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Comments 151 to 175 out of 388:

151. RW1 at 06:49 AM on 28 February, 2011
     Actually the Archer model you reference seems to be consistent with White's numbers: 385 W/m^2 x .566 (% absorbed clear sky) = 217.9; 217.9 W/m^2 x 0.333 (% clear sky) = 72.6 W/m^2 385 x .857 (% absorbed cloudy sky) = 329.9; 329.9 W/m^2 x 0.666 (% cloudy sky) = 219.7 W/m^2 219.7 W/m^2 + 72.6 W/m^2 = 292.3 W/m^2; Archer = 287.8 W/m^2, which is awfully close.
152. hank at 06:57 AM on 28 February, 2011
     http://www.google.com/search?q=top+of+atmosphere+255K First hit: meteo04.chpc.utah.edu/class/1020/Lecture2.201009.pdf Start around slide 31.
153. Tom Curtis at 10:55 AM on 28 February, 2011
     RW1 @150, if you care to look at the settings, Modtran is run for specific typical locations, with the default being the tropics. It does not produce a globally averaged result. Because the tropics is warmer than the global average, OLR at the tropics is warmer than the global average of approx 240 w/m^2. As set for default, it also does not include the effect of clouds. Further, and for the umpteenth time (as this is just your same question in a different guise) the OLR from the atmospheric window is included in the calculation. This can clearly be seen in the graph of the emissions for each model run. It can also be seen with line by line detailed data by viewing the whole output file. If you want a closer approximation to the global average, use the 1976 US Standard atmosphere (effective brightness temperature = 259 K). Alternatively, use the ground temperature offset to either set the surface temperature at 288 K (effective brightness temperature = 257 K), or adjust it to match an output of 240 w/m^2, and then run the doubling of CO2 experiment. As previously indicated, this is an obsolete model. And as implemented on the net, it does not even allow us to control all parameters so that you cannot set up a globally averaged surface temperature plus cloud cover. It's use is to show you quite clearly that the 3.7 w/m^2 is the difference in total OLR from doubling CO2. If you want a more up to date model, you'll have to pay the licensing fees.
154. Tom Curtis at 11:27 AM on 28 February, 2011
     RW1 @151, it is no surprise that the figures are close to George White's in that GW used a hi-tran model to get his figures. He then misinterpreted the Outgoing Long-wave Radiation as being the the total energy emitted from the top most layer of the atmosphere, and divided it by two to get what he believes to be the OLR. Looking at the Modtran model you can clearly see that that is a mistake. That model calculates the IR radiance at a given location that is either out going, or incoming. You can set the altitude to 0 and Look Up to calculate the back radiation. Or you can use the default to set the altitude to 70 km and look down to model what a satelite at 70 km altitude would detect. Clearly that satellite is not going to detect the radiation that is returning to the Earth, it will only detect the OLR. So, the I(out) of the model with that setting is the OLR. There is no need to divide it by two, and doing so shows complete incompetence on this subject. (Not a problem in somebody who is trying to learn, but a huge problem in someone like George White who purports to lecture.)
155. RW1 at 12:27 PM on 28 February, 2011
     "Looking at the Modtran model you can clearly see that that is a mistake." I've spent the last hour or more looking at the detailed line by line output and don't see the 'mistake' you're referring to. I do see that the atmospheric window is included in the data though (more on that in my next post). Also, I know this model is out of date, but for the purposes of understanding what these numbers mean, let's break them down as they are. Here are the inputs I'm using: CO2 (ppm) 375 & 750 CH4 (ppm) 1.7 Trop. Ozone (ppb) 28 Strat. Ozone scale 1 Ground T offset, C -1 hold water vapor pressure Water Vapor Scale 1 Locality 1976 US Standard Atmosphere No clouds or rain Sensor Altitude km 70 Looking Down This is the data output I'm looking at.
156. Tom Curtis at 13:04 PM on 28 February, 2011
     You did not show the data output. However, with settings as indicated and 375 ppm CO2, the base output is: I, W / m2 = 255.565 Ground T, K = 287.20 For 750 ppm, the output is: I, W / m2 = 252.801 Ground T, K = 287.20 The difference in I is 2.764 w/m^2. That is the difference, according to this model, between the IR leaving the atmosphere with 350 ppm and with 750 ppm. Plainly, if that is the IR leaving the atmosphere, it is incorrect to divide it by two to determine the difference in the IR energy leaving the planet in the two cases. But that is exactly what George White does with his equivalent calculation.
