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Location: Tennessee, USA | It is 1/4 x , based on 341.5 avg. and is based on clear skies. What's more, 102 w/m^2 is directly due to water vapor at an average density of 0.00225 (quarter %) concentration. There may be a problem there as some of the h2o vapor has to be in the form of clouds which are a different creature, leaving somewhat less than that averaged over clear skies. This paper concludes that column water vapor is lower for clear skies than cloudy skies. The effect is small in the tropics and increases as you move away from the equator. Most of the water vapor is also in the first couple of kilometers of altitude. I think you may be double counting the effect of water because you may have some of the same water reflecting incoming solar energy as clouds but also absorbing energy as water vapor. Considering that most of the water vapor is below the cloud tops, do you subtract most of the area covered by clouds from the total used to calculate absorption? Assume that there is no further absorption of incoming energy than what is aborbed by the clouds themselves. Clouds radiate a lot of energy. The cooling at the cloud top is large enough to create convection in the cloud.
So what's your current allocation of incoming solar radiation between reflected, absorbed by the atmosphere and absorbed by the surface? Can you plot a spectrum of incoming radiation at the surface compared to the top of the atmosphere? |
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| DeWitt - 8/11/2007 19:04
It is 1/4 x , based on 341.5 avg. and is based on clear skies. What's more, 102 w/m^2 is directly due to water vapor at an average density of 0.00225 (quarter %) concentration. There may be a problem there as some of the h2o vapor has to be in the form of clouds which are a different creature, leaving somewhat less than that averaged over clear skies. This paper concludes that column water vapor is lower for clear skies than cloudy skies. The effect is small in the tropics and increases as you move away from the equator. Most of the water vapor is also in the first couple of kilometers of altitude. I think you may be double counting the effect of water because you may have some of the same water reflecting incoming solar energy as clouds but also absorbing energy as water vapor. Considering that most of the water vapor is below the cloud tops, do you subtract most of the area covered by clouds from the total used to calculate absorption? Assume that there is no further absorption of incoming energy than what is aborbed by the clouds themselves. Clouds radiate a lot of energy. The cooling at the cloud top is large enough to create convection in the cloud.
So what's your current allocation of incoming solar radiation between reflected, absorbed by the atmosphere and absorbed by the surface? Can you plot a spectrum of incoming radiation at the surface compared to the top of the atmosphere?
Hi Dewitt,
Common sense causes me to assume that. If there's a calculation that averages all h2o in the atmosphere and determines xxx average, and there are clear and cloudy skies, then there must be some in clouds that isn't present in clear. The common sense doesn't quite carry into lattitude dependence though. I've down loaded the file and will read it soon.
The last numbers and report is based on the 1976 std atm numbers (averaged over the column). Some of it undoubtedly must be in clouds. However, for this I was dealing with clear sky only as I was attempting to validate the basics before proceeding.
Clouds may be radiating better as they have particulate sizes not just gas molecules but clouds also are highly reflective - limiting what is expected to radiate at least in the visible. If one assumes the cloud to be like liquid h2o or snow, that reflectivity plumets as wavelengths increase so snow is highly reflective high Bond albedo while being an excellent IR radiator as it's in the high 90% s in absorptivity there.
I'll be getting back towards this stuff as I get comfortable with the results for clear sky again.
At present, my rework is progressing fairly reasonably. I'll be trying to make another run tonight. It takes about 1 hour of continual processing to create the table of Tau or emission/absorption values /unit length for a spectrum that is 1nm resolution between 200nm and 65000nm with 40 or 50 vertical layers. It almost worked last time except for a minor error which caused it to keep repeating the last entry and adding it to the table multiple times.
After that, it remains to be seen just what will happen with my postprocessing Excel spreadsheet efforts. The table may exceed the memory capabilities of computer/excel - forcing additional preprocessing or other variations to the effort.
The model then becomes a 1 dimensional model rather than a 0 dimensional one.
