What Causes the Greenhouse Effect?

June 13th, 2015 by Roy W. Spencer, Ph. D.

I’ve had a request to (once again) go through an explanation of the (poorly-named) Greenhouse Effect (GHE). Hopefully there is something which follows that will help you understand this complex subject.

The greenhouse effect usually refers to a net increase in the Earth’s surface temperature due to the fact that the atmosphere both absorbs and emits infrared radiation. (Our miniscule enhancement of the natural greenhouse effect with carbon dioxide emissions, and its possible role in global warming, is a separate issue).

This GHE temperature increase is frequently quoted as being around 60 deg. F, thus keeping the Earth from being an ice planet, since its average surface temperature is somewhere around 59 or 60 deg. F.

This 60 deg. F warming attributable to the GHE is actually incorrect; the greenhouse effect on surface temperature, if left to its own devices, would actually be at least twice that strong…more like 140 deg F average surface temperature…but most of that theoretical surface temperature rise is short-circuited by convective heat loss from the surface caused by convective air currents, in turn caused by the greenhouse effect, which also largely creates the weather we experience.

That’s right – without the greenhouse effect, we would not have weather as we know it. The greenhouse effect, energized by solar heating, creates weather.

The GHE is somewhat controversial among some skeptics, probably because we can’t “see it” the way we can see visible sunlight and the resulting heating of surfaces sunlight falls upon – a rather non-controversial cause-and-effect process. It instead involves infrared (IR) light, which we cannot see, but which is an essential part of the energy flows in our climate system…and in most other systems that generate heat. You can actually feel if it is sufficiently strong (e.g. radiant heat from a stove or fire).

I must preface the following discussion with this: The temperature of any object represents a balance between energy gained and energy lost by that object. Temperature is an energy balance issue. Unless phase changes are involved (e.g. melting ice), if more energy is gained than lost, temperature goes up. If more energy is lost than gained, temperature goes down. Understanding this is fundamental to understanding weather and climate, as well as the following discussion.

The atmosphere contains “greenhouse gases” (GHGs), which means gases which are particularly strong absorbers and emitters of IR radiation. In the Earth’s atmosphere, the main GHGs are water vapor and carbon dioxide. Absorption and emission of IR go together because anything that is a good absorber of IR is also a good emitter, although in general the rates of absorption and emission are not the same since absorption is mostly temperature-independent but emission is very temperature-dependent.

In the classical Kiehl-Trenberth global energy budget diagram, the energy flows I have marked with an “X” would not exist without GHGs:

greenhouse-gas-and-KT-diagram

Now, recall I said that temperature is a function of rates of energy gain and energy loss. Thus, those energy flow arrows marked with an “X” in the above diagram represent huge flows of energy which can affect temperature, if they really exist.

So, let’s now think through what happens as sunlight enters the climate system. As the Earth’s surface absorbs sunlight it warms up. As it warms up, it emits more and more IR energy, limiting its temperature rise (remember “energy balance”?).

If the atmosphere could not intercept (absorb) any of that surface-emitted IR energy, the energy would readily escape to outer space and as a result it has been estimated that the Earth’s average surface temperature would be only about 0 deg. F. But we really don’t know exactly because there would be a lot more ice, which would reflect more sunlight, which would make temperatures even colder. Also, we have no idea why kinds of clouds would exist under those conditions. Suffice it to say the Earth would probably be too cold for most life as we know it to survive.

But the atmosphere DOES absorb IR energy. The IR absorption coefficients at various wavelengths, temperature, and pressures have been measured for water vapor, CO2, etc., in laboratories and published for decades.

This absorption means the atmosphere also EMITS IR energy, both upward and downward. And it is that DOWNWARD flow of IR energy (sometimes called “back radiation”) which is necessary for net warming of the surface from the greenhouse effect.

(Technical diversion: This is where the Sky Dragon Slayers get tripped up. They claim the colder atmosphere cannot emit IR downward toward a warmer surface below, when in fact all the 2nd Law of Thermodynamics would require is that the NET flow of energy in all forms be from higher temperature to lower temperature. This is still true in my discussion.)

Now, some will claim the atmosphere’s decreasing temperature with height is also necessary for the greenhouse effect to occur. While this is true, the decrease in temperature with height in the troposphere is ultimately caused by the greenhouse effect itself.

You see, as long as an atmosphere (it doesn’t matter from which planet) has greenhouse gases, the temperature will decrease with height. Without convection, the temperature would decrease drastically with height…the so called “pure radiative equilibrium” case, first demonstrated by Manabe and Strickler (1964). The net effect of GHGs is to strongly warm the surface lower atmosphere temperature, and strongly cool the upper atmosphere temperature, compared to if those gases did not exist. The GHE makes the atmosphere so unstable that convection – weather – results, which restores the atmospheric temperature lapse rate to somewhere between dry adiabatic and moist adiabatic.

Remember, without greenhouse gases, the upper atmosphere could not lose the energy it accumulates from all sources, and would stay warm, and the atmosphere would not destabilize and cause convective overturning (weather).

This net result is not intuitively obvious. I sometimes use the (admittedly imperfect) analogy of insulation in a house in winter (even though heat conduction is a different physical process from radiation). Given the same rate of energy input into the home by its heating system, addition of insulation slows the net rate of heat flow from the warmer interior to the cold exterior, causing higher temperatures inside and lower temperatures outside, compared to if the insulation did not exist.

Again, temperature is the result of energy gain AND energy loss. If you reduce the rate of energy loss, temperature will rise…even if the energy input is the same. Extremely high temperatures can even be created with very little energy input…even from a tiny battery…if you can reduce the rate of energy loss to near zero. You cannot say anything about temperature based upon the rate of energy input alone, any more than you can say what the average level of a lake will be based upon the rate of water input alone. It just ain’t physically possible.

Analogous to insulation in a heated home, greenhouse gases reduce the net rate of infrared energy transfer from the surface and lower atmosphere to outer space, causing the surface and lower atmosphere to be warmer, and the upper atmosphere to be colder, than if greenhouse gases did not exist.

Since the effect is not entirely intuitive, years ago we programmed up the equations ourselves in a 1-D radiative-convective model for me to be convinced this is what actually happens. When the model is initialized with global average sunlight and atmospheric greenhouse gas concentrations, from any initial temperature profile you want (even absolute zero), it eventually equilibrates to the observed average vertical temperature structure of the atmosphere.

And I suppose it is the non-intuitive nature of the process (I required a model demonstration to finally believe it) that breeds so much controversy and alternative ideas. I get that.

Now, I know that this post will cause a few people (you know who you are) to object with hand-waving arguments involving technical jargon that what really happens is something different. But until they put their ideas in the form of physical equations based upon known (and laboratory-measured) processes, which conserve energy, in a time-dependent model that also produces the observed average temperature profile of the atmosphere, I will not believe them.

What those people need to do is go read a book on atmospheric radiation, say Grant Petty’s A First Course In Atmospheric Radiation. I know Grant, and he is a brilliant and careful scientist. If you disagree with him (and the many other experts who agree with him), you’d better have some pretty good evidence to back your case up.

The bottom line, then, is the Greenhouse Effect, due mostly to greenhouse gases, is largely caused by the fact that the atmosphere emits IR energy downward, the so-called “back radiation”. This single component of the whole GHE process basically then determines all of the other features of the greenhouse effect and leads to net GHE warming of the Earth’s surface.

You can measure the greenhouse effect yourself with a handheld IR thermometer pointed at the sky, which measures the temperature change caused by a change in downwelling IR radiation. In a clear sky, the indicated temperature pointing straight up (“seeing” higher altitudes) will be colder than if pointed at an angle (measuring lower altitudes). This is direct evidence of the greenhouse effect…changes in downwelling IR change the temperature of a surface (the microbolometer in the handheld IR thermometer). That is the greenhouse effect.

If I’ve make a mistake in the above, I’ll fix it. I realize some might not like the way I’ve phrased certain things. But I’ve been working in this field over 20 years, and the above is the best I can do in 1-2 hours time. From some of the objections you will see in the comments, you will find it is a complex subject, indeed.


524 Responses to “What Causes the Greenhouse Effect?”

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  1. angech says:

    Feel your comment about the upper atmosphere being cooler somehow misleads an understanding of the energy processes occurring
    It suggests that the outer atmosphere is cold implying that the earth is not radiating as much heat out as your explanation otherwise asserts.
    The outer atmosphere may be colder and the lower atmosphere warmer but the amount of heat radiated out, the amount of heat passing through this colder area is still the same is it not?
    I presume without knowing that this can only be the case if the upper atmosphere is deficient in GHG like water vapor and CO2 hence the outgoing radiation passes easily through the thinner and colder outer atmosphere without inducing any extra heating up.
    If this is indeed the case , and the lower more GHG rich atmosphere is absorbing and emitting more IR in all directions, then the atmosphere is effectively a shell of hotter air whose heat emmissions outwards equals that incoming. Why dos this have to cause a TOA imbalance if as you have already said the heat has just been distributed between the different layers differently, implying there is no need to shift the heat into the oceans as it has already been moved down into the lower atmosphere?

    • Given the same amount of absorbed solar energy coming in, the amount of IR escaping to space at the top of the atmosphere will indeed be the same no matter how many greenhouse gases there are (assuming the system is in equilibrium).

      What instead changes is how energy is exchanged between layers, which then alters the vertical temperature distribution.

      I’m not sure what you asking about the oceans. If the surface is undergoing net heating from increasing CO2, the ocean will as well…just more slowly due to vertical mixing.

      • Mike Flynn says:

        Dr Spencer,

        The surface of the Earth has cooled since its creation. That seems obvious. Therefore, the surface has lost more energy than it has received from all sources.

        This includes heat from the Sun. If there was energy balance, the Earth would never cool.

        Geophysicists have different figures, but around one to three millionths of a degree per year seems to cover it.

        No energy balance at all. The surface heats during the day, cools at night. Winter is cooler than summer, daytime clouds cool things, nighttime clouds slow the rate of cooling.

        I won’t ask you what you think, obviously. Sorry to keep plugging away.

        • jerry l krause says:

          Hi Mike,

          Believe we have touched bases before. And I do not mean to start any disagreement. But it is not obvious to me why the earth has cooled since its creation. Is it because of the volcanic activity (and hot springs) which still occurs, bringing sensible heat to the surface where it is radiated to space?

          You also wrote: “No energy balance at all. The surface heats during the day, cools at night. Winter is cooler than summer, daytime clouds cool things, nighttime clouds slow the rate of cooling.” What intrigues me is that you referred to clouds. The word cloud, clouds, cloudy were only mentioned 11 times in Roy’s post (once) and the comments (10, two of which were yours).

          Because I have no formal training in meteorology and climatology, I prefer to quote the knowledge of those who are recognized authorities in these sciences. I am sure I have shared the following several times during the past 10 or so months, but I again will share it because it is my opinion that clouds are the earth’s thermostat. For R. C. Sutcliffe in his 1962 book, Climate & Weather, wrote: “Clouds which not give rain, which never even threaten to give rain but which dissolve again into vapour before the precipitation stage is ever reached, have a profound effect on our climate. This is obvious enough if we only think of the difference between a cloudy and a sunny day in summer or between an overcast and a clear frosty night in winter. Taking an overall average, about 50 per cent of the earth’s surface is covered with cloud at any time whereas the precipitation is falling over no more than say 3 per cent. Non-precipitating clouds are thus the common variety, rain clouds are the exception.

          “The climatic importance of clouds lies in their effectiveness in reflecting, absorbing, transmitting, and emitting radiation, … . To the sun’s visible radiation they are efficient reflectors, throwing up to as much as 80 per cent back to space, and so shining white in the eyes of the space traveler. What is not reflected mostly penetrates and is absorbed in clouds of sufficient vertical depth so that the amount of light reaching the earth is then quite small, as every photographer knows. Long-wave radiation from the earth, the invisible heat rays, is by contrast totally absorbed by quite a thin layer of clouds and, by the same token, the clouds themselves emit heat continuously according to their temperatures, almost as though they were black bodies. In this way clouds by day keep much of the sun’s heat away and in the nighttime too they return to the earth much of the heat that would have been lost. A completely cloudy day may be close and humid but never exceptionally hot, whereas during a clouds night the temperature may hardly fall from its day-time value.”

          You wrote: “ … , daytime clouds cool things, nighttime clouds slow the rate of cooling.” So it seems that you and Sutcliffe agree as to clouds’ thermostat actions. And mpainter (6/14/15 at 6:23am) wrote: “Satellite imagery confirms that most radiation is emitted at cloud tops, providing observations which confirm the error of such diagrams.” If we change most to much this statement is consistent with what Sutcliffe wrote. But the most amazing fact is that searching cloud led to me to: Global Warming as a Natural Response to Cloud Changes Associated with the Pacific Decadal Oscillation (PDO) by Roy W. Spencer, Ph.D. October 20, 2008 (updated December 29, 2008).

          Because some comment about my long comments, I will come back to what Sutcliffe has written about clouds in another response.

          Have a good day, Jerry

          • Mike Flynn says:

            Hi Jerry,

            As I understand it, the near vacuum of space allows radiation to travel freely.

            Most of the Earth is a glowing molten mass. The core is hotter than the mantle, which is hotter than the crust, which is “hotter” than the 3K or so of outer space.

            Therefore the core loses heat to the mantle, the mantle to crust, the crust to space. After four and a half billion years, nothing has managed to stop the crust solidifying and thickening as it cools. Fairly obviously, without a perfect insulator intervening, there will be a temperature gradient from say 6000K to 3K or so.

            So the Earth cools, and will presumably continue to do so until it is isothermal at depths beyond that of the Sun’s influence – maybe 100 m or do.

            I agree with Sutcliffe it seems. I might go a little further, and say that the reflective properties of cloud extend to more than just the visible. Cloud base measuring equipment has been made using the principle of bouncing UV from the surface to the cloud, and measuring the return time.

            Likewise, IR of various wavelengths is reflected quite nicely by clouds. Radar wavelengths up to 10 cm are used in weather radars, both ground and aircraft based. These things are easily observed, but apparently overlooked by climatologists at time, if they don’t seem to fit preconceived ideas.

            I don’t need a GHE, when normal physics seems to explain every observation I know about.

            Maybe a lot of time, effort, and money has been wasted in pursuit of shadows.

            Have fun.

          • jerry l krause says:

            Hi Mike,

            I keep finding that I have not kept current on your comments. And I do not know where this comment will end up relative to your comments of 6/15 at 8:39pm. Relative to your comment: “As I understand it, the near vacuum of space allows radiation to travel freely.”, I am comfortable with stating: As I understand it, the cloudless (a haze is a cloud in my thinking) atmosphere also allows radiation to travel freely beyond the scattering of the shorter wavelengths of solar radiation. And I ignore the important ozone system of the stratosphere.

            I asked you about your statement about the perpetual cooling of the earth. And you have replied.

            I am a believer in the Creator God referred to in Genesis 1. One reason I am a believer is because of what I read in Genesis 1. From the beginning water was a central issue. It is said the man Moses is the human author of Genesis 1. I think I can say that we know that Moses lived at a time when volcanic activity (consistent with your comment: “Most of the Earth is a glowing molten mass.”) was very evident. So, why did Moses begin with water, which is not compatible with a glowing molten mass, if his writing was not inspired by the Creator God? I do not believe the six days of creation are literal days because our human definition of the period of time which we call a day is based upon the earth’s 24 hour period of rotation as the earth very slowly revolves about the sun. However, we find that, according to what is written, that the sun, moon, and stars were not created until the fourth day. But guess what was created on the third day. I suspect you do not need to guess because you know that plants were created before there was a sun to drive the photosynthesis process. Another fact that makes no human sense. I hope you see my point.

            In your comments (6/15 at 9:06pm) you wrote: “I would prefer Feynman to Sutcliffe if Sutcliffe claims total absorption rather than partial absorption and partial reflection.” I have previously encountered the proposal of partial absorption and partial reflection and the substitution of reflection for the mechanism of scattering which Feynman taught. Sutcliffe is not the only one who claims total absorption of LWIR radiation by quite thin clouds; Roy and NASA do also. At least I have not read that they propose the partial absorption and partial reflection to which you refer. But of course, I cannot claim to have everything. The issue of the substitution of reflection for the scattering which Feynman taught seems a serious problem because scattering and reflection are clearly different physical phenomena. Different physical laws apply to each phenomenon. The reflection of light from condensed matter (either solid or liquid) is usually partial. In Feynman’s teaching of scattering, I read nothing to suggest that scattering can be partial. I believe you can accept that the scattering which Feynman taught is not a part of conventional (traditional) physics.

            Can you agree with Sutcliffe that ‘one cannot explain the broad features of world climate if one does not know the actual mechanisms involved’? I do and that is the importance of resolving the question as to the validity of what can be read about the scattering which Feynman taught. For it (Feynman’s scattering) certainly is not being commonly considered.

            Have a good day, Jerry

        • VikingExplorer says:

          Mike, you are correct. There is no radiative balance on Earth or any other planet. Mercury is still warming, and Jupiter has been cooling.

          In the latest Judith Curry explanation of the GHE, it rests entirely on the fictitious idea of energy balance.

          • Curt says:

            No one, not even the worst of the alarmists, thinks the earth is out of energy balance by more than about 1.0 W/m2, averaged (integrated) over the earth’s surface and over a year.

            So the earth’s power inputs and outputs must balance to within this amount.

            The energy imbalance caused by the long-term cooling of the earth over billions of years is so minuscule compared to the radiative fluxes and their possible variations that it is not worth including them in the calculations.

          • VikingExplorer says:

            >> So the earth’s power inputs and outputs must balance to within this amount

            Mike, my point is that they don’t have to balance. That’s the flaw. The input and outputs are largely independent of each other. Whatever they are, they are. It happens locally, and there is no physical law that is integrating over time and space. Like Jupiter, Earth is happy to be exothermic, bleeding energy into space.

          • Curt says:

            VE:

            You miss the key point. To the extent that the energy inputs and outputs for the earth are out of balance, the energy level of the earth must change. That is what the 1st LoT demands. This would manifest itself as temperature changes and/or phase changes of water.

            If the earth were significantly exothermic, we would see substantial temperature drops, and possible freezing and condensation. We are not seeing this.

            Nobody who has seriously looked at the issue thinks the earth is out of energy balance by more than 1.0 W/m2, averaged over the surface and a year, in either direction. Most estimates are that it is gaining a fraction of a W/m2. The arguments are really whether this is less than 0.5 or greater.

          • VikingExplorer says:

            >> out of balance, the energy level of the earth must change.

            Yes, the energy level of earth would change, and that would be reflected in some internal temperature or phase change or work performed.

            >> That is what the 1st LoT demands

            The point is that the 1st law of Thermo does not require or imply radiative balance. The assumptions of the 1st law are completely violated.

            Also, don’t put words in my mouth about the level of the imbalance. I’ve said nothing about that, so I’m not sure why you keep talking about that.

    • Without Greenhouse gases (GHGs) the temperature gradient of the atmosphere from the surface to the top of the atmosphere (TOA) would be almost linear. With GHGs the lower levels of the atmosphere are warmer than they would be otherwise and the upper levels are cooler than they would be otherwise. It’s no longer a straight line relationship of altitude vs temperature.

      Regardless of the makeup of the atmosphere, energy in will equal energy out. The complexity of atmospheric conditions and climate in general is what happens in between.

      • Toneb says:

        Diane:
        “Without Greenhouse gases (GHGs) the temperature gradient of the atmosphere from the surface to the top of the atmosphere (TOA) would be almost linear.”

        No it wouldn’t (if by linear you mean isothermal).
        It would if there were no incoming SI.
        Even without GHG’s there would be a lapse rate. GHG’s only serve to raise the level of the -18C (Earth’s BB temp) level to ~7km from the surface, where it would be without.
        With incoming solar without GHG’s, a totally dry planet with the same orbital/rotational characteristics as Earth would still have a LR – this is because the atmosphere would still be dynamic. Convection, turbulence, winds/jet-streams formed via deltaT between equator and poles. The atmosphere would still be in strong motion. This rise/fall of air creates the LR. Air ascending cools (reduction of internal energy) and the consequent descending air warms as it’s pressure rises.
        Constant motion -> constant LR (but with local variation). A “heat pump” in effect. That’s NOT to say that the GHE does not effect the LR however.

      • VikingExplorer says:

        Dianne,

        You are exactly correct about this. GHG change the shape of the curve from a steady slope from bottom to top into one that is less steep at the bottom, yet steeper at altitude. You said it better than I just did.

        The point is that it doesn’t change overall energy flow. It only affects the transient timing.

        However, this effect is made irrelevant because there are multiple other energy transfer mechanisms acting in parallel. Like with Kirchhoff’s current law, when one increases the resistance of one path, more current will flow in the parallel circuits.

  2. Curious George says:

    I don’t buy the measurement of the greenhouse effect with an IR thermometer pointed at the sky. A thermometer pointed straight up would read very low on a planet with no atmosphere, or on the Moon. Low reading has little to do with a greenhouse effect.

    • then you missed my point. It is the increase in indicated temperature as you scan down from the zenith that is the demonstration of the greenhouse effect.

      • Curious George says:

        True. But it is difficult to tell when your scan touches the ground. Thanks for the explanation.

        On a related theme, the Earth’s albedo is sometimes measured – rather indirectly – by an Earthshine, a light reflected from the Earth and then reflected again from an unlit part of the Moon. Would a direct measurement by a Moon-based instrument (measuring a spectral and angular distribution of a radiation coming from the Earth) add anything to a satellite-based measurement?

        • The instrument would have to be well calibrated and measure over a fairly small angular distance. The CERES instruments already do this from 700 km orbit, but if there is ever another lunar base that might be one of the instruments proposed to put to there. Who knows?

    • …or, I just took my FLIR I7 and measured the inside of my freezer (around 5 deg F), then measured the blue sky above me outside (air temperature 86 and dewpoint of 71 F), and the indicated IR temperature was about 25 deg F.

