: this is an example of a problem that does actually require calculus). No, this is a problem that is pointless to try to solve. The Parrot Killer Debunked in my sig. - tmiddles Google keeps track of paranoid talk and i'm not on their list. I've been evaluated and certified. - keepit nuclear powered ships do not require nuclear fuel. - Swan While it is true that fossils do not burn it is also true that fossil fuels burn very well - Swan
30-03-2017 19:40
Surface Detail ★★★★☆ (1673)	Into the Night wrote: Surface Detail wrote: Leitwolf wrote: Well .. we would need to read the book. But there is good reason why do not rely on such data. There is plenty of non sense data out there. Like putting emissivity to 1, when calculating the GHE, or claiming the cloud albedo effect was 44 - 50W/m2 (while it is about 2/3 of a 106W/m2 for the total albedo), or elevating clear sky albedo for earth 0.15 and beyong. All that is non sense, but apparently claimed to support the theory of the so called GHE. However, if it done correctly, the result is as above. You can read some of the book here: Thermal Radiative Transfer and Properties My guess (you've not shown it, so it remains a guess) is that your calculation of hemispherical emissivity from the emissivity as a function of polar angle is wrong. How did you work it out? Did you integrate the function over the surface of surface of a hemisphere to find the total emissivity as shown in the book? Did you use an analytical or a numerical approach? (Note to ITN: this is an example of a problem that does actually require calculus). No, this is a problem that is pointless to try to solve. For you, certainly
30-03-2017 19:53
Into the Night★★★★★ (21588)	Surface Detail wrote: Into the Night wrote: Surface Detail wrote: Leitwolf wrote: Well .. we would need to read the book. But there is good reason why do not rely on such data. There is plenty of non sense data out there. Like putting emissivity to 1, when calculating the GHE, or claiming the cloud albedo effect was 44 - 50W/m2 (while it is about 2/3 of a 106W/m2 for the total albedo), or elevating clear sky albedo for earth 0.15 and beyong. All that is non sense, but apparently claimed to support the theory of the so called GHE. However, if it done correctly, the result is as above. You can read some of the book here: Thermal Radiative Transfer and Properties My guess (you've not shown it, so it remains a guess) is that your calculation of hemispherical emissivity from the emissivity as a function of polar angle is wrong. How did you work it out? Did you integrate the function over the surface of surface of a hemisphere to find the total emissivity as shown in the book? Did you use an analytical or a numerical approach? (Note to ITN: this is an example of a problem that does actually require calculus). No, this is a problem that is pointless to try to solve. For you, certainly So what is the point of trying to solve such a problem? You LIKE ignoring cloud cover, wave action, etc. You LIKE attempting to use the Stefan-Boltzmann law improperly. You LIKE feeling that your contribution is somehow relevant to the actual Earth. The Parrot Killer Debunked in my sig. - tmiddles Google keeps track of paranoid talk and i'm not on their list. I've been evaluated and certified. - keepit nuclear powered ships do not require nuclear fuel. - Swan While it is true that fossils do not burn it is also true that fossil fuels burn very well - Swan
30-03-2017 19:57
Surface Detail ★★★★☆ (1673)	Into the Night wrote: Surface Detail wrote: Into the Night wrote: Surface Detail wrote: Leitwolf wrote: Well .. we would need to read the book. But there is good reason why do not rely on such data. There is plenty of non sense data out there. Like putting emissivity to 1, when calculating the GHE, or claiming the cloud albedo effect was 44 - 50W/m2 (while it is about 2/3 of a 106W/m2 for the total albedo), or elevating clear sky albedo for earth 0.15 and beyong. All that is non sense, but apparently claimed to support the theory of the so called GHE. However, if it done correctly, the result is as above. You can read some of the book here: Thermal Radiative Transfer and Properties My guess (you've not shown it, so it remains a guess) is that your calculation of hemispherical emissivity from the emissivity as a function of polar angle is wrong. How did you work it out? Did you integrate the function over the surface of surface of a hemisphere to find the total emissivity as shown in the book? Did you use an analytical or a numerical approach? (Note to ITN: this is an example of a problem that does actually require calculus). No, this is a problem that is pointless to try to solve. For you, certainly So what is the point of trying to solve such a problem? You LIKE ignoring cloud cover, wave action, etc. You LIKE attempting to use the Stefan-Boltzmann law improperly. You LIKE feeling that your contribution is somehow relevant to the actual Earth. Please could you piss off and leave this discussion to the grown-ups, ITN? You clearly have nothing to contribute to it.