157. hank at 13:28 PM on 28 February, 2011
     "he then misinterpreted ... that is a mistake." The misinterpretation.
158. RW1 at 13:31 PM on 28 February, 2011
     The first thing I notice in the data is that at 375 ppm, the average transmittance is 0.2526, and the average transmittance at 750 ppm is 0.2465 (a reduction of 0.0061 or about 2.4%). At temperature of 287.2K, the earth's surface emits about 385 W/m^2. 385 W/m^2 x .2526 = 97.251 W/m^2 passing through the atmospheric window at 375 ppm, and 385 W/m^2 x .2465 = 94.903 passing through at 750 ppm for (a reduction of 2.348 W/m^2). 385 W/m^2 - 97.251 W/m^2 = 287.749 W/m^2 absorbed by the atmosphere at 375 ppm. 385 W/m^2 - 94.903 W/m^2 = 290.907 W/m^2 absorbed at 750 ppm. Here is what I can't figure out: The output of the data is showing 255.565 W/m^2 leaving at 375 ppm and 252.801 W/m^2 leaving at 750 ppm (a reduction of 2.764 W/m^2). If I divide 287.749 W/m^2 (375 ppm) by 2, I get 143.8745 W/m^2. 143.8745 + 97.251 = 241.1255 W/m^2 leaving (255.565 W/m^2 needed to match the data?). If I divide 290.907 W/m^2 (750 ppm) by 2, I get 145.0485 W/m^2. 145.0485 W/m^2 + 94.903 = 239.9515 W/m^2 leaving (252.801 W/m^2 needed to match the data?). 145.0485 W/m^2 - 143.8745 W/m^2 = 1.174 W/m^2, which is exactly half of the 2.348 W/m^2 reduction in the atmospheric window. To match output of the data exactly, at 375 ppm there needs to be 158.314 W/m^2 from the atmosphere (158.314 + 97.251 = 255.565 W/m^2). For 750 ppm there needs to be 157.898 W/m^2 from the atmosphere (157.898 + 94.903 = 252.801 W/m^2). The difference between 158.314 W/m^2 and 157.898 W/m^2 is 0.416 W/m^2, which is the exact difference between 2.348 W/m^2 reduction in the window and the reduction in the data output of 2.764 W/m^2). What accounts for the missing 0.416 W/m^2???
159. KR at 13:37 PM on 28 February, 2011
     RW1 - Why are you dividing by 2? Seriously, why? What Modtrans outputs is the total outgoing IR, and hence the 2.764 change on doubling CO2 is the entire, whole, complete difference between outgoing IR. Not half the amount, not twice the amount, but the whole amount. Dividing by 2 is wholly unphysical and wrong. This is the basic mistake that GW makes, and that you have repeated. It is wrong.
160. RW1 at 14:06 PM on 28 February, 2011
     KR, Then why does the 2.764 W/m^2 difference outputed NOT match up to the difference in the transmittance data outputed? Are you saying it shouldn't? Explain why. What accounts for the difference? All I've done is run some calculations showing what the numbers would be dividing by 2. Those calculations using the exact transmittance data provided at least yield about 240 W/m^2 (255K) leaving.
161. Tom Curtis at 14:15 PM on 28 February, 2011
     RW1 @158: 1) You did not account for the emissivity of 0.98 for the Earth's surface. That means Surface Radiation (SR) = 385.8 * 0.98 = approx 378 w/m^2 2) The average transmittance, ie, the sum of each line's transmittance divided by the number of lines, cannot be used as you have done it. The energy emitted at each line is not constant, so the distribution in variation in transmittance relative to the distribution in emitted energy can make very large differences in the net transmission. Therefore using a simple average of transmission will give invalid results. 3) Total radiance obviously includes values for emissions by the atmosphere, as for example at line 400: Surface Transmission: 3.18E-29 Total Radiance: 1.42E-03 Transmittance: 0.00000 Clearly with a transmittance of 0, Total Radiance would be 0 if radiation emitted from the atmosphere was excluded. To conduct the analysis you wish to make, you need to go through line by line, and sum the total of surface radiation * transmittance to get the amount of radiation from the surface that escapes to space unabsorbed. You then need to go through line by line and sum (total radiance - (surface radiation * transmittance)) to get the amount of radiation emitted from the atmosphere to space. You will then be in a position to do what you are trying to do in 158. Have fun.