      In both cases the handheld IR thermometer was essentially the same temperature.

      So, what source of energy caused the sky to indicate a temperature 20 deg. warmer than the freezer? The only answer is downwelling IR from the sky warming the surface of the microbolometer inside the instrument.

      That’s the greenhouse effect.

      • gbaikie says:

        “So, what source of energy caused the sky to indicate a temperature 20 deg. warmer than the freezer? The only answer is downwelling IR from the sky warming the surface of the microbolometer inside the instrument.”

        What if air temperature was 86 F without any greenhouse gases in the air.
        One could have 1 atm of N2 and with air being 86 F and in such as atmosphere what do think the FLIR I7 would indicate in terms of temperature.

        Anyhow to answer the question what are the the sources?
        It seems like it could be water droplets, or dust.

        It seems you think it can only be water vapor or CO2.
        And within 1000 meter, it seems there would more water vapor
        than CO2.

        My cheap hand held infrared thermometer doesn’t seem to measure liquid water very well. If pointed thru steam raising from a boiling pot, it seems to mostly measure the wall behind it.
        Perhaps since air vapor could interfere with measurements it’s designed not measure water vapor.
        And if your FLIR I7 was designed not to measure water vapor, then it would not be able to measure most of greenhouse effect?

        Do you think you are only measuring IR from CO2 gas molecules?

        It doesn’t like much CO2 in “line of sight” of the FLIR 17

        Say there 1 atm of CO2 and one point the FLIR 17 so looks thru meter of this pure CO2 at 1 atm, does it measure this amount of CO2. Or if had column of 1 by 1 meter by 1000 meter of air which one added 1 cubic meters of CO2, that is
        1 part per 1000 or 1000 ppm.

        • I’m not claiming it’s all CO2, I’m claiming the clear sky emits IR downward. In the winter in Alabama, when the humidity is low, my FLIR routinely bottoms out at -40 deg F., even when the air temperature is in the 60s. Yesterday, in a humid atmosphere, is was +25 deg. F. So that’s a 65 deg. difference, far larger than the air mass temperature difference.

          There are upward-viewing radiometers making measurements at hundreds of wavelengths, clearly showing CO2 emission and water vapor emission at the wavelengths more sensitive to these gases. There is no other possible explanation for these things.

          • gbaikie says:

            –I’m not claiming it’s all CO2, I’m claiming the clear sky emits IR downward. In the winter in Alabama, when the humidity is low, my FLIR routinely bottoms out at -40 deg F., even when the air temperature is in the 60s. Yesterday, in a humid atmosphere, is was +25 deg. F. So that’s a 65 deg. difference, far larger than the air mass temperature difference. —
            My question was what your FLIR was measuring. But it appears from what you say, that since drier air measures is much colder that can’t be just measuring CO2.

            And next question does your FlIR measure a small quantity of CO2 gas- does it measure 1 meter length of pure CO2 gas at 1 atm pressure. Because if it can’t then we conclude that the FlIR can’t be measuring the CO2 gas in the atmosphere.

            So have deep freezer, large bag 1 meter long. Pure CO2 gas which is 30 C.
            Wrap and seal FLIR to open end of bag. Put bag in freezer. If it measure the back of plastic bag or freezer, then it’s not measuring the CO2.

        • Infrared radiation (IR) is electromagnetic radiation in a specific frequency range, regardless of the source. It takes other, more precise instruments to pinpoint the actual absorption spectrum of various molecules.

          The sun itself actually emits very little radiation in the IR frequency spectrum and, of course, in the dark there is no other possible source for IR than re-emission from the atmosphere itself.

          When a molecule absorbs a quanta of radiative energy it vibrates in various combinations of motions. When it settles back down that quanta is re-emitted. The direction of re-emission is random … up, down, left, right. Ignoring the fact that the Earth is a sphere (a very small added geometric complexity) that means half goes away from the Earth’s surface and half goes towards it.

          It’s the half that goes towards the surface that you are measuring with the IR meter. The closer you get to pointing at the surface itself, the more atmospheric re-emission sources you have (because there’s more atmosphere between you and space) and so the higher the reading.

          • gbaikie says:

            You mean to say that Sun emits very little Longwave IR,
            because more than half the energy of the sun is Shortwave IR
            [which also not visible to human eyes].

      • VikingExplorer says:

        Dr. Spencer,

        Everything with a temperature emits radiation. There is nothing profound about this.

        • Tim Folkerts says:

          That is true VE, but two things at the same temperature can emit VERY different amounts of radiation. For example, a column of N2 @ 300 K will indeed emit some IR, but it will be a miniscule amount. A column of CO2 @ 300 K will emit WAY more IR (orders of magnitude more). A column of H2O droplet (ie a cloud) @ 300 K will emit more yet – maybe 3 times more than CO2.

          There IS something profound about *this*!

          • VikingExplorer says:

            Tim,

            That nature of atomic radiation or “black body” radiation is that it has a specific spectrum and intensity that depends only on the temperature of the body, not the material itself.

            An electron causes an electric field. A moving electron causes a magnetic field. EM fields are such that a changing E field will create a B field, while a changing B field will create an E field. The happy result is the EM fields are self propagating.

            A vibrating atom (with an electron) creates the changing electric (E) and magnetic (B) fields necessary to cause EM radiation.

            Thus, everything with a temperature (which is measuring kinetic energy of atoms) is also radiating.

            Measuring air with an IR meter neither proves not disproves the GHE. It merely proves that air has a temperature, which was never in much doubt.

          • Tim Folkerts says:

            “Atomic radiation” and “blackbody radiation” are two very different things. Blackbody radiation is an ideal, approximated quite well by cavity radiators and approximated less well by various materials. Your high school description of EM waves — while basically correct — adds nothing to this discussion. You need to dig into quantum mechanics to learn why simple gases (like Ar or N2) hardly radiate any IR, why more complex gases (like CH4 or CO2) radiate in specific band of IR, and why some solids radiate almost like ideal blackbodies.

            So pointing an IR thermometer at the sky does indeed tell you quite a bit if you have the scientific sophistication to grasp the importance.

          • VikingExplorer says:

            Tim,

            I’ve got a BS in Electrical Engineering. I got an A in Electromagnetics. I learned from one of the top EM consultants in the US.

            There is a lot I don’t know, but I am quite familiar with harmonics and Fourier transforms. In my work with aerospace generator design, I did extensive work designing the windings to meet a customer specification for harmonics, and then measuring the results.

            I also understand that molecules, like most physical structures, have various resonant or characteristic frequencies. For example, my van has a resonant frequency that shows up around 64 mph.

            If I were blindfolded, and driven around in two vehicles, with the driver calling out the MPH, I would feel/hear the resonant frequency when the driver said “64”, and I would be able to tell which one of these vehicles was my van, and which one wasn’t. It would tell me nothing about the temperature of the car body going down the road.

            That’s what spectroscopy is all about. I’ve also worked on a project that interacts with instruments for Raman, UV, NIR, and Mid-IR. Basically, by hitting an unknown substance with a certain frequency, and measuring the response, we can get a clue about what the substance is. When we change the input frequency from 0 to N, we can plot a spectrum of response values.

            However, this should not be confused with EM radiation that every object is radiating, or the dominant frequency of that radiation. A molecule has multiple modes of vibration. The dominant mode is associated with the kinetic energy of the atom or molecule. The level of kinetic energy is what temperature is a measurement of. The wavelength of the peak of the blackbody radiation curve gives a measure of the temperature. This is why I said previously that the dominant harmonic depends only on the temperature of the body, not the material itself.

            An IR meter designed to measure temperature is detecting this dominant harmonic. This is why it generally works on any object you point it at. This is why I said Everything with a temperature emits radiation. There is nothing profound about this..

            Now, if one carefully looked at the radiation waveform, one would see other minor harmonics that would give us a clue as to what kind of thing we’re looking at. In the case of CO2, we would see harmonics in the IR range, but with Nitrogen, we wouldn’t. You seem to be confusing these secondary harmonics with the dominant primary harmonic.

            The casual reader (who perhaps isn’t suffering from a superiority complex and assuming people with engineering degrees have only a high school level understanding) might be wondering: What is the frequency of the primary harmonic at various temperatures?

            2 um => 1450 K (2150F)
            4 um => 728 K (850F)
            9 um => 322 K (120F)
            10um => 289 K (62F) 263 K (14F)
            12um => 241 K (-26F)
            13um => 223 K (-58F) 193 K (-112F) <——- average upper troposphere temp at the equator

            The wavelength of Infrared goes from about 1 um (micro meter) to 1000 um. As you can see, the entire range of normal temperatures from the surface to the top of the troposphere are in the IR range.

            This means that a Nitrogen atom next to a CO2 molecule, in thermal equilibrium, will both radiate at the SAME dominant frequency. If the temperature is within the range above, then both will radiate infrared radiation. The IR meter is simply measuring that and displaying the associated temperature.

            Again, this doesn’t change the fact that Nitrogen is generally transparent to IR radiation, while CO2 is not transparent to certain frequencies of IR radiation. As explained, this has to do with the fact that CO2 has resonant frequencies associated with it’s molecular structure.

            I’m left wondering what level of education you have to make this statement: “a column of N2 @ 300 K will indeed emit some IR, but it will be a miniscule amount. A column of CO2 @ 300 K will emit WAY more IR”.

            ref
            ref
            ref

          • VikingExplorer says:

            Data didn’t seem to come through:

            2 um => 1450 K (2150F)

            4 um => 728 K (850F)

            9 um => 322 K (120F)

            10um => 289 K (62F) 263 K (14F)

            12um => 241 K (-26F)

            13um => 223 K (-58F) 193 K (-112F) <——- average upper troposphere temp at the equator

          • VikingExplorer says:

            Let’s try one more time:

            2 um => 1450 K (2150F)
            4 um => 728 K (850F)
            9 um => 322 K (120F)
            10um => 289 K (62F) (average lower troposphere temperature)
            11um => 263 K (14F)
            12um => 241 K (-26F)
            13um => 223 K (-58F) (average upper troposphere temp at middle latitudes)
            15um => 193 K (-112F) (average upper troposphere temp at the equator)

          • gbaikie says:

            –This means that a Nitrogen atom next to a CO2 molecule, in thermal equilibrium, will both radiate at the SAME dominant frequency.–

            Or a CO2 molecule surrounded by about 2000 nitrogen molecules.

            But anyhow,
            would what you saying would only apply if the Nitrogen and CO2 molecules were solid, liquid, or Plasma state?

            Are certain gases behaves this way.

            Or do at least acknowledge, that there would be some kinds differences in this regard between gases and other states of matter?

          • Tim Folkerts says:

            VikingExplored,

            It is helpful to know your background to know how to respond. And my background is a PhD in physics. It sounds like you have some good practical knowledge of resonance, but your knowledge of quantum mechanics may be not quite so good.

            Quantum mechanics limits the energies and wavelengths that a material can emit. For solids, pretty much any wavelengths can be emitted. For monatomic gases, almost no frequencies can be emitted. For polyatomic gases, there are an intermediate # of possible wavelengths.

            It turns out that symmetric diatomic gases basically do not have energy levels available that fall in the IR bands, so they neither absorb nor emit IR (at any appreciable amount). More complex molecules (like CO2 or CH4) can absorb & emit IR.

            For a couple quick examples, I would suggext the following.
            1) Spectracalc is used for many practical engineering applications of IR spectra. If you compare N2 & CO2, you will find that CO2 lines cover much wider bands and are ~ 10^7 stronger. In engineering, if one effect is 1 / 10,000,000 as large, it can pretty much be ingnored.
            http://www.spectralcalc.com/spectral_browser/db_intensity.php

            2) ModTran is another engineering application. In particular, it can be used to calculate the spectra expected from the atmosphere. Try setting the simulation to look upward in a cloudless sky from 0 km altitude. Then remove all the GHGs and watch the spectrum. The radiation dissappears, indicating that the N2 tht makes up the bulk of the atmosphere really doesn;t radiate.
            http://climatemodels.uchicago.edu/modtran/

          • VikingExplorer says:

            Tim,

            Picture a gas with no external radiation applied to it in any way. Those atoms are vibrating according to their temperature.

            That vibration causes radiation. If the temperature is in the range 2500 K (4040 F) to 3 K (-454 F), then that radiation is in the Infrared range.

            That’s why I said:

            that the dominant harmonic depends only on the temperature of the body, not the material itself.

            An IR meter designed to measure temperature is detecting this dominant harmonic. This is why it generally works on any object you point it at. This is why I said Everything with a temperature emits radiation. There is nothing profound about this.

            gbaikie, this is true regardless of the state of the substance.

            Now, in the ideal, simplest case, the resulting EM waveform is a perfect sinusoid. As the atomic configuration becomes more complex, the structure has other modes of vibration happening simultaneously (see here).

            These other modes of vibration introduce harmonics to the original sinusoid. This means that a Fourier spectrum will show a primary peak at the frequency associated with the substance temperature and multiple smaller harmonics at various other frequencies. The more complex the atomic structure, the more complex the waveform harmonics.

            Saying that a Nitrogen atom doesn’t radiate in the IR range, just because it’s too simple to have secondary harmonics in the IR range is like saying that a well built car can’t be a certain temperature because it has no resonant frequencies.

            As for quantum mechanics, it doesn’t change anything I’ve said. QED basically recognizes that energy is not a pure floating point number. It can only take on discrete values. This is easily understood because we can’t have 1.5 atoms, or 1.5 electrons. The elementary charge q is the electric charge carried by a single proton or electron. This requirement that we can’t have half an electron means that all derived values appear to be floating point values, but in fact can only have discrete values. This is why molecules can only be at discrete energy levels.

            It is often said that Nitrogen absorbs no IR. This is clearly false or overstated. It should be changed to Nitrogen is relatively transparent to IR. Nitrogen absorbs infrared radiation between 3.8 and 5 um (micrometers).

          • gbaikie says:

            –gbaikie, this is true regardless of the state of the substance.–

            Ok.
            Why I asked is because I think there is a big difference and for various reasons. But I suspect most climate scientist would basically agree with you.
            Anyhow I wrote wordy reply but decided not to post it.
            Then found something interesting.
            So question is what happens when you put a cup of liquid nitrogen in a microwave oven.

            We know microwaves are designed to just heat water [and I would say mostly liquid water rather than water vapor].

            Anyways someone had this question, and asked it, the answers
            surprised me, mostly because people seemed afraid to do it.
            Because I generally would imagine it would have little effect.
            So, here:

            ” Dear Straight Dope:

            I have a question that neither of my high school science teachers could answer for sure, including my physics teacher, who scored 1590 on his SATs (he also found an error on the test) and is a member of Mensa. If I were to put liquid nitrogen in my microwave, what would happen? Would it superheat and instantly explode or would it just boil off like if it was in regular air? By the way, my teacher suggested for me to try this experiment. ”

            “SDSTAFF Alphaene replies:

            Here at work I am within three feet of both a microwave and 240 liters of liquid nitrogen. However, even if I leave my job I’d like my boss to be a reference. And if he said “everything was fine until he superheated liquid nitrogen and destroyed the lab microwave,” it may not reflect positively on me. ”
            And another reply:

            ” SDSTAFF Chronos replies:

            When I saw this question, I was going to answer based on the resonance properties of water and nitrogen molecules. Microwaves work by exciting resonances in water molecules, and the water, once excited, heats up everything else in the frozen dinner you’re cooking. Since water is an asymmetric triatomic molecule, it has three distinct rotational modes and three more vibrational modes, making it easy to find resonances. Nitrogen, on the other hand, is a symmetric diatomic molecule, so it only has one distinct rotational mode and one vibrational mode. That makes it much harder to match a resonance for nitrogen, so the microwave probably wouldn’t do anything to it.

            That’s how I was going to answer. But that’s boring. Here at the Straight Dope, we do experiments.

            I went to the instructional lab supervisor here and asked if I could have some liquid nitrogen for, uh, “instructional purposes.” Having obtained the substance, I then proceeded to pour equal quantities into two identical paper cups. One was left outside as a control, and the other was placed in the microwave, which was set for one minute.

            It turns out that my prediction was half-correct. Liquid nitrogen is, as I suspected, transparent to microwave radiation. The effect of this transparency, however, is not that “nothing happens.” The effect is the same as when a microwave oven is run containing only air: a spectacular light show of sparks and flashes. Although my view of this phenomenon was somewhat hindered by the skull-and-crossbones sign on the door of the microwave, I decided that it would be nonetheless prudent to cut short the experiment after approximately one second. Afterward the liquid level in both cups remained the same, to within the limits of measurement. When an empty cup of the same sort was placed in the oven, similar pyrotechnics were produced. We may therefore conclude that, so far as the microwave oven is concerned, liquid nitrogen behaves in the same way as the atmospheric nitrogen normally present in the cooking chamber.

            Later it occurred to me that there was no reason to halt the experiment prematurely. A cup of liquid nitrogen would hardly be expected to catch fire, so the only concern would be damage to the magnetron tube. Since the next experiment proposed for this microwave was, “What happens to a microwave oven dropped from the ninth story?” the fate of the tube was likewise not an issue. Unfortunately, this thought did not occur to me until after my office-mates had already scattered the liquid nitrogen on the floor to see the Leidenfrost effect, so further tests were impossible.

            — Chronos”
            http://www.straightdope.com/columns/read/2064/what-would-happen-if-you-put-liquid-nitrogen-in-a-microwave

            Now, for me, I never had sparks fly from running microwave empty. And being forgetful creature I have done this [with no fire shows ensuing], and I can’t imagine a microwave being sold to the public [when one has warning labels, such as “This is not a step”] which would have sparks if the oven were to be run empty].

            But I was surprised liquid nitrogen would cause spark- until remembered that nitrogen is reflective- so that made it seem more likely that such event could actually occur.

            Since microwave are so cheap, not sure why anyone would afraid of some sparks for the sake of science.
            I simply don’t keep any liquid nitrogen. But anyhow, my question is, first, do you think that liquid nitrogen actually does cause sparks, and if it does, why does it?

          • wayne says:

            gbaikie, with that evidence do you suspect it might have been the cup?

            I have had that very same thing happen many times, by not noticing that the coffee cup or a paper cup has decorative foil stamped to add a little flair… also adds an impressive fire and sparks show every time. Or an unnoticed small screw if it has a wooden handle. Try it with a 1/16″ x 2″ aliminum foil strip and expect the smell of vaporized aluminum!

            On the nitrogen in the microwave. Think that may have a little to do with water being a polar molecule and nitrogen not?
            http://hk-phy.org/energy/domestic/cook_phy/images/emwave_h2o.gif
            (notice the ‘cook’ in the link? the wave is a microwave and the oscillation causes molecular friction that actually does the heating and cooking, not EMR absorption)

          • gbaikie says:

            — wayne says:
            June 29, 2015 at 8:07 PM

            gbaikie, with that evidence do you suspect it might have been the cup?–

            It seems possible. Here, are some braver creatures who used what appears to be plastic cups:
            https://www.youtube.com/watch?v=Bzr_Akg7WJ0
            But they also put in a cup of water with the cup of liquid nitrogen.
            They concluded the liquid nitrogen didn’t warm from microwave- which is not surprising.

            Anyways I didn’t see any sparks- or hear anything.
            Though it’s possible it could related to having the cup water in microwave.

            This gives general explanation and apparently you can make hot dogs spark, by cutting the so have sharp edges:
            http://engineering.mit.edu/ask/why-can%E2%80%99t-we-put-metal-objects-microwave
            Link given from others explaining why microwave oven sparks from metal:
            http://physics.stackexchange.com/questions/67880/why-do-metal-objects-in-microwaves-spark

            –I have had that very same thing happen many times, by not noticing that the coffee cup or a paper cup has decorative foil stamped to add a little flair… also adds an impressive fire and sparks show every time. Or an unnoticed small screw if it has a wooden handle. Try it with a 1/16″ x 2″ aliminum foil strip and expect the smell of vaporized aluminum!

            On the nitrogen in the microwave. Think that may have a little to do with water being a polar molecule and nitrogen not?
            http://hk-phy.org/energy/domestic/cook_phy/images/emwave_h2o.gif
            (notice the ‘cook’ in the link? the wave is a microwave and the oscillation causes molecular friction that actually does the heating and cooking, not EMR absorption)–

            Yes, and I don’t think water vapor would warm by moving the molecules back and forth. Or the effect would be so small that it would be hard to measure.
            But agree this whole microwave thing is mostly unrelated to greenhouse gases being affected by sunlight or IR of earth.
            What is kind of interesting, people worrying about it possibly exploding.

          • VikingExplorer says:

            >> I think there is a big difference and for various reasons. But I suspect most climate scientist would basically agree with you.

            It’s not a matter for subjective opinion. Temperature is defined to be a measurement of atomic kinetic energy. It is exactly that kinetic energy that causes radiation. If the radiation from a liquid @ 300 K had a different frequency from a gas @ 300 K, then they would be at different temperatures, which is contradicted by the premise.

            >> So question is what happens when you put a cup of liquid nitrogen in a microwave oven. We know microwaves are designed to just heat water

            What on earth would make you say that microwaves are designed to just heat water? Microwaves heat because a) electromagnetic waves forces act on charges and food has lots of polarized (+) and (-) particles, and b) the EM waves have the right frequency and amplitude to cause eddy currents.

            This second mechanism is due to Faraday’s law of induction which says that a changing magnetic field will induce an electric field to counter it. It’s like nature would like the magnetic field at any given point to remain constant, so an E field forms which causes current to flow that attempts to cancel the applied changing magnetic field.

            It flows in water because it conducts electricity. Many foods have some salt in them, which helps conductivity.

            As I stated previously, Nitrogen absorbs infrared radiation between 3.8 and 5 um (micrometers). You are talking about applying microwaves, which are defined to have a wavelength of between 1,000 um and 1,000,000 um. So, no direct absorption.

