Do GCMs Model a Flat Earth?

May 8th, 2014 by Roy W. Spencer, Ph. D.

I keep getting dragged into e-mail threads dealing with the claim that climate models (General Circulation Models, or GCMs) don’t even model a spherical Earth….that they apply to a flat Earth where the sun shines 24 hours a day.

I really don’t know where this idea ever came from. It might be because of the Kiehl-Trenberth diagram which illustrates the global-average and time-averaged major energy flows in the climate system, which is often presented over a flat cartoon representation of the Earth’s surface:
kiehl-trenberth

But the 3D computer models of the climate system most assuredly address a spherical, rotating earth, with one side illuminated by the sun.

For example, here’s a YouTube video of one of the GFDL models outgoing infrared radiation, and you can see (1) the land warm and cool during the day/night cycle, as well as (2) rotating weather systems (caused by the Earth’s rotation), and (3) the mid-latitude westerlies which develop in response to the temperature gradients resulting from the curved Earth receiving less sunlight at high latitudes. These are all the result of modeling the system on a spherical, rotating Earth.

Or, if you prefer a map projection that shows the spherical Earth, here’s a different GFDL model showing the ocean surface salinity evolving over time (the model also contains atmospheric circulation systems, which are partly driving the surface currents you see here, but you really can’t effectively visualize more than one or two variables at a time):

(see the very impressive full-res version here).

So, PLEASE, folks…there is plenty to criticize about the climate models. But let’s stick to the stuff that is true, rather than stuff made-up and repeated by people who apparently mistake a cartoon labeled with averages for a full-blown, 3D coupled ocean-atmosphere climate model containing a half-million lines of computer code.

Now, I really loathe having to defend climate models, because I believe they are not yet useful for making climate predictions. But let’s give credit where credit’s due.

The simplest way I’ve found to express climate model shortcomings is this:
Today’s climate models can be tuned to reasonably represent the *average* climate system. But they are, so far, largely useless for what we *really* want to know, that is, how will the climate system change over time?


55 Responses to “Do GCMs Model a Flat Earth?”

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  1. David Gray says:

    Thanks Roy. As usual, an excellent summary of reality.

  2. stephen richards says:

    Today’s climate models can be tuned to reasonably represent the *average* climate system. But they are, so far, largely useless for what we *really* want to know, that is, how will the climate system change over time?

    OR MORE PRECISELY!

    Today’s climate models cannot be tuned to accurately represent the earth’s climate system and they are, so far, utterly useless for what we have to know. That is, how will the climate system change over time, precisely?

  3. Today’s climate models can be tuned to reasonably represent the *average* climate system. But they are, so far, largely useless for what we *really* want to know, that is, how will the climate system change over time.

    And we need to add the reasons why which are the models have incomplete data, lack of accurate data and lack of the data needed to show the proper state of the climate to begin with.

    Until that is dealt with they will never be able to forecast the climate going forward.

    • …but even if we had perfect data and perfect knowledge of the forcings (which we don’t), we STILL might not be able to produce useful predictions. How the system RESPONDS is still largely unknown for some important variables that will determine the outcome(e.g. cloud feedback, precipitation efficiency changes).

      • rossbrisbane says:

        Rubbish Roy, that comment is plain rubbish. See the following uTube. You like sitting with the scorners. I personally would not. http://youtu.be/JrJJxn-gCdo

        • KuhnKat says:

          rossbrisbane,

          when are you going to show us a high fidelity backcast that acutally worked?? Until then don’t even bother giving us any more arm waving about forecasts.

          • rossbrisbane says:

            Sir with due respect the following interview should answer you directly and My reasoning behind calling out Roy Spencer on this: “It should be noted that people do depend on models throughout their lives. Airplanes are designed using models, flown using models, and the weather analyses they use are based on the models that are very much akin to our climate models. The models we use in study of the Earth’s climate system are basically the integrators of our knowledge of how the atmosphere and Earth’s climate work, and they do quite a good job of explaining it.

            “We depend heavily on observations in our analyses. The well-measured CO2 increase exactly fits with our understanding of the carbon cycle as being associated with the burning of fossil fuels and land use change. There is NO other explanation for the increase in CO2. You can’t just say it is natural cycles when the ice core observations show that CO2 levels haven’t been as high as they are now (or even close to as high as they are now) for over 800,000 years.

            “Going further, we have observation-based analyses of the past 2000 years that tell us the changes in climate we have seen in the last 50 years are way outside the norm. Plus they fit with our basic understanding of the greenhouse effect that go back nearly two centuries. There are no natural cycles that can explain all of the observed changes we are seeing on Earth — the atmospheric is warming, the oceans are warming, the land is warming, and the ice is melting. Climate does not change randomly, it changes beyond the normal natural variability due to external forcings. In the past, nature played the major role in those through changes in the solar flux or through large volcanic eruptions. But nature cannot explain what we have seen the last 50 years (the Sun has actually decreased in flux slightly over that time). But the greenhouse effect and the increasing greenhouse gases do fit.

            “What we are experiencing is outside of anything humans have seen on our planet and the only explanation that makes any real sense is that it is due to human actions.”

            Taken from a interview with:

            Donald Wuebbles, the Harry E. Preble Professor of Atmospheric Science at the University of Illinois

          • Dr. Strangelove says:

            “What we are experiencing is outside of anything humans have seen on our planet and the only explanation that makes any real sense is that it is due to human actions.”

            It’s a shame the professor has never heard of the Medieval Warm Period or the Holocene climatic optimum, or he is just in denial of their existence. I’m pretty sure the Vikings farming in Greenland were humans, and there were plenty of humans sweating in the warm climate of 8,000 years ago.