30-03-2017 21:24
Leitwolf ★☆☆☆☆ (117)	Ok .. I need to take back the 280K figure. I addepted a text bood formula with the initial parameter 5770 for T, 700 for R and 150 for a. I will hold on to my correction with regard to a, but I should also have fine tuned T and R. If I do so, I get to 344.6W/m2 and 279.2K respectively. I should also note, that 341W/m2, a figure oftenly quotet, would only result in 278.48K, being obviously closer to 278 than 279. Anyhow.. Calculating hemispheric A and E is not really a big deal. The trouble is only in the details of the synthetic functions with regard to inclination. I'll attach the excel file (sheet9). Attached file: klimax.xls Edited on 30-03-2017 21:25

30-03-2017 21:48
Surface Detail ★★★★☆ (1673)	Leitwolf wrote: Ok .. I need to take back the 280K figure. I addepted a text bood formula with the initial parameter 5770 for T, 700 for R and 150 for a. I will hold on to my correction with regard to a, but I should also have fine tuned T and R. If I do so, I get to 344.6W/m2 and 279.2K respectively. I should also note, that 341W/m2, a figure oftenly quotet, would only result in 278.48K, being obviously closer to 278 than 279. Anyhow.. Calculating hemispheric A and E is not really a big deal. The trouble is only in the details of the synthetic functions with regard to inclination. I'll attach the excel file (sheet9). Thanks. I'm genuinely curious - it's not obvious to me whether a theoretical water-covered world (with no atmosphere) would be warmer or cooler than a black body Earth. My hunch is cooler, but I'm not at all sure about that. I'll get back to you, but not until tomorrow. Guests coming round in a bit...
31-03-2017 00:56
Into the Night★★★★★ (21588)	Surface Detail wrote: Into the Night wrote: Surface Detail wrote: Into the Night wrote: Surface Detail wrote: Leitwolf wrote: Well .. we would need to read the book. But there is good reason why do not rely on such data. There is plenty of non sense data out there. Like putting emissivity to 1, when calculating the GHE, or claiming the cloud albedo effect was 44 - 50W/m2 (while it is about 2/3 of a 106W/m2 for the total albedo), or elevating clear sky albedo for earth 0.15 and beyong. All that is non sense, but apparently claimed to support the theory of the so called GHE. However, if it done correctly, the result is as above. You can read some of the book here: Thermal Radiative Transfer and Properties My guess (you've not shown it, so it remains a guess) is that your calculation of hemispherical emissivity from the emissivity as a function of polar angle is wrong. How did you work it out? Did you integrate the function over the surface of surface of a hemisphere to find the total emissivity as shown in the book? Did you use an analytical or a numerical approach? (Note to ITN: this is an example of a problem that does actually require calculus). No, this is a problem that is pointless to try to solve. For you, certainly So what is the point of trying to solve such a problem? You LIKE ignoring cloud cover, wave action, etc. You LIKE attempting to use the Stefan-Boltzmann law improperly. You LIKE feeling that your contribution is somehow relevant to the actual Earth. Please could you piss off and leave this discussion to the grown-ups, ITN? You clearly have nothing to contribute to it. Fine. Then clearly you have nothing to contribute by specifically naming me in your hate posts. The Parrot Killer Debunked in my sig. - tmiddles Google keeps track of paranoid talk and i'm not on their list. I've been evaluated and certified. - keepit nuclear powered ships do not require nuclear fuel. - Swan While it is true that fossils do not burn it is also true that fossil fuels burn very well - Swan
31-03-2017 17:04
Surface Detail ★★★★☆ (1673)	Leitwolf wrote: Ok .. I need to take back the 280K figure. I addepted a text bood formula with the initial parameter 5770 for T, 700 for R and 150 for a. I will hold on to my correction with regard to a, but I should also have fine tuned T and R. If I do so, I get to 344.6W/m2 and 279.2K respectively. I should also note, that 341W/m2, a figure oftenly quotet, would only result in 278.48K, being obviously closer to 278 than 279. Anyhow.. Calculating hemispheric A and E is not really a big deal. The trouble is only in the details of the synthetic functions with regard to inclination. I'll attach the excel file (sheet9). Just had time for a quick look at this. Regarding your synthetic function for emissivity as a function of polar angle (angle from vertical): Am I correct in thinking that you have attempted to numerically produce an approximation of the theoretical variation of emissivity with polar angle given by the Fresnel equations? If so, my first question would be: why not just use the Fresnel equations themselves to compute your values?