            Water is a polar molecule, which means it has a positive and a negative end. This helps greatly for the first mechanism of microwave oven heating. This isn’t the case for Nitrogen. It’s demonstrated here. I would speculate that the microwaves might make some difference that wouldn’t show up in such a short experiment. Nitrogen in liquid form at about 101 kPa would be boiling, so no temperature change would occur, but I suspect that if they microwaved it for longer and took more accurate measurements, they might see a difference in volume.

          • gbaikie says:

            — VikingExplorer says:
            June 30, 2015 at 11:10 AM

            >> I think there is a big difference and for various reasons. But I suspect most climate scientist would basically agree with you.

            It’s not a matter for subjective opinion. Temperature is defined to be a measurement of atomic kinetic energy. It is exactly that kinetic energy that causes radiation. If the radiation from a liquid @ 300 K had a different frequency from a gas @ 300 K, then they would be at different temperatures, which is contradicted by the premise.–

            The difference would the liquid would have fuller spectrum
            of radiation. And a liquid has a surface and gases do not.

            But also other differences, such as dynamics of gas in an atmosphere with gravity. Though gravity also effects liquids
            in terms of how they warm and cool.

            So big difference is both radiant and non-radiant heat transfers.

          • VikingExplorer says:

            >> The difference would the liquid would have fuller spectrum of radiation.

            Only if the atoms are at different temperatures!!

            I’ve already carefully explained the source of the radiation, so I will not repeat it. If you understood that, you wouldn’t have said any of the rest.

            >> radiant and non-radiant heat transfers

            Huh? We’re specifically talking about radiation, and what an IR meter is measuring. An IR meter pointing up is confirming that the atmosphere exists, which as I said, is not very profound.

          • wayne says:

            Hi VikingExplorer, you seem to be a new voice here, welcome.

            You will find an entire group of commentators here that follow what the likes of Willis Eschenbach, WattsUpWithThat, Tim Folkerts, etc that have claimed for years, many years, that if the atmosphere were just of nitrogen and oxygen that if you were to point an IR thermometer toward space it should read basically a bit above the 3K of the universe no matter of the atmospheres temperature, being closer to somewhere around 235 K. No emission, no absorption (or miniscule amount as Tim allows). They seem to base their belief on very rarified gas FTIR spectrums of the gases where collisions are rare and then massaged by programs (MODTRAN, HITRAN) supposedly following theory adding the broadenings and that is what is real. What I think are the flaw is in a near collisionless environment gathering per-line data how can you then manipulate that data when the values are nearly zero to begin with because of being rarified. I have never bought that entire line of thought and think much like you seem to.

            I am thinking those programs are accentuating only the vibration modes and suppressing any collision induced bb emission/absorption. probably because the military use that they were created for in the first place (missle tracking, infrared sensing/locking, etc) is where these have a real use.

            Your thoughts?

            Glad to have another voice around.

          • Tim Folkerts says:

            VikingExplorer says: “What is the frequency of the primary harmonic at various temperatures?

            2 um => 1450 K (2150F) …”

            No. This is the peak wavelength from Planck’s Law and/or Wein’s Law. This is the wavelength of the “brightest” emission from a blackbody — ie a body that is free to oscillate and emit at any frequency.

            The frequency of the “primary harmonic”, on the other hand, is determined by the mass of the atoms in the molecule and the “stiffness” of the bonds holding he molecule together. You can read up up these “normal modes” various places likehttps://en.wikipedia.org/wiki/Rotational%E2%80%93vibrational_spectroscopy

            For solids, there are generally a huge number of possible modes with huge options for frequency. Not so for gas molecule — they have very limited modes and frequencies they can vibrate. So for example, the “primary harmonic” for the CO2 bending mode is 667 cm-1 (the famous 15 um band for CO2). This is true independent of temperature. The molecule can bend at twice the amplitude (ie twice the energy = twice the frequency = 1/2 the wavelength).

            “This means that a Fourier spectrum will show a primary peak at the frequency associated with the substance temperature and multiple smaller harmonics at various other frequencies.”
            Again, no. The frequencies of the harmonics are associated with the STRUCTURE of the molecule, not the TEMPERATURE.

            It is often said that Nitrogen absorbs no IR. This is clearly false or overstated. It should be changed to Nitrogen is relatively transparent to IR.
            1) few people say N2 is *completely* transparent to IR. The term “transparent” is usually mean in the same vein as “glass is transparent to light” — ie the vast majority gets through.
            2) This is exactly what I said: “For example, a column of N2 @ 300 K will indeed emit some IR, but it will be a minuscule amount.” N2 will both absorb and emit very little IR compared to GHGs like CO2 and H2O and CH4.

            PS “QED” has to do primarily with photons, not with integral numbers of electrons, or the quantized energy levels of electrons around atoms.

          • VikingExplorer says:

            wayne,

            Thanks for the welcome. I have a long history on CA and WUWT. I have not commented here much, but felt compelled to correct the record on what an IR meter is measuring, and the source of that radiation.

            Tim, your response is very weak and basically consists of correcting my usage of the word “harmonic”. Technically, I should have called it the “peak wavelength” instead of “primary harmonic”. Harmonic refers to the secondary peaks.

            There is no substance as the rest is all nitpicks. I had professors in college with PhDs that when you asked them a question, they would answer “I’ll get back to you on that”. Even though I’ve had teachers like you in college, I’m still surprised that someone with a PhD in Physics could say:

            For example, a column of N2 @ 300 K will indeed emit some IR, but it will be a miniscule amount. A column of CO2 @ 300 K will emit WAY more IR (orders of magnitude more).

            It’s absurd, and easily tested with an experiment. This is one experiment that I’m not going to bother to perform however. If they are both at 300 K, an IR meter will read…

            wait for it… drum role please… 300 K

            As for the total energy, that depends on the number of molecules of each. There is one CO2 molecule for every 10,000 Nitrogen and Oxygen molecules. If all are at the same temperature, then the amount of energy radiated from the CO2 molecule is about 1 / 10,000 of the energy radiated from Nitrogen + Oxygen.

            Now, that’s not to say that CO2 doesn’t absorb a whole lot more IR being transmitted through it. However, CO2 is absorbing IR on the way down as well as on the way up.

            However, Nitrogen and Oxygen do absorb incoming IR. To claim otherwise is to claim that dry air isn’t heated by direct sunshine. That is obviously false from everyone’s personal experience, and easily falsified.

            As can be seen here and here, there is solar radiation in the range 3.8 – 5 um. More than half the Sun’s power output is in the form of infrared light, and much of it is absorbed by the Earth’s atmosphere. If you look at the first link, O2 and N2 are busy beavers at absorbing energy.

          • wayne says:

            V.E., thank you for that link at Columbia and yes I do see what you are speaking of within that lower %absorbance spectrum. I could (and will) spend and hour scrutinizing that graph. Notice how the w.v. lines start off in the visible low and thin, as you move toward infrared they keep getting taller and taller (approaching 100%) and wider and wider. That is how I remember it.

            Yet the bottoms (the minimum absorbance) bottoms out at about 20-25% and that is exactly why you cannot have an infrared telescope sitting at sea level, there is simply too much attenuation, all frequencies. Also notice what is the very lowest in absorbance? 10.5 μm and that is exactly the primary ‘window’ that they use in such infrared telescopes sitting about 13,750 feet ASL. There are no ‘clear’ windows and that is the underlying black-body (gray-body really) of our atmosphere in general, not just GHG lines and continuum. Must be from all atoms and molecules… don’t leave argon out, poor sole.

            You and I seem very parallel and I bet you began your studies back in the early 70’s or even before, books from the 60’s, when they actually taught physics, all of it, without having passed through this co2/gw meme filter.

            To me the world is more like this in a graph, with all of the man-made adjustments backed out and I will always stand that this is a more correct view.
            http://i39.tinypic.com/1118rnl.png
            Just a rough back out, I admit, just enough for me to get a glimpse of reality.

            Thanks VE.

          • Tim Folkerts says:

            It’s absurd, and easily tested with an experiment. This is one experiment that I’m not going to bother to perform however. If they are both at 300 K, an IR meter will read…

            wait for it… drum role please… 300 K

            And I have had students like you, who THINK they know what they are talking about, but reality their intuition is way off.

            This IS easily tested with an experiment — one that is performed essentially every time an IR thermometer is used!

            Stand near a hot wall (of maybe over a large sheet of steel fresh from a furnace) @ 1000 K. Measure the temperature up close, so there is almost no atmosphere between you and the wall. The IR thermometer will read the temperature of the wall. Now stand far from the wall, so that you are looking through a large length of 300 K air. The IR meter will read …
            wait for it … drum please … 1000 K. (NOT 300 K)

            You see, the IR meter is designed to look THROUGH the atmosphere, especially in the “atmospheric window” — IR meters typically measure from ~ 8-14 um where even H2O and CO2 are fairly transparent. This means that even at large distances, the thermometer does not (and cannot) measure the temperature of the air. A simple image like this shot from an airplane (http://www.imagingnotes.com/ee_assets/Figure1_EarthScope.jpg) should convince you that the IR meter is measuring the ground and ignoring several 100 m of air. If there was a big sheet of dry ice on the ground, the thermometer would read ~ 200 K. If a similar column of air lead to space, the thermometer would read ~ 3 K.

            Why do you suppose good IR thermometers include an adjustment for emissivity if any object 300 K would read the same?

            “As can be seen here and here, there is solar radiation in the range 3.8 – 5 um. More than half the Sun’s power output is in the form of infrared light, and much of it is absorbed by the Earth’s atmosphere. If you look at the first link, O2 and N2 are busy beavers at absorbing energy.”
            Again, this is only partly right, and misses the main ideas. The very webpages you link to clearly indicate how you are wrong.

            Yes, about half of the sun’s output is IR — but that is almost entirely from 0.7 – 4 um. Yes, a significant part of that IR is absorbed coming in (which makes up much of the “78 W/m^2 absorbed by atmosphere” in the KT diagram). But now much of that is due to N2 or O2?

            If you look at the SECOND link, you will see which gases are “busy beavers at absorbing energy”. The busy beavers in the IR range are H2O, CO2, N2O, O3 and CH4 (ie the standard GHGs). These gases are what is absorbing that incoming IR (and in fact for solar IR, it is almost entirely due to H2O). O2 absorbs a little visible light and most of the UV below 0.3 um, which is not that strong to begin with). N2 is NOWHERE on the graph. Across this entire range, N2 absorbs so little that they didn’t even bother including it anywhere. Across the thermal IR range, neither N2 nor O2 absorb enough to be significant. Hence they also do not EMIT enough to be significant.

          • wayne says:

            Tim : N2 is NOWHERE on the graph.

            So? Why would N2 be singled out? Because ALL matter radiates and absorbs in a gray body manner due to interaction/collisions with the atoms and molecules around it Tim, not as much, but still, that is unless gases are so rarified in spectrometers (which I have read, they do) and interaction/collisions approach zero to create your belief system.

          • Tim Folkerts says:

            Wayne says: “Why would N2 be singled out?”
            Because that is what the graph is showing! Specific molecules are singled out when they have observable effects. If there was some observable impact of N2, then it would have been included.

            That graph is the measured absorption for the entire atmosphere.

            “Because ALL matter radiates and absorbs in a gray body manner due to interaction/collisions with the atoms and molecules around it Tim”
            Reference??? Where can you find anything that backs up your “belief system”?

            All matter radiates/absorbs in ways specific to that material. For visible wavelengths, surely you are familiar with line spectra of gases. Here the light emitted is NOT at all continuous, but consists of very sharp lines as electrons jump from one quantum level to another. For IR, there are lines due to specific vibrations/rotations of the molecules. For a variety of reasons these IR lines tend to be broader (ie “bands”) compared to fairly sharp lines of visible lights. But they do not smear out into “graybody” radiation.

            Furthermore, satellites routinely measure the surface temperature, which means the radiation is coming FROM the surface THROUGH the atmosphere basically unimpeded. If the atmosphere were absorbing/emitting significantly across the whole IR spectrum, then there would be no way to measure the radiation from the surface. (And of course, they can only measure in IR windows).

            There is a whole field of “Infrared spectroscopy” that studies exactly these highly-non-graybody IR spectra.

            “… unless gases are so rarified in spectrometers (which I have read, they do) and interaction/collisions approach zero to create your belief system.

            Ah … so I provide detailed theory, backed by experimental measurements of the IR spectrum of earth’s actual atmosphere, supported by detailed and validated engineering software used by NASA and the US Air Force, confirmed by satellite surface temperature measurements.

            You provide ‘I read somewhere’. And *I* am the one operating on a “belief system”? 🙂

        • VikingExplorer says:

          >> All matter radiates in ways specific to that material

          You are changing the subject. This isn’t about absorption, it’s about radiation.

          The dominant frequency is associated with the kinetic energy of the atom or molecule. The level of kinetic energy is what temperature is a measurement of. The wavelength of the peak of the blackbody radiation curve gives a measure of the temperature. This is why I said previously that the dominant frequency depends only on the temperature of the body, not the material itself.

          This is directly opposite of your statement. You’re either lying, seriously confused, or pushing some political agenda. You simply can’t deny this basic scientific truth and get away with it.

          The entire range of normal air temperatures from the surface to the top of the troposphere are in the IR range.

          Measuring that air temperature, regardless of how, tells us nothing profound.

          • Tim Folkerts says:

            “You are changing the subject. This isn’t about absorption, it’s about radiation.”
            Read up on Kirchhoff’s Radiation Law. “With this definition, a corollary of Kirchhoff’s law is that for an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium, the emissivity is equal to the absorptivity.”
            https://en.wikipedia.org/wiki/Kirchhoff's_law_of_thermal_radiation

            “The dominant frequency is associated with … is a measurement of … ”
            These fuzzy words seem to be masking some fuzzy thinking (or perhaps simply miscommunication).

            There TWO sorts of “dominant frequency”.
            1) The peak of the blackbody radiation curve — the curve described by Planck’s Law. This peak does depend only on temperature. (But does not have any “harmonics” or “resonance”).
            2) There are resonance peaks associated with the mass & bonding of atoms. For example, CO2 resonates to create/absorb 15 um radiation. But it doesn’t create (significant) 12 um or 20 um IR — there are no resonances in those ranges. No matter what the temperature, CO2 will not resonate to create these wavelengths. There ARE harmonics for these resonances.

            The fact that you seem to be talking about #1 but discussing resonance suggests you are not clear on these concepts. I would encourage you to play with this MODTRAN site: http://climatemodels.uchicago.edu/modtran/

            It shows the shifting peak for the ideal blackbody curves at various temperatures. It shows the frequencies/wavelengths where specific gases absorb/radiate. Play around with gas concentrations and clouds and looking up/down and altitude. When you can intuitively predict what the spectrum will look like and why, then you will be much closer to understanding all this.

            For example, try looking up from the ground at subarctic summer conditions. An IR thermometer tuned to a band around 15 um would read a temperature of ~ 285 K. An IR thermometer tuned close to 12 um would read ~ 200 K. Remove all the GHGs and the measured “temperature” would be well below 100 K at pretty much all wavelengths.

            PS you still haven’t been able to address the challenge to your “300 K” reading from pure N2 gas. It will NOT radiate anything like a BB ~ 300 K and the IR thermometer will basically read the tempeerature of whatever is BEYOND the N2.

          • VikingExplorer says:

            No, I’m not fuzzy, but you have been. If you look at what I’ve been talking about the whole time, it is #1 and that you have been obsessed with #2. #1 makes your statement that N2 doesn’t radiate in the IR range non-sense.

  3. jimc says:

    Perhaps if DC actually reads the whole thing he will understand. On the other hand, I suspect he will respond with the usual bombast and bluster – leading to another 1000 comment post complete with planetary angular momentum and a tour of Venus and Mars – proving his home is Cruithne, not Oz.

    • yeah, I know. I’m not out to change his mind. That’s not possible. But it occurs to me that “Wizard of Oz” would be a good nickname for him. 😉

    • Slipstick says:

      Unfortunately, that is extremely unlikely. The “Wizard”‘s (an apt moniker, Doctor) mistaken belief that the 2nd law applies to individual photons is unshakable.

  4. Massimo PORZIO says:

    Hi Dr. Spencer,
    you wrote “When the model is initialized with global average sunlight and atmospheric greenhouse gas concentrations, from any initial temperature profile you want (even absolute zero), it eventually equilibrates to the observed average vertical temperature structure of the atmosphere.”

    This is the second time I read you write about that.
    What was the TOA altitude for that 0 K run?
    I’m really surprised reading that a 0 K planet has any atmosphere on its surface.
    Do I misunderstand something?

    Have a great day.

    Massimo

    • The top layer of the atmosphere was very high as I recall, above 0.1 mb pressure altitude, so that would be at least 99.99% of the atmosphere.

      In the model run you assume an average atmospheric composition as a function of pressure altitude, and just assume it is initially at 0 K…or any other temperature you want, say 500 K. The model then warms or cools all layers of the atmosphere independently until energy balance is reached throughout. It creates a tropopause and stratosphere, too, since those are largely a result of the assumed vertical profile of ozone and the sunlight it absorbs.

      • Massimo PORZIO says:

        Ok, maybe I get your point.
        Do you started from 0 K and raised the simulated surface temperature to a defined temperature much higher than 0 K.
        The atmosphere build-up as the temperature raise, is it so?
        I mean at 0 K should not exist any gas at all and the atmosphere gases should be all collapsed to the ground most in form of solid and helium probably in form of liquid because of gravity, isn’t it?

        Massimo

        • sigh….yes, the 0 K assumption is physically inconsistent with the physical state of the gases.

          But you are missing my point.

          You can initialize the model at ANY temperature, say isothermal at 200 K throughout the whole depth. It will still change the temperatures, based upon the energy imbalances between layers, to what is observed in the real world.

          • Massimo PORZIO says:

            Uhmmm… Is that whole initial condition of 200 K isothermal atmosphere which doesn’t convince me at all.

            I really have very difficulties to understand how could that atmosphere really exists under a gravitational field, since the outer molecules should run in every direction at the same speed of the ones at the ground level except for the vertical direction where they must be at 0 m/s (otherwise they where not at the TOA, they were still running away, do you get it?).

            I feel that situation very improbable.

            It’s 1:15AM here in Italy, tomorrow I have to attend a military veterans meeting in the early morning, much better I go to sleep now

            Have a great Sunday.

            Massimo

    • wayne says:

      “… it eventually equilibrates to the observed average vertical temperature structure of the atmosphere.”

      It has to Massimo, the Earth like all of the rocky planets and gas giants in our solar system are close to 1.5 polytropes (Earth is right at 1.5) and they must have the temperature profiles (ELR) that they must have in the lower portions or their tropospheres. These equations take into account the graviational field strength, the molecular mass of the components and the specific heat capacity for each different atmosphere. The placement of the intesection of the surface depends of course on the solar radiation that can even get down to the surface (TSI minus albedo and direct atmospheric absorption). Through this polytropic path you can calculate the lapse of each other atmosphere and they do match the data from the probes.

      Took me quite a while learning a bit of general atmospheric physics and thermodynamics of all atmospheres to get to the bottom of what is always being discussed… our atmosphere.

      That is what I have been digging into during the last couple of years. Opened my eyes.

      • Massimo PORZIO says:

        Hi wayne.
        I agree with your argumentation, a lapse rate should exist anyways without GHGs.

        Being Italian I missed the meaning of “It has to Massimo”.

        What does it means?

        Have a great day.

        Massimo

  5. Notagain says:

    Roy, firstly the “classical Kiehl-Trenberth global energy budget diagram” is a mistake or worse, since surface there emits more energy than it gets from the sun, which is impossible. Unless you believe that energy can be creating out of nothing just like that, by writing down some numbers.

    Secondly, generally, the idea that the surface can be warmer than the sun can it possibly make (in absence of a more powerful source of heat, of course) is nonsense for the same reason. The surface would then radiate more energy away than it gets. There is simply no such a “greenhouse effect”, for this pure arithmetical reason. Just count the energy, that is all.

    • …Jetzt geht das schon wieder los…

      actually, no. The diagram shows the net IR loss by the surface is only 63 w/m2. All energy losses by the surface equal the net energy inputs to the surface (except for their assumed 0.9 imbalance caused by humans).

      Just count the energy.

      And it appears you really didn’t even read my post, since I already addressed these issues. You are just spouting the standard SDS talking points.

    • Bob says:

      Re Notagain:

      You wrote:
      since surface there emits more energy than it gets from the sun, which is impossible.

      You need to measure at the right place. Look at the outflowing energy while standing on the moon (or anywhere else outside the atmosphere. That figure shows 341 going in and 341 going out. The energy also flows around inside the earth system. When I run my car, some of the waste heat of combustion goes out the tail pipe in hot exhaust gasses. Other waste heat is transported to the radiator and heats the air flowing past the radiator. If my engine does not manage to get rid of 99.99+% of the energy it generates, it will either boil over or melt (assuming it’s aircooled like the old VW bug). The figure discusses the transport mechanisms—that’s all.

      Buy an IR thermometer and measure the temperature of the sky at various angle. Tell me why the sky is warmer than outer space—but usually not as warm as the ground.

      Bob

  6. KevinK says:

    Dr. Spencer, with respect, it sure sounds like you are trying to convince yourself more than anyone else ?

    One small correction; In the K&T energy “budget” diagram the surface radiation would exist in all cases above absolute zero. It makes no difference whatsoever if the atmosphere contains radiative gases or not. All surfaces (above absolute zero) emit radiation all the time without any regard for the optical properties of any gases that may surround them.

    You really should read up a bit about optical integrating spheres and how they operate. The mystery of why the “Radiative Greenhouse Effect” has no “Effect” on the average temperature at the surface of the Earth is clearly understood once one examines what happens to photons inside an integrating sphere. If one places an illuminated light bulb (with an electrical power supply attached) inside an integrating sphere the continuous stream of photons travel (at the speed of light) towards the interior surface of the sphere. A few “lucky” photons exit immediately through the “exit port” (a relatively small hole in the reflective surface of the sphere). But most photons “bounce” off of the interior surface. Some of these “unlucky” photons do in fact get absorbed by the filament of the light bulb and do in fact warm the light bulb. This is exactly the posited result of the “GHE”. And in fact this “warming” of the filament from backradiation changes the efficacy (or efficiency) of the filament. A warmer filament will produce more photons from the same number of electrons and the light bulb will appear “brighter/warmer/assume a higher temperature”. BUT (and this is a very large BUT) it requires the complete system of a filament and an electrical power supply for this effect to happen.