          • rossbrisbane says:

            Sir (addressing alias Strangelove), You make an assumption here. We understand the Medieval Warming Period and LIA only too well. With due respect to the historical evidence: There is absolutely no evidence whatsoever that the warming in this present period or cooling is consistent with the global wide spread phenomena of warming we are seeing today swamping our climate systems. The global scale requires one to get of their fat bums and do some on ground examination. I have and I am stunned, absolutely stunned by the continual denial of global warming on such a scale as we see here on this blog. We will see it if we dare by BEING THERE. I suggest every denier of Climate Global warming get off their bums (stop being armchair ruler and judge of everything) and see the world for THEMSELVES. You do not need a handful of inept sceptics telling you everything is okay. Now sleep easy.

          • Jake says:

            Ross,

            I fail to see your point.

            I don’t dispute that models are used extensively in many applications. But, the models WORK. If someone designed a model to simulate an airplane, and it failed … and they then built the airplane for commercial use and it crashed, what would the expected outcry be? Please answer me that question!

            BUT, we have climate models that don’t back-predict, have failed in their predictions over the past 20 years, and we’re supposed to bow to their validity? To question the models effectiveness also leaves an individual effectively ostracized, wandering the scientific community with a scarlet D across their chest.

            I think that climate models should be an area of study, and that they can continue to improve in their ability to improve our understanding of variables which may (or may not) drive climate variability. To use our current versions to drive policy is on the edge of madness.

          • Dr. Strangelove says:

            Sir ross, you are the denier of MWP who need to get off your fat ass and see the world instead of preaching AGW bullshit. Greenland is one of the coldest place on earth. If Greenland was warm during the MWP, the rest of the world must be warmer. Do you realize how dumb your claim is that MWP was not global? That means the rest of the world was colder than Greenland or did not warm. Why? Where is your evidence to that absurd claim?

  4. But at least if the data were perfect it would be a step in the correct direction.

    However, what you say in the above post is hitting the nail on the head.

    Unknowns all over the place.

  5. Svend Ferdinandsen says:

    Land use and temperature.
    I took a walk with an IR termometer in my garden on a day with 12C air temperature and clear sky.
    Lawn in shadow 10C
    Lawn in sun 25-30C
    Bare and rather dry soil 35-40C

    It shows that land use, vegetation, humidity and changes in it can make big changes in temperature and radiation.
    You can use a radiation change of 5W/m2/K and it gives 150W/m2 in difference between the lawn in shadow and the bare soil.

  6. gbaikie says:

    It seems if you concerned/interested in topic of climate of the world, one use map which focus on the tropics, rather than Mercator projection map.

  7. geran says:

    “Cartoon” is a perfect description. Kinda like the “silly Wabbit” had something to do with it.

    How many things bring a laugh? For example, the “396” from surface radiation goes to the atmosphere. And, “333” gets returned to Earth from “back radiation”. That’s a 94% return when molecules actually radiate in all directions.

    But, in cartoons, you can fall off a 1000 ft cliff and survive.

    Beep-beep!

  8. RW says:

    Roy,

    “I keep getting dragged into e-mail threads dealing with the claim that climate models (General Circulation Models, or GCMs) don’t even model a spherical Earth….that they apply to a flat Earth where the sun shines 24 hours a day.

    You really get a lot of emails complaining about this? That’s curious, as of all the things that might be considered wrong with Trenberth’s depiction, that’s one I wouldn’t expect.

  9. Thanks, Dr. Spencer. An interesting post.
    But, come to think about it, I never understood the Kiehl-Trenberth cartoon. 😉

    • RW says:

      What’s not to understand? It’s pretty straightforward.

      • gbaikie says:

        I can understand 169 watt per square meter average passing thru an atmosphere. Just as I can understand the with sunlight at zenith and clear day one can get 1050 W/m2 of direct sunlight, but include indirect light one gets total
        of 1120 W/m, or another 170 watts of indirect light.
        And in comparison very little IR radiation from Earth could be considered directed light- or one could expect a lot of indirect IR light leaving earth. Whether it is absorbed and re-radiate or simply diffused light.
        But can’t see how the atmosphere is a source of 169 watts of energy. Or it seems clouds can be a source heat which is radiate and the surface has something which has heated up and radiating, but not air, nor a very small percentage gases cooling so as to emit energy

  10. Streetcred says:

    No credit for a perfectly shaped ‘stool’ when what you want is a chocolate cake. You know what I mean?

  11. lewis says:

    RE: Rossbrisbane.

    I consider myself a denier because 1 – I hope the earth does get warmer. Five months in the south – North Carolina – of winter is too long. I prefer a longer growing season. 2 – as is often pointed out – the recent history of the world is one of ice-ages and intermittent warm periods. Which leads to 3 – 20,000 years ago the oceans were much lower than they are today because so much water was tied up in ice on land.

    So the most recent history of the world, prior to man, especially prior to industrialization, is one of climate change – warming. Prior to that – ice.

    Now – suddenly – actually simply because man has devised a way to measure various inputs – some (the AGW scare the world crowd) have decided man is a bad actor and is ruining the world. Bah.

    The climate changes due to various inputs: See Salvatore or volcanoes or asteroids or Milankovitch or Spencer. We are not yet able to determine exactly how or why, but some are trying. In the meantime, I prefer warm over more ice. The important question is: why do you want more ice?

    Also, the fact man is stupid enough to build in the way of mudslides, hurricanes (see Miami and New Orleans or New Jersey), floods or in earthquake zones is only a tribute to his stupidity – it is not an indictment of the climate or the use of hydrocarbons to run an internet.

  12. David South says:

    Question… Regarding the top Youtube video above, here is some text that gives me the impression that the NOAA GCM model is not a sphere…. but instead is a cube!!!