02-04-2017 18:06
IBdaMann★★★★★ (14389)	Leitwolf wrote: I tried to figure out what the total hemispheric emissivity of water would theoretically be, based on the named measured data and graphs. ... I would love to have some educated opinions on this issue. You can't do that. Black body radiation pertains to bodies. Bodies are the atomic unit of black body radiation. You don't get to break it down beyond that. Emissivity applies to a body, not to sub-parts of a body. The earth is the body in question. The hydrosphere is a subpart of the body. It does not have it's own emissivity. . I don't think i can [define it]. I just kind of get a feel for the phrase. - keepit A Spaghetti strainer with the faucet running, retains water- tmiddles Clouds don't trap heat. Clouds block cold. - Spongy Iris Printing dollars to pay debt doesn't increase the number of dollars. - keepit If Venus were a black body it would have a much much lower temperature than what we found there.- tmiddles Ah the "Valid Data" myth of ITN/IBD. - tmiddles Ceist - I couldn't agree with you more. But when money and religion are involved, and there are people who value them above all else, then the lies begin. - trafn You are completely misunderstanding their use of the word "accumulation"! - Climate Scientist. The Stefan-Boltzman equation doesn't come up with the correct temperature if greenhouse gases are not considered - Hank :sigh* Not the "raw data" crap.* - Leafsdude IB STILL hasn't explained what Planck's Law means. Just more hand waving that it applies to everything and more asserting that the greenhouse effect 'violates' it.- Ceist
03-04-2017 22:20
Leitwolf ★☆☆☆☆ (117)	why not just use the Fresnel equations themselves to compute your values? Well quite simply because I have no clue what a Fresnel equation is (I am not physicist), or rather did not have before dealing with the subject. As I wanted to have results, I took the quick and dirty approach. I will deal with it later on.. Right now I was writing some programs to automate the retrieval of temperature data (in larger quantities) as I need them. They are up and running now and will allow me to aquire deeper insights into the role that clouds play.
04-04-2017 04:58
Surface Detail ★★★★☆ (1673)	Leitwolf wrote: why not just use the Fresnel equations themselves to compute your values? Well quite simply because I have no clue what a Fresnel equation is (I am not physicist), or rather did not have before dealing with the subject. As I wanted to have results, I took the quick and dirty approach. I will deal with it later on.. Right now I was writing some programs to automate the retrieval of temperature data (in larger quantities) as I need them. They are up and running now and will allow me to aquire deeper insights into the role that clouds play. I didn't know what the Fresnel equations were either, until I investigated your work! It looks like the graph of the emissivity of water vs angle that you linked to comes from here: https://scienceofdoom.com/2010/12/27/emissivity-of-the-ocean/ According to that site, the different graph colours represent theoretical variation in emissivity with measurement angle, versus "refractive index" as computed by the Fresnel equations. So the graph that you have attempted to recreate using an ad-hoc function is, in fact, just a theoretical graph computed from the Fresnel equations. Didn't you think that it might be a good idea to find out where the graph came from before attempting to replicate it? You could have replicated the graph exactly by plotting the Fresnel equations, rather than trying to replicate it with your own "synthetic function"!
04-04-2017 06:53
Leitwolf ★☆☆☆☆ (117)	Didn't you think that it might be a good idea to find out where the graph came from before attempting to replicate it? You could have replicated the graph exactly by plotting the Fresnel equations, rather than trying to replicate it with your own "synthetic function"! For the named reasons.
06-05-2017 17:31
Leitwolf ★☆☆☆☆ (117)	Ok, I redid the whole calculation based on fresnel equations. As to be expected, there is no fundamental change in the outcome. Also I corrected the previously assumed 280K for a PBB to 279.2. Yet off course I do not exactly know the specific parameters for n2 with regard to E and A. Simulations gave me 1.27 for n2 for E, which seems the best fit in comparison to the measured data here: https://scienceofdoom.com/2010/12/27/emissivity-of-the-ocean/ For A things seem a way more uncertain. With various sources claiming total R (reflectivity = albedo) of (ocean) water to be 0.06, and A henceforth 0.94, this would be in line with n2 = 1.3. On the other side there are indications that vertical R could be as high as 0.04, which was true for n2 = 1.5, and would go in line with a total hemispheric R of 0.092. For this reason I think this would be a rather unlikely scenario, drawing the upper limit for the highest thinkable value for n2. Anyhow, we can draw different scenarios based of these assumptions. While holding n2 = 1.27 for E, we can put n2 for A to.. 1.3 resulting in 287.1K 1.4 resulting in 285.9K 1.5 resulting in 284.7K So, after all, the ocean will drive earths climate to about 286K PS. again calculations are to be found in sheet 9, Fresnel equations are on the two succeeding sheets. Attached file: klimaxx.xls
07-05-2017 18:14
Leitwolf ★☆☆☆☆ (117)	Ok, finally I could find out more precise information on the refractive index of water, which apparently is n = 1.33 for visible light. I found this little site highly usefull in that regard (however there are a couple of other sources naming the 1.33 as the refractive index for water).. https://refractiveindex.info/?shelf=main&book=H2O&page=Hale Now we can simply enter those 1.33 for n2 for absorptivity into the spread sheet provided and get to read 286.7K for the natural temperature of water. Which of course makes liquid(!) water by far the most significant "greenhouse gas" of all. Science settled, GHE nullified.