Initially, in the 0 dimensional model, Clouds were dealt with very simply. The model assumed uniform temperature pressure and concentrations. Clouds were assumed to be 0 thickness objects at 1/2 pathlength - essentially 4km up. They reflected 80% of incoming sw solar and absorbed 20%. They were assumed to absorb 100% of LW upward radiation and have perfect BB emission. I'm not sure just what the proper method to estimate the BB curve as the particles are going to doubtlessly have diameters in the middle of the range of IR wavelengths which will screw that up somehow. Also, 62 % cloud coverage was used. An additional assumption could also be made as to permitting 10% of incoming radiation to be allowed through.
For it's simplicity, I think the cloud parts were quite reasonable in the early modeling although some numbers doubtlessly need adjusting. I still need to find some reasonable number for h2o vapor concentrations for clearsky which has to be were the biggest problem was. Since cloudy skies are treated differently inside the cloud layer, I probably don't need to do a great deal to mimic it other than to subtract out path effects above and below.
Another factor of importance is to know cloud temperature vs atm temp. The SW absorption of 20% or so has got to warm it above the atm and the perfect BB spectrum vs total line emission/absorption limitations has got to improve the overall emissions.
The corrected values indicated about 120 W /m^2 absorbed I think (it's in an above post somewhere). About 102W/m^2 of that was h2o vapor based on 1/4% concentration. Surface albedo is very low as 75% is ocean and below 5% with virtually none reaching 20%.
I should be able to plot a spectrum when the program gets put back together. My biggest problem is that the excel doesn't like plots with over 32000 points. I need to come up with something more like 500-1000 points across the spectrum to make it practical. Since I've not worked on that, I'm not sure other than by brute force, how it might be easily accomplished.
I've got an occultation to deal with very early sat morning after midnight and an all day event sat. and preparations for some stuff next week as well so my available timing is becoming quite limited again for several days. I'll be around here plenty of time I'm sure but I wont be able to just let the computer run the caculations to the exclusion of all else after tonight.
best regards,
cba
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Location: Tennessee, USA | Clouds aren't completely opaque. There is scattering as well as absorption and some of the scattered light gets through. It's still a lot brighter on a cloudy day than at night. In fact, for relatively thin clouds, you can have 80% diffuse transmittance with zero direct transmittance. In the visible, there's essentially no absorption, it's all scattering, otherwise clouds wouldn't look white. That also means, of course, that clouds don't emit in the visible since they don't absorb. The problem with zero thickness clouds is that implies that top and bottom are at the same temperature, which isn't true. Cloud tops are much colder than cloud bottoms. Cloud bottoms absorb radiation from the surface, which keeps the bottom warm and more buoyant than the colder air above. That generates convection. Convection transfers heat from the bottom to the top, not to mention the latent heat from water vapor condensing at the cloud top. Clouds are pretty good absorbers of near IR (wavelengths longer than 1 micrometer) from solar radiation. But a lot of that is going to be radiated back to space at thermal wavelengths rather than transferred through the cloud. Still, clouds transfer heat much less efficiently than direct radiation, so the surface doesn't cool as fast on a cloudy night. Note that the clouds don't warm the surface, they keep the surface from cooling as fast as it otherwise would. |
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| DeWitt - 9/11/2007 01:52
Clouds aren't completely opaque. There is scattering as well as absorption and some of the scattered light gets through. It's still a lot brighter on a cloudy day than at night. In fact, for relatively thin clouds, you can have 80% diffuse transmittance with zero direct transmittance. In the visible, there's essentially no absorption, it's all scattering, otherwise clouds wouldn't look white. That also means, of course, that clouds don't emit in the visible since they don't absorb. The problem with zero thickness clouds is that implies that top and bottom are at the same temperature, which isn't true. Cloud tops are much colder than cloud bottoms. Cloud bottoms absorb radiation from the surface, which keeps the bottom warm and more buoyant than the colder air above. That generates convection. Convection transfers heat from the bottom to the top, not to mention the latent heat from water vapor condensing at the cloud top. Clouds are pretty good absorbers of near IR (wavelengths longer than 1 micrometer) from solar radiation. But a lot of that is going to be radiated back to space at thermal wavelengths rather than transferred through the cloud. Still, clouds transfer heat much less efficiently than direct radiation, so the surface doesn't cool as fast on a cloudy night. Note that the clouds don't warm the surface, they keep the surface from cooling as fast as it otherwise would.