    “To improve this aspect of the model, for this and for
    related projects at relatively high resolutions, we have
    made the following changes to AM2.1: the finite-volume
    dynamical core (Lin 2004) on a latitude–longitude grid
    has been replaced by a finite-volume core using a cubed-sphere
    grid topology (Putman and Lin 2007);”

    “…and the polar Fourier filtering for the fast waves, resulting in improved computation and communication load balance, using 2D domain decomposition on each of the six faces of the cube.”

  13. P. Tuvnes says:

    You are right Dr. Spencer. It is the Kiehl-Trenberth diagram that is the reason why some climate realists say that some climate researchers models the earth as flat and the sun “cold” (sun radiation reduced by a factor of 4 from ca 1360W/m2 at the top of the atmosphere (TOA) to ca 340 W/m2, because the area of a sphere is 4 times the sphere circle). The earth is made flat as a map projection.
    A slightly modified version of the K-T diagram is also included in the IPCC AR5 WG 1 report as fig 2.11, adapted from Wild et al. (2013).
    It is called an “energy budget”, but the energy unit is J, not W/m2 (W = J/s). Length is also not measured as m/s, but as m.
    The KT-figure is actually an attempt to make a snapshot of the radiation intensity flows taking place in one second!
    The sun do not shine on all of the earth in one second, only half of it. Therefore, it is wrong to divide the sun radiation at TOA by 4 in order to even out the radiation in one second all over the earth.
    This simple and fundamental error of the K-T “energy budget” makes the calculation of the effect of “Greenhouse Gases” on “Back Radiation” completely wrong.
    The correct calculation of energy would be to integrate over time and a semi sphere, with maximum sun radiation intensity at the zenith and gradually reduced intensity until zero at the tangent for the radiation, taking into the calculation a round rotating earth with day and nigth.
    Outgoing radiation, however, takes place from the whole sphere.
    This erroneous K-T budget is used to explain the average temperature of the earth to be + 15 C, due to the “Greenhouse effect”.
    Any “imbalance” of this calculation is used to explain change in global temperature and climate.
    Advanced GCMs are not programmed in this way, as you have correctly pointed out. They have other errors, revealed by prof. Judith Curry and Dr. Tim Ball among others, with a much too high sensitivity for CO2 (linear sensitivity to CO2).
    A curiosity is that Anthony Watts has posted that this “flat-earth-cold-sun” concept of climate alarmism is supported by the US leader of the “Flat Earth Society”.

    • MikeB says:

      Idiot

      Did you not read the article?

      • P. Tuvnes says:

        MikeB, you must be a member of the Flat Earth Society. Sorry for having insulted you. Reality can be hard to accept.

    • Go Whitecaps!! says:

      Wouldn’t it be easier to just use the satellite data to get the solar TOA flux as K&T do. Do you have an aversion to data?

      I see that MikeB is his usual charming self.

      • P. Tuvnes says:

        From Wikipedia:
        http://en.wikipedia.org/wiki/Solar_constant
        The solar constant, a measure of flux density, is the amount of incoming solar electromagnetic radiation (the solar irradiance) per unit area that would be incident on a plane perpendicular to the rays, at a distance of one astronomical unit (AU) (roughly the mean distance from the Sun to the Earth). The solar constant includes all types of solar radiation, not just the visible light. It is measured by satellite to be roughly 1.361 kilowatts per square meter (kW/m²) at solar minimum and approximately 0.1% greater (roughly 1.362 kW/m²) at solar maximum.[1]

  14. Go Whitecaps!! says:

    Satellites measure more than just the solar constant.
    http://en.wikipedia.org/wiki/Earth_Radiation_Budget_Satellite

  15. Curt says:

    In thermodynamics, the steady-state case is so vastly easier to analyze than any dynamic case that it is the focus of any introductory course. Handling the dynamic case can be orders of magnitude more complex.

    So if you want to show the major power flows of the earth/atmosphere system in a single cartoon, you really have no choice but to display average values.

    Of course, since key attributes of this system are strongly non-linear, the temperatures resulting from constant power flows at the average values are significantly different from those where there are variations about the average.

    One thing that continually amuses/exasperates me about the Slayers’ objections to the KT diagram is that as soon as you introduce variations of these power flows about the average, the lower the average temperature (and energy level) of the earth would be without greenhouse gases, so the larger the power gap that must be made up.

    • geran says:

      Curt, your rambling incoherent comment would be great fodder for an artist. A good cartoon needs both an artist and hilarity.

      If you can’t draw, just start with “stickmen”. Also, there are courses in basic drawing techniques.

      But, don’t quit spouting your comedy. My favorite line:

      “…as soon as you introduce variations of these power flows about the average, the lower the average temperature (and energy level) of the earth would be without greenhouse gases, so the larger the power gap that must be made up.”

      The dangling science phrases, along with the lack of syntax, adds to the hilarity.

      More, please!

      • Curt says:

        So I dangled a preposition (but English isn’t Latin). My underlying point was obviously too subtle for you, so I’ll break it down into baby steps for you.

        A planetary surface at a constant and uniform -18C (255K) with near-unity emissivity would emit about 240 W/m2 over the surface, just enough to balance the 240 W/m2 absorbed from the sun, if the atmosphere were fully transparent to the far infrared emitted from the surface.

        But, the Slayers object, this is a “flat earth” model, where the earth’s surface receives constant and uniform insolation and is the same temperature over the entire surface. So let’s look at non-uniform cases.