07-05-2017 22:20
Wake ★★★★★ (4034)	Leitwolf wrote: Ok, finally I could find out more precise information on the refractive index of water, which apparently is n = 1.33 for visible light. I found this little site highly usefull in that regard (however there are a couple of other sources naming the 1.33 as the refractive index for water).. https://refractiveindex.info/?shelf=main&book=H2O&page=Hale Now we can simply enter those 1.33 for n2 for absorptivity into the spread sheet provided and get to read 286.7K for the natural temperature of water. Which of course makes liquid(!) water by far the most significant "greenhouse gas" of all. Science settled, GHE nullified. I am quite impressed. As someone that knows spectometry I approached it from angle of an almost complete lack of energy in the spectral absorbance range of CO2. That the amount of energy in this band is so low that the entire absorbance band even at very low levels absorbs all there is. With this lack of energy additional CO2 simply has no effect in regards to radiation. So the effect of additional CO2 is no more than other molecules in that it is part of the conduction/convection cycle that moves the Earth's energy into the upper atmosphere where it is radiated away almost entirely by H2O.
12-05-2017 00:51
Wake ★★★★★ (4034)	Leitwolf wrote: Ok, finally I could find out more precise information on the refractive index of water, which apparently is n = 1.33 for visible light. I found this little site highly usefull in that regard (however there are a couple of other sources naming the 1.33 as the refractive index for water).. https://refractiveindex.info/?shelf=main&book=H2O&page=Hale Now we can simply enter those 1.33 for n2 for absorptivity into the spread sheet provided and get to read 286.7K for the natural temperature of water. Which of course makes liquid(!) water by far the most significant "greenhouse gas" of all. Science settled, GHE nullified. By the way - if it matters (I can't find the paper now) a study showed that that absorptivity of the ocean didn't seem to be effected by any wave actions other than extreme cases such as great storms.
12-05-2017 02:09
Leitwolf ★☆☆☆☆ (117)	Wake wrote: By the way - if it matters (I can't find the paper now) a study showed that that absorptivity of the ocean didn't seem to be effected by any wave actions other than extreme cases such as great storms. Right! Initially I assumed that waves might increase emissivity, as obviously that will interfere with the very low emission angles. But ultimately, on a microlevel, the watersurface remains flat, which will not change anything. On the other side, a small fraction of the emissions will hit water again, where it gets either reflected, or again absorbed. When I ran some simulations I soon saw there was no way to get to a higher emissivty. The only thing really making a difference is the cool skin effect, which is heavily depending on wind. Because of constant vaporization the "skin" of water is always somewhat colder than water underneath, but that delta is very hard to measure. It interferes however with radial emissions relative to normal temperatures. This factor also increases with the angle of emissions, which may be the reason why measurements show a slight deviation from an ideal Fresnel curve. But all that can only reduce emissivity.
26-05-2017 22:08
Leitwolf ★☆☆☆☆ (117)	Calculating the water sphere.. Attached file: wasser.xls
26-05-2017 22:55
Wake ★★★★★ (4034)	Leitwolf wrote: Calculating the water sphere.. Again what I most noticed was the rather substantial difference between absorption and emission. This is because most of this energy is extracted via conduction and convection. The majority of the emissions is also trapped rapidly in the humidity of the air directly above the ocean's surface and it too becomes part of the conduction and convection cycle. So virtually all but a tiny portion of the energy in the lower atmosphere is transported into the tropopause by conduction and convection. We know that there is a 40 uM hole in the atmosphere. This is, what, 25 khz or virtually no energy per cycle - very close to absolute zero - 1 millionth of a degree kelvin per cycle. So the thermal energy that goes into the conduction must be stupendous. There are something like four pretty sharp absorption lines in CO2 and H2O cover all of them but the (from memory) 6.8 uM line. This one isn't very wide and there simply isn't energy in that area either. Some 7 millionths of a degree kelvin per cycle. These extremely low IR frequencies do not carry energy via radiation and so there leaves the direct contact conduction and convection. The radiation that does occur is immediately absorbed by the gases and moved into the tropopause where trigonometric analysis shows that the majority of it is then radiated into space. But at damned low amounts. Looking through absorption and emission lines I find a great deal of confusion so my numbers might be marginally wrong. But the fact is that at low frequencies (very low temperature differentials) there is very low radiation and the bulk of the heat is move through conduction.