That's a nice description of cloud effects. For a relatively simple model I was figuring on using some average effective value rather than a combination of cloud types. That's where the number 10% comes in at as that should give a fairly good start to overcast conditions. I don't recall exactly the source but it's associated more with photovoltaic systems and overcast conditions than a direct measurement.
Remember too, at present, this is a radiative only model. There is currently not even scattering added. The original model approach was a blackbox (not a black body). No attempt was being made to determine a lapse rate. It was assumed there was an equilibrium lapse rate that at least balanced for the default atm by undefined means that included radiation.
The computer churned away into the night and I may have actually obtained an absorption dataset for the 1976 std atm concentrations/conditions. I was also able to import the file into an empty spreadsheet. It was too late to look at the numbers for any blatant problems or compare them to previous runs of a single T and p atmosphere. The next step is to try to determine if there's any blatant problems caused by software bugs. Calculations are mostly done with the existing code, except for the T correction. Hopefully, there will be some time over the weekend to get a handle on this and compare values to my previous runs for ballpark reasonableness and ultimately with the existing measurements/calculations.
I still need to find that average clear sky column values for h2o vapor to plug in as I really suspect it's too high as provided, including that portion which would be in the clouds.
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Location: Graham, North Carolina | Hey Charles;
Just a heads up incase you missed it in the earlier post. The reference on Page 2 of this thread has this link: http://plot.dmf.arm.gov/PLOTS/TWP/twpskyrad60s/20061018/twpskyrad60... labeled as a near cloudless sky in The Western Pacific near the Equator actually I think it is on an island about 10 Deg. East of Brisbane Queensland and about 10 Deg. Southeast of New Guinea, between New Caledonia and Fiji. If you were to draw up the satellite hybrid map you will see that most land area in this region is over covered with large stratocumulus clouds. If you look at the attached image you will note that the temperature being sensed is the bottom of these low clouds which my experience has shown shades the earth during the day and acts like a blanket at night. If you look at the opposite type of cloud which is more in character with the more Northern variety, you will see cirrocumulus clouds. These clouds generally are associated with a trapping radiant energy in the day and night. Meaning they have temperatures that are slightly higher then the surrounding air and act as a sort of blanket and low radiant angle reflector, reflecting radiant energy (both visible and high frequency) back towards the earth's surface at a hight latitude.
You have to keep in mind that the Equatorial or ITCZ band is more characterized by the stratocumulus and the near Arctic clouds have more of a cirrocumulus character and hence their radiant/reflective functions will be different. I do not know that any one model actually discusses this character very well and hence it may be likely that many of the variations we see in the modeling data could be related to the need to look at this effect much better. A professor that pops in from time to time here, Professor Tom Choularton might be able to add a bit to this discussion and to suggest some of the participation of aerosols if you were able to track him down.
In the meantime, try taking a second look at the sky temperatures at the ARM.gov site. I believe if you look carefully over the SKYRAD data sets you might be surprised. I wait to see what the clear sky data looks like from your model. I am curious if you can get a daytime between the 10AM and 2PM local and a night time between 11PM and 3AM local run across several parameters such as winter/summer high pressure and winter/ summer low pressure with a regional run for 10-30 Deg North/South and another for 45-65 Deg. North/South. In the meantime, have lots of fun!