    In precision radiometry applications this error source is well understood and termed “self absorption”. Generally light bulbs used as precision radiometric sources are powered with a constant current electrical source. A constant voltage source makes this well understood error larger. In the case of the surface of the Earth no equivalent of the electrical power supply exists.

    An optical integrating sphere (while exhibiting nearly 100% back radiation “forcing”) merely delays the stream of photons from the filament to the exit port, same as the “Greenhouse Gases” do in the atmosphere. This delay is not long enough to have any effect on the average temperature of the Earth.

    With all due respect, the fact that you needed a “model” to convince you of a non-intuitive “effect” means you should be way more curious about what this “effect” really does.

    None of the predictions (Arrhenius, Callendar, Hansen) about temperature versus CO2 postulated over the last 100 plus years has come true, how much longer must we wait to declare this “conjecture” DOA ?

    Thanks for taking the time to explain the postulated “Greenhouse Effect” one more time, it is a truly elegant explanation, but sadly the observations do not bear out.

    Cheers, KevinK.

  7. Mike Flynn says:

    Dr Spencer,

    A demonstration of the greenhouse effect under laboratory conditions would settle the matter.

    However, experimenters from John Tyndall onwards, show that the supposed effect does not exist. Tyndall’s work “Heat as motion” demonstrated by experiment what does happen, and uses an example, accompanied by diagram, of an IR absorbing glass fire screen (popular in Victorian times).

    The reason he did this was to overcome the natural reluctance of people to appreciate the heat absorbing and radiating abilities of invisible gases.

    I am reasonably sure that people who quote Tyndall in support of the GHE have not actually read his works.

    By the same token, people using Arrhenius may not like what Svante actually wrote –

    “We often hear lamentations that the coal stored up in the earth is wasted by the present generation without any thought of the future, and we are terrified by the awful destruction of life and property which has followed the volcanic eruptions of our days. We may find a kind of consolation in the consideration that here, as in every other case, there is good mixed with the evil. By the influence of the increasing percentage of carbonic acid in the atmosphere, we may hope to enjoy ages with more equable and better climates, especially as regards the colder regions of the earth, ages when the earth will bring forth much more abundant crops than at present, for the benefit of rapidly propagating mankind.”

    So Arrhenius thought increasing CO2 levels was desirable, leading to “more equable and better climates”, but appeared concerned by his generation burning all the coal, and leaving none for us.

    But if you still think reducing the level of CO2 in the atmosphere is good, please don’t reduce it below the level needed to maintain plant life. We’ll all die then, won’t we?

    • Notagain says:

      You might be surprised to hear that, Mike, but the picture of most of us dead as soon as possible is what some people have been dreaming of.

      This is what Jacques-Yves Cousteau said in an interview published by The UNESCO Courier, November 1991, p.13: “It’s terrible to have to say this. World population must be stabilized and to do that we must eliminate 350,000 people per day. This is so horrible to contemplate that we shouldn’t even say it.”

      http://unesdoc.unesco.org/images/0009/000902/090256eo.pdf

      • Many environmentalists look at the lives of those people as the price we must pay to approach some imaginary Eden, living in equal partnership with a Gaea that never was and never will be.

  8. Volokin says:

    Here is an interesting recent paper that discusses the magnitude of Earth’s atmospheric GHE. As it turns out, GHE has been incorrectly calculated for over 35 years. Our atmosphere raises Earth’s surface temperature by about 90 K, not 33 K as currently assumed:

    http://www.springerplus.com/content/3/1/723

    • From your source:

      “It is concluded that the contribution of greenhouse gases to Earth’s ATE defined as GE = ATE – TE might be greater than 33 K, but will remain uncertain until the strength of the hereto identified TE is fully quantified by future research.”

      GE (Greenhouse Effect) is not the same as ATE (Atmospheric Thermal Enhancement), which is what they are pegging (rightly or wrongly) at 90°K.

      • Volokin says:

        Read the paper. There is much more to the story than revealed by the Abstract.

        The effective radiating temperature (Te = 255 K) used so often by the Greenhouse Theory is a non-physical (abstract mathematical) quantity that has no relationship to actual (measurable) physical temperatures at the surface. There has been a major confusion about how to measure the size of the atmospheric thermal effect. The 33 K (60 F) GE discussed by Dr. Spencer is simply an erroneous number devoid of any physical meaning. The paper provides a very good explanation as to why this is the case …

        • The context of the discussion is not the same. The paper is comparing a theoretical Earth without an atmosphere, oceans, ice or vegetation to Earth with an atmosphere and all its other features, whereas this discussion here is an atmosphere with GHGs to atmosphere without GHGs.

          • This discussion is basically about whether GHGs are required to achieve a decline in temperature with height.

            All else follows from that.

            Roy and many others think that with no GHGs the atmosphere would become isothermal with no temperature decline with height.

            I say that is nonsense because density decreases exponentially with height around a sphere due to the distribution of atmospheric mass tending to be greater nearer a centre of gravity and less the further away one moves.

            All one then needs to cause convective overturning is uneven surface heating (inevitable for a rotating sphere illuminated by a point source of light)which causes density variations in the horizontal plane.

            GHGs have nothing to do with the greenhouse effect which is simply a surface temperature enhancement above the S-B prediction beneath the mass of an atmosphere which insulates the surface by conducting and convecting.

          • VikingExplorer says:

            many others think that with no GHGs the atmosphere would become isothermal with no temperature decline with height.

            I say that is nonsense because density decreases exponentially with height around a sphere due to the distribution of atmospheric mass tending to be greater nearer a centre of gravity and less the further away one moves

            Both view points are incorrect.

            Without GHG (and convection), the slope of temperature vs. altitude would be closer to a straight line. GHGs serve to warm up the lower altitude, making the decline shallower, then steeper.

            Your viewpoint misses the fact that PV = nRT. The lapse rate is NOT due to decreasing P. P decreases, but that is compensated for by decreasing n (the number of air molecules).

            The lapse rate is NOT caused by gravity. The lapse rate IS caused by being heated from below and cooled from above. The energy level of the earth IS caused by a previous gravity collapse.

            The previous gravity collapse (e.g. a planetoid crashing into the earth, liquefying it, and resulting in our rather large moon) and NOT gravity, is responsible for setting the T at the surface.

            This large reservoir of energy is reflected in the average temperature of the surface.

          • Tim Folkerts says:

            VikingExplorer,

            Unfortunately, your attempts at ‘correcting’ Stephen’s post introduces many new errors.

            “The previous gravity collapse … is responsible for setting the T at the surface.”
            HARDLY! Temperature is determined by the combined effects of all energy inputs and outputs. The initial collapse was responsible for much of the initial warming, its true. But consider a “twin” of earth, with the same in-falling materials 4.5 billion years ago. Only this twin is out a bit past Pluto. This twin would have a surface MUCH colder than the earth due to the lack of sunlight.

            It is sunlight & albedo that primarily determines the surface temperature of planets (ie the closer to the sun, the warmer). This temperature is modified by atmospheres and GHGs (more atmosphere & more GHGs = warmer). Geothermal heat flows are a MINOR player in the surface temperatures.

            ********************************************

            I agree the lapse rate is CAUSED by uneven heating. However, the fact the the lapse rate is nearly the same over wide ranging conditions is enforced by convection and g/Cp. And since gravity plays a key role in convection and in ‘g’, then clearly gravity is very important in setting the observed laspe rates.

            ******************************************

            I do like what you said about the ideal gas law. 🙂

          • gbaikie says:

            -VikingExplorer,

            Unfortunately, your attempts at ‘correcting’ Stephen’s post introduces many new errors.–
            🙂

            –“The previous gravity collapse … is responsible for setting the T at the surface.”
            HARDLY! —
            I will not miss the rare opportunity, to agree with Tim.

            –Temperature is determined by the combined effects of all energy inputs and outputs. The initial collapse was responsible for much of the initial warming, its true. But consider a “twin” of earth, with the same in-falling materials 4.5 billion years ago. Only this twin is out a bit past Pluto. This twin would have a surface MUCH colder than the earth due to the lack of sunlight.–

            Not sure why one wants to go along with idea of 1000+ km diameter planetary body smashing into a larger 1000+ km diameter at same speed of around 20 km/sec, should called
            a gravity collapse.
            Though I suppose if one imagined one of them being hollow it might makes some sense.
            As for a Earth twin beyond Pluto being cooler, this would be unwise to disagree. But much colder, would depend what you meant. I would agree that Earth interior would be a bit cooler. A bit. And major part of the bit, would related to tidal heat related to Sun distance to the tidal interaction of Earth and the Moon. Or if had some other body [like Jupiter size planet] providing same kind of tidal as the sun does with Earth-Moon system. Then it would less than a little bit.
            But I assume you talking about the twin Earth’s surface temperature being Much Colder. Well I would say the nearest amount of large amount of liquid water would be about 1 km below the frozen surface. Or perhaps most to Earth ocean would liquid. But certainly too cold to grow turnips.

            –It is sunlight & albedo that primarily determines the surface temperature of planets (ie the closer to the sun, the warmer). This temperature is modified by atmospheres and GHGs (more atmosphere & more GHGs = warmer). Geothermal heat flows are a MINOR player in the surface temperatures.

            ********************************************–
            More to agree with Tim, about.
            Volcanic activity on Earth doesn’t have much affect upon surface temperatures. And bigger events like trillion of tonnes of volcanic ejecta sent into stratosphere would tend to cause a significant amount of cooling. And the even rarer continental type scale of lava flows may add as much a 10 C to air temperature for short time periods [million years]. Though less large scale lava flows in deep ocean could have longer term and more significant effect upon ocean temperature and air temperatures.

            –I agree the lapse rate is CAUSED by uneven heating. However, the fact the the lapse rate is nearly the same over wide ranging conditions is enforced by convection and g/Cp. And since gravity plays a key role in convection and in ‘g’, then clearly gravity is very important in setting the observed laspe rates.

            ******************************************–
            Mostly agree. I would say the radiant effect of greenhouse gases [not counting the condensation effects of H20 [H2O is main greenhouse gas on Earth] has a minor effect upon the lapse rate.
            So 1/2 gravity and you roughly half lapse rate.
            And one has no convection without gravity.

            As far as uneven heating, roughly that could related to rotational effect of a planet, so doubling or halving rotation rate would have measurable regional effects in terms it’s effect upon lapse rate.

            Also connected to rotation:
            “Mars is subject to strong thermal tides produced by solar heating rather than a gravitational influence. These tides can be significant, being up to 10% of the total atmospheric pressure (typically about 50 Pa). Earth’s atmosphere experiences similar diurnal and semidiurnal tides but their effect is less noticeable because of Earth’s much greater atmospheric mass.”
            https://en.wikipedia.org/wiki/Climate_of_Mars
            So as understand this has minor effect of “acting like”
            doubling rate of rotation.

            And it seems radiant effect upon lapse rate would a less of an effect.

            –I do like what you said about the ideal gas law. 🙂 —

            Well I think Pressure times Volume equals amount and mass of molecules times their velocity squared [velocity: 400 to 500 m/s for earth atmosphere troposphere] is important concept related to Earth climate.

            But it should kept in mind it applies to pressure vessels. Or it applies to small pressure vessel as compared to the scale of Earth Atmosphere.
            Or Earth has pressure because of gravity. Pressure vessels have pressure due the strength of the container walls to withstand the pressure. And because they tend to be smaller than cubic km, gravity is little effect upon the pressure in typical pressure vessels.

          • dvolikin says:

            gbaikie,

            You said: “…I think Pressure times Volume equals amount and mass of molecules times their velocity squared [velocity: 400 to 500 m/s for earth atmosphere troposphere] is important concept related to Earth climate.

            Where did you study physics and the Gas Law, my friend? Pressure times Volume equals Joule (kinetic energy), i.e. kg*m^2/s^2. It does NOT equal “mass of molecules times their velocity squared” !!!

  9. justaguy says:

    I don’t understand why this is so hard for people to understand, and I am probably wasting my time, but I am going to try this again.

    The IR activity exhibited by the atmosphere isn’t a warming effect at all, it’s a cooling effect! If the atmosphere was completely transparent to IR energy and completely unable to absorb or emit IR energy, the atmosphere would still warm via conduction with the planets surface, but would not possess any means of cooling to space.

    Under those conditions, near-surface air temperatures would be extremely hot 24/7, year-round. There wouldn’t be any ice anywhere on the planet.

    No, the IR activity that is exhibited by the atmosphere isn’t a greenhouse effect at all, it is a refrigeration effect, keeping our planet much cooler than it would be otherwise. It’s like this: the sun heats our planets surface, the surface heats the atmosphere, the atmosphere radiates that energy to space. It’s just that simple.

    Not to mention that atmospheric gases do not play any significant role. The IR activity exhibited by the atmosphere is completely driven by the latent energies of water changing phase in the atmosphere, condensing, fusing and sublimating. The role of gases is barely measurable.

    Why is that so difficult for people to understand? Too many years of indoctrination in malarkey? Probably.

    • wayne says:

      Cause most people can’t keep their eyes on the pea. Either too specialized in their area or scientifically ignorant. Both seem to get fooled.

      I agree with your view of a hot world without ir active gases, a quite hot atmosphere actually. Inversions at the surface everywhere but in the tropics would rule out even conduction from the atmosphere to the surface to radiate of any size to cool it.

      • Norman says:

        wayne,

        Why would inversions rule out even conduction? Conduction moves in any direction and an inversion would not stop its flow. It will always flow from a hotter body to a cooler one. If the atmosphere above gets warmer than the surface this energy will conduct downward to the surface.

        I am unsure of why you believe people to be scientifically ignorant if they do not accept GHG’s as coolants. What is your scientific evidence that your view is correct? Is it your belief, opinion or do you have some actual evidence to support your thought process?

        • wayne says:

          Norman, please don’t take the word ignorant as some put down, the word means someone does not know or has never thought on it and I know a lot of people that know little of science and spend no time trying to, no big deal. That is all I meant. I myself am ignorant in many areas of art, music, etc that others are fully versed or even experts. The word doesn’t bother me, sorry it bothered you, not intentional.

          As for the inversion, the tropics now have the full brunt of the 1361 Wm-2 which should bring the very surface to around 390 K and only able to convect if all of the column is not at the same temperature, cooler above. Like in your room all warm air would rise to the TOA with no way to cool. Initially any cooler air will subside to be warmed to this temperature or close to that. Near vertical stagnation would form there, convection grinds to nearly a halt. In the beginning air would travel to the poles to cool but the only way is by direct contact and slow conduction, the surface would be frozen solid and this creates an inversion with cold air right at the surface unable to convect, the hot air above not able to subside or radiate. That cooler surface air would slowly migrate toward the equator but there would be little of it compared to all of the entire atmosphere, conduction is linear by the temperature difference of every one meter layer, no fourth power. The air everywhere would be very hot yet the solid surface closer toward the poles would be very cold radiating any energy it does receive directly out to space.

          Basically a hot isothermal atmosphere with the surface air slightly cooler but only cooler by what direct conduction could perform.

          I guess it matter much whether you are thinking of the solid surface temperature, or the air’s temperature, or somekind of mean that makes little sense.

          • Norman says:

            wayne

            Hi and thanks for the reply. I am not bothered by your use of the term ignorant. I read lots of posts with opinions and no links to actual science that I tend to question things of that nature.

            Your evolution of the atmosphere without GHG could be a possibility but I do not think it is factual enough one way or the other to make a positive claim about it.

            The surface will radiate like a beast at night without GHG’s according to Roy Spencer (in a previous thread).
            http://www.drroyspencer.com/2015/04/why-summer-nighttime-temperatures-dont-fall-below-freezing/

            Plug some values into his model and see what you get.

            Here is a link from a paper by Judith Curry. Look at figure 2 of her paper. If you look at the temperature of the polar air 4 km above the surface it is at around -25C on day one. Even though the air is moving toward isothermal on the 14th day the air at this level has still dropped to -35C so conduction is slow but will still occur.

            http://curry.eas.gatech.edu/currydoc/Curry_JAS40.pdf

            At night without the GHG’s Roy Spencer demonstrated the temperarture of the surface would get quite cold and would also cool the lower levels of air. Until you can provide some really good support of your idea I think the overall Earth system would be as cold as claimed.

            Even though the moon gets very hot during its day it gets very very cold at night and its average temperature is much lower than Earth’s.

            I am not saying you are not correct but I would like to see some support material that verifies your claim. Keep it science and not belief on what might happen.

            Remember that is what Roy Spencer is fighting against. The IPCC uses computer models to project this massive heating that will occur in just a few decades melting all the polar ice and flooding most coastal cities. It is a belief and not a science. It is my opinion that Dr. Spencer is trying to bring science back into climate studies. One belief over another will not lead to a more scientific understanding. Hard evidence is a much better approach.

          • Norman says:

            wayne,

            For real world items. A car sitting in the sun gets very hot inside but at night it cools down to ambient conditions. I think once convection stopped in the atmosphere (and the upper atmosphere would still be much cooler than the surface because of the lapse rate) only a few meters of atmosphere would warm during the daytime and cool back down at night and with no GHG present the surface would get much colder and so would the air above via conduction.

          • Objectivist says:

            There’s a simple real-world example of how greenhouse gasses trap heat.

            Go visit any nice, hot, dry desert. It may well exceed 100 F during the day, but at night the low humidity and lack of cloud cover allow energy to escape like nobody’s business. It often grows uncomfortably cold, even with extremely high daytime temps.

            In the Sahara, a swing from 100 F to 31 F was observed in a single day…

            CO2 is a weak greenhouse gas, and has a low concentration in the atmosphere. Water vapor dominates it by orders of magnitude.

    • Norman says:

      justaguy

      Seems like in your post you are neglecting a very powerful cooling effect.

      You state: “It’s like this: the sun heats our planets surface, the surface heats the atmosphere, the atmosphere radiates that energy to space. It’s just that simple.”

      You are not considering the surface is radiating and doing so at a much higher rate than the atmosphere with GHG present. You should include that the sun heats the Earth’s surface, the heated surface radiates away heat based upon its temperature, the surface does heat the atmosphere but conduction is a very slow process in air (air is a very good insulator, heat moves slowly in air via conduction alone). The atmosphere radiates in both directions so some will radiate to space and some will return to the surface (the GHE).

      If you have no GHG in the atmosphere what is to stop the surface from radiating away heat much faster than can be moved into the atmosphere by conduction or even convection (much faster but very slow compared to radiant heat flow).

      I am not sure how you logically conclude that GHG act as coolants and the Earth would be warmer without them. Do you have more than your opinion to back up your statements?

      • Mike Flynj says:

        Norman,

        I know you didn’t ask me, but – maximum surface temperature Earth (with GHG) – 85-90C.

        Maximum surface temperature Moon (without GHG) – 107C?

        Moon heats a little quicker, of course. No atmosphere to prevent radiation reaching surface.

        I believe these are facts, not opinions. What do you think? GHGs cool or heat?

        • Norman says:

          Mike Flynj

          Mike thanks. The surface of the moon gets hotter not just because of the GHG preventing some IR from reaching the surface of Earth. It is also because the atmosphere on Earth allows for convection which moves heat rapidly away from a surface and keeps from getting to the higher 107 C that the moon experiences. The moon’s surface does say something about radiation though. It has a two week day and only gets to 107C because this is the equilibrium temperature for radiation in vs radiation out. If you slowed the radiation out the moon’s surface would get even warmer.

          • Mike Flynn says:

            Norman,

            I don’t think you are right.

            A greenhouse which prevents convection does not reach 107 C.
            A non convecting solar pond does not reach 107C.
            A maximally absorptive surface at the bottom of a vacuum chamber does not reach 107 C.

            My solar hot water system designed to as efficient as possible gets to maybe 85 C, even here at 12 S, on a really hot day, with no observable atmospheric convection here at ground level. Calm. Still. Extremely uncomfortable.

            With respect to the Moon, the temperature reaches its peak slightly quicker than the Earth. Please check the instrumental record if you wish. You may stand in front of an incandescent lamp with a filament temperature above 2000C. You will not get any hotter after a year than a day, or even an hour. The Moon and the Sun obey the same physical laws.

            With regard to slowing the rate of heat loss leading to warming, this is incorrect. Take a small hot potato. Allow it to cool to ambient. It cools. Reheat the potato, and place it in a vacuum flask, or other insulated container. It cools. It doesn’t get hotter, does it? It cools to ambient temperature, albeit more slowly than when not insulated. That’s what insulators do.

            Read the works of John Tyndall if you wish. The 1905 edition of Heat as Motion covers the experimental and personal accounts of this sort of thing rather nicely. Subsequent experiments have not managed to improve on Tyndall’s in any substantial way.

    • The surface of the Earth absorbs sunlight in its natural spectrum and emits energy in the IR spectrum – black body radiation. If there were no GHGs all of that IR energy would go directly into space. With GHGs some of it is absorbed and re-emitted back towards the surface. This causes the lower atmosphere to be warmer than it would be otherwise and the upper atmosphere to be cooler than it would be otherwise.

      Energy in equals energy out, regardless of the composition of the atmosphere. GHGs just change the distribution of that energy along the path between the surface and space. Phase changes between water vapor, liquid water and ice constitute the latent energies in that part of the atmospheric conditions, but they do not negate the GHG effect. In fact, water vapor itself is, by far, the strongest GHG in the atmosphere.

      The Earth’s energy balances at any given time are about as stochastic a system as you’re likely to ever encounter. While simple answers may be emotionally satisfying, they don’t describe the reality.

      • GHGs just change the balance between energy to space from the surface and energy to space from within the atmosphere and in the process change the rate of convective overturning and the atmospheric circulation.