26-05-2017 23:33
Into the Night★★★★★ (21588)	Wake wrote: Leitwolf wrote: Calculating the water sphere.. Again what I most noticed was the rather substantial difference between absorption and emission. There is no difference. They are always the same value. Wake wrote: This is because most of this energy is extracted via conduction and convection. The majority of the emissions is also trapped rapidly in the humidity of the air directly above the ocean's surface and it too becomes part of the conduction and convection cycle. You cannot trap light. You also cannot trap heat. Wake wrote: So virtually all but a tiny portion of the energy in the lower atmosphere is transported into the tropopause by conduction and convection. This part is actually correct. Wake wrote: We know that there is a 40 uM hole in the atmosphere. Did you go up there with a micrometer and measure it? Wake wrote: This is, what, 25 khz or virtually no energy per cycle - very close to absolute zero - 1 millionth of a degree kelvin per cycle. Temperature is not measured in frequency. Neither is amplitude. Wake wrote: So the thermal energy that goes into the conduction must be stupendous. Stupendous! Wake wrote: There are something like four pretty sharp absorption lines in CO2 and H2O cover all of them but the (from memory) 6.8 uM line. There are hundreds of absorption lines in CO2 and again in H2O. Wake wrote: This one isn't very wide and there simply isn't energy in that area either. All absorption lines are at frequencies of light above zero Hz. That means they all have a level of energy associated with them. Wake wrote: Some 7 millionths of a degree kelvin per cycle. Temperature isn't measured in frequency. Wake wrote: These extremely low IR frequencies do not carry energy via radiation All light is energy, whether that light is 1Hz, 1Mhz, infrared light, visible light, or X-Rays. All light is electromagnetic radiation. Wake wrote: and so there leaves the direct contact conduction and convection. All three forms of heating are taking place all the time. Wake wrote: The radiation that does occur is immediately absorbed by the gases You are now violating the Stefan Boltzmann law. You are reducing radiance while keeping the same temperature. Wake wrote: and moved into the tropopause Light doesn't stop at the tropopause. Neither does heating by conduction or convection. Wake wrote: where trigonometric analysis Buzzword. Wake wrote: shows that the majority of it is then radiated into space. Heating from the surface throughout the entire atmosphere is by conduction, convection, and radiation. You seem to be confused by air density. Wake wrote: But at damned low amounts. ALL the energy absorbed from the Sun is radiated out again. They are equal. The mass of the Earth does not change significantly, and the Sun has not changed it's output much. Wake wrote: Looking through absorption and emission lines I find a great deal of confusion Not surprising. Wake wrote: so my numbers might be marginally wrong. They are completely wrong. Mostly due to incompatible units. Wake wrote: But the fact is that at low frequencies (very low temperature differentials) there is very low radiation and the bulk of the heat is move through conduction. Frequency is not amplitude. The Parrot Killer Debunked in my sig. - tmiddles Google keeps track of paranoid talk and i'm not on their list. I've been evaluated and certified. - keepit nuclear powered ships do not require nuclear fuel. - Swan While it is true that fossils do not burn it is also true that fossil fuels burn very well - Swan

26-05-2017 23:52
Wake ★★★★★ (4034)	Into the Night wrote: Wake wrote: Leitwolf wrote: Calculating the water sphere.. Again what I most noticed was the rather substantial difference between absorption and emission. There is no difference. They are always the same value. Wake wrote: This is because most of this energy is extracted via conduction and convection. The majority of the emissions is also trapped rapidly in the humidity of the air directly above the ocean's surface and it too becomes part of the conduction and convection cycle. You cannot trap light. You also cannot trap heat. Wake wrote: So virtually all but a tiny portion of the energy in the lower atmosphere is transported into the tropopause by conduction and convection. This part is actually correct. Wake wrote: We know that there is a 40 uM hole in the atmosphere. Did you go up there with a micrometer and measure it? Wake wrote: This is, what, 25 khz or virtually no energy per cycle - very close to absolute zero - 1 millionth of a degree kelvin per cycle. Temperature is not measured in frequency. Neither is amplitude. Wake wrote: So the thermal energy that goes into the conduction must be stupendous. Stupendous! Wake wrote: There are something like four pretty sharp absorption lines in CO2 and H2O cover all of them but the (from memory) 6.8 uM line. There are hundreds of absorption lines in CO2 and again in H2O. Wake wrote: This one isn't very wide and there simply isn't energy in that area either. All absorption lines are at frequencies of light above zero Hz. That means they all have a level of energy associated with them. Wake wrote: Some 7 millionths of a degree kelvin per cycle. Temperature isn't measured in frequency. Wake wrote: These extremely low IR frequencies do not carry energy via radiation All light is energy, whether that light is 1Hz, 1Mhz, infrared light, visible light, or X-Rays. All light is electromagnetic radiation. Wake wrote: and so there leaves the direct contact conduction and convection. All three forms of heating are taking place all the time. Wake wrote: The radiation that does occur is immediately absorbed by the gases You are now violating the Stefan Boltzmann law. You are reducing radiance while keeping the same temperature. Wake wrote: and moved into the tropopause Light doesn't stop at the tropopause. Neither does heating by conduction or convection. Wake wrote: where trigonometric analysis Buzzword. Wake wrote: shows that the majority of it is then radiated into space. Heating from the surface throughout the entire atmosphere is by conduction, convection, and radiation. You seem to be confused by air density. Wake wrote: But at damned low amounts. ALL the energy absorbed from the Sun is radiated out again. They are equal. The mass of the Earth does not change significantly, and the Sun has not changed it's output much. Wake wrote: Looking through absorption and emission lines I find a great deal of confusion Not surprising. Wake wrote: so my numbers might be marginally wrong. They are completely wrong. Mostly due to incompatible units. Wake wrote: But the fact is that at low frequencies (very low temperature differentials) there is very low radiation and the bulk of the heat is move through conduction. Frequency is not amplitude. Since you do not understand the difference between emission and absorption there's no use in discussing it. "You cannot trap light. You also cannot trap heat." Tell me - what do you think absorption lines are? Oh wait - you think that absorption and emission are the same thing. Moronic comments showing that you know nothing about transfer of energy are not funny - they just show you as stupid. 100's of absorption lines in CO2 huh. Why don't you give us a reference for that. "Temperature is not measured in frequency. Neither is amplitude." Now what was the Stefan-Boltzmann equation again? Hmm, any temperature has a specific flux or amplitude of energy and visa versa. Aren't you the one arguing that the entire universe is a blackbody obeying the Stefan-Boltzmann theory sans emissivity?
