8 Comments

  1. I may ask some questions later, but the main problem I see with this presentation is that it assumes cirrus clouds are a good average for the planet. The problem is that these very likely model very low humidity, so the 1.024 scale value still leaves you with almost negligible water in the atmosphere. Because of that, the first incremental temp increase is only .06, leading to overall low temp increase (after the infinite sum).

    Not surprisingly, then, this low water feedback temp increase is way below the world-wide consensus and what complex computer models and much more basic models predict. Another clue that this humidity level is way too low is that if we look up from .001km using modtran, we observe 260-ish IR value rather than the Trenberth diagram 333 average global back radiation IR value.

    In contrast, if we use any other cloud model except the cirrus ones we end up with back radiation around 360 (much more reasonable). Further, if we keep the cirrus but raise the water vapor scale to around 4 (400% of normal, rather than 1.024!), we then get close to the 333 back radiation level. It makes sense that the water humidity level (which is what causes a lot of the back radiation seen on the ground and the consequent higher temp) should be much higher on average world-wide than what cirrus provides. Even when it isn't cloudy, many places are humid, eg, in the tropics.

    The 1976 US std atmosphere doesn't model humidity, and I'd wager the cirrus models have very low humidity values. Anyway, I'll re-calculate some of the key numbers (assuming most of the formulas are good enough for an estimate) and post again.

  2. 1. I think we might use modtran (or better, hitran data) to calculate many different values around the planet with x and y percentages of various clouds and then we would integrate all of that data to get the actual global averages we want. To look at just one setting (of temp, cloud type, humidity, etc) is not modeling the planet, and I don't think can be made to match the relevant values from the Trenberth diagram.

    However, we might try to do an estimate using the single set of modtran parameters used in this presentation (eg, Appendix B). That seems like a reasonable approach that might yield a ballpark result.

    For example, let's make 2 estimates.

    A) 1976 us std atmosphere, noaa cirrus model lowtran 6, ground temp=288.2K, water vapor scale=1.

    i) 400ppm, 70km look down: 242.691
    ii) 800ppm, 70km look down: 240.241

    So, using (i) to fix the sun's effective intensity, we can adjust (offset) the ground temp at 800ppm so that when we look down we also get 242.691. An offset of +0.81 does the trick.

    So by this estimate, a doubling of CO2 from today's level will lead to a .81 C increase in the ground temp when we ignore all water and other feedbacks.

    B) 1976 us std atmosphere, cumulus cloud base .66km top 2.7km, ground temp=288.2K, water vapor scale=0.0001.

    i) 400ppm, 70km look down: 242.031
    ii) 800ppm, 70km look down: 240.116

    So, using (i) to fix the sun's effective intensity, we can adjust (offset) the ground temp at 800ppm so that when we look down we also get 242.031. An offset of +0.64 does the trick.

    So by this estimate, a doubling of CO2 from today's level will lead to a .64 C increase in the ground temp when we ignore all water and other feedbacks.

    In conclusion, while there are more accurate procedures to use with modtran for estimating the no feedback 2x CO2 temp raise at ground and while there are more accurate tools to use than modtran itself, we still can get a rough idea of what 2x CO2 does using a one shot modtran calculation and a simple model like 1976 US std atmosphere model.

    2. Is equation 1 useful? I don't think much.

    Energy entering atmosphere – leaving atmosphere:

    [78 (radiation from sun) +
    17 (thermal, contact from earth) +
    80 (evaporation) +
    356 (radiation from earth)]

    [169 (radiation from air towards space) +
    30 (radiation from clouds to space) +
    333 (radiation from air and clouds towards ground)]
    =
    -1

    So basically the atmosphere is close to equilibrium (since we also have rounding errors). If not it would be getting hotter (or colder) at a noticeably fast rate.

    In contrast, equation 1 along with the calculated values of the various H suggests the atmosphere is gaining huge amounts of net radiation flux (energy per second per area) and thus makes no sense.

    3. Later it's stated that using 200 meter ocean depth makes modtran useless. That could be a clue that the methodology might have problems. Instead the value 20 is used. Question, why not keep going and use just 1 meter or .001 meters? How would that impact the data? According to the logic employed, would that lead to a lot of heating? So why cherry-pick 20? [I am not sure because it wasn't clear to me how Table 1 was used, if at all.]

    4. Putting aside the problems with equation 1 for a moment, Table 1 suggests that the "heat" in the atmosphere will be averaged into the ocean, but that doesn't make much sense since the sun is heating the oceans and the atmosphere at the same time. It isn't an either or proposition. What is Table 1 about?