        Compared to natural variation caused by sun and oceans we could never measure the changes from our emissions.

        GHGs warm the troposphere in ascending air but cool it in descending air for a zero net effect at the surface.

        They have no effect on the stratosphere because the temperature there is determined by the balance of ozone destruction / creation processes and they appear to vary differently above equator and poles in response to solar variability.

      • Mike Flynn says:

        Diane Merriam,

        I understand when a surface’s temperature is increasing, it is absorbing energy faster than it can radiate it away. And vice versa, of course. The Earth’s surface is in balance at most twice each 24 hours, at temperature maxima and minima.

        Only at these points is the temperature unchanging.

        With regard to energy absorbed from the Earth’s surface, you may have overlooked the fact that when the surface loses energy, it cools. GHGs cannot return all the energy to the surface from whence it came.

        An extreme example of this is noticed in tropical arid deserts at night. Temperatures can drop in excess of 50 C, due to the rapid loss of of EMR to the cold sink of outer space.

        This is reality.

  10. pochas says:

    Roy, I agree with what you say, but I’d like to inject an element of controversy. Yes, you can calculate an energy balance at the surface, and downwelling and upwelling fluxes, but they have nothing, nothing to do with surface temperature. Because the instant the lapse rate is infringed by the tiniest amount, the molecules in the air organize themselves like ranks and columns of tiny soldiers and perp walk those overheated molecules directly to the radiating zone, to be replaced by cooler troops from above. Surface temperature is determined by the mass of the atmosphere and the radiant flux from the sun. Everything else is a quibble.

    • You’re confusing convective and conductive heat transfer with radiative heat transfer. The mechanisms are totally different. It’s the interplay among them that causes weather.

      • Yes.

        Climate and weather are the consequence of conduction and convection removing radiative imbalances.

        http://www.public.asu.edu/~hhuang38/mae578_lecture_06.pdf

      • pochas says:

        “You’re confusing convective and conductive heat transfer with radiative heat transfer.”1

        No, I’m not. The lower 10,000 feet of the atmosphere is the convective zone. Above that is the radiative zone. The adiabatic temperature profile applies in both regions up to the tropopause.

        Without convection the convective zone would have superadiabatic daytime temperatures as high as 70C. Convection functions to reduce daytime temperatures to comfortable levels (for us) and recover the adiabatic temperature profile. It also removes any important effect from CO2.

        Amazing how our planet seems designed to support life. We are provided with a near-surface convective zone that stabilizes surface temperatures and supports the weather systems we depend on.

  11. donb says:

    Jeremiah 5:21

  12. geran says:

    Well, it’s good to see Dr. Roy did not use the phrase “traps heat”, or the word “blanket”, or state that “CO2 produces warming”. So, there is a slight movement away from the IPCC/GHE/CO2/AGW nonsense, and that is good.

    I think what many of the remaining “Lukewarmers” are being confused by are the heat transfer mechanisms of the atmosphere. As Dr. Roy mentioned before, he believes the fact that the Earth does not go to absolute zero at night is due to the GHE. That is NOT why the Earth does not go to absolute zero at nighttime. Heat transfer is not instantaneous. Heat transfer involves a “lag time”, or response time.

    But, there are so many things wrong with this latest post, I’ll just say respectfully that Dr. Roy and I “agree to disagree”. And hopefully, he will admit that the “science is NOT settled”, regardless of his beliefs.

  13. The hardest part of understanding heat transfer on a gut level is that there are three methods to it. Conduction and convection, based solely on physical movement, and radiation, unseen and unfelt. The first two are intuitively obvious from direct and constant personal physical sensation. The latter may be sensed at a sufficiently high level, but it’s easiest for the mind to simply hold that as an exaggerated case of the first two, when it’s actually totally different.

    Even when you know and can follow the math and intellectually understand the process, it’s hard to emotionally accept what it means. To prove to yourself that what is actually happening really is what’s actually happening.

    The difficulty in learning is not so much learning what you don’t know, it’s unlearning what you think you do know that just ain’t so.

    • Kelvin Vaughan says:

      Light a candle. See how close you can hold your finger to the side (radiation). Then see how close you can hold your finger to the top. ( radiation plus convection). I wouldn’t advise trying out conduction though.

      • Lick your fingers and pinch out the wick. That slight moisture barrier is enough to protect your fingers long enough to douse the flame by depriving it of oxygen. 🙂

        What I’m saying though is on an emotional level … for people who don’t understand the science, meaning the vast majority of the population. Everyone can understand convection and conduction. We experience convection every second of every day throughout our lives and conduction on a very regular basis on a purely physical sensation level.

        IR radiation is something we don’t see, can’t feel at all unless it’s very strong, and even then it’s easy to think it’s just more of the same. Gases are extremely poor conductors of heat, but that fact isn’t obvious. Your candle analogy would be explained as conduction to the sides and conduction plus convection above. That’s a good enough explanation for people to use as a practical guide for what they do or avoid doing and that’s all they want. No need for further analysis.

        Unfortunately that means they have no other basis on which to actually understand the greenhouse effect which makes them vulnerable to all kinds of emotional manipulation. Few people analyze. It’s too much trouble and would take learning a lot more than they are willing to put in the effort to do. Man is far more often a rationalizing animal than a rational one.

  14. mpcraig says:

    I have a fairly naive question. What would the surface temperature of Venus be if the atmospheric mass was similar to Earth and thus the surface pressure also close to one bar?

    • Notagain says:

      Venus having a higher surface temperature due to pressure is pure nonsense, unfortunately supported by some so called skeptics. For the same reason I mentioned in my previous comment. Not to mix up high pressure with compression. If it is true that the temperature there is that high, meaning higher than it can be possibly induced by the Sun, it can only be because of heat coming from underneath the surface.

    • gbaikie says:

      Hmm. Interesting question.
      Couple things Venus atmosphere is very dry, but it’s huge, in in total it’s a lot water. But assume it’s same dry air a Venus [or one removes 91 atm of CO2 with it’s water, rather than leave the water].
      So:
      “96.5% Carbon Dioxide (CO2), 3.5% Nitrogen (N2) ”
      http://nssdc.gsfc.nasa.gov/planetary/factsheet/venusfact.html

      At Venus distance from Sun it has 2700 watts per square meter.

      Venus day is longer than year, if same rotation as current
      Venus, the CO2 atmosphere would freeze out.

      But I assume you wanted that atmosphere, so give it 24 rotation and throw in Earth’s tilt of 23 degrees. And since CO2 doesn’t freeze on Earth we can easily assume it will not freeze if Venus has rotation [Btw Venus atmosphere upper rotates the planet every few days].

      CO2 is more transparent than nitrogen, so at noon at equator
      at surface I would guess one get 2400 to 2500 watts per square. And such a planet could have a lot of dust and could have a lot of wind- which complicated it. So assume no dust- or same amount dust as earth.
      So should heat surface to around 460 K [186 C]. And air temperature around 150 C.
      Now at equator at 4 pm or sun is at 30 degree angle, the sunlight hitting surface at angle and one get 1200 to 1250 watts per square meter. Or surface would warm to 90 C- or air could still be warmer than 90 C. Or could assume the air would lower than 100 C by the time the sun sets. And assume night time low could reach around 50 C [or around hottest day time air temperature which has occurred on Earth.
      So could have high swing of 100 C [or K] in day and night temperature at the equator.
      Now say equinox, and 45 degree latitude [North or South]
      So noon, sun at 45 degrees, and similar to equator- maybe 10 C cooler. So more than 1/2 world has average air temperature of around 100 C.
      Now Equinox at 60 degrees latitude [Oslo Sweden and below Cape Horn towards antarctic in south].
      Sun at noon is 30 degree above horizon. So like 4 pm ground
      does not warm above 90 C and might have air temperature high of 60 C and night might cool below 40 C [because of average air temperature of half the world.
      And say polar regions might get as cool as 20 C and could get above 50 C in summer.

      But then you to account for lapse rate, which should about 9 C per 1000 meter elevation. So if on 2000 meter high mountain at say 60 degree latitude, it’s on warmer side of warmest temperatures on Earth. But UV would be very harsh, long unprotected exposure is in lethal range- and/or very quick sun tans [or it would seem it would be brighter and stronger than tanning booths].

      • Good grief!!! There’s essentially NO water vapor in Venus’s atmosphere. The clouds are sulfuric acid, almost opaque to incoming solar radiation. No free hydrogen. No free oxygen. It has no magnetic field, so the solar wind has long since blown most of the lighter gases away. What little solar energy that makes it to the surface from the sun, combined with subsurface heat (it does still seem to have a molten core) doesn’t make it back out until it reaches equilibrium at over 450°C.

        The basic Ideal gas law is PV = nRT. Pressure times volume equals the amount of gas times the ideal gas constant times the temperature. All else being equal, as pressure goes up, temperature goes up. That’s basic thermodynamics. When you compress a gas (increase the pressure) it gets hot. Venus’s surface pressure is about 90 times that of Earth.

        The energy profile of Venus is far different from that of Earth in many ways. Aside from size and mass, they have very little in common. Almost nothing when it comes to atmospheric conditions.

        • Notagain says:

          Diane, there is no PROCESS of compression on Venus. Convection is not compression. There is high pressure though, compared to Earth. As for compressed gas in a bottle e.g., once compressed it will cool to the temperature of the environment.

          Even if you do not understand that, and you are not alone there, think of it in terms of energy. This theory is as nonsensical as GHE, since the “warmer” surface will produce energy out of nothing, which is absurd.

          • Convection involves simultaneous decompression in ascent and compression in descent and both are equal when the atmosphere is in hydrostatic balance which is when the upward pressure gradient force is equal to the downward gravitational force.

            The atmosphere as a whole never loses the potential energy created from kinetic energy by convection.

            It is simply recycled up and down constantly and requires an ‘extra’ 33K at the surface to drive the process.

            That ‘extra’ 33K comes from conduction and convection reducing the rate of radiation to space. For Earth the mass of the atmosphere requires a surface temperature 33K above S-B to maintain the hydrostatic balance.

            If GHGs upset the radiative balance then the rate of convective overturning simply changes to negate any thermal effect at the surface

            The enhancement of surface temperature above S-B is entirely a product of atmospheric mass, held withi a gravitational field and subjected to external insolation. and not the radiative capability of the constituent gases.

            That is why, after adjusting for distance from the sun, the atmsphere of Venus and the atmosphere of Earth are roughl;y the same at the same pressure.

            This is old but forgotten knowledge.

          • Notagain says:

            Stephen, there can not be any “‘extra’ 33K” since it would mean the “warmer” surface will radiate more energy away than it gets, which is physically impossible.

            It does not matter how plausible the explanations for that “‘extra’ 33K” might appear, they are all false. Because they “explain” something that is physically impossible. It’s a pity that quite a few people are trapped within that.

        • Are you suggesting that the surface is not warmer than the S-B equation would predict ?

          • Notagain says:

            I have made it clear what I am suggesting and why, no need to go in circles.

          • Mike Flynn says:

            Stephen Wilde,

            The Earth has a core of around 6000 K, I believe.

            It is surrounded by an environment of around 3 K.

            The Sun, effective temperature of around 5800 K is about 150,000,000 km distant.

            Now calculate the temperature of the surface, with and without the Sun. Even without the Sun, the Earths surface temperature must be above 3 K, unless it is a perfect insulator. This is obvious after a moment’s thought.

            So, S-B is irrelevant, wouldn’t you agree? It would only apply to a body with no initial temperature, ie at 0 K. Not the molten Earth, of course.

        • gbaikie says:

          — Good grief!!! There’s essentially NO water vapor in Venus’s atmosphere. —

          Yes, apparently that is true. Or said differently:
          Venus:
          Total mass of atmosphere: ~4.8 x 10^20 kg
          20 ppm water vapor
          http://nssdc.gsfc.nasa.gov/planetary/factsheet/venusfact.html

          Which would be about 10 times 4.8 x 10^14 kg of water.
          So, ~4.8 x 10^12 tons of water [4.8 trillion tonnes]
          Or 4800 cubic km of water.
          So Vevus is drier than Mars which has ~210 ppm of water vapor, but there is also is lot atmosphere on Venus.

          Another way to express it:
          Venus has 460 million square km and
          4800 cubic km divide by 460 million is .01 meters deep.
          So water of Venus would cover 1 cm of the surface.
          As compared to Earth:
          “The amount of water in the atmosphere at any moment in time is only 12,900 cubic
          kilometers, a minute fraction of Earth’s total water supply: if it were to completely
          rain out, atmospheric moisture would cover the Earth’s surface to a depth of only
          2.5 centimeters. ”
          http://earthobservatory.nasa.gov/Features/Water/page2.php

          So according to this, Earth has 2.5 times more water in it’s entire atmosphere as compare to the entire atmosphere of Venus.

          • gbaikie says:

            “So according to this, Earth has 2.5 times more water in it’s entire atmosphere as compare to the entire atmosphere of Venus.”
            Note I mean per square meter of surface area- or amount vapor
            above you.

            So I excluded this water additional water, or kept it so there was only 20 ppm water vapor. And though Venus has vast clouds of acid, it’s also small portion of the mass of entire atmosphere, so likewise kept it in portion, and so there would not be much acid either.

  15. Kevin Hearle says:

    I was recently at the New Zealand Governments consultation process for setting targets for Paris. The publics level of scientific understanding of the green house effect and the role of CO2 was virtually non existent, if not frightening. People actually believed that the temperature will rise 12 degrees C and that their grandchildren will fry.

    What is more important than the overall process is the miniscule effect of increasing CO2 in that process.

    I am sure that your repeated efforts to dispel this ignorance are sorely needed. The people at the consultation were in the main politically part of the NZ Green Party who had a political agenda. Where were the average NZer’s, home watching TV while the rabid green blob determine their fate for Paris and that is again frightening.

    Given that our Politicians are on average as scientifically illiterate as most of the population we are doomed not by increasing CO2 but by the policies to be initiated by the scientifically illiterate politicians.

  16. gbaikie says:

    –The greenhouse effect usually refers to a net increase in the Earth’s surface temperature due to the fact that the atmosphere both absorbs and emits infrared radiation.–

    Greenhouse is suppose to increase average temperature.
    The moon is much colder than Earth during the night but much hotter than the Earth during the day.
    Mars is similarly as cold in it’s night and has little greenhouse effect. And since Mars has 24 hour day it’s like Earth but Mars orbital distance has about half the solar flux as compared to Earth. And therefore Mars cooler day, but Mars temperature during the day does reach to about 20 C- and so Mars daylight can warmer than in parts of Earth during it’s it’s daylight.

    –This GHE temperature increase is frequently quoted as being around 60 deg. F, thus keeping the Earth from being an ice planet, since its average surface temperature is somewhere around 59 or 60 deg. F. —

    Earth is a planet with ice on it. It should noted that F conversion to C is +32 times 5 and divided by 9 and that 33 C is not equal to 59 to 60 F. And GHE is suppose to cause 33 C [or +33 K] to Earth average temperature.
    And if Earth average temperature was freezing [0 C or 32 F]
    Most of Earth would not be ice nor freezing.
    Earth would always have a warmer tropics than the rest of the world. Or non tropics will always be colder than the Tropics. And colder Earth, as during glacial periods was mostly about a cooler non tropical areas.

    And warmer periods in Earth history, say when Earth’s average temperature was about 25 C, it was the non tropical regions which became warmer [or nearer to Tropical average temperature].
    And though I don’t Earth has ever had an average surface temperature of -18 C, but even at that average temperature the tropics would not be frozen. Earth is simply too close to the Sun and the Earth surface gets about 70% of the sunlight in area of 40% of the total surface area, which lie within 23 degree north and south latitude.

    –This 60 deg. F warming attributable to the GHE is actually incorrect; the greenhouse effect on surface temperature, if left to its own devices, would actually be at least twice that strong…more like 140 deg F average surface temperature…–
    There many places on Earth which the surface ground temperature exceed 140 F.
    But Earth never had an air temperature of 140 F.
    And GHE could not have an effect upon causing the ground temperature to increase. And little effect upon causing a higher daytime or highest air temperatures.

    • At least two or three times in Earth’s history it has been completely, or almost completely, iced over. Google snowball earth for more information.

      Our orbit changes, our axial tilt changes, continents move around, ocean currents change, atmospheric composition changes, volcanic activity changes, airborne particulates change, solar activity changes.

      The Earth is a very dynamic place, just not on the scale of a human lifetime. And bless our little hearts, if we don’t actually see something we would consider unusual in the short span of our personal existence with our own eyes, it just doesn’t seem real.

      Perfect example … The California drought. What is it, four years now? Oh – it’s the worst drought in recorded history!!

      That depends on what you’re recording with because that have been at least two extreme drought periods in parts of California that have lasted over 200 years, just in the last millennia. And many, many of them that have lasted over 20 years. This may have been to worst in the last 100 years (actually, an unusually wet period for the region) but it’s entirely normal for the area. Nothing to be surprised by at all. It was just before any of us saw it so it isn’t quite real and so we feel that this must be something completely different and new.

      • gbaikie says:

        “At least two or three times in Earth’s history it has been completely, or almost completely, iced over. Google snowball earth for more information.”

        Snowball earth is a hypothesis.
        It’s not even considered a theory. Greenhouse Effect is regarded as theory. One could argue whether Greenhouse effect should be regarded as hypothesis rather than theory. Or since so vague and people are so certain of it, one could call it non science- pseudoscience or a rather silly belief.

        There are at least 6 hypothesis of how the Moon formed, so at least 5 are wrong, the the most favored hypothesis is the giant impact hypothesis:
        https://en.wikipedia.org/wiki/Giant_impact_hypothesis
        I also tend to agree that it the most likely, though accept the possibility it’s wrong, or basically I tend to regard as a work in process. So the giant impact hypothesis is correctly called an hypothesis which could develop someday into a theory.

        • jerry l krause says:

          Hi gbaikie,

          “There are at least 6 hypothesis of how the Moon formed, so at least 5 are wrong” Very, very, important point. In science there is only one right answer which we can never be sure of. But in science we have the possibility of eliminating the wrong hypothesis by observation. This is hard because we cannot perform laboratory type experiments with natural systems. But we often ignore what we do know. Case in point: We know the moon has a 1-2cm extremely good insulating layer on its surface. But few who compare the temperature of the moon to the earth consider the obvious consequences of it.

          Have a good day, Jerry

      • jerry l krause says:

        Hi Diana,

        While your comment has nothing to do with the GHE, it has everything to do with the GHE. Irrigation changes climate on the solid earth part of this planet. And I just came back from a part of Oregon which is only about fifty miles north of California and it is greener (natural rain) than I have seen it since 2005.

        Have a good day, Jerry

  17. jerry l krause says:

    Hi Roy and others,

    In his previous post, Roy in his response to my request for his accurate definition of the greenhouse effect wrote: “Of course it’s [greenhouse effect] been defined. It’s the downwelling IR flux at the surface. More importantly, it is measured continuously at many sites around the world.” To which I replied: “I agree fully with this definition for I too have always observed the result of this downward IR flux from a clear sky during the nighttime with my ‘modified’ Suomi, Staley, Kuhn net radiometer.”

    Now you (Roy) write: “The greenhouse effect usually refers to a net increase in the Earth’s surface temperature due to the fact that the atmosphere both absorbs and emits infrared radiation. … This GHE temperature increase is frequently quoted as being around 60 deg. F, thus keeping the Earth from being an ice planet, since its average surface temperature is somewhere around 59 or 60 deg. F.” If this is your accurate definition of the greenhouse effect, I cannot agree and I have brought observations to your attention which refute this proposed (hypothesized) consequence of the downward radiation from a clear sky. But you did not stop here with a 59 or 60 deg. F colder average earth’s atmospheric temperature observed about 1.5m above its surface. You continued: “This 60 deg. F warming attributable to the GHE is actually incorrect; the greenhouse effect on surface temperature, if left to its own devices, would actually be at least twice that strong…more like 140 deg F average surface temperature …but most of that theoretical surface temperature rise … .”

    Observation is my thing because Galileo’s science was built on actual observations which refuted several well accepted theories. Early in 2014 I saw that a very common observation seemed to simply refute the greenhouse effect as just stated by Roy. It was that the atmosphere’s temperature had never been observed to cool below the atmosphere’s dewpoint temperature. This regardless of location, altitude, or time. Hence, the low temperature of the diurnal atmospheric temperature oscillation could never cool below the atmosphere’s dewpoint temperature. Hence, the greenhouse effect could not be a factor in limiting the nighttime cooling. For various reasons, one of which I am a nobody, I was not successful in generating any interest in my ‘discovery’. So, I finally turned to responding to whatever of Roy’s posts because I was sure that he was a scientist who did not embrace the hypothesis, just stated by him, of the greenhouse effect. I was successful in attracting his attention as he replied: “yes, the dewpoint temperature is always below the temperature. Not sure what you are claiming that proves, Jerry.” (Climate Polling Results Lead to Weird Press Coverage August 13, 2014).

    In a response to Roy’s April 10, 2015 (Why Summer Nighttime Temperatures Don’t Fall Below Freezing) I called his attention to a portion of the historical weather record (available at wunderground.com) observed May 10 and 11, 2015 at the nearest airport to Huntsville AL. For the period (a little more than 10hrs) between sunset on the 10th and sunrise on the 11th, the sky was observed to be continuously clear. The recorded observation just before sunset for the atmosphere’s temperature was 82.9 (degrees Fahrenheit) and that for the atmosphere’s dew point temperature was 65.3. After sunset the first recorded observation for the temperature was 80.2 and for the dew point was 65.1. Before sunrise on the 11th the temperature was 69.6 and the dew point was 65.3. After sunrise the temperature was 67.6, the minimum temperature of the 11th, and the dew point was 65.1. About these observations I made the following comment.