        Because the radiative losses are proportional to the 4th power of absolute temperature, an earth with some temperatures above the average of -18C and some below would radiate more power away than an earth with a uniform temperature of -18C. This would put it out of power balance with the incoming power absorbed from the sun. The result is that the average temperature would be lower than this -18C for a planet without a radiatively active atmosphere, and the bigger the variation over time and distance, the lower the average temperature (and average energy content) would be.

        We only have to look at the moon to see this. Despite its lower Bond albedo, which means it absorbs more solar power, its average temperature is far lower than that of earth, even though it is the same distance from the sun.

        So by objecting to the (over) simplified steady-state analysis as depicted in diagrams like the KT picture Roy shows above, Slayers are digging themselves into a deeper hole. An earth with a transparent atmosphere would be far colder than the steady-state analysis would predict, and a radiatively active atmosphere is the only thing that can explain the difference between that and what we observe.

        • geran says:

          The problem is you start with an assumption–255K.
          Then, you confuse atmosphere temp with surface temp.
          Then, you bring in the Moon for comparison, which has NO oceans, and has no meaningful comparison.
          Then, you conclude that any rebuttal to what you have indicated is just “digging themselves a deeper hole.”

          You must be a climate “scientist”….

          • Curt says:

            Yikes! I overestimated you. I’ll have to break it down into even smaller steps.

            The 255K is no assumption. It is the result of a calculation that any undergraduate in an introductory engineering course is expected to be able to make very quickly. That is: what is the steady-state temperature of a surface with near-unity emissivity radiating to deep space while receiving a constant 240 W/m2 input?

            The next thing you fail to understand is the trivial point that a transparent atmosphere has no capability to remove energy from (or add energy to) such a surface on an ongoing basis. This is because it has no other source or sink for energy. In the steady-state (“flat earth”) case, the atmosphere at the surface would be at the same temperature as the surface. In a day/night case, it could absorb some energy during the day and return some energy at night.

            The moon is a very good comparison to the hypothetical uniform-temperature body. The fact that it has no oceans makes it even better. The moon rotates slowly enough, and the depth of the lunar surface directly affected by the day/night cycle is shallow enough, that the measured temperatures, especially during the day time, to what would be expected from the solar power flux input and the resulting radiation to deep space. And on average, this is much lower than 255K.

  16. Nicias says:

    The surface of the earth in GCM are made of flat tiles with different altitude.
    There’s no mountain, no tree etc … no vertical structures (and yet they still cool the earth to …).

    It’s a flat earth in a sense.

  17. geran says:

    Curt says: “The 255K is no assumption. It is the result of a calculation that any undergraduate in an introductory engineering course is expected to be able to make very quickly. That is: what is the steady-state temperature of a surface with near-unity emissivity radiating to deep space while receiving a constant 240 W/m2 input?”
    >>>>
    Uh, Curt, 255K is the correct answer for the situation you describe, but that situation does not compute with Earth. “They” divide Earth’s insolence by 4 to achieve the “240”. That is one of the errors. (They confuse photon flux with energy–not the same animals.)

    Curt says: “The next thing you fail to understand is the trivial point that a transparent atmosphere has no capability to remove energy from (or add energy to) such a surface on an ongoing basis. This is because it has no other source or sink for energy. In the steady-state (“flat earth”) case, the atmosphere at the surface would be at the same temperature as the surface. In a day/night case, it could absorb some energy during the day and return some energy at night.”
    >>>>
    I think you got so confused that you just made one of my points for me.

    Curt says: “The moon is a very good comparison to the hypothetical uniform-temperature body. The fact that it has no oceans makes it even better. The moon rotates slowly enough, and the depth of the lunar surface directly affected by the day/night cycle is shallow enough, that the measured temperatures, especially during the day time, to what would be expected from the solar power flux input and the resulting radiation to deep space. And on average, this is much lower than 255K.”
    >>>>
    If you seek a hypothetical uniform-temperature body, with no oceans, why not just use a hypothetical “ideal” absorber. A flat “ideal” absorber, receiving the same solar insolence as Earth (after albedo, approx. 960 W/m2) would produce a temp of 361K. But, that would be hotter than Earth, huh?

    In the big picture, if you contend that the K-T energy diagram (cartoon) is valid science, then we will just have to “agree to disagree”. Have a great week!

  18. Curt says:

    Good grief! You don’t even understand basic high school geometry! The earth intercepts the sun’s radiation over a circular area of Pi*R^2, where R=6400km is the radius of the earth. At the earth’s distance from the sun, the power flux density of this radiation is about 1370 W/m^2.

    But the spherical earth’s surface area is 4*Pi*R^2, with 2*Pi*R^2 facing toward the sun (day) and 2*Pi*R^2 facing away from the sun (night). So it is a simple matter of high school math to compute that, with an albedo of 0.3 (which you use), the average power flux density received on the earth’s surface, over time and area, is:

    1370 * 0.7 * (Pi*R^2) / (4*Pi*R^2) = 240 W/m^2

    Why is this number important? Because if you multiply this number by the area of the earth (4*Pi*R^2), that is the power input to the earth. (Not 480 or 960.) To stay at a constant energy level, the earth must reject the same total amount of power to space, which it can only do by its thermal radiation. And no one, alarmist or skeptic, thinks the earth is out of balance by more than about 1 W/m^2.

    Now, is this a claim that the effect of this average insolation (not insolence…) is the same as if all of the earth’s surface received this constant average value? Not at all! But the interesting thing is (and this was my original point) that this hypothetical constant case leads to the highest average temperature/energy level. Any variations lead to lower average results, and the bigger the variations, the lower the resulting temperatures/energy levels.

    You bring up the case of the surface receiving 960 W/m^2 and calculate a resulting temperature of 361K. But this insolation level only occurs for a small percentage of the time in the equatorial region at noon. It is not a steady-state value.