    5. How was humidity calculated and how does that translate to a water vapor scale value?

    6. Measurements and theory suggests that water will result in a significant amount of temperature rise. This is inconsistent with the r result of .08. Instead a value of about 2/3 C is what leads to a 2 C rise beyond the 1 C anticipated for CO2 working by itself.

    7.
    (a1) "A, L, C, and F are the CO2 quantities in the atmosphere, land, ocean and fossil fuels, respectively. The time dependent transfer rate from atmosphere to ocean is Ac; Ca is from ocean to atmosphere, etc. …For this analysis it is assumed that the transfer rate out of a region is proportional to its concentration."

    (a2) "La = 0.05L(t)"

    (a3) Eqn 8: "A_n+1 = A_n*(1-0.244) + 0.05*L_n + 0.002451*C_n + 10*(1+0.005*n)"

    (b1) "Therefore La(0)*L(t) = 100"

    (b2) "A’ (t) = Fa(t)*F(t) + La(t)*L(t) + Ca(t) –Al(t)*A(t) –Ac(t)*A(t)*A(t)"

    a1, a2, and a2 are consistent with each other. b1 and b2 are consistent with each other. a1, a2, a3 are inconsistent with b1, b2.

    I'll assume b1, b2 were effectively typos and that a1, a2, a3 was intended.

    For example, La(0) = 100 = .05*L(0).

  3. Thanks to Jose for his thorough read of the document and his comments. I hope I can answer these to his satisfaction.

    Response to paragraph 1 of Jose's first comment:

    I quoted ModTran's writeup as NOAA cirrus clouds and U. S. Standard atmophere as being a world average (which was quoted in Appendix B). But I didn't base any conclusion or results on that particular climate region and weather condition. In fact, I tried to span the full range of weather conditions and climate regions. I sometimes used one example calculation to display typical results and I sometimes used what seem to be a reasonable number of calculations. If there was one specific example that did not fit the conclusions of the several calculations, it was noted and displayed. In fact the higher humidity samples had the lowest temperature increase. That can be seen in Figure 11. I will be glad to supply the results for any climate region, weather condition, or humidity condition of Jose's choosing. I DID NOT SHORT-CHANGE HUMIDITY IN THESE CALCULATIONS.

    2nd Paragraph. If I had been in agreement what "world-wide consensus, and what complex computer models and much more basic models" predict, I would have had no reason to publish the paper. What an example of a much more basic model? I do't believe that a concatenation of multiple computer models with many guessed parameters adds up to precision. I, too noticed that greater or less than 333 back radiation could be generated with Modtran. I used Figure 2 to illustrate the components of atmospheric heat balance and to derive the second term of Equation 1, otherwise I relied on Modtran's ranges and parameters for temperature calculations.

    3rd paragraph. See the last sentence on the first paragraph.

    The 1976 U. S. Standard Model can be modeled with any humidity level you choose and you have 14 weather conditions to chose from in Modtran

    Response to paragraphs of Jose's second set of comments.

    1. I was not trying for an average of the planet and I was not trying to match Trenberth's figure. I was trying to do a broad enough sweep of all the world's available climate regions and weather conditions to ensure a valid conclusion of CO2's world-wide effect on temperature and on man's ability to control it. I I have left a major stone unturned in this quest, please inform me of what it was.

    With your two estimates please note from my Figure 7 that your doubling of CO2 with NOAA cirrus clouds and with cumulus clouds and my results shown in Figure 7 are very close to the same. I don't really understand what point you are making. Your method may be better than mine but their divergence is minuscule.

    In your method did you just do temperature increases by trial and error with the doubling until you matched the non-doubled exit radiation?

    2. First, if my use of Equation 1 produced the same answer as your method did and you (I assume) consider your method to be correct, why is mine not much useful?

    The reason for not including the back radiation as part of the atmospheric energy balance is difficult to grasp. I wrestled with that for some time myself. It doesn't permanently escape the atmosphere as those which exit to out space do. It comes right back again through the IR of the increased temperature of the earth. But that's not too satisfying either. But Modtran does not use it, otherwise there would be no escape of radiation to outer space and Modtran shows a lot of that. But the explanation I find most satisfying is that it is totally created within the atmosphere, so when the atmosphere loses that there is a net 0 in energy balance. In Figure 2, for all other energies into the atmosphere their source can be traced to outside the atmosphere. Back radiation's source cannot be traced to outside the atmosphere. I never found any way of using back radiation that gave anything but an irrational answer.

    The atmosphere is, indeed, gaining huge amounts of net radiation flux. That is, after all, what global warming is all about. However, it is also losing a huge amount to outer space. The loss being proportional to absolute temperature to the 4th power, it doesn't required a large increase to get rid of a lot of energy. Global warming is a rather slow, steady process as more IR energy is dumped into the atmosphere, the hotter temperature quickly gets rid of it to outer space, but it has to have a higher temperature to do that.