Dave Cooke
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Location: Tennessee, USA | David,
The ARM.gov site is interesting. I was wondering about the details of the downwelling IR spectrophotometer. The AERI instrument pages were quite enlightening, particularly the AERI Handbook linked in the upper right corner. It doesn't say, but I'm assuming that the solid state detector is chilled to minimize Johnson noise if nothing else. I was surprised that it could be calibrated with just two blackbody sources at ambient temperature and 330 K. Apparently, the beam splitter is the major source of stray light. Not only do the measured spectra match line-by-line calculated spectra, but you can invert a spectrum using radiation transfer codes to profile the atmosphere above the instrument. The close match between calculated profiles and those measured by independent methods are yet another proof that atmospheric radiation transfer codes do model reality reasonably accurately.
The plots of total diffuse vs. direct solar radiation were interesting as well and demonstrate that total irradiance can be relatively constant even when a cloud blocks direct radiation from the sun. |
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| Attached is a transmission chart from my 1 dimensional model. It is using the 1976 std atm concentrations, temperatures and pressures. The model is broken down by 1km vertical resolution from 0-25km, 2.5km resolution 25km-50km and 5km resolution from 50km through 100km. This is a clear sky situation and it is an outbound energy. 197W/m^2 is the power reaching the TOA from the surface and 187.5W/m^2 being absorbed somewhere in the atmosphere. There are two lines on the chart. The top shows the transmission value and the lower line shows power /m^2 /nm radiated outward in all outward directions. Note that this doesn't show energy being radiated by the atmosphere or where in the atmosphere any energy is being absorbed.
Edited by cba 10/11/2007 23:29
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| cba - 11/11/2007 00:15
Attached is a transmission chart from my 1 dimensional model.
Sorry to jump into the discussion so late like this, but, Im preparing for a long trip beginning tomorrow and don't have time to go back and review all previous postings in this thread (I will, in about 2 wks time, when I return, do just that though!)
cba - is this model or the code publicly available?
I would do this myself given the ingrediants: over lay the equivalent black-body curve for power on the radiance spectra in the bottom plot, yielding a clearer picture of where the H2O and CO2 suck-out filter -er- absorption bands lie. In particular, I am interested in what is going on past the peak to the left at the shorter wavelengths.
Nice plot BTW.
Edited by _Jim 11/11/2007 19:56
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| _Jim - 11/11/2007 19:53
cba - 11/11/2007 00:15
Attached is a transmission chart from my 1 dimensional model.
Sorry to jump into the discussion so late like this, but, Im preparing for a long trip beginning tomorrow and don't have time to go back and review all previous postings in this thread (I will, in about 2 wks time, when I return, do just that though! )
cba - is this model or the code publicly available?
I would do this myself given the ingrediants: over lay the equivalent black-body curve for power on the radiance spectra in the bottom plot, yielding a clearer picture of where the H2O and CO2 suck-out filter -er- absorption bands lie. In particular, I am interested in what is going on past the peak to the left at the shorter wavelengths.
Nice plot BTW.
Attached are the file with the BB curve plus a zoom in of the left side. Data resolution in my model is 1 nm which is significantly beyond the resolution of computer screens for charting.
The code is my proprietary code written in c++ for the bulk of the processing and the final processing and display uses excel spreadsheet.
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Location: Tennessee, USA | cba,
That doesn't look like anywhere near enough absorption on the long wavelength side of the curve. The CO2 band looks about right, though. What about line broadening at low altitude? Also, spreading the water vapor over the whole column would tend to minimize continuum absorption from water vapor, which is a function of the square of the partial pressure of water vapor. Everything I've seen says that the atmosphere is completely opaque for wavelengths longer than 13.5 micrometers. You must be leaving something out. Continuum absorption seems the likely candidate. |
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Location: Sydney Australia | Charles,
The code is my proprietary code written in c++ for the bulk of the processing and the final processing and display uses excel spreadsheet.
I'm impressed. I do think you should acknowledge the data source along with above statement. |
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| DeWitt - 13/11/2007 17:20
cba,
That doesn't look like anywhere near enough absorption on the long wavelength side of the curve. The CO2 band looks about right, though. What about line broadening at low altitude? Also, spreading the water vapor over the whole column would tend to minimize continuum absorption from water vapor, which is a function of the square of the partial pressure of water vapor. Everything I've seen says that the atmosphere is completely opaque for wavelengths longer than 13.5 micrometers. You must be leaving something out. Continuum absorption seems the likely candidate.