    Now, on the moon, this surface temperature is reached in the equatorial region at lunar noon. But the moon is rotating relative to the sun at 3% of the rate of the earth, and it has no oceans or atmosphere to move energy over the globe. This high temperature is not steady state. Even on the moon’s equator, the nighttime temperature falls well below 100K in the lunar night, as it radiates to deep space. If you calculate the average surface temperature of the moon, it is well below 255K, even though its albedo is lower than that of the earth.

    You seem to think that my point that a transparent atmosphere cannot steadily transfer power to or from the surface is ridiculous, but you offer no counter-analysis. If such an atmosphere is steadily absorbing power from the surface, where does it reject that power?

    In your “big picture”, what good is the K-T diagram? It does give reasonable approximations of globally averaged power flux density values. But these are not constant values over time and area (it does not claim they are), and analysis that assumes these values are constant will lead to significantly wrong results. But one key way these results based on constant power flux density values will be wrong is that they will result in temperature values that are too high.

    • geran says:

      Curt, you make all the same mistakes as the IPCC. Clear your mind of the climate “science” confusion, go back to the books, and start all over.

      You can NOT average the solar insolation (and thank you for correcting my mistake, it was late night!). The cartoon divides 960 by 4 to get the 240. They indicate, as do you, that this is because of the ratio of area of circle to area of sphere. Their geometry is correct, but their quantum physics is WRONG. The 960 is a “photon flux”, not energy. It is confusing if you are rusty on quantum physics. Think of it this way: A photon is NOT energy, it only transports energy. Upon impact with an absorber, the energy is released, and the photon is annihilated. (A photon can be created or destroyed, but energy cannot.)

      And, that is why the 255K is WRONG. They (you) use the right equation, but the wrong value for photon flux. GIGO!

      Your other points I have already addressed, but it must have been late at night for you also and you did not understand them.

      This is over for me, so you get the last word.

      • Curt says:

        geran: I am using completely standard engineering thermodynamics and heat transfer analysis, plus standard statistical mechanics.

        And you are completely wrong here. We are talking about POWER flux and POWER flux density here. The units are WATTS and WATTS per square meter. Watts are units of power, which is the rate of energy transfer. That should be a clue.

        This radiative energy is carried by photons, and if a photon is absorbed, the internal energy of the absorber is increased by the energy carried in the photon.

        We know the energy quantum carried by each photon: e = h * v. The power flux values cited are computed by integrating over all of the relevant wavelengths and over the relevant area.

        These are very standard calculations used every day in engineering heat transfer analysis. Until you understand these very basic concepts, you will remain in total confusion.

        • geran says:

          (Sheesh, I am trying to get out of this, but I keep getting drawn back in!)

          Curt, I will try one more time. Consider a flat plate that is one sq meter in area. The plate is a perfect absorber. Let solar flux of 960 W/m2 hit the plate perpendicularly. I am convinced you can calculate the temp of the plate once equilibrium is reached. You will get 361K.

          Now, divide the plate into 4 equal areas–not touching each other. Now, would you divide the solar flux by 4? That would give you the IPCC’s 255K. And, that would be WRONG. (Of course, the flux does not change, so the 361K would be reached REGARDLESS of the change in geometry.)

          You cannot arbitrarily divide a proton flux. It affects the result, and is NOT properly applying the S-B equation.

          If you think algebra works with proton flux, consider multiplying by 4. (Just increase the plate to 4 sq. meters.) That would give you a flux of 3840 W/m2, which calculates to a temp of 510K!!! (Which is WRONG.)

          Hope this helps.

          • Curt says:

            This is beyond ridiculous. You don’t even understand the issues at play. You obviously have never taken a heat transfer course.

            In your first example, given your information, all we have established is that the plate is absorbing 960W/m^2 * 1m^2 = 960W. We cannot calculate the steady-state temperature until we know the details of how it can lose power.

            You are completely missing the point of the “divide by 4” concept, which is that 3/4 of the surface is not receiving this solar flux. So take your 1m^2 plate at make it into a tetrahedron, with each of the 4 surfaces having 0.25m^2 surface area. One surface is receiving the 960W/m^2 solar flux perpendicularly, which means the other three surfaces are receiving none.

            So we have established that the body is receiving 960W/m^2 * 0.25m^2 = 240W of power from the sun. To set further conditions that allow us to solve for a steady state temperature, let’s also say that this body is in the vacuum of space, and away from any other power sources. The body is of a highly conductive metal, so it is a uniform temperature throughout.

            For the body to be at steady state, it must emit 240W from its 1m^2 surface area. It is radiating to deep space, which can be considered to have an effective blackbody radiative temperature very close to 0K. So to solve:

            240 = 5.67×10^-8 * T^4 * 1.0

            T = (240 / 5.67×10^-8)^(1/4) = 255K

            Now take a sphere with 1m^2 surface area under the same conditions. Its cross-sectional area is 0.25m^2, so it absorbs 240W of power from the sun. For the steady-state temperature, the calculations are identical to those above for the tetrahedron.

            Your examples miss the whole point that on some of the surfaces, the sun doesn’t shine. I can’t believe that I have to point out that the sun doesn’t shine at night, so half of the earth’s surface is not receiving any solar input.

            (Oh, in your first problem, the most reasonable solution is to compute that the plate absorbing 960W on its front surface is radiating from both front and back – 2m^2, so radiating 480W/m^2, yielding a steady state temperature of

            T = (480 / 5.67×10^-8)^(1/4) = 303K

  19. Kermit42 says:

    Well, I wonder.
    From my point of view the K/T cartoon has some really strange physics.