    These observations do not refute the greenhouse effect. In fact they might seem to support it because the atmosphere’s temperature never decreased to the atmosphere’s dew point temperature. A critical observation of this night and morning is never reported because it is not directly related to the atmosphere. But it is very easy to observe if it exists. It is the dew which is commonly observed to form on surfaces, totally exposed to the clear sky, when their temperatures do cool to the dew point temperature. Blades of grass, roofs of cars and houses are some of these common exposed surfaces. The sensor of the atmospheric temperature is not directly exposed to the clear sky because the temperature it would measure during the daytime, if directly illuminated by direct solar radiation, would not begin to reflect that of the atmosphere. I often have observed that dew does not form on the top of my van parked under a tree at the same time dew has formed on the tops of nearby cars fully exposed to the clear sky. This explains the temperature difference observed between the temperature of the atmosphere (measured under a roof 1.5m above the surface) and the observed dew point temperature of the atmosphere when dew has clearly begun to form on exposed surfaces well before sunrise.

    It’s getting late and I know some do not appreciate my long comments. If interested you can find my continued comments in Roy’s May update and his other posts to this one.

    Have a good day, Jerry

    • Of course the temperature can never fall below the dew point. The definition of the dew point is that temperature where, given a certain amount of water vapor in the air, it begins to condense out. The water that condenses out reduces the amount of water that’s in the air, so the dew point at that concentration is lower still. A natural process is what it is. Identifying it doesn’t suggest anything new.

      Nor is the temperature of a given volume of air the same as the temperature of another volume of air at a different but close point. It’s cooler in the shade than in the sun. The reverse is true at night. Then you have to figure in latent heat from the immediate surroundings – whether that’s heated buildings or other enclosed areas such as vehicles, the bigger, the more heat retained or biological processes such as decay or any of a number of other sources.

      • gbaikie says:

        — Diane Merriam says:
        June 14, 2015 at 2:58 PM

        Of course the temperature can never fall below the dew point. The definition of the dew point is that temperature where, given a certain amount of water vapor in the air, it begins to condense out. The water that condenses out reduces the amount of water that’s in the air, so the dew point at that concentration is lower still. A natural process is what it is. Identifying it doesn’t suggest anything new.–

        I would say there is a lot which known, but that doesn’t mean it’s not ignored.
        Everyone knows that CO2 is trace gas. Everyone knows that CO2 doesn’t warm very much.
        Everyone knows there is more than trillion tons of CO2 in the Atmosphere. Everyone knows 1 billion is 1/1000th of a trillion. Everyone knows that seasonal variation in amount of CO2 in the atmosphere is larger than amount of CO2 added from the burning of hydrocarbons.
        But this does stop idiots from being afraid of human CO2 emission.

        The issue is how much effect occurs to average temperature related to the dew point.
        Keep in mind the claim of greenhouse effect theory is that only the radiant effects of greenhouse gas increase the average temperature.
        And dew point has *nothing* to do with radianr effects of greenhouse gases.
        So if dew point is related to increasing the average temperature of earth by 1 C, it refutes the greenhouse effect theory- and this scale of warming is greater than what could be caused by increase in CO2 levels on Earth in the last 100 years.

  18. gbaikie says:

    –…more like 140 deg F average surface temperature…but most of that theoretical surface temperature rise is short-circuited by convective heat loss from the surface caused by convective air currents, in turn caused by the greenhouse effect, which also largely creates the weather we experience.

    That’s right – without the greenhouse effect, we would not have weather as we know it. The greenhouse effect, energized by solar heating, creates weather.–

    I would say with water becoming a gas- therefore becoming a part of the atmosphere- that this factor is related to a lot weather we see.
    But I think a rotating planet and any atmosphere will cause wind- and wind normally considered part of weather.
    And one would get convective if a Planet only had a nitrogen atmosphere and sunlight reaching and warming the surface.
    And convection of gases is directly connected to gravity. Or in the international space station [which lacks gravity] you don’t get convection- and the addition of greenhouse gases won’t create convection for ISS.

    –The GHE is somewhat controversial among some skeptics, probably because we can’t “see it” the way we can see visible sunlight and the resulting heating of surfaces sunlight falls upon – a rather non-controversial cause-and-effect process. It instead involves infrared (IR) light, which we cannot see, but which is an essential part of the energy flows in our climate system…and in most other systems that generate heat. You can actually feel if it is sufficiently strong (e.g. radiant heat from a stove or fire).–
    The heat of fire or stove is Near Infrared or shortwave IR,
    And the heat from the sun is almost entirely from visible light and Near Infrared light part of electromagnetic spectrum. The GHE is dealing with is longwave infrared, of very low intensity.

    –The atmosphere contains “greenhouse gases” (GHGs), which means gases which are particularly strong absorbers and emitters of IR radiation. —

    Per definition, any gas which are particularly strong absorbers and emitters of IR radiation, are called greenhouse gases. But liquids and solids are stronger absorbers and emitters of IR radiation. Or gas tend not to be absorbers and emitters of most electromagnetic energy- and gases generally tend to be transparent.

    –Now, recall I said that temperature is a function of rates of energy gain and energy loss. Thus, those energy flow arrows marked with an “X” in the above diagram represent huge flows of energy which can affect temperature, if they really exist.–

    As said above most gases are transparent, but this doesn’t mean that all light passes thru something which is transparent and things which are transparent tend to reflect and scatter light.
    If passing vertical thru the atmosphere, light would go thru about 10 tons of transparent gases per square meter.
    Normal window glass has density of about 2600 kg per cubic meter, so if have window glass about 4 meter thick that is roughly like earth atmosphere- both are transparent and both you could see thru. Or one could compare it to 10 meter depth of water- water is also transparent to visible light. But if leaving earth not vertically, you go thru more atmosphere.
    So if leave at 45 degree angle, you go thru
    about 1.4 times more atmosphere, and at 30 degree angle is twice as much atmosphere.
    And similar to 40% of earth surface is the tropics, if leaving surface in random direction 40% of the direction would below 23 degrees, or about 1/2 would below 35 degrees
    or 45 to 90 degree is about 25% of the direction something going at random direction would go.

    And sunlight is reflected and scattered going thru earth atmosphere. Or some of spectrum of sunlight absorbed by gases but all of the spectrum has some portion which is reflected/scattered by our atmosphere- blue light does this more strongly- and hence the blue skies.
    And so the IR light also would need to get thru this transparent barrier.
    But anyhow, Mars has 25 trillion tonnes of CO2, and I do think that all this CO2 does slow the cooling, but it doesn’t slow it by much.
    “There is always water in the atmosphere. Clouds are, of course, the most visible manifestation of atmospheric water, but even clear air contains water—water in particles that are too small to be seen. One estimate of the volume of water in the atmosphere at any one time is about 3,100 cubic miles (mi3) or 12,900 cubic kilometers (km3). ”
    http://water.usgs.gov/edu/watercycleatmosphere.html
    1 billion tonnes times 12,900 is 12.9 trillion tonnes, hmm less tonnage of liquid water than Mars CO2.
    I wonder how many trillion of tons of water vapor there is in Earth’s atmosphere. Anyhow, thought there was more water
    than Mars CO2. But it seems as a guess, that clouds would better at delaying radiant heat from leaving as compared to Mars atmosphere.

  19. coturnix says:

    I find it still a little bit confusing as far as the surface temperature goes, because it is not the surface tenperature that the meteorologist measure but the boundary layer air temperature. Like, just few days ago i measured, the air temperature was +29C but meanwhile the dry soil grass or rocks were between 45C and 55C.

  20. Cathy S. Lewis says:

    Poor Dr. Spencer. What a tragically lonely man. If he tries to convince scientists that his personal theories are correct, he is met with scorn and criticism and ridicule. If he tries to convince his blog readership that standard atmospheric physics are correct, he is met with scorn and criticism and ridicule.

    I must confess that I admire your endurance.

    CS

    • Kelvin Vaughan says:

      I don’t see scorn criticism and ridicule. I just see a lot of alternative theories and a lot of interest in this web site.

  21. RW says:

    Roy,

    “The bottom line, then, is the Greenhouse Effect, due mostly to greenhouse gases, is largely caused by the fact that the atmosphere emits IR energy downward, the so-called “back radiation”. This single component of the whole GHE process basically then determines all of the other features of the greenhouse effect and leads to net GHE warming of the Earth’s surface.”

    I would say this is not correct. First of all, you need to differentiate between ‘back radiation’, i.e. downwelling IR emitted from the atmosphere to the surface, and just downward emitted IR towards the surface. They are not the same thing, though of course some IR ultimately being passed to the surface is a requirement of downward IR emission at all levels of the atmosphere.

    The GHE is first due to the fact that some of constituents of the atmosphere, i.e. GHGs and clouds, are largely opaque to surface emitted IR acting to cool the surface. That is, they absorb or ‘block’ most of the surface emitted IR from being transmitted into space (this amount is quantified by the spectral absorptivity evaluated at the temperature of the surface). The GHGs and clouds then re-radiate this absorbed IR energy both up towards space and down towards the surface. The atmosphere must make the push toward radiative balance with the Sun at the TOA by re-emitting absorbed surface IR up towards space (i.e. it must ultimately pass the required amount of IR into space that wasn’t already transmitted to space from the surface), but in order to do that it must also push back the other way, because absorbed IR is re-radiated by the atmosphere both up and down. The downward IR push towards the surface which must occur in addition to the upwards push towards the TOA to achieve balance is the fundamental underlying driving mechanism of the GHE. Everything else is 2nd order or a subsequent mixed effect, including downward IR emitted from the atmosphere to the surface (which has multiple sources in addition to surface IR absorbed by the atmosphere and re-radiated back to the surface).

    “You can measure the greenhouse effect yourself with a handheld IR thermometer pointed at the sky, which measures the temperature change caused by a change in downwelling IR radiation. In a clear sky, the indicated temperature pointing straight up (“seeing” higher altitudes) will be colder than if pointed at an angle (measuring lower altitudes). This is direct evidence of the greenhouse effect…changes in downwelling IR change the temperature of a surface (the microbolometer in the handheld IR thermometer). That is the greenhouse effect.

    If I’ve make a mistake in the above, I’ll fix it.”

    Yes, I think you have made a mistake and this is not correct either. Downwelling IR at the surface is really just the consequence of there being downwelling emitted IR at all levels of the atmosphere and some of that downward emitted IR is transparent to the constituents of the atmosphere (and passes through to the surface). Dowwelling IR to the surface itself doesn’t say anything about or establish the existence of the GHE. I remain perplexed why you think it does.

    • RW says:

      Roy,

      I should add that because of the underlying driving mechanism I laid out, the net combined result of everything is the surface and lower atmosphere must be at a higher temperature and be emitting IR at a higher rate in order to achieve balance at the TOA. That is, they must emit at a higher rate in order to achieve radiative balance at the TOA, or they must emit at a higher rate in order to ‘push through’ and ultimately pass the required IR flux into space in order to achieve balance with the Sun. If the constituents of the atmosphere were not opaque to surface IR acting to cool the surface, or the atmosphere only re-radiated absorbed IR up towards space, there would be no GHE due to GHGs in the atmosphere.

      Downwelling IR at the surface has little if anything to do with the GHE. It’s more a consequence of the underlying physics of the GHE than a driver or indicator.

    • RW says:

      Roy,

      “What those people need to do is go read a book on atmospheric radiation, say Grant Petty’s A First Course In Atmospheric Radiation. I know Grant, and he is a brilliant and careful scientist.”

      Maybe he is, but he has a kind of hyper and eccentric personality that makes having a logical and systematic discussion really difficult!

      Really smart and first rate scientist though, I agree.

  22. If Roy defines the greenhouse effect as the rise in surface temperature caused by greenhouse gases then he is wrong and the rest of his narrative becomes flawed.

    The rise in surface temperature is caused by the tendency of conduction and convection through the mass of an atmosphere to slow down radiative loss to space.

    Greenhouse gases are not rquired for a decline in temperature with height. That decline is caused by the density gradient with height which allows expansion of gases with height and thus the creation of potential energy, which is not heat and does not radiate, at the expense of kinetic energy.

    Convection would still occur even without GHGs because uneven surface heating would cause density differentials in the horizontal plane which inevitably leads to convective overturning so no isothermal atmosphere is possible.

    The space surrounding a spherical object increases exponentially with height. That is quite unlike the space in a vertical column so the experimenmts involving vertical columns are invalid.

    Once surface kinetic energy becomes entrained in convective overturning it becomes unavailable for emission to space and remains locked into that convective overturning in a never ending cycle for so long as the atmosphere remains suspended off the surface in hydrostatic balance.

    That is old science as set out in the Gas Laws and the concept of hydrostatic balance whereby an atmosphere can only be held off the surface to the extent that the upward pressure gradient force is balanced with the downward force of gravity.

    The upward pressure gradient force is fuellled by kinetic energy at the surface. That kinetic energy (33K for Earth) is engaged in holding the atmosphere up and is not available for radiation to space.

    That is the true Greenhouse Effect and it is independent of GHGs.

    If a radiative imbalance occurs as a result of GHGs then convective adjustments negate it.

    http://www.public.asu.edu/~hhuang38/mae578_lecture_06.pdf

    Every ascending or descending column of air is a reservoir of potential energy that was initially drawn from kinetic energy at the surface by conduction and convection. It is the return of potential energy to kinetic energy during the descent phase that warms the surface 33K above S-B.

    That applies to every planet with an atmosphere regardless of the amount of GHGs which is why the temperature within the atmospheres of Venus and Earth are approximately the same at the same pressure after adjusting only for distance from the sun.

    That was all established science before the radiative theory came to the fore about 20 to 30 years ago

    • Bob says:

      S. Wilde wrote:
      The space surrounding a spherical object increases exponentially with height.

      Are you sure? I thought it expanded quadratically. X**2 grows much more slowly than exp(X).

      Bob

    • Tim Folkerts says:

      Stephen says:
      “Greenhouse gases are not rquired for a decline in temperature with height. ”

      and earlier:
      “This discussion is basically about whether GHGs are required to achieve a decline in temperature with height.
      All else follows from that.”

      Stephen, most will agree that uneven heating (both poles vs equator and day vs night) will result in convection even in the absence of GHGs. I think this is one place where Dr Spencer overstated his case a little. I agree with him that there would not be “weather as we know it”, but there would still be some significant wind, etc in the atmosphere.

      OTOH, you vastly overstate your side when you say “all else follows from that.” Convection & the subsequent lapse rate is NOT all that matters. Your GHG-free atmosphere would have convection and would have a lapse rate, but it would be doing it from a surface that has an average temperature of ~ 255 K rather than ~ 288 K. (And yes, I know those numbers are very broad approximations, but the GHGs WILL cause a higher average surface temperature.)

      • gbaikie says:

        “OTOH, you vastly overstate your side when you say “all else follows from that.” Convection & the subsequent lapse rate is NOT all that matters. Your GHG-free atmosphere would have convection and would have a lapse rate, but it would be doing it from a surface that has an average temperature of ~ 255 K rather than ~ 288 K. (And yes, I know those numbers are very broad approximations, but the GHGs WILL cause a higher average surface temperature.)”

        GHGs will cause a slighter higher average temperature.
        GHGs do not cause a higher surface temperature.

        The lunar surface warms to about 120 C, GHGs will not cause
        a higher surface temperature than 120 C.
        With a greenhouse on Moon one could get an air temperature of around 120 C- is really hot- highest sauna temperatures approach about 100 C air temperature.

        And on Earth a parked car have air temperature of about 70 C. A parked car on Moon: 120 C, Earth: 70 C.
        The high air temperature is due to inhibiting convection, and the Moon one gets as much 1400 watts and on Earth one gets about 1000 watts per square meter of direct sunlight at noon and clear skies.

        Anyways also it should be noted that Earth more than 1000 watts of sunlight because it also gets about 70 watts of indirect sunlight at noon [and Moon does not get such addition of indirect sunlight].
        Or wiki:
        “If the extraterrestrial solar radiation is 1367 watts per square meter (the value when the Earth–Sun distance is 1 astronomical unit), then the direct sunlight at Earth’s surface when the Sun is at the zenith is about 1050 W/m2, but the total amount (direct and indirect from the atmosphere) hitting the ground is around 1120 W/m ”
        https://en.wikipedia.org/?title=Sunlight

        And it appears to me as a guess that the 161 watt absorbed by earth surface in chart of global heat flows, does not include this indirect sunlight.
        The Moon does not have an additional component of indirect sunlight because the Moon lacks an atmosphere to scatter the sunlight.

        And on Earth at noon with sun at zenith with clear skies the sunlight is passing thru the least amount of atmosphere- or when it’s 4 pm, the sunlight passes thru more atmosphere and this reduces the amount of direct sunlight, but would also cause more than 70 watts of indirect sunlight.
        So that chart isn’t merely lacking a ratio of 1000 vs 1100 watts [10% off] but rather it’s probably closer to 20% or more.

  23. TB says:

    Roy,
    thanks so much for the time you spend on this. You may be able to explain something for me.
    Whenever I see articles on the GHE as in this one the emphasis is always on the RADIATIVE effects of the GHG. In my mind a CO2 or H2O molecule absorbing IR energy would vibrate more vigorously (ie its temp. would rise). This extra molecular energy would be dissipated by collisions with other molecules nearby (ie heat energy transfer by CONDUCTION). But this is never talked about. Is there something I’m missing?
    Thanks again for your time

    • Sometimes that energy is transmitted conductively, but gas molecules are so far apart they are a very poor conductor of heat, even when taking convection into account. That’s why double pane windows are such good insulators. Generally, however, the photon of absorbed energy is re-emitted and the molecule drops back down into a more stable lower energy state.

      When you’re talking about greenhouse gases specifically, the only thing that differentiates them from any other gas is their radiative absorption spectrum, so the conduction/convection modes of heat transfer aren’t pertinent.

      • They are pertinent because any energy absorbed radiatively results in faster convection.

        Faster convection negates the potential thermal effect of GHGs at the surface by creating more PE (not heat)within the vertical column at the expense of KE (heat).

        Convection adjusts to correct for radiative imbalances:

        http://www.public.asu.edu/~hhuang38/mae578_lecture_06.pdf

      • TB says:

        Thanks Diane and I see what you mean about double pane glass. I would note that a vacuum flask has air removed between the two sides of the flask to increase efficiency so I would infer there would be SOME energy transfer by direct conduction from molecular collisions. Yes they are relatively far apart but they move fast – the speed of sound in fact.
        I would feel it is a bit relevant as any energy conduction would heat the layer of air that the GHG molecule was in and not immediately radiate it away. And I would wonder how the proportion of radiant/conductive heat transfer varies with altitude and temp. Maybe someone knows of a study that has been done?
        Perhaps I’m wrong but if a GHG molecule immediately sheds its high energy state through any means it is then ready to absorb the next IR photon coming its way which would suggest that a small amount of a GHG would be capable of absorbing a lot of energy in its wavelength range. Am I wrong to think that this implies a GHG would reach a saturation level in the atmosphere?
        Thanks again for your time

  24. gbaikie says:

    –So, let’s now think through what happens as sunlight enters the climate system. As the Earth’s surface absorbs sunlight it warms up.As it warms up, it emits more and more IR energy, limiting its temperature rise (remember “energy balance”?).–

    The most common place on Earth is the ocean. And it’s surface does not warm up much, and it’s temperature is governed by evaporation.

    –If the atmosphere could not intercept (absorb) any of that surface-emitted IR energy, the energy would readily escape to outer space and as a result it has been estimated that the Earth’s average surface temperature would be only about 0 deg. F–

    Most of area is ocean and not warmed as much as a land surface or land surface would radiate far more radiant energy
    during day from sun shining on it- though land also evaporate
    a significant amount of water and if ground is wet will remain cooler until it’s surface dries up.

    But say we talking unusual spot of Earth which is dry and it’s noon with sun high in sky, with about 1000 watts per square meter. So surface might warm as high as 70 C.
    If you could stop all conduction and convection losses it might reach 90 C. And difference of 70 to 90 C is about
    200 watts.
    Now part of being able to reach ground surface temperature of 70 C [158 F, would be warm air, say around 50 C [120 F]. And air were less than 40 C [100 F] the cooler air may take away too much heat to reach 70 C,
    [[Btw the 70 C [158 F] is suppose to be temperature needed to fry eggs [on sidewalk].]]
    So ground is dry, and not any wind, air is near 50 C, with 1000 watts of sunlight can reach 70 C and would convecting about 200 watts per square meter of heat.
    Leaving you with say 600 watts of radiant energy.
    And as understand it, of 600 watts, about 300 watts goes directly into space. And so the most amount radiant energy from the surface which can be blocked or absorbed by the atmosphere [preventing it going directly into space].

    And that chart says 396 watts is emitted from surface and it
    is getting 333 watt from the sky.
    Now the sky would not add any heat to 70 C surface.
    One could bounce the sunlight down underground and if the air was 50 C, and sunlight was 1000 watts per square meter
    it should reach 70 C. Or if on the moon and have sunlight striking at angle so it’s 1000 meter per square meter and
    it should warm a surface to 90 C.

    So on average according to chart the sky absorbing 356 watts from the surface- and average, day and night, ocean and land, clear and cloudy [etc].
    Returning ocean, it’s getting most of solar heating below the surface [and does not radiant heat under the surface].
    And as said the ocean surface temperature [like a wet ground] is governed by evaporation. Or why ocean surface never gets above 35 C.
    So something on the moon which was 35 C would radiate about 510 watts per square meter. [Though 35 C water on the moon would explode and instantly become ice.]
    And look at a satellite map of IR from Earth, tropic is bright red.
    So seems only possible way to get an average number of 356 or the 396 watts is because tropics and tropic ocean.
    Now if tropic ocean surface temperature was cold, say 10 C [363 watts] it’s surface not going warm up in one day.
    Or put ice in tropical swimming pool so it’s 10 C, and it will not warm up within one day.
    Or one cubic meter of water require 84 million joules to warm 20 C- we can assume the swimming pool is deeper than 1 meter.
    Or if mixed the top 100 meters of tropic water with 1000 to 1500 meter tropical water, one immediately cools the ocean and it takes decades to warm up again. And since it seems
    atmosphere require ocean heat to come anywhere near close.
    So mix entire ocean- have surface to 100 meter 4 C [333 watts].
    And also of course there no way that back radiation could warm ocean [Ocean not transparent to IR whereas sunlight is]. So ocean heat is unrelated to this back radiation. So got ocean which allows the large number of 396 watts and then got the 333 number disconnected to it, doesn’t cause ocean to warm nor does it heat in term heating land area so the land area radiated more energy to the sky.