Dewitt,
These graphs are using the 1 dimensional model approach where the atmosphere is broken up by 1km increments up to 25km, 2.5km to 50km and 5km to 120km. The concentrations, temperatures, and pressures are straight out of the 1976 standard atmosphere tables as published by GATS. Line broadening is fully invoked again (at least I think I turned it back on) according to the methodology in Hitran's 1996 documentation paper along with temperature and pressure corrections. It should be the full implementation including pressure shifting. Quantum state populations of Q(T) are also implemented using the Hitran data. The only thing missing from what I can gather at present is is particles aerosols and clouds.
Note that for these graphs, only every 65th nm point is sampled and for the blow up, only about one point every 10 nm is displayed. Also, my total absorption amount is probably higher than the commonly accepted values so I'm concerned I might be absorbing too much.
I'm not sure how continuum absorption fits into the hitran approach. It is not specifically added at the moment.
have to prepare for class and hopefully the other computer is finished 'thinking' about this stuff as it is really getting loaded down by all the data.
cba
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| Jan Pompe - 13/11/2007 20:31
Charles,
The code is my proprietary code written in c++ for the bulk of the processing and the final processing and display uses excel spreadsheet.
I'm impressed. I do think you should acknowledge the data source along with above statement.
It's been mentioned in the past and the question was the programming (which is processing of data). I am using the Hitran 2004 absorption line database with the Hitran 1996 documentation for temperature and pressure adjustments of the line data including pressure shifting, intensity correction with T, pressure broadening etc. Currently, the atmospheric input data is the 1976 Standard Atmosphere data obtained from the GATS website. I'm using Excel for final processing and a proprietary program written in microsoft c++ to create the wavelength spectrum out of the Hitran and std atm data. Partition function data values Q(T) also comes from the Hitran package.
gotta run
cba
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Location: Tennessee, USA | Note that for these graphs, only every 65th nm point is sampled and for the blow up, only about one point every 10 nm is displayed. Also, my total absorption amount is probably higher than the commonly accepted values so I'm concerned I might be absorbing too much.
For some reason, I have been under the impression you were calculating the absorption of incoming solar radiation. Assuming that your x scale is in nanometers, you have actually been calculating the absorption of Outgoing Longwave Radiation (OLR). In that case, your absorption is nowhere near enough, not too much. Absorption of OLR is about 70% to 85%. Only the window centered at 10,000 nm has any significant transparency. Compare this lower resolution transmission spectrum:
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Location: Tennessee, USA | Why we can calculate the approximate value of greenhouse warming.
It has been stated elsewhere that we can't calculate the value of greenhouse warming because the simplistic calculation uses an isothermal body and a non-isothermal body will have a lower average temperature on an area weighted calculation. That is indeed true, but the question is how much less? Well, it's not all that difficult to model a non-isothermal body. If there is no conduction, but infinite heat capacity, there will be no diurnal or longitudinal temperature variation, but the temperature will vary with latitude because insolation decreases as the cosine of the latitude. If the body has the same dimensions, shortwave albedo (0.31), and solar constant as the Earth, but with the axis of rotation perpendicular to the plane of the ecliptic, the temperature will vary from 2.7 K at the poles (cosmic microwave background temperature) to 269.8 K at the equator, assuming a longwave albedo of 1. The global average temperature is then 250.8 K compared to 254 K for an isothermal body.