    3. I didn't say Modtran was useless. I just said it may be beyond its resolution capability. That, in itself is good information about its extremely small value.. Your question about why not keep going to 1 or .001 meters is a good one. Initially I was using ten meters then I discovered the curve shown as Figure 9 which indicated excellent mixing (i.e., constant temperature) down to 200 meters which means there is little error in using that full depth as the recipient of the atmospheric heat, Then I discovered the heat increase was maybe too small to calculated with Modtran and certainly too small to show up on a reasonable sized graph that showed the range needed. I opted for 20 as reasonable conservative, yet still demonstrable.

    4. It doesn't matter than the sun is heating the oceans and the atmosphere at the same time. I can separately estimate that which is due solely to created water vapor and treat that as addition. I am only after the addition.
    Table 1 is a tabulation of the results of the Clausius-Clapeyron equation. The last column in the table gives the fractional increase in vapor pressure per degree increase in water temperature. So multplying that fraction by the degrees of temperature rise of the water
    gives the total fractional increase in vapor pressure. Modtran works on water vapor pressure, so that can be just added to the vapor pressure previously used for the modtran calculation. Vapor pressure of water doesn't depend on how much vapor is already above the water, it depends only on the water temperature and atmospheric pressure. Conveniently atmospheric pressure is given in atmospheres. I believe Modtran uses 1 atmosphere in all its calculations.

    6.One of the three main goals of the paper was to refute your paragraph 6. The other two were to show that CO2 can never reach a harmful value and that man couldn't do anything about it if it did get that high.

    7. I think your assumption is correct. I have checked my many excel calculations on this and they seem to be correct. Plus. their plots are good at displaying errors.

    Again, Jose, I appreciate your read-through and your questions. You may have provided the most thorough review I have had.

    Thanks,

    Bryce

  4. I should make it clear that I am new to modtran use (through the free web service or otherwise) and also have not used any other similar tools. I try to research online and appeal to past discussions I have read on its use. Also, while I think point 7 is no big deal, I was a little harsh on earlier points as I best was able to understand the paper. Not only could I be wrong, but I can certainly not be recognizing the value of the analysis.

    Further, while I am not a climate scientists and understand the field and topics in pieces, I think anyone getting significantly different results than most other experts in a field should take that as a warning and really try to make sure the logic and research are sound (and this is part of what I am trying to prod in this particular case). It seems also in order in time in order to establish a new set of significantly different results would be a critique of many key papers that have established a foundation of modern climate estimates of 2x CO2. The computer models from which temperatures may have slightly slipped off the edge (depending on interpretation of its assumptions made by the computer predictions) is not an automatic condemnation of the believed science but perhaps at most on its incompleteness, especially since the IPCC took pains to clarify that it presents the predictions from models that are known to be incomplete in a few important ways but offer our best estimates. Keep in mind the error bars cited are a confidence range of no more than 2 sigmas and have never claimed 100% absolute coverage (eg, as noted clearly in IPCC summary sections of their main reports). New papers attribute some of the lower temps than expected over this past decade to a large number of cyclical cooling la nina action. One way to attack the status quo would be from an analysis of the open source climate programs and/or from appealing for the non-open-source models to have their legitimacy knocked down a notch for their lack of full transparency. It's a tough job for any new player to analyze so much computer code, so it may help to promote open discussion forums for such an endeavor or to participate in any that might already exist. The models use more extensive physics (with many fewer approximations and greater appeal to fundamental physical equations), so it is easy to see why their results would carry extra weight in the minds of most experts. At the same time, there is greater room for error in greater complexity.

  5. Bryce Johnson @3

    >> In fact the higher humidity samples had the lowest temperature increase.

    I noticed that as well after writing that first comment (and to my best understanding of what Modtran does).

    >> The 1976 U. S. Standard Model can be modeled with any humidity level you choose and you have 14 weather conditions to chose from in Modtran

    That was the assumption I made for comment #2. I am glad you agree since that increases the odds I am understanding how Modtran works.

    >> I don’t really understand what point you are making.

    I was implicitly agreeing that your .75 estimate of 2x CO2 seems as reasonable as anything I as able to pull from Modtran (if I understand that tool).

    I was also emphasizing that these Modtran estimates are among a toolbox that includes more accurate techniques. [I guess you did use a larger sample size with more sensible statistics than I did. Again, I haven't spent all the time required to understand every part of your paper.]

    This also means I was implicitly supporting your approach to use conservative boundaries.

    So pt 1 agrees with this paper's general attitude and agrees with the ballpark of the 2x CO2 values calculated.