  25. Svend Ferdinandsen says:

    This paper can put it all in perspective
    http://www.geocraft.com/WVFossils/greenhouse_data.html
    It compares the total effect, and not just the differential changes, but anyway.

  26. dave says:

    Dr Spencer says:

    “convective currents…caused [sic] by the greenhouse effect…”

    The main cause of convection over the oceans is the fact that water vapor from evaporation displaces denser dry air, and destabilizes the atmosphere.

    • jerry l krause says:

      Hi Dave,

      You wrote: “The main cause of convection over the oceans is the fact that water vapor from evaporation displaces denser dry air, and destabilizes the atmosphere.”

      This is a very, very, insightful observation and very, very, important if we are to understand the world’s atmospheric circulation. I would change a few words because I do not like the word: displaces. I would edit your statement to read: The main cause of convection over the oceans is the fact that water vapor from evaporation dilutes denser, drier, air which destabilizes the atmosphere by creating low atmosphere pressure area.

      I describe it as insightful because I have never found an author who has written something similar to your observation. For the best example of where this occurs is the tropical ocean region known as the doldrums, of which the ancient mariners were well aware. Because I have thought at length about this I will take this opportunity to share what I have considered.

      It seems best to begin where Richard Feynman began The Feynman Lectures On Physics as he taught physics students at Caltech.

      “If, in some cataclysm, all of scientific knowledge were to be destroyed, and only one sentence passed on to the next generations of creators, what statement would contain the most information in the fewest words? I believe it is the atomic hypothesis (or the atomic fact, or whatever you wish to call it) that all things are made of atoms—little particles that move around in perpetual motion, attracting each other when they are a little distance apart, but repelling upon being squeezed into one another. In that one sentence, you will see, there is an enormous amount of information about the world, if just a little imagination and thinking are applied.” (pp1-2)

      From ‘little particles that move around in perpetual motion’ came the kinetic molecular theory of gases from which we learn the relationship between the average kinetic energy of the atoms (molecules) motions and that observation we call temperature. But if the little particles have different masses, we have also learned, based on the definition of kinetic energy, that the different particles have different average speeds. Water molecules have a molecular mass of 18amu, nitrogen molecules a molecular mass of 28amu, and oxygen molecules a molecular mass of 32amu. So, to have the same average kinetic energy, the particles having the lesser mass must have an average greater speed. It is commonly stated that, if it were not for the convective mixing of the atmosphere, the atmosphere would be a layered cake with the bottom layer being composed of the most massive particles and the top layer containing the least massive particles, which in the case of the three atmospheric molecules being considered would be the water molecules.

      The doldrums can be defined as large, humid, tropical ocean regions having a stagnant (little convection) atmosphere for weeks at a time. One might think that once the surface atmosphere becomes saturated with water vapor (molecules), further evaporation of molecules would cease. But the faster moving water molecules are diffusing upward into drier, denser, atmosphere above. Hence, more water molecules evaporate to replace those diffusing upward. Normally, diffusion as a transfer process is considered to be a slow process, relative to atmospheric convection.

      Now, I begin to imagine. However, many, many, water molecules can be added to and distributed upward by diffusion because the atmosphere is stagnant for weeks. Hence, a large volume of this stagnant atmosphere becomes nearly saturated with water molecules, thus creating what is termed a tropical low pressure region. This large tropical low pressure region (doldrum) is surrounded by atmospheres having higher pressure at the ocean surface. So at the edges the higher pressure atmosphere begins to flow under the lower pressure atmosphere, which is a huge volume of atmosphere nearly saturated with water molecules. And as the drier, more dense atmosphere flows under the lower pressure atmosphere, this nearly water vapor saturated atmosphere is lifted and adiabatically quickly cooled to its dew point temperature so the condensation of the water vapor begins. Starting the atmospheric heat engine, fueled by water vapor, which convectively lifts the formerly nearly saturated surface layer to great heights in what is commonly called a thunderstorm. Now be reminded I am thinking a little and imagining a lot as I imagine that this thunderstorm could be the beginning of a tropical storm which might grow into a hurricane or typhoon.

      And this is why I commented: This is a very, very, insightful observation and very, very, important if we are to understand the world’s atmospheric circulation.

      Have a good day, Jerry

  27. CoRev says:

    Dr Roy and anybody, I have been trying to get a reasonable approximation of the residence time for the IR photon. Dr Roy says: “… although in general the rates of absorption and emission are not the same since absorption is mostly temperature-independent but EMISSION IS VERY TEMPERATURE DEPENDENT.” If the mechanism for emission is temp change how long before the change (residence time), how much change, and what are the mechanisms for the temp change.

    Please, this has been a black hole in my understanding.

    • dave says:

      CoRev says:

      “…residence time for IR photon…”

      There is a distinction between the residence time for “a particular joule” of added heat, and the residence time of the photons which successively carry that energy – ultimately out to space.

      If, to calculate the latter time, one says, rather arbitrarily, that each photon travels one hundred meters from emission to absorption, then some ten thousand photons will be involved before final escape, they will travel a total of one million meters at the speed of light, and the total accumulated residence time of all the photons will add up to one three-hundredth of a second. Therefore, the energy in the atmosphere actually “locked up” in IR photons at any time is negligible.

      The times BETWEEN the successive photon emissions will be much greater than the times of actual existence and flight.
      Absorbed photonic energy “relaxes” in molecules almost instantly, so that general heat is always reborn. The time before re-admission will depend on the temperature of the atmosphere containing the greenhouse gas. But, essentially, one is asking how long a body*, cooling by radiating its heat and starting at 301 K, will take before closing in to within 1/2 K of a similar body starting at 300 K. That is easy calculus, using the Stefan-Boltzmann law as an approximation for calculating the governing rates. You will recall that this law indicates that the difference between 301 K and 300 K is that the hotter body radiates more than 1% more energy per second. Anyway, the answer here is “a few minutes”. Now, if this has to happen vertically with a couple of hundred slices of atmosphere you have to multiply a few minutes by 200 and you get “a day or so,” as the total residence time of modest excess heat in the atmosphere. If you wish, model the atmosphere as a couple of thousand slices and you will get “a week or so,” as an indication of the upper limit on residence time.

      Of course, none of this is applicable in the troposphere where convection is far more important than radiation in moving heat upwards.

      *in this context, a gas having some of the character of a black-body

      • CoRev says:

        Thank you. I feel fulfilled. My black hole of knowledge is filled. 😉 (smiley face)

  28. mpainter says:

    The Kiehl, Trenberth diagram does not properly add up. It gives 333W/sq. m of DWLWIR “absorbed by the surface”. Add the 161W/sq. m of insolation, and the earth’s surface absorbs a total of 494 W/sq. m radiation.
    Some simple math:
    Assume sea surface cooling is 60% via evaporation. Ocean=71%of earth’s surface, hence:
    .60×.71×494= 207 W/sq. m. Compare this to the figure of 80W/sq. m of evaporation in the Kiehl, Trenberth diagram.
    These sort of diagrams show how badly distorted is the GHE by the present body of “consensus” science.
    The GHE is real, but alarmist doctrine has generated a science of dubious value.
    IMO, the back radiation “absorbed” at the earth’s surface is egregious and the total radiation emitted at the earth’s surface is wrong, as given by such diagrams.
    Satellite imagery confirms that most radiation is emitted at cloud tops, providing observations which confirm the error of such diagrams.

    • MikeB says:

      It does add up, unless of course you want to assume something unwarranted, as you do.

      Where is your evidence that 60% of sea surface cooling is by evaporation? A number picked out of the air with absolutely no justification (and totally wrong).

      The KL figure of 80 watts per square metre can be justified. Your made up figures can’t.

      Unfortunately, this is the sort of nonsense that gives sceptics a bad name. It is why they are ignored.

  29. mpainter says:

    Mike B ,
    A number of studies have partitioned ocean cooling into its various components. These give slightly different figures, but the figure of 60% via evaporation, 28% via radiation, and 12% via conduction and sensible heat loss would be representative. Some put the heat loss via evaporation at over 65%.
    There is no question sea surface cooling is predominately via evaporation. And no, I made nothing up and I suggest that you do a little research on the issue before you fling about wild accusations.

  30. Ulric Lyons says:

    I think it is confusing when it is said that without a greenhouse effect but still with 0.3 albedo that the planet would be only 255K, when the 255K is actually the approx temperature of the greenhouse effect.
    239W/m2 (255K) downwelling IR, plus 159W/m2 solar shortwave, makes 398W/m2, giving a surface temperature of just over 289K:
    http://www.spectralcalc.com/blackbody_calculator/blackbody.php

    • MikeB says:

      You can’t do it like that, Ulric. The Stephan-Boltzmann Law, which I assume you have used, is not based on the ‘inputs’ to a body, but solely on the radiation ‘emitted’ by the body.

      A blackbody at 15 deg. C (288K) will emit 391 watts per square metre, irrespective of any heat losses from convection, evaporation, conduction or whatever.
      The amount a body emits depends only on its own temperature and its emissivity. The ‘inputs’ to the system do not enter the calculation.

      The 255K is the blackbody temperature of the Earth, i.e. the temperature it appears to be when viewed from space. The greenhouse effect is the difference between 255K and the actual SURFACE temperature of the planet.

  31. mpainter says:

    MikeB
    To put it another way,
    .16 × 494W/ sq. m =
    80 W/ sq. m
    To claim that 80 W/ sq. m is the correct figure for evaporation is to claim that evaporative heat loss from the ocean is less than 16% of total loss from the ocean. Do you adhere to that position?

    • MikeB says:

      Ok, work it out yourself, like this:
      What is the global annual rainfall? – LOOK IT UP

      How much latent heat of condensation does this represent? – WORK IT OUT

      Do you get 80 watts per square metre? – If not, work it out again CORRECTLY

  32. mpainter says:

    Mike Flynn,
    One degree per million years means that the earth’s surface has cooled 65 degrees since the beginning of the Tertiary and 600 degrees since the beginning of the Cambrium.

    • Mike Flynn says:

      a painter,

      No it doesn’t. There are factors at work which mean that the rate of cooling since the presumably glowing white hot surface of the Earth is not linear.

      1. Radiation intensity is proportional to the fourth power of the absolute temperature. Temperature will fall faster at first.

      2. Heat from radioactive decay series. Initially high, this will decrease logarithmically depending on the half lives of the initial elements, their concentrations and the same factors applied to their decay products.

      3. Heat from the Sun, which varies moment to moment. It is quite possible that at one time, the Sun’s output was considerably lower than today – the “faint Sun”.

      4. Changes to the conductivity and surface emissivity of the ever changing mantle, crust, and surface, aquaspheric, biosphere and so on.

      An example, albeit very simple, is observing a white hot chunk of steel being worked by a blacksmith. It loses heat very rapidly at first. Quite quickly, it ceases to appear even a faint red, and achieves a “black heat”. Depending on size and ambient temperature, I would suggest you check carefully before picking up an apparently cold chunk of metal with your bare hands.

      It will continue to cool, ever more slowly, until it reaches ambient temperature.

      Even when heating steel for annealing purposes, it takes a while before the centre is at the outside temperature. A rule of thumb, at around 900 C, is to allow 2 hours for every 25 mm of thickness, and hold this for a while longer.

      I could go on, but I’ll quit, and give you a rest.

  33. Bryan says:

    The Greenhouse Effect?
    Which Greenhouse Effect?

    Various version’s of the ‘effect’ claim 33K or even 90K increase in earth surface temperatures’.
    This increase is claimed to be mainly a radiative effect.

    Willis with his Steel Greenhouse gives an extreme example in a model that explains this extraordinary radiative phenomena.
    The planet surface temperature ends up being 1.9 times the initial surface temperature.

    However can any of them be called an ‘effect’ in the same sense as say the Photoelectric Effect?

    I doubt it.

    When examined carefully it turns out that its almost entirely an insulation effect.
    Various complicated and interrelated interactions involving conduction, convection, radiation and latent heat contributing.

    What about the Expanded Polystyrene model?

    With a constant heat source like the Willis model the final surface temperate reached is almost infinite.
    Is the atmosphere more like expanded polystyrene than a vacuum with a thin absorbing and radiating shell?

    In reality neither model is in any way realistic or even claim to be.
    So there is no mystery or effect when an absorbing medium like the atmosphere raises the temperature of the surface.

  34. mpainter says:

    Diane Merriam,
    actually, TSI is roughly one half IR, but this is mostly in the SWIR spectrum.

    • Right. The effect of the greenhouse gases is in the longer wavelengths.

      • dave says:

        Dianne Merriam says:

        The effect of the greenhouse gases [sic,plural] is in the longer wavelengths [of IR].

        Not true for the main “greenhouse gas” – water vapor.

  35. jimc says:

    One common mistake people make when making projections into the future is to assume a static situation. Luddites, Hobesians, Marxists, Progressives, and Environmentalists seem especially vulnerable to this malady.

    Are we to assume there will be no scientific and technological progress in this century – that there will be no realization of fusion energy, better batteries, and things not even imagined? Not according to what I’ve seen in the past. The above may seek to stymie progress by stifling economic growth, but hopefully, they won’t succeed.

    RWS posits that 1 deg C may be realistic by double CO2. I don’t think it will be the same world then though.

  36. Fred Miller says:

    One thing that must be remembered in the discussion of “back radiation”, is that the molecules of carbon dioxide and water in the atmosphere are randomly oriented. This is based on fundamental understanding of gases. Thus, for any given region of the atmosphere, there is an equal probability of these molecules emitting IR radiation in all directions. On a macro scale this translates into equivalent emission towards space and towards the ground. Thus “back radiation” is real.

  37. dave says:

    “re-admission” should be re-emission”.

  38. wayne says:

    Diane Merriam says:

    “The sun itself actually emits very little radiation in the IR frequency spectrum … ”

    Diane, you seem to try to speak with some authority but in that statement you are not correct. By Planck integrated blackbody irradiance with a cutoff between shortwave (visible/UV) and infrared at 760 nano-meter wavelength and sun temperature of 5777 K you will find that out of the 63.2 MW/m² that there is 28.6 MW/m² of infrared and 34.6 MW/m² of shortwave and I personally would never call that “very little”, how about “somewhat smaller” of the ir that the sun emits.

    However, if you isolate out the UV that is absorbed in the far upper atmosphere and the high degree of visual frequency albedo then you will find that there is actually quite a bit more infrared than the shortwave frequencies entering Earth’s lower climate system.

    Just trying to not have readers leave here with some incorrect assumptions. If you were instead just ignoring near infrared, please state it so.

    • Yes, I was leaving the shorter wavelengths out since the discussion is about the greenhouse effect and that occurs in the longer wavelengths. I do need to remember to put ALL of my conditions in. Thanks.

  39. It comes down to this which is, is the strength of the GHG effect the result of the climate/environment or is the GHG effect the cause of the climate.

    Thus far when viewing the data presented at the recent Heartland Climate Conference which you Dr. Spencer were part of, the evidence they presented ,along with even more sources of data beyond what they presented all strongly suggest that the GHG effect is in response to the climate/enviroment not the cause.

    Until that changes the relative strength of the GHG effect is really not materially important to the climate for if the climate cools the GHG effect will diminish.

    In addition to much emphasis is being put on the CO2 aspect of the GHG effect and not enough on water vapor. If water vapor has a negative feedback with CO2 then AGW theory is toast.

    The lack of a lower tropospheric hot spot ,the fact that water vapor in the upper atmosphere seems to be correlated to ENSO, the fact that temperature always and is still leading CO2 concentration changes, the fact that as CO2 concentrations increase the saturation factor lessens CO2’S effects, the fact that CO2 concentrations were much higher in the geological past and Ice Ages took place all make a very strong case that AGW theory based on CO2 increases amplifying the GHG effect and thus driving the climate is weak to say the least.

    • geran says:

      And, in case you don’t have time to be “indoctrinated”, here’s just one important quote:

      “But some of the heat is trapped by the greenhouse gases in the atmosphere.”
      ++++++

      There’s that “trapping” thing again.

      Does it get any funnier?

      • And somehow they don’t actually give the conclusion that’s obvious from their last point. They just gloss right over it. That being that since we don’t know whether clouds are a net greenhouse enhancer or moderator, we really don’t know, when it’s all said and done, what the net effect of any changes in the other greenhouse gases will actually be.

  40. In support of what I said in my previous post. Time will tell.

    Presentation of Evidence Suggesting Temperature Drives Atmospheric CO2 more than CO2 Drives Temperature

    Guest Blogger / 19 hours ago June 13, 2015

    Note: I present this for discussion, I have no opinion on its validity -Anthony Watts

    Guest essay by Allan MacRae

    Temperature, among other factors, drives atmospheric CO2 much more than CO2 drives temperature. The rate of change dCO2/dt varies ~contemporaneously with temperature, which reflects the fact that the water cycle and the CO2 cycle are both driven primarily by changes in global temperatures (actually energy flux – Veizer et al).

    To my knowledge, I initiated in January 2008 the hypothesis that dCO2/dt varies with temperature (T) and therefore CO2 lags temperature by about 9 months in the modern data record, and so CO2 could not primarily drive temperature. Furthermore, atmospheric CO2 lags temperature at all measured time scales.

  41. Erik Magnuson says:

    Roy,

    Nice write-up and appreciate the information on Grant Petty’s book. I have a bit of experience with radiation transport involving neutrons and high energy photons, but only a cursory exposure to IR.

    The whole brouhaha over sky temperature is kind of funny for me, one problem in an engineering heat transfer class from 38 years ago was about how much above freezing the ambient air temperature could be and still allow frost to form. One assumption given for the problem was, IIRC, a sky temperature of -140F (obviously clear skies).

  42. ossqss says:

    Physics is, well, physics. Nice job Doc.

    Per your request, first sentence last paragraph.

    • FTOP says:

      Yes, physics is physics.

      Which is why the K&T energy budget is nonsensical. It attributes greater forcing to the atmosphere than the sun.

      I appreciate the efforts and belief of Roy, but the GHE effect fails to explain so much.

      LWIR is unable to heat water, yet the earth is 70% water surface
      All objects release IR, but the hotter the object the broader and more complete the spectrum. Lava rock glows from its heat. Is it 33C warmer from CO2
      Thermal capacity plays a huge part. Air vs. water or air vs. rock carry fundamental heat flow challenges based on thermal capacity.

      Measuring IR establishes an object is above absolute zero. It is not a given that it will warm things around it.

      A couple of thought experiments.

      1) if the atmosphere was separated from the earth so it existed in a ring 100 meters above the surface holding all the same properties as the current atmosphere but condensed to a thin layer at its average temperature of -18C Eliminating convection and conduction. This would not effect IR exchange. Would one expect the earth surface to be 33C higher?

      2) we exhale air with a higher CO2 content than the atmosphere. If CO2 forcing is so high how can blowing on an object cool it?

  43. Norman says:

    Mike Flynn,

    Not sure where this post will end up in the thread as the webpage is in the weird mode. The difference between Earth and moon surface temperature can be explained by the energy budget Roy Spencer has embedded in his post. The atmosphere reflects and absorbs 157 watts/meter^2 that do not make it to the surface where on the moon they would.

    Mike I do respect your posts but your potato example is not a valid analogy to the Sun/Earth system. For a valid potato example you would have to add energy to the potato constantly (some portions of the Earth are always receiving energy input). Put a heater in a vacuum flask of hot liquid and see what happens. Will the liquid now cool? If you are continuously adding energy to the Earth system and you slow the rate of cooling the end result will be a warmer final equilibrium temperature than if you did not slow the rate of energy loss. Energy in constantly at a certain rate, slow rate of energy loss and the system ends with more energy and when this energy is randomized it will exhibit itself as a higher temperature. I think so many on this sight do not understand this fundamental. They are of the false understanding that Carbon Dioxide and water vapor warm the surface and use many examples to show how absurd this is. The carbon dioxide and water vapor are slowing the loss of radiation from the system (not the only way the surface can cool but the only way the Earth system as a whole can cool) and with a constant input of energy, if you slow the output you end with more energy. The concept really cannot get much easier than this. No further complications are needed. Energy In-Energy out will determine the net energy of the system. Change either one (more energy in same out) and you have more energy in the system and will have a warmer temperature. Same energy in and less out and you still end with more energy in the system.

  44. KuhnKat says:

    Respectfully Dr. Spencer, you have never answered my question. We are at the bottom of the atmosphere with the highest density of air. The IR thermometer measures the temperature of whatever is immediately in front of it. Why does it show the temperature of the air molecules thousands of feet up, according to you, rather than the temperature of the molecules actually banging against the thing when you point it straight up??

    Until you can reasonably explain that to me I cannot believe in the worth of the instrument for measuring anything to do with the atmosphere.

    • Dr. Strangelove says:

      Gas molecules are not banging the IR thermometer. IR photons are doing the banging. Some photons from distant stars are also banging but mostly from atmosphere.

      • kuhnkat says:

        Actually the gas molecules and atoms ARE banging the freaking IR thermometer as it is right there in your hand in the middle the densest part of the earth’s atmosphere. They are NOT banging the actual sensor probably.

        So, you are saying the stars photons are hitting it. Yup. Now explain to me why the largest number of photons coming from right in front of the sensor is NOT causing it to give a reading for the temp of the gasses right in front of the sensor as opposed to an average of ever fewer molecules as you go up including those few stars photons and maybe the moon…

        If you cannot give a solid physical explanation for this reading then the IR thermometer is NOT telling you what you think it is and this type of sensor system is NOT reliable for measuring the temp of the atmosphere.