    First, if the values of absorption at surface were right the surface has to have an average temperature of about 305 Kelvin.
    (161W/m² short wave + 333W/m² long wave = 494W/m² total –> 305K)

    Second, the three kinds of heat transport (radiation, evapotranspiration and conduction) share the same energy pool. To get a total of 493W/m² output requires these 305K (or more, assumed an emission coefficient slightly smaller than 1). Therefore the cartoon would only be of importance if radiation would be the only kind of heat transport.

    Third, even if one assumes the values of energy transmission were correct, the cartoon is only valid for about 29% of the Earth’s surface. For the remaining 71% it cannot be valid due to the very different radiative behaviors of water regarding short wave and IR. Here another physical quantity has to be taken into account: time. Time required for the transport of absorbed solar heat to the water surface.

    Finally, due to the absorption by plants and the transformation to chemical energy a balanced energy budget would cause cooling.

    So, what have I missed?

    • Curt says:

      Kermit: You have missed a fundamental understanding of what the SB equation tells you. It tells you for a given temperature, how much a blackbody (or if you add an emissivity term, you make it more general) EMITS in terms of power flux density. So a blackbody at 305K EMITS a power flux density of about 494 W/m^2.

      You have simply tallied up how much the surface is absorbing. Different animal. Now, IF the surface could only lose power by radiative means, then in the steady state, it would have to radiatively emit 494 W/m^2, and this would result in a steady-state temperature of 305K.

      But the earth’s surface also loses power by conductive/convective means and by radiation, as shown in the K-T diagram. It shows these at 97 W/m^2, so when added to the 396 W/m^2 of radiative losses, it (almost balances the incoming power flux). A blackbody at 289K would emit this much (or a graybody of slightly higher temperature.

      If the surface were at 305K in this scenario, it would radiatively emit 494 W/m^2, plus lose over 97 W/m^2 through non-radiative means. This is more than the incoming 494 W/m^2, so it would cool.

      As to land versus ocean, there really is very little difference. LWIR is absorbed right at the surface of both solid and liquid objects, as there are virtually no natural substances that are transparent to these wavelengths.

      Solar radiation can penetrate ocean water to a depth of a few meters, occasionally tens of meters, but not that significant on a global scale. The thermal mixing in the oceans reduces transients, but does not affect the average much.

      Photosynthesis converts a few percent of the sunlight hitting a plant into chemical energy, but this energy is released when the resultant substances decompose. Over a year, these two effects are in very close balance.

    • RW says:

      Kermit42,

      The key point you are missing is — for a state of energy balance — all power in excess of 396 W/m^2 entering the surface must be exactly offset by power in excess of 396 leaving the surface, and that the surface radiates about 396 W/m^2 as a consequence of its average temperature (about 289K). In reality the surface is only gaining joules at a net rate of about 396 W/m^2 to replace the 396 W/m^2 radiated away. The joules in excess of 396 W/m^2 entering and leaving the surface are perpetually ‘in limbo’, i.e. they are neither adding joules or taking away joules from the surface — they are just joules circulating from the surface to the atmosphere and back to the surface.

  20. RW says:

    BTW, few ever mention what is arguably the biggest flaw in the Kiel-Trenberth diagram. That is, the non-radiant flux of 97 W/m^2 from the surface to the atmosphere is total flux leaving the surface, which is then exactly offset at the surface by 332 W/m^2 of so-called ‘back radiation’ (really just downward LW). There is no up minus down component depicted for the non-radiant flux from the surface to the atmosphere. The diagram suggest an amount equal to it (97 W/m^2) is returned to the surface solely as direct radiant power from the atmosphere.

  21. geran says:

    Curt says: This is beyond ridiculous. You don’t even understand the issues at play. You obviously have never taken a heat transfer course.
    >>>>>
    WRONG, both thermo and heat transfer courses, as well as QUANTUM PHYSICS.

    Curt says: In your first example, given your information, all we have established is that the plate is absorbing 960W/m^2 * 1m^2 = 960W. We cannot calculate the steady-state temperature until we know the details of how it can lose power.
    >>>>>
    WRONG, I specified “at equilibrium”.

    Curt says: “You are completely missing the point of the “divide by 4″ concept, which is that 3/4 of the surface is not receiving this solar flux. So take your 1m^2 plate at make it into a tetrahedron, with each of the 4 surfaces having 0.25m^2 surface area. One surface is receiving the 960W/m^2 solar flux perpendicularly, which means the other three surfaces are receiving none.
    >>>>>
    WRONG, you are missing the point that photons do not heat surfaces that they do impact.

    Curt says: So we have established that the body is receiving 960W/m^2 * 0.25m^2 = 240W of power from the sun…(blah, blah, blah about the IPCC nonsense)…
    >>>>>
    WRONG, I already gave a simple example why the divide-by-4 thingy is WRONG. You choose to obfuscate my example.

    Curt says: Your examples miss the whole point that on some of the surfaces, the sun doesn’t shine. I can’t believe that I have to point out that the sun doesn’t shine at night, so half of the earth’s surface is not receiving any solar input.
    >>>>>
    WRONG, my example was clearly about explaining S-B calculation to you. You choose to bring in the “red herring” about the false notion that I don’t know about night and day.

    Curt says: (Oh, in your first problem, the most reasonable solution is to compute that the plate absorbing 960W on its front surface is radiating from both front and back – 2m^2, so radiating 480W/m^2, yielding a steady state temperature of
    T = (480 / 5.67×10^-8)^(1/4) = 303K
    >>>>>
    WRONG, you WANTED the plate to be able to radiate from both front and back, to avoid the obvious result that you are WRONG. (Try the calculation again, if no radiation leaks out the back.) Once again, you try to confuse a simple example.