If the heat capacity is zero, then temperature varies with both latitude and longitude at any given time. Peak temperature at the equator for the lit side is then 359.2, the temperature of the dark side is 2.72 and the average temperature is 142.4 K or -130.7 C. This is interesting, but irrelevant to the real Earth which has a finite heat capacity. As the heat capacity increases, the minimum temperature increases rapidly while the peak temperature decreases less rapidly. The peak time shifts to later in the day and the minimum shifts to just after sunrise. For a heat capacity that gives a diurnal temperature range of 49.6 degrees at the equator, the average at the equator is 268.3 K compared to 269.8 K for the infinite heat capacity case. The actual diurnal temperature variation of surface temperature is quite small because most of the Earth is covered by water and diurnal variation in sea surface temperature is less than one degree.
What else do we know? The modeled non-isothermal bodies emit the same amount of energy at each latitude as they receive. The isothermal body, because it conducts heat from the equator to the poles, will emit more radiation than it receives at the poles and less than it receives at the equator. So does the Earth. For an isothermal body, the excess emitted at the poles is 236 W/m2 and the deficit at the equator is 64.5 W/m2. This compares to the measured deficit of about 70 W/m2 at the equator and excess at the poles of 100 W/m2 for the earth. Considering that the Earth's axis is tilted so the poles actually receive about 50 W/m2 average annual insolation at the South Pole and nearly 80 W/m2 at the North Pole , this makes the Earth much closer to an isothermal body than a non-isothermal body.
If the average temperature of the Earth is 288 K that puts a lower bound on size of the greenhouse effect of 34 degrees and an upper bound of 37.2 degrees. Given the amount of heat actually transferred, the true value is almost certainly very close to the lower bound.
Edited by DeWitt 14/11/2007 17:55
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Location: Graham, North Carolina | Hey Dewitt;
Okay, I'll byte, first you may need to reduce your equatorial input as the LW albedo is much higher then 1 within +/- 12 Deg. of the equator Next you will need to adjust your nightside value as nearly 30% between 12 Deg. and 62 Deg. is covered by clouds offering an increase of albedo on the dayside and in essence preventing the radiant and convective transfer of stored LW ground temperature as the cloud bottoms run at about 270-290 Deg. K. This means that for 13% of the surface, the high temperature near the equator the value should be closer to 291 Deg. K and not 306 Deg. K, then we have the balance average of between 12 and 62 Deg of 70% at 302 Deg. for the average summer high or 279 Deg. K for the average winter high with 30% running at about 294 Deg. K for the summer high and about 267 Deg. K for the winter high. During the evening or nightside where the temperatures ran with a summer range between the high and low of 22-28 Deg. have recently started running as high as 32-36 Deg. in regions were the air is dry and as low as 18-22 Deg. where there is high soil moisture.
As to the idea of non-isothermal transport you really need to review the character of the actual current data and not rely on the generalizations or the former standard tables, IMO. (Even the stuff I am suggesting above is incorrect in that it is not based on an actual measured data base and is instead reflecting known local outliers.) What makes things even slightly worse is that the orbit places the earth closer to the sun during the NH winter and further away during the SH winter and when coupled with the higher albedo of land in the NH and lower albedo of the ocean in the SH will cause uneven heating of the two hemispheres. You may also need to calculate in your same cosine rule for East and West values greater then +/- 23 Deg E/W of the line perpendicular between the core of the sun and the core of the earth, otherwise you will over state the zonal energy content. Also according to an old discussion point on RealClimate the suggested solar insolation value should be around 240 - 242 Deg. K if I remember correctly, then again it might be 255 Deg. K, (I can not remember if the suggested difference due to atmosphere was 33 or 18 Deg. C.). I think part of the issue was trying to define whether the calculation was supposed to address the average or the seasonal extremes.