162. RW1 at 14:22 PM on 28 February, 2011
     Tom, The average transmittance for '100 TO 1500 CM-1' is given at the bottom. Also, I tried using an emissivity of .98 and it didn't make much difference (2.3058 W/m^2 instead of 2.348 W/m^2).
163. Tom Curtis at 14:22 PM on 28 February, 2011
     Addendum to 161: Looking at the values, it appears quite probable that "Surface transmittance" is the surface radiation that escapes to space at each line, with transmittance rounded to five significant figures, thus showing 0 in this case. In that case, to get the transmittance you would have to calculate independently the surface radiance at the surface for each line. However, it would save you a step in integrating determining the total emissions from the atmosphere. You may need to find a manual to clarify this. Of course, the sensible thing to do would probably be to assume that no fundamental errors slipped into the programing based on the fact that a large number of independently programed models yield essentially the same result.
164. Tom Curtis at 14:24 PM on 28 February, 2011
     @162, I know the average is given at the bottom. That does not mean you can use it as you are doing.
165. RW1 at 14:27 PM on 28 February, 2011
     Tom, 'INTEGRATED ABSORPTION FROM 100 TO 1500 CM-1 = 1054.84 CM-1 AVERAGE TRANSMITTANCE =0.2465' You're saying this doesn't account for the differences in energy emitted at each line? How do you know this?
166. RW1 at 14:28 PM on 28 February, 2011
     Tom, "I know the average is given at the bottom. That does not mean you can use it as you are doing." How do you know? Have you added all the lines up and divided?
167. KR at 14:30 PM on 28 February, 2011
     Tom Curtis, RW1 - do you have a link to the Modtran model you are using? I'm not seeing the same freedom of parameters you seem to have discussed at the model here.
168. RW1 at 14:33 PM on 28 February, 2011
     Tom, Most of the radiance is in the window, so if anything that would seem like it would make the number much higher than only about 0.25?
169. RW1 at 14:36 PM on 28 February, 2011
     Tom, Should I add up all individual transmittance lines and divide?
170. RW1 at 14:50 PM on 28 February, 2011
     RE: my 168 Actually most of the energy is not really in the window.
171. Tom Curtis at 15:00 PM on 28 February, 2011
     KR @167, that is the model. If you run it there will be a link to "View the whole output file" which shows a large number of additional values. RW1 @168, as can be seen in this image, the peak of the surface transmission is in the 400 to 800 wavenumber band, ie, in the band with a deep trough due to CO2 and a number of troughs due to H2O. The peak radiance is at wavenumber 592, inside the left hand side of the CO2 trough. Therefore if the average transmittance is the mean, it would definitely underestimate the reduction in outgoing IR from the surface. @164 and 165, I don't know, but that seems the most natural reading to me. You should always take average to mean "mean", not "weighted mean" (or median or mode) unless there are clear contextual reasons to think otherwise. There are no such contextual reasons here; and furthermore, your discreprancy gives weight to that interpretation. Which is more likely, that a program developed by the air force for research and which has been used in various incarnations since 1988 with good correspondence to observational results has an error that produces up to 20% errors in its output? Or that you are simply mistaken in your interpretation of average? Regardless, if you disagree with me, you do the LBL integration. I am not the one chasing windmills here. @169, no, you just add up the individual lines. Any divisions (if necessary, see 163) shoud be done for each line only.
     Response: Fixed image width.
172. RW1 at 15:11 PM on 28 February, 2011
     I'm downloading the manual.
173. RW1 at 15:20 PM on 28 February, 2011
     wrong version.
174. RW1 at 15:25 PM on 28 February, 2011
     Tom, Do we agree that the reduction in the window should be twice the 2.764 w/m^ outputed (or about 5.528 W/m^2)? If not, why not? You don't think that all the infrared the atmosphere absorbs is directed toward the surface? Clearly it's not.
175. KR at 15:38 PM on 28 February, 2011
     RW1 - "Do we agree that the reduction in the window should be twice the 2.764 w/m^ outputed" Ummm, absolutely not. It's both a reduction in the atmospheric window and a deepening in the GHG emission bands. As we have said repeatedly. Not just a single effect, but two different ones that make up the total reduction in emissions.

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