    Well, you were WRONG on each of your points. And, I have made a sincere effort to make the examples clear and simple, so now it is definitely time for me to move on. Have a great weekend.

  22. Curt says:

    You continue to sink lower and lower.

    “You are missing the point that photons do not heat surfaces that they do impact.”

    My jaw just dropped at that one. You cannot be serious. If you have taken courses, you have not understood anything in them. What happens to the e=h*v energy in the photon when it is absorbed.

    In your example, I stated that you did not provide enough information to calculate a temperature. “At equilibrium” is not enough information.

    There is only one very particular set of circumstances under which the answer you want is obtained:

    * All other surfaces are completely insulated
    * There are no conductive, convective, or evaporative transfers
    * The surface is radiating to absolute zero

    The bigger point is that you are interpreting the SB equation exactly backwards. It calculates the radiative flux of a blackbody given its temperature at any time, whether at equilibrium or not. There is only one very special set of circumstances (listed above) where it just so happens that the equilibrium temperature as a function of received radiative flux has the same mathematical relationship.

    Your “examples” of cutting the plate in 4 are complete strawmen, as I showed. You have all the plates still facing the sun. That is not what the divide-by-4 issue is about. I repeat that it is a trivial matter of high school geometry, which you are incapable of understanding even after it is repeatedly pointed out to you, to see that if only 1/4 of the surface is facing the source of the supplied radiation, the flux density, and therefore the power received, averaged over the entire surface, is 1/4 of the flux density received by the surface directly facing the source.

    I thought the tetrahedron example would make it easy to understand. What part of my analysis do you disagree with? (“Blah, blah, blah” doesn’t tell me anything.) Do you think that I am wrong that the 0.25m^2 facing the source absorbs 960*0.25=240W of power? Do you think that the total 1.0m^2 of surface area radiating to the ~0K of deep space (the implicit condition for your example) would reach a different temperature than what I calculated?

    If your heat transfer professor gave you this problem, what would your solution be for the equilibrium temperature of the tetrahedron? This is just a slight modification of the problem you gave me. Show your work.

    • Don says:

      “You are missing the point that photons do not heat surfaces that they do impact.”

      If a photon is absorbed then the energy it contained is incorporated into the chemical bond involved and raises that electron-bond to the next permissible energy level. In the IR, for gases like CO2 and H2O, these are so-called v-level transitions, or vibrational energy levels. For CO2, the vibration involved is asymmetric stretching, which produces a moving charge in the quantum field. Because the time constant for that gas molecule to strike another (at STP) is orders of magnitude faster than the time for quantum relaxation, it is highly probably that that CO2 molecule will transfer that new photon energy to another molecule through kinetic collisions. There is a distribution of kinetic energies among molecules at any temperature. And those molecules are constantly transferring energy among kinetic motion and bond energies. Only if/when that original C-O bond regains the original energy it acquired from the IR photon can it emit an IR photon.
      This is how photon absorption heats materials.

      In the case of a solid (or liquid like the ocean), any O-H bond has H-bonding with many surrounding molecules. This gives greater flexibility to distribute energy acquired by absorbing an IR photon and even makes that absorption more likely.

    • geran says:

      (Okay, this really is my LAST effort. Curt does not want to understand. He keeps twisting, confusing, and distorting what I am trying to explain.)

      geran said: “You are missing the point that photons do not heat surfaces that they do impact.”

      Curt says: My jaw just dropped at that one. You cannot be serious. If you have taken courses, you have not understood anything in them. What happens to the e=h*v energy in the photon when it is absorbed.
      >>>>>

      Curt, see how you try to distort and confuse what I am trying to explain? My wording was “IMPACT”, not absorb. You tried to insinuate that I did not know that an absorbed photon releases energy. You have to twist, distort, and confuse what I write in your effort to mislead yourself, and others, for some unknown reason. I am trying to help you out of your confusion, and you are trying to purposely confuse my efforts.

      My point was that not all photons in a photon flux are converted into heat energy upon impact. Some are REFLECTED. You do not understand photons. You think that the S-B equation is just another algebra problem. I tried to explain it is not that simple. Quantum physics is not always intuitive. You have to know how to deal with the physics of photons. Before you do the math, you have to understand the physics, or you will get the wrong answers. (For interested readers, see example below.)

      I am convinced that you are not here to improve your understanding of science, but rather, you have another agenda. My examples are clear and simple. You attempt to confuse them. But, if I am wrong, and you really want to increase your understanding, I suggest you study some of the early experiments in photons. The science is over 100 years old, so it is not something new. However, it is not intuitive, and it is somewhat obscure, as people cannot “see” photons. The early experiments are easy to understand and they help explain how photons behave. Hey, if I can understand the experiments, you can too!

      >>>>>>>>>>
      Example of how you must understand the science BEFORE applying the math.

      Problem: What is the average length of the sides of 3 perfect cubes of volumes 8, 27, and 125?

      Solution 1: Average of 8, 27, and 125 is 53.3. Taking the cube root of the average yields 3.76.

      Solution 2: Take the cube roots of each cube — 2, 3, 5. Now, average the sides–3.33

      The correct answer, of course, is Solution 2. Solution 1 is WRONG because taking the cube root of a number (or raising it to a power, as in the S-B equation) is a NONLINEAR operation. You cannot work a nonlinear equation arbitrarily.

      HINT: The S-B equation is NONLINEAR.

      (Will Curt understand the point of this, or will he, again, try to confuse my simple example?)