Dave |
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Location: Tennessee, USA | ldavidcooke - 14/11/2007 19:18
Hey Dewitt;
Okay, I'll byte, first you may need to reduce your equatorial input as the LW albedo is much higher then 1 within +/- 12 Deg. of the equator Next you will need to adjust your nightside value as nearly 30% between 12 Deg. and 62 Deg. is covered by clouds offering an increase of albedo on the dayside and in essence preventing the radiant and convective transfer of stored LW ground temperature as the cloud bottoms run at about 270-290 Deg. K. This means that for 13% of the surface, the high temperature near the equator the value should be closer to 291 Deg. K and not 306 Deg. K, then we have the balance average of between 12 and 62 Deg of 70% at 302 Deg. for the average summer high or 279 Deg. K for the average winter high with 30% running at about 294 Deg. K for the summer high and about 267 Deg. K for the winter high. During the evening or nightside where the temperatures ran with a summer range between the high and low of 22-28 Deg. have recently started running as high as 32-36 Deg. in regions were the air is dry and as low as 18-22 Deg. where there is high soil moisture.
As to the idea of non-isothermal transport you really need to review the character of the actual current data and not rely on the generalizations or the former standard tables, IMO. (Even the stuff I am suggesting above is incorrect in that it is not based on an actual measured data base and is instead reflecting known local outliers.) What makes things even slightly worse is that the orbit places the earth closer to the sun during the NH winter and further away during the SH winter and when coupled with the higher albedo of land in the NH and lower albedo of the ocean in the SH will cause uneven heating of the two hemispheres. You may also need to calculate in your same cosine rule for East and West values greater then +/- 23 Deg E/W of the line perpendicular between the core of the sun and the core of the earth, otherwise you will over state the zonal energy content. Also according to an old discussion point on RealClimate the suggested solar insolation value should be around 240 - 242 Deg. K if I remember correctly, then again it might be 255 Deg. K, (I can not remember if the suggested difference due to atmosphere was 33 or 18 Deg. C.). I think part of the issue was trying to define whether the calculation was supposed to address the average or the seasonal extremes.
Dave
LW albedo higher than one? Do you mean emissivity? That would explain why the OLW at 330 W/m2 acccording to the graph in Petty's book is higher than the theoretical insolation. Of course a simpler and more physically realistic explanation is that the albedo at the equator is less than 0.31 plus the day is slightly longer than 12 hours because of atmospheric refraction. If there is more energy coming in at the equator than in my models, and there probably is, it just means that advective heat transport is even larger, and it has no effect at all on the isothermal model. Anything like clouds that makes the diurnal temperature variation lower by cooling during the day and warming at night or axial tilt that makes the poles warmer than for a perpendicular axis makes the planet more isothermal on average, not less. Orbital eccentricity combined with axial tilt that causes different insolation at the poles is a second or third order effect at best. Huge diurnal temperature differences don't have much effect, why do you think that a tiny effect like that would make a significant difference. Do you mean 240-242 W/m2 rather than K? 255, OTOH, probably is K because it's the temperature of an isothermal blackbody with an average energy flux of ~240 W/m2. So the lower bound is 33 degrees, not 34, not significant. In fact, let's just say it's 34 plus or minus 2.
As far as your comments on varying surface temperatures, I'm modeling TOA brightness temperature, not surface temperature. In spite of Tom Vonk and G&T, TOA OLW radiation has to be in close balance to incoming SW radiation over time. The 288 K global average surface temperature, OTOH, is measured, not calculated by me. Whether the measurement is correct is another matter, but I don't really think it's too far off.
Edited by DeWitt 14/11/2007 21:04
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Location: Graham, North Carolina | Hey DeWitt;
First, I need to make the point that part of what I am trying to recommend is staying away from the average or median values. There is nothing natural in median values, if you want to model you have to model the edges not the middle. Changes at the edges are the changes that matter not the middle and two changes at the edge are the issues that need to be measured and addressed.
Second actually I was using your value for albedo not really knowing the scale. The point is that the cloud density is so great and the clouds are so tall actually the only region that is not strongly covered by clouds in the InterTropical Convergence Zone is over the Sahara Desert Other then at the Eastern and Western extremes. If anything the emissivity is very low in the region as nearly all the energy transport is virtually pure convection at this latitude. However, in the region just above/below the ITCZ you get about 1000-1200-1000 watts/meter^2 peak overhead for about 4 hours/day and about 750-1000 watts/meter^2 for about 2 hours on either side. The balance is about 2 hours on either side ranging from about 46-52 watts/meter^2 up to 750 watts/meter^2. As you move further north/south you will see a decrease of about 100 watts / 6 degrees up to about 75 Deg. N/S.