27-05-2017 00:11
Into the Night★★★★★ (21588)	Wake wrote: Since you do not understand the difference between emission and absorption there's no use in discussing it. There isn't any difference in their values. Emissivity and aborption ability are the same value. Wake wrote: "You cannot trap light. You also cannot trap heat." Tell me - what do you think absorption lines are? Oh wait - you think that absorption and emission are the same thing. This stems from your inability to distinguish between energy, heat, and light. Wake wrote: Moronic comments showing that you know nothing about transfer of energy are not funny - they just show you as stupid. Ah. Again with trying to say I'm stupid, because YOU don't understand the difference between heat, light, and energy. Wake wrote: 100's of absorption lines in CO2 huh. Why don't you give us a reference for that. Go look it up. The spectral absorption lines of CO2 are well known. So is H2O. Wake wrote: "Temperature is not measured in frequency. Neither is amplitude." Now what was the Stefan-Boltzmann equation again? radiance = SB constant * emissivity * temperature ^ 4 Do you see a frequency in there at all? Is it because you don't understand the units used in this equation? Wake wrote: Hmm, any temperature has a specific flux or amplitude of energy and visa versa. Do you see a frequency in there? Wake wrote: Aren't you the one arguing that the entire universe is a blackbody obeying the Stefan-Boltzmann theory sans emissivity? No. The Stefan-Boltzmann law incorporates the effect of emissivity into the equation. This is the same as absorption ability. The Parrot Killer Debunked in my sig. - tmiddles Google keeps track of paranoid talk and i'm not on their list. I've been evaluated and certified. - keepit nuclear powered ships do not require nuclear fuel. - Swan While it is true that fossils do not burn it is also true that fossil fuels burn very well - Swan
27-05-2017 00:21
Wake ★★★★★ (4034)	Into the Night wrote: Wake wrote: Since you do not understand the difference between emission and absorption there's no use in discussing it. There isn't any difference in their values. Emissivity and aborption ability are the same value. Wake wrote: "You cannot trap light. You also cannot trap heat." Tell me - what do you think absorption lines are? Oh wait - you think that absorption and emission are the same thing. This stems from your inability to distinguish between energy, heat, and light. Wake wrote: Moronic comments showing that you know nothing about transfer of energy are not funny - they just show you as stupid. Ah. Again with trying to say I'm stupid, because YOU don't understand the difference between heat, light, and energy. Wake wrote: 100's of absorption lines in CO2 huh. Why don't you give us a reference for that. Go look it up. The spectral absorption lines of CO2 are well known. So is H2O. Wake wrote: "Temperature is not measured in frequency. Neither is amplitude." Now what was the Stefan-Boltzmann equation again? radiance = SB constant * emissivity * temperature ^ 4 Do you see a frequency in there at all? Is it because you don't understand the units used in this equation? Wake wrote: Hmm, any temperature has a specific flux or amplitude of energy and visa versa. Do you see a frequency in there? Wake wrote: Aren't you the one arguing that the entire universe is a blackbody obeying the Stefan-Boltzmann theory sans emissivity? No. The Stefan-Boltzmann law incorporates the effect of emissivity into the equation. This is the same as absorption ability. Thanks for demonstrating that you are a complete idiot. Back to high school chimp. Try looking up Wien's displacement law. Edited on 27-05-2017 00:27
27-05-2017 01:01
Into the Night★★★★★ (21588)	Wake wrote: Into the Night wrote: Wake wrote: Since you do not understand the difference between emission and absorption there's no use in discussing it. There isn't any difference in their values. Emissivity and aborption ability are the same value. Wake wrote: "You cannot trap light. You also cannot trap heat." Tell me - what do you think absorption lines are? Oh wait - you think that absorption and emission are the same thing. This stems from your inability to distinguish between energy, heat, and light. Wake wrote: Moronic comments showing that you know nothing about transfer of energy are not funny - they just show you as stupid. Ah. Again with trying to say I'm stupid, because YOU don't understand the difference between heat, light, and energy. Wake wrote: 100's of absorption lines in CO2 huh. Why don't you give us a reference for that. Go look it up. The spectral absorption lines of CO2 are well known. So is H2O. Wake wrote: "Temperature is not measured in frequency. Neither is amplitude." Now what was the Stefan-Boltzmann equation again? radiance = SB constant * emissivity * temperature ^ 4 Do you see a frequency in there at all? Is it because you don't understand the units used in this equation? Wake wrote: Hmm, any temperature has a specific flux or amplitude of energy and visa versa. Do you see a frequency in there? Wake wrote: Aren't you the one arguing that the entire universe is a blackbody obeying the Stefan-Boltzmann theory sans emissivity? No. The Stefan-Boltzmann law incorporates the effect of emissivity into the equation. This is the same as absorption ability. Thanks for demonstrating that you are a complete idiot. Back to high school chimp. Try looking up Wien's displacement law. What about it? State your argument. So far you are using this as a buzzword. The Parrot Killer Debunked in my sig. - tmiddles Google keeps track of paranoid talk and i'm not on their list. I've been evaluated and certified. - keepit nuclear powered ships do not require nuclear fuel. - Swan While it is true that fossils do not burn it is also true that fossil fuels burn very well - Swan