    >> In your method did you just do temperature increases by trial and error with the doubling until you matched the non-doubled exit radiation?

    Yes, and, partly because of that simple approach, I perhaps used fewer data trials than you did.

    Again, my point 1 offers no real complaints except a note of caution that we are playing with estimates inferior to other approaches in using this tool to get these results.

    >> First, if my use of Equation 1 produced the same answer as your method did and you (I assume) consider your method to be correct, why is mine not much useful?

    In my mind, this paper's divergence with status quo results hinges significantly around the very low value obtained for the first iteration of the water (feedback) temp contribution. If we are to believe the status quo for a moment, that feedback iteration should be nearer to a 2/3rd C. You got around .06, which is only about 10% of what I think would be expected.

    >> But Modtran does not use [back radiation], otherwise there would be no escape of radiation to outer space and Modtran shows a lot of that.

    >> But the explanation I find most satisfying is that it is totally created within the atmosphere, so when the atmosphere loses that there is a net 0 in energy balance.

    I agree that the back radiation is created within the atmosphere. My view of that is that it leads to a higher temp ("average energy") within that region. All forms of insulation create that effect, where the internal enclosed area resides at a much higher temp than the boundary. This is probably unintuitive at first sound (it certainly was for me), yet it doesn't seem you can prod the internal temps from the outside (at least not with a simple thermometer) but must infer it or otherwise place a thermometer inside. An oven or a sweater give only indirect clues of the temp inside. Note as well that the hottest point on the planet+atmosphere (statistical measurements and theory) is supposedly the surface, so I don't think there is any inconsistency with thermodynamic expectations. What we should remember is that the sun keeps adding more and more energy to the earth's ground level surface. Hypothetically, if the planet were to keep that energy that was added years and months back by the sun, that effectively today goes into space quickly (especially if the planet were to have no atmosphere, as is basically the case for the moon), instead fully accumulating within the earth's atmosphere by delaying its exit into space indefinitely (a fictional and boundary case, of course), then there is reason to expect that average temperatures in the atmosphere will rise "stratospherically" while a sensor looking from space to the planet would observe almost no energy coming out. Another example might be using mirrors (or something like a laser mechanism) to aggregate a lot of energy within a narrow enclosure in order to produce a very high temperature, rather than letting that energy (photons) continue along the path without mirrors and end up elsewhere. Energy that would otherwise be available to aliens quickly would instead be used at home.

    >> The atmosphere is, indeed, gaining huge amounts of net radiation flux. That is, after all, what global warming is all about.

    I would change "radiation flux" (the rate of energy movement) in that sentence to say "radiation energy". The latter allows for the accumulation of lots of total energy when looking at the entire planet, even if the rate is very small locally in time and space. The Trenberth values are the rates for an average second and meter square averaged over the entire planet's surface over an entire year, more or less. The average is over a large set of points in time and space, but it still represents the value of a single second and meter squared.

    These things are not something I understood except after lots of online debate and thought.

    >> Initially I was using ten meters then I discovered the curve shown as Figure 9… Then I discovered the heat increase [for 200 m] was maybe too small to calculated with Modtran and certainly too small to show up on a reasonable sized graph that showed the range needed. I opted for 20 as reasonable conservative, yet still demonstrable.

    No disrespect wished, but that is why I used the word "cherry-pick". Your logic suggests that any value between 0 and 200 would do, so why not do the math when using 1 meter? If the method is robust, then you should get a result that still fits within the conclusions.

    There are math steps missing in the paper related to the use of Table 1 that I have not yet figured out, so I can at most only perhaps suspect that if you use a value of .001 meter, that you might end up with a totally bogus result inconsistent with the conclusions of the paper.

    >> 4.

    I will read that more carefully later today/tonight, but try to see in particular if what I said above about "radiation flux" vs "radiation energy" makes sense.

  6. >> I was implicitly agreeing that your .75 estimate of 2x CO2 seems as reasonable as anything

    Oops, I should clarify that I am talking strictly about the non-feedback theoretical result. I think the traditional value is around 1 C, which is close to .75 C.

  7. Any updates to this computer model in last 4 years given that the speed of climate science is constantly leaving most of us behind the curve?

  8. I would also recommend that you incorporate the following Chapter 3 of a science textbook. It is titled "Absorption,emission,reflection and scattering".

    http://www.ssec.wisc.edu/library/coursefiles/03_a

    After reading it and proofing it yourself for any errors, if you add that chapter or the equivalent info, to the above presentation, it will provide a more complete explanation of the whole absorption debate which you must agree is the crux of the matter in the false theory of AGM. This would be especially helpful to the non nuclear scientists like myself who found the explanation in your above presentation to be a tad skimpy. Thank you very much for helping to expose the AGM scandal.

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