  23. Tim Folkerts says:

    geran says: “Curt does not want to understand. He keeps twisting, confusing, and distorting what I am trying to explain.”
    You two seem to be talking past each other quite a bit. However, for the most part, I am finding Curt easier to understand and more accurate. My two cents …

    *** geran says: “My point was that not all photons in a photon flux are converted into heat energy upon impact. Some are REFLECTED.”
    This seems an odd point to focus on at this point. The entire discussion up to this point was assuming that 960 W/m^2 was indeed the absorbed power (and something like 1360-960 = 400 W/m^2 was the reflected portion.)

    *** Both of you are talking about “at equilibrium”, but none of these systems are truly at equilibrium. All involve objects at different temperatures with heat flowing in and heat flowing out, which could not happen at equilibrium. You both really mean “steady-state”, with equal heat in and out of the system.

    *** geran says: “Problem: What is the average length of the sides of 3 perfect cubes …
    I think everyone involved at this point realizes there are many possible solutions for the temperature around an object that is absorbing 960 W/m^2 over some part of its entire surface.

    Even assuming no conduction or convection, and assuming the object is a blackbody radiator, there are still many steady-state solutions. At the risk of “confusing” people with concrete examples with specific conditions …

    * A flat plate with one side facing the sun with no thermal conductivity would be ~ 361 K on the front and ~ 0 K on the back (for an average of ~ 180 K).
    * A flat plate with perfect thermal conductivity so that the front and back are the same temperature would be ~ 303 K on both surface (for an average of 303 K).
    * A flat plate with intermediate thermal conductivity would be somewhere in between (for an average between 180 K and 303 K)

    * a tetrahedron with one side facing the sun with no thermal conductivity would be ~ 361 on the front face and ~ 0 K on the other 3 surfaces (for an average of ~ 90 K)
    * a tetrahedron with perfect thermal conductivity would be 255 K on every surface (for an average of 255 K)
    * a tetrahedron with intermediate thermal conductivity would be somewhere in between (for an average between 90 K and 255 K).

    * I don’t feel like calculating the actual temperature distribution for a non-rotating sphere with no thermal conductivity. The point facing the sun would be 361 K; the entire back hemisphere would be 0 K. The average would be below (360/0)/2 = 180 K, but above 90 K).
    * perfectly conducting sphere would be 255 K everywhere.
    * A real, non-rotating sphere would have an average between 90 K and 255 K).

    * a real *rotating* sphere would have a higher average than the non-rotating sphere, but it cannot be warmer than 255 K.
    * a real *rotating* *non-blackbody* sphere would be a bit warmer. (Roughly, a drop of 1% in the emissivity will raise the temperature about 0.25%).

    geran’s flat, one-sided, 361 K example is indeed “simple”, but is so overly simple as to be useless. Rather than Curt’s example being “wrong” because he wants the object to radiate from all surfaces, his two-side plate (or his tetrahedron) is a much more appropriate example for understanding a sphere like the earth.

  24. Curt says:

    Geran: You have an amazing combination of a lack of ability to apply the knowledge you have to new situations and lack of ability to express yourself.

    On your “impact” statement: We were talking about idealized blackbodies, which by definition absorb all photons that “impact” them. And in all real substance, which are between perfect absorbers and perfect reflectors, some of the photons will transfer heat to the body. You made no such qualification. So I still maintain that statement as you made it is just wrong. (And yes, I understand the the energy in reflected photons is not absorbed – but you did not say that.)

    And I still believe you do not really understand the issues in play. The whole point of my original post was that, BECAUSE OF the non-linearity of the S-B equation, a body with varying temperatures, in order to radiate away the same total amount of power (to stay in steady state with the constant power received) must have a lower average temperature (and therefore lower energy level) than a comparable body with uniform temperature.

    But we really have not been able to discuss that point because of some of your fundamental misunderstandings and your strawman examples. To wit:

    You cannot just plug in the incoming radiative flux density to a body (even if that is the only power input) into the SB equation and solve for the steady-state temperature of the body. You solve for this temperature by using the energy balance equation including all power inputs and power outputs. As I pointed out, the only case where the answer is the same is the case where the only power output is radiation from the receiving surface to the absolute zero of deep space.

    As I pointed out before, your example of splitting up a 1m^2 blackbody plate into 4 pieces, all still receiving the 960W/m^2 input, is completely irrelevant to the discussion at hand. (For the record, I agree that the temperature would not change.) As I said, the example that is relevant is the one where 3 of these pieces are receiving no power input from outside the system, but are in thermal contact with the 1 pieces that does receive this power.

    I will take an additional step this time, first considering the case where the pieces are not in thermal contact with each other. The one piece receiving 960W/m^2 and radiating from that surface only to absolute zero would have a steady-state temperature of

    T = (960 / 5.67×10^-8)^(1/4) = 361K

    The other three pieces would have:

    T = (0 / 5.67×10^-8)^(1/4) = 0K

    In the next case, which I presented earlier, the four pieces are in thermal contact, and we’ll consider that the conductivity is good enough that the whole body is isothermal. So now, the body takes in 960*0.25=240W from its “front” and has to radiate this away from all 4 surfaces with a total area of 1m^2. In the uniform case, this means 240W/m^2 from the whole surface, and we get:

    T = (240 / 5.67×10^-8)^(1/4) = 255K

    Note very carefully that 255K is NOT 1/4 of 361K — because of the non-linearity of the S-B equation.

    And, to go back to my original point, if the body were not uniform in temperature, to reject the same total amount of power, the average temperature would be lower than 255K, again because of the non-linearity of the S-B equation.

    I believe your fundamental error comes from the fact that you don’t realize that the 1st Law conservation of energy equations are mathematically linear, unlike the S-B equation. Once you get to energy values, you can add and subtract them and even average them.

    • Tim Folkerts says:

      Kinda cool how we get the same basic results using similar logic when starting from the same fundamental science. 🙂

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