27-05-2017 01:14
Wake ★★★★★ (4034)	Into the Night wrote: Wake wrote: Into the Night wrote: Wake wrote: Since you do not understand the difference between emission and absorption there's no use in discussing it. There isn't any difference in their values. Emissivity and aborption ability are the same value. Wake wrote: "You cannot trap light. You also cannot trap heat." Tell me - what do you think absorption lines are? Oh wait - you think that absorption and emission are the same thing. This stems from your inability to distinguish between energy, heat, and light. Wake wrote: Moronic comments showing that you know nothing about transfer of energy are not funny - they just show you as stupid. Ah. Again with trying to say I'm stupid, because YOU don't understand the difference between heat, light, and energy. Wake wrote: 100's of absorption lines in CO2 huh. Why don't you give us a reference for that. Go look it up. The spectral absorption lines of CO2 are well known. So is H2O. Wake wrote: "Temperature is not measured in frequency. Neither is amplitude." Now what was the Stefan-Boltzmann equation again? radiance = SB constant * emissivity * temperature ^ 4 Do you see a frequency in there at all? Is it because you don't understand the units used in this equation? Wake wrote: Hmm, any temperature has a specific flux or amplitude of energy and visa versa. Do you see a frequency in there? Wake wrote: Aren't you the one arguing that the entire universe is a blackbody obeying the Stefan-Boltzmann theory sans emissivity? No. The Stefan-Boltzmann law incorporates the effect of emissivity into the equation. This is the same as absorption ability. Thanks for demonstrating that you are a complete idiot. Back to high school chimp. Try looking up Wien's displacement law. What about it? State your argument. So far you are using this as a buzzword. So far you have shown that you know absolutely NOTHING. You've even gone so far as to tell us all Wikipedia is wrong because it contradicts your poor stupid ass. When asked for your absolute ignorant statement that CO2 has "hundreds" of absorption lines you don't actually know anything about it and tell me to look it up. I think that you need your diapers changed but you probably had your mother leave town a long time ago to get away from you. Edited on 27-05-2017 01:14
27-05-2017 01:29
Into the Night★★★★★ (21588)	Wake wrote: Into the Night wrote: Wake wrote: Into the Night wrote: Wake wrote: Since you do not understand the difference between emission and absorption there's no use in discussing it. There isn't any difference in their values. Emissivity and aborption ability are the same value. Wake wrote: "You cannot trap light. You also cannot trap heat." Tell me - what do you think absorption lines are? Oh wait - you think that absorption and emission are the same thing. This stems from your inability to distinguish between energy, heat, and light. Wake wrote: Moronic comments showing that you know nothing about transfer of energy are not funny - they just show you as stupid. Ah. Again with trying to say I'm stupid, because YOU don't understand the difference between heat, light, and energy. Wake wrote: 100's of absorption lines in CO2 huh. Why don't you give us a reference for that. Go look it up. The spectral absorption lines of CO2 are well known. So is H2O. Wake wrote: "Temperature is not measured in frequency. Neither is amplitude." Now what was the Stefan-Boltzmann equation again? radiance = SB constant * emissivity * temperature ^ 4 Do you see a frequency in there at all? Is it because you don't understand the units used in this equation? Wake wrote: Hmm, any temperature has a specific flux or amplitude of energy and visa versa. Do you see a frequency in there? Wake wrote: Aren't you the one arguing that the entire universe is a blackbody obeying the Stefan-Boltzmann theory sans emissivity? No. The Stefan-Boltzmann law incorporates the effect of emissivity into the equation. This is the same as absorption ability. Thanks for demonstrating that you are a complete idiot. Back to high school chimp. Try looking up Wien's displacement law. What about it? State your argument. So far you are using this as a buzzword. So far you have shown that you know absolutely NOTHING. So now you have reduced yourself to insulting and telling people they know nothing. Not an argument. Insults are a fallacy. Wake wrote: You've even gone so far as to tell us all Wikipedia is wrong because it contradicts your poor stupid ass. No, Wikipedia is wrong because it is biased, is written badly, and often describes physics very poorly or incorrectly. It is not an authoritative source. The authors of a theory are the ONLY authoritative source of a theory. There is no other. References to Wikipedia are summarily discarded. I do not accept them as a source. There are too many things wrong with too many articles at Wikipedia. I am not going to go through and filter out the good stuff that's in that crap. I am not going to defend against their crap. They are no substitute for an argument. Since you have decided to try to defend 'sources', be aware