After living at 25 Degrees N for a long period it becomes obvious that the transition between light and dark is nearly like a knife the closer to the equator you are. Generally, I would suggest that you are more likely to see light longer at the upper latitudes then you would ever see near the equator.
The orbital and tilt differential is actually well distributed in the ocean via the THC, meaning that though there may be a difference as far as the land versus sea albedo and surface temperature the actual atmospheric heating which is based on the surface LW emissivity and convection are fairly well distributed in with the SH providing more to the THC then the NH. At the same time the albedo in the NH is generally going to be in the higher frequency band to the tune of nearly 2/3rds more then the SH. This seems to suggest that the NH must contribute much lower LW to the atmosphere then the SH as the vegetation land area has almost three times the albedo of the ocean and sand covered should be slightly less then clouds.
As to the isothermal temperature without atmosphere at this distance from the sun my memory suggested that the temperature ranged from about 240 to 255 Deg. K. As for the 34 versus 33 degrees really doesn't matter if you are trying to replicate/model the thermal processes that is about the level of precision that could be expected at best from the temperature record prior to calibrated digital thermometers.
As for your attempt to work at the TOA probably the best references may be the older Nimbus and ACRIMM packages on page 6 of the following: http://mac01.eps.pitt.edu/courses/GEOL0030/Solar_Energy_Heat_Budget... .
Additional reference might be found in the data used by Dr.Scafetta at Duke University:
http://www.fel.duke.edu/~scafetta/pdf/2006GL027142.pdf
http://www.fel.duke.edu/~scafetta/pdf/2006GL025668.pdf
http://www.fel.duke.edu/~scafetta/pdf/2005GL025539.pdf
I like your approach in that you would desire to stand at the TOA and measure the input on the dayside and then the output on the nightside. The problem is that the energy has been changed almost like a doppler narrowing of the emissivity from a black body. The problem is that different spectral energy comes out at different points of the planets rotation requiring up and down welling measure throughout the rotation.
Otherwise you and Charles appear to be doing great work. I am interested in what the outcomes will be all in all.
Dave
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Location: Sydney Australia | Charles,
One of the reasons I mentioned acknowledging the data source is that not everybody will go back over to find it and some even forget if they had seen it.
Your model output shows banding where the charts produced by DeWitt show none after the 15 micron CO2 absorption band.
So who is right?
If you have left nothing out from the HITRAN data and I read the Boisoles pater correctly that area does in fact show banded transmission. |
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| Dewitt,
I'm doing both up and down radiation. I was concentrating on downward radiation while attempting to correct problems but both are equally important. There is still more absorption on the downward insolation than the accepted standard values, something like 120W/m^2 for clear sky versus a 70 to 90 W/m^2. Also, the values for surface radiated clear sky - as shown in my graph above total just under or over 50% absorption, something like 190 W/m^2. The values shown on your chart indicate transmission totals of between 15% and 30% for outward LW. Assuming a mix of clouds invoked, the total average absorption would rise significantly so it would depend upon the conditions assumed for that chart as to whether I'm above or below the absorption level it presents. Also, my graphs are using the exp(-tau)extinction rather than taking the optical depth Tau directly and lopping it off at 1.0. I'm not sure what your provided chart is doing or what was the resolution of the chart data used.
Looking at raw numbers, my line data is indicating there are tiny windows of several nm widths all around, surrounded by serious extinction areas.
Referencing Keihl and Trenberth's 1997 paper, they offer clear sky LW radiation absorption to the TOA as being 125 W/m^2 versus my 187 W/m^2 absorption. Their paper is a conglomeration of measured data from other researchers and calculated data they performed.
Charles
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Simple Radiative Models of the Atmosphere