of this: The use of 'sources' as your argument only shows your inability to think for yourself. It shows you must borrow upon the arguments of others to think for you. It blinds you to see what may be wrong with their argument. Name dropping doesn't work with me either. It does not matter who someone is or what organization it is, nothing about it confirms or 'blesses' the argument. The argument is the key, not where it comes from. Depending on where an argument comes from is Bulverism, a fallacy. Wake wrote: When asked for your absolute ignorant statement that CO2 has "hundreds" of absorption lines you don't actually know anything about it and tell me to look it up. It's not hard to find. Are you really that lazy? People have measured the spectral lines of CO2 and water for a long time now. It's practically a student practice project these days. Wake wrote: I think that you need your diapers changed but you probably had your mother leave town a long time ago to get away from you. Ah...dropping to insults again. Insults are a fallacy, they are not an argument. They just waste space. While we're at it discussing what I accept and what I don't accept, you should be aware that I have a very high standard for any data presented. For details on what those standards are and why, see the first article in the 'Data Mine' thread. The Parrot Killer Debunked in my sig. - tmiddles Google keeps track of paranoid talk and i'm not on their list. I've been evaluated and certified. - keepit nuclear powered ships do not require nuclear fuel. - Swan While it is true that fossils do not burn it is also true that fossil fuels burn very well - Swan Edited on 27-05-2017 01:34
27-05-2017 05:52
Leitwolf ★☆☆☆☆ (117)	I am feeling very, very sorry about re-starting this discussion. I mean I love to discuss it back and forth, but this time I was only using the board as a convenient file host. I have posted this excel file here before, I just took out all the other sheets to avoid confusion. @wake These data really only refer to liquid water, as the specific geometry of a water hemisphere is pivotal. H2O molecules as vapour may behave very differently, in fact I never even considered it. And again, I did not and will not, because there are not GHGs and H2O is neither one. And btw. guys.. relax! I mean, if you knew what I have been trough, and what I am still in, you would not take the discussion here so personal and / or serious. There are idiots everywhere, some even have nobel prizes. If you want to fight them all, you'll be very busy, for the rest of your lifes.
29-05-2017 17:44
Wake ★★★★★ (4034)	Surface Detail wrote: Leitwolf wrote: Well .. we would need to read the book. But there is good reason why do not rely on such data. There is plenty of non sense data out there. Like putting emissivity to 1, when calculating the GHE, or claiming the cloud albedo effect was 44 - 50W/m2 (while it is about 2/3 of a 106W/m2 for the total albedo), or elevating clear sky albedo for earth 0.15 and beyong. All that is non sense, but apparently claimed to support the theory of the so called GHE. However, if it done correctly, the result is as above. You can read some of the book here: Thermal Radiative Transfer and Properties My guess (you've not shown it, so it remains a guess) is that your calculation of hemispherical emissivity from the emissivity as a function of polar angle is wrong. How did you work it out? Did you integrate the function over the surface of surface of a hemisphere to find the total emissivity as shown in the book? Did you use an analytical or a numerical approach? (Note to ITN: this is an example of a problem that does actually require calculus). After seeing that this man is carefully calculating all of the variables down to the fine details what do you say? "My guess" So your religion overrides absolutely anything. You yourself are incapable of calculating any of this but you are willing to "guess" that others that can are wrong. I suggest you grow up.
29-05-2017 19:34
Wake ★★★★★ (4034)	Surface Detail wrote: I didn't know what the Fresnel equations were either, until I investigated your work! It looks like the graph of the emissivity of water vs angle that you linked to comes from here: https://scienceofdoom.com/2010/12/27/emissivity-of-the-ocean/ According to that site, the different graph colours represent theoretical variation in emissivity with measurement angle, versus "refractive index" as computed by the Fresnel equations. So the graph that you have attempted to recreate using an ad-hoc function is, in fact, just a theoretical graph computed from the Fresnel equations. Didn't you think that it might be a good idea to find out where the graph came from before attempting to replicate it? You could have replicated the graph exactly by plotting the Fresnel equations, rather than trying to replicate it with your own "synthetic function"! That was rather insulting of me and I apologize. But I would rather you actually look at all of the facts than at those that agree with your own belief system. Now that research grants aren't automatically being gifted to people willing to agree with a position that would give more power to politicians we will start seeing more honest research. It is ALWAYS well to remember what the Chairman of the IPCC said: "This is not about global warming, it is about redistribution of the world's resources".

Join the debate Emissivity of the ocean:

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