        • Dr. Strangelove says:

          The photons in front of you are from one-inch thick air. The atmosphere is 20 km thick. More photons.

          • Kristian says:

            “More photons.”

            The more photons you get in, the higher the temperature reading, one would expect. So measuring the entire 20 km of generally cold atmospheric column would give a higher reading than only measuring the one-inch warm air layer right next to the instrument, is that what you’re saying? If so, the instrument is not doing what it’s supposed to do, now is it?

  45. Dr. Strangelove says:

    “You can measure the greenhouse effect yourself with a handheld IR thermometer pointed at the sky, which measures the temperature change caused by a change in downwelling IR radiation.”

    There’s a loophole is this argument that Slayers can exploit. Look at your chart. The atmosphere absorbs 78 W/m^2 from incoming solar radiation, 80 W/m^2 in latent heat and 17 W/m^2 from thermals. Even without greenhouse gases, there would be downwelling IR radiation to balance the heat inflow in the atmosphere but not 333 W/m^2. The Slayers would say the atmosphere absorbs more than 333 W/m^2 from the three mechanisms mentioned to explain away the observed back radiation.

    • Ernest Bush says:

      But there are CO2 and water vapor in the atmosphere working as absorbers and emitters and they must be accounted for in your reasoning.

  46. SteveB says:

    You don’t think your FLIR can measure outbound LW radiation? Do you think it can only read IR coming towards the surface?

    • Ernest Bush says:

      It should be obvious that the sensor in the instrument must be struck by a photon (or energy wave) to register its presence. How, then, does a photon (or energy wave) speeding away from the instrument register its presence. Instruments that produce infrared video (this is where I have experience) have clusters of sensors that are sequentially scanned. To produce video each sensor must be struck by more or less photons to produce an image. The bigger the emitter, the lighter the spot in the processed image.

      • SteveB says:

        Sorry, I said FLIR when I meant to say IR thermometer. The IR thermometer measures temperature, not downward radiation….IR moving out of the atmosphere has a temperature and pointing an IR thermometer at it will register a temperature whether the radiation is moving away or coming towards you.

  47. Ernest Bush says:

    Dr. Spencer, rest assured most of us follow and accept what you are saying based on the evidence presented. There are other skeptics who have tried to explain the “greenhouse effect” with similar results in the comments.

  48. http://icecap.us/images/uploads/TomQuirkSourcesandSinksofCO2_FINAL.pdf

    CO2 GHG EFFECT A CONSEQUENCE OF CLIMATE/ENVIROMENT.

    If so AGW theory is history.

  49. jerry l krause says:

    Hi Mike,

    Because I have no formal training in meteorology and climatology, I prefer to quote the knowledge of those who are recognized authorities in these sciences. I am sure I have shared the following several times during the past 10 or so months, but I again will share it because it is my opinion that clouds are the earth’s thermostat.

    R. C. Sutcliffe in his 1962 book, Climate & Weather, also wrote: “All this may seem a far cry from the general circulation of the world’s atmosphere but the detail serves to point the moral, that one cannot explain the broad features of world climate if one does not know the actual mechanisms involved.” (pp 138) after he had written: “The climatic importance of clouds lies in their effectiveness in reflecting, absorbing, transmitting, and emitting radiation, … .” (pp 34). While I am not certain what Roy’s position is on the mechanism of reflection by clouds, he has, in a reply to my question, clearly stated that clouds absorb the long-wave radiation from the earth as Sutcliffe wrote: “Long-wave radiation from the earth, the invisible heat rays, is by contrast totally absorbed by quite a thin layer of clouds, … .”

    While I have some formal training in physics, I certainly do not claim to have the theoretical understanding behind many of its theories. So I turn to a physicist who was an authority: Richard Feynman. The following is an excerpt about light scattering, from The Feynman Lectures On Physics, which Feynman taught physics students at Caltech.

    “One interesting question is, why do we ever see clouds? Where do the clouds come from? Everybody knows it is the condensation of water vapor. But, of course, the water vapor is already in the atmosphere before it condenses, so why don’t we see it then? … We have just explained that every atom scatters light, and of course the water vapor will scatter light, too. The mystery is why, when the water is condensed into clouds, does it scatter such a tremendously greater amount of light?” I omit the next paragraph (pp 32-8) because it is the theoretical reasoning, which I do not understand, and go to the next paragraph that I claim to understand.

    After this omitted paragraph Feynman taught: “If we have N atoms in a lump, which is a tiny droplet of water, then each one will be driven by the electric field in about the same way as before (the effect of one atom on the other is not important; it is just to get the idea anyway) and the amplitude of scattering from each one is the same, so the total field which is scattered is N-fold increased. The intensity of the light which is scattered is then the square, or N2-fold, increased. We would have expected, if the atoms were spread out in space, only N times as much as 1, whereas we get N2 times as much as 1! That is to say, the scattering of water in lumps of N molecules each is N times more intense than the scattering of the single atoms. So as the water agglomerates the scattering increases. Does it increase ad infinitum? No! When does this analysis begin to fail? How many atoms can we put together before we cannot drive this argument any further? Answer: If the drop get so big from one end to the other is a wavelength or so, then the atoms are no longer all in phase because they are too far apart. So as we keep increasing the size of the droplets we get more and more scattering, until such a time that a drop gets about the size of a wavelength, and then the scattering does not increase anywhere nearly as rapidly as the drop gets bigger. Furthermore, the blue disappears, because for long wavelengths the drops can be bigger, before this limit is reached, than they can be for short wavelengths. Although the short waves scatter more per atom than the long waves, there is a bigger enhancement for the red end of the spectrum than for the blue end when all the drops are bigger than the wavelength, so the color is shifted from the blue toward the red.”

    While Feynman never refers to the invisible wavelengths that are longer than those of visible solar radiation, it seems plain that what Feynman taught would apply to these longer, invisible if the diameter of the cloud droplets were 20µm, which according to Sutcliffe is the approximate diameter of ordinary cloud droplets. And if ordinary cloud droplets have this approximate diameter, it would seem this scattering mechanism should also apply to the long, invisible, heat rays being emitted from the earth’s surfaces. And I see that Feynman probably misspoke, as mortals tend to do from time to time, when he taught: “So as we keep increasing the size of the droplets we get more and more scattering, until such a time that a drop gets about the size of a wavelength, and then the scattering does not increase anywhere nearly as rapidly as the drop gets bigger.” It seems, to make sense, the last three words were intended to be: as the wavelengths get longer.

    I have yet to read anyone writing about what Feynman taught his students about the scattering of light by cloud droplets. So I do not know if what I understand about it is valid. But I know there are problems with Sutcliffe’s and Roy’s mechanism that even thin clouds nearly totally absorb the longwave radiation being emitted from the earth’s surface. Before I review this problem I must state that I accept that clouds in general “emit heat continuously according to their temperatures, almost as though they were black bodies” as Sutcliffe wrote. I accept it because I know this statement is based upon the observation of the emission from the cloud top and the temperature of the cloud top. And because of these two observations it must be concluded that the cloud is blocking the transmission of the longwave radiation being emitted from the much warmer earth surface beneath a high, thin, cold, cirrus cloud. Hence, there is the question: By which mechanism does the high, thin, cold cirrus cloud block the transmission of these ‘invisible heat rays’ and return then back to the earth’s surface where they can be observed by radiometers?

    And, of course, I have not found anyone who questions how these ‘invisible heat rays’ are able to pass, unhindered, from the earth surface to the bottom of these high, cold, cirrus clouds and then back to the earth’s surface where they can be observed by radiometers.
    Have a good day, Jerry

    • Mike Flynn says:

      Hi Jerry,

      Like you, I’m a Feynman reader (of sorts).

      I don’t know whether you have read a short Feynman book, QED the strange theory of light and matter.

      You might find, as I do, sufficient explanations of why, for example, clouds can reflect light of wavelengths from UV, through visible and IR, to the much longer wavelengths of weather RADAR systems. Observable fact and experiment support Feynmans writings.

      I would prefer Feynman to Sutcliffe if Sutcliffe claims total absorption rather than partial absorption and partial reflection. Of course, what happens to the absorbed energy is a whole ‘nother question.

      Have fun.

  50. Derek Alker says:

    Excerpt –
    “This GHE temperature increase is frequently quoted as being around 60 deg. F, thus keeping the Earth from being an ice planet, since its average surface temperature is somewhere around 59 or 60 deg. F.”

    Interesting way to describe the bare earth model Roy, without actually calling it that. 0 degrees Fahrenheit being minus 17.77778 degrees Celsius.
    http://www.metric-conversions.org/temperature/fahrenheit-to-celsius.htm

    Roy, would you say that the massive heat capacity of the oceans is a negative or positive feedback? I ask because they are obviously not present in the bare earth model, but they are present on earth’s surface with an atmosphere. Have you taken the effect this has into account at all? If so, how much would it be? Would it be more or less than the supposed “greenhouse effect”?

    ALSO, you write –
    “(Technical diversion: This is where the Sky Dragon Slayers get tripped up. They claim the colder atmosphere cannot emit IR downward toward a warmer surface below, when in fact all the 2nd Law of Thermodynamics would require is that the NET flow of energy in all forms be from higher temperature to lower temperature. This is still true in my discussion.)”

    PLEASE stop misrepresenting what the Slayers said. What the Slayers stated (I was involved in the email discussions) is that there is no observable WARMING EFFECT at earth’s surface by atmospheric back radiation. NOTE that is NOT saying there is no atmospheric back radiation, but it is saying it has no observable warming effect at earth’s surface. AND, that as the atmosphere at altitude is far colder than the surface then such a proposed “mechanism” is in violation of the second law of thermodynamics that states, colder (or the same) can not heat hotter (or the same).

    Also, in regard of the global energy budgets today on facebook you have written –
    “The only component of the energy fluxes directly related to temperature is outgoing IR from the surface (396), which through the Stefan-Boltzmann equation gives 15 deg. C (assuming a broadband IR emissivity near 1.0)”

    YET, most reasonable estimates of actual surface temperature, NOT near surface air, or just under the water surface ocean temperature readings, or measured by satellite atmospheric temperature, suggest that the average surface temperature is nearer 27C to 28C (roughly 81F). Quite a difference, most are not aware of…

    • Mark Stoval says:

      “… PLEASE stop misrepresenting what the Slayers said. …”

      Good luck on that! The luke-warmers are never going to address the real argument against their belief in back-radiation warming the earth’s surface. Never.

    • Tim Folkerts says:

      I’m not sure what your point was, Derek.

      Roy claimed that ‘slayer’ think back radiation violates 2nd Law of Thermodynamics.

      You (a self-admitted ‘slayer’) say “such a proposed “mechanism” is in violation of the second law of thermodynamics” — which is exactly what Roy said! Rather than ‘misrepresenting’, he exactly stated what you yourself say IS the slayer position!

      Furthermore, you claim “the second law of thermodynamics, that states, colder (or the same) can not heat hotter (or the same).”
      Well, there is no heating going on here, so your objection is invalid. The atmosphere and ‘back radiation’ merely LIMIT the ability of heat to leave the surface. The back radiation does NOT heat the surface (in the strict definition of thermodynamic heat).

      During he winter, the insulation around my house is cooler than the interior of the house. By your warped thinking, the insulation would be forbidden to make the interior warmer, since it is cooler that the interior.

      “Roy, would you say that the massive heat capacity of the oceans is a negative or positive feedback? “
      It is neither. Heat capacity merely serves to even out temperature swings; the days are cooler and the nights warmer than without this large thermal reservoir. [This smoothing actually has a minor warming effect, but this has to do with the T^4 temperature dependence of radiation and is not germane this this particular discussion).

      … suggest that the average surface temperature is nearer 27C to 28C (roughly 81F)
      Really? Do you have a reference for this claim? I can believe that the tropics have such an average for the year. I can believe much of the mid-latitudes have such an average for the summer. But as an annual global average this just doesn’t pass the ‘sniff test’.

      • gbaikie says:

        — Tim Folkerts says:
        June 25, 2015 at 9:38 AM

        I’m not sure what your point was, Derek.

        Roy claimed that ‘slayer’ think back radiation violates 2nd Law of Thermodynamics.

        You (a self-admitted ‘slayer’) say “such a proposed “mechanism” is in violation of the second law of thermodynamics” — which is exactly what Roy said! Rather than ‘misrepresenting’, he exactly stated what you yourself say IS the slayer position!

        Furthermore, you claim “the second law of thermodynamics, that states, colder (or the same) can not heat hotter (or the same).”
        Well, there is no heating going on here, so your objection is invalid. The atmosphere and ‘back radiation’ merely LIMIT the ability of heat to leave the surface. The back radiation does NOT heat the surface (in the strict definition of thermodynamic heat).

        During he winter, the insulation around my house is cooler than the interior of the house. By your warped thinking, the insulation would be forbidden to make the interior warmer, since it is cooler that the interior. —

        The insulation around your house prevents conductive and convective heat losses, a paper thin wall blocks radiant energy.
        If one had a paper thin house, the outside heat easily transfers to inside and heat inside easily transfer outside,
        but it this isn’t about a radiant transfer, it’s about inhibiting the convection and conduction of heat.

        If there not an atmosphere- say on the moon, a paper thin house could work fine in terms of insulation. [Though if want a pressurized interior the walls have to be strong enough to withstand this pressure- just as you would if you wanted the interior on Earth to be say 12 psi higher than outside pressure.]

        Any warm object radiates indirect radiation.
        The forms of direct radiation one can find on Earth is sunlight and man made things which are designed make directed radiation.
        Sunlight is directed light because it traveled tens of millions of km [Earth on average is 149.6 million km from the Sun].
        The Sun is enormous ball of hot plasma which emits indirect
        light in all direction. Not only does it radiate light in all directions because it’s spherical, but every square meter [or square cm] of it’s surface radiate from a flat surface in all direction [radiates towards the sun, but only radiates away for the sun if in a direction away from the Sun.
        So does back radiation warm the sun. No.
        Or the lack of the radiation not emitting away from the sun does not warm the sun. Or it’s a silly way to look at it.
        Or chunk of metal radiates at 100 watts per square meter, we don’t consider it radiating 100 watts per square meter at itself.
        So only if you want to over complicate something [and confuse yourself {and every other idiot].

        Now if an object is in vacuum. Say 1 cubic meter of iron.
        So six sides with 1 meter square. And it’s heated within it so it’s radiating 100 watts per square meter, and has 6 square meter [total of 600 watts radiating from it, into space]. Now put another 1 cubic meter box, so it “floats” 1 meter away from it, have one of it’s sides facing the other side.
        Because it’s one meter away from other box and the heated box is emitting indirect light, the other box is not going to be heated by 100 watts. Or any radiant heat leaving heated box at a less than 45 degree, will miss hitting non heated cube-except for the corners.
        And most of the radiant energy leaves a surface at less than 45 degrees.
        It made the heated box emit more directed light, one could have heat the heat cube more. So simply putting golf ball dimples on heated cube would increase amount of radiate light which reached the other cube.
        Of course at microscopic level surfaces are not made completely flat- and on gets variation of polished surfaces [and etc].
        So radiant light can be represent as emitting into a hemisphere- the flat surface cuts a sphere in half.
        And if you cut earth in half at the equator, about 40% of the surface area of that hemispheric dome is below 23 degrees [tropical line of Capricorn or Cancer] and at say 40 degree latitude is more than half of that Earth’s hemisphere.
        So one could start by having 1 meter diameter hemisphere [which has height of 1/2 meter] and putting in the middle of the 1 meter square side of box. So on 1 cm square area in middle of square 1 meter box, more than 1/2 of the energy does not reach the other cubic box [even if the other box were instead of 1 meter away it was 1/2 meter away].
        So one could calculate how much of the 100 watt reached the other box side which was 1 meter away, but we could make the box in such way that 20 watt of the 100 watts reaches the other box. Now this 20 watt per square meter at this point could be called directed light.
        Or roughly speaking, it will follow the rules of directed light [if it traveled a million km it would more “directed”].
        So we got 20 watts of directed light hitting the other box 1 meter away. That light could be reflected. If had some mirror for that kind of radiant light we could reflect
        it back at the cube which warming. Which essentially reduces the amount the 600 watt cube emits by 20 watts.
        And one could call that backradiation, if it made you happy. But it does not warm the warming cube side facing the other cube, but rather prevents entire cube from radiating as much as it would if not for the other cube which is reflecting it’s radiant “back” at it. Or how it affect the cube will depend on how the cube manages to uniformly heat all sides of the cube [there many ways to do this].
        But I wasn’t planning on second cube reflecting the directed radiant light, but plan was it was going to absorb the radiant energy. So adsorbs the 20 watts and heats the rest of this unheated box. And saying with the directed light of 20 watts we follow rules regarding it. So if move the cubes further apart so it’s twice the distance, instead absorbing 20 watts at 1 meter distance, at 2 meter distant it absorbs 1/4 of the energy- so 5 watts rather than 20 watts. And rule following is called Inverse-square law:
        https://en.wikipedia.org/wiki/Inverse-square_law
        Oh, and getting back to point of reflecting it if 20 watts of directed light were reflected it would travel another meter, giving total distance of 2 meters and would arrive with 5 watts.

        As said one could make boxes which direct the light better.
        Or if dealing with sunlight or lasers one dealing with light which far more like most directed light one can get.

        But anyhow with climate except in regards to Sunlight one is dealing with very diffused/indirect radiant energy.
        Or some have said, not all watts are the same.

        • Tim Folkerts says:

          If there not an atmosphere- say on the moon, a paper thin house could work fine in terms of insulation.

          This is way off topic, but no, it wouldn’t. For simplicity, consider the night time. A given heater will cause the outside of the ‘paper house’ to reach a give temperature. If there is no other insulation, the interior will also be that temperature.

          If, however, there is insulation inside the paper, the EXTERIOR surface of the insulation will still be the temperature of the paper, but the INTERIOR surface of the insulation will be warmer. And hence the interior of the ‘house’ will be WAY warmer!

          I got lost in the rest of what you were trying to say.

          If your cube is a blackbody, it will be 205 K when floating is space, far from any other warm object. ANY warm object nearby — even one cooler than 205 K — will cause the cube to warm above 205 K. That is really all we need to know.

          • gbaikie says:

            — Tim Folkerts says:
            June 25, 2015 at 9:22 PM

            “If there not an atmosphere- say on the moon, a paper thin house could work fine in terms of insulation.

            This is way off topic, but no, it wouldn’t. For simplicity, consider the night time. A given heater will cause the outside of the ‘paper house’ to reach a give temperature. If there is no other insulation, the interior will also be that temperature. —

            One make the paper thin material reflective, which in a vacuum work as insulation.

            –I got lost in the rest of what you were trying to say.

            If your cube is a blackbody, it will be 205 K when floating is space, far from any other warm object.
            ANY warm object nearby — even one cooler than 205 K — will cause the cube to warm above 205 K. That is really all we need to know.–

            Space is big and has wide variety of thermal environments available. And particularly in context of space, it’s not clear to me what you mean by nearby.

            But a summary of what I said is that all heated matter emits indirect electromagnetic radiation.

            And sunlight near the sun’s surface is also indirect light and therefore it spread out into spherical volume. But at distance of say, Earth from the Sun, all the radiant energy from the sun is heading in near parallel direction- or what remains at such a distance is directed light.

            If one use say a parabolic reflector, you can make directed light- such as flashlights or car headlights do.

  51. mpainter says:

    FTOP: ” LWIR is unable to heat water, yet the earth is 71% water surface”

    Many people do not seem able to grasp the implications of these simple facts. In fact, some AGW types insist that LWIR heats the ocean, in their infatuation.

    • FTOP says:

      I followed a link to Skeptical Science “explaining” how LWIR heats the ocean at depth. I left the page begging for my time back and looking for the jaws of life to remove my forhead from my palm.

      Thes gyrations to defend a flawed premise put most fiction writers to shame.

  52. Genghis says:

    Nice write up and correct, but I have a small quibble.

    If the earth was a perfect blackbody, which it isn’t, it would be ~5˚ C according to the S-B calcs.

    But, the oceans are a pretty good conductor and their average temperature is very close to 5˚ C So I think a pretty good case can be made that the green house affect is only good for ~ 10˚ C or so. Assuming that the average temperature is 15˚C.

    However, the average temperature of the surface of the ocean is ~21˚C, 16˚C warmer than the S-B calcs. This is where the Greenhouse effect really is and the oceans primarily cool by evaporation, not radiation.

    In agreement with your treatise above I would say that the greenhouse gases downwelling long wave radiation almost equals the radiation from the surface of the ocean meaning that the ocean “only” cools via evaporation.

    Here is the interesting thing, long wave downwelling radiation only penetrates the first couple of microns, which stimulates evaporation. The surface of the ocean is therefore cooler because of the radiation from the green house gases.

  53. Rich says:

    @Dr, Spencer.

    Those that claim that there is NO, ZERO, IR coming back to the earth should review some of the explanations and documents describing Near Vertical Incidence Skywave (NVIS) Communications. Google NVIS Communications and there are pages of web sites and PDF documents.

    The NVIS antenna radiates the radio signal from the transmitter into the atmosphere. This radio signal can be collected on an antenna and thus a communications link established.

    IR, like radio waves, are Electromagnetic radiation. The IR from the earth (land/water) will be radiated into the atmosphere, just like the transmitted signal from the NVIS Communication system. This IR can then be detected by an IR detector.

    However, Here is where I (like some others) have a problem. How can the “experts” claim that only GHG will reflect this IR back to the earth? Surely some of the other gases, dust particles, whatever, in the atmosphere is going to reflect this energy back to the earth and not e IR signal that Dr. Spencer measures is reflected in exactly the same way as the radio signal and how much is added from Water vapor, CO2 and the other recognized GHGases? Or, Why do the non GHG not reflect any IR whatsoever back to earth? How much is from the Ionosphere doing the exact same thing it does to Radio signals?

    http://www.w8ne.com/Files/NVIS%20nvis_AI0W.pdf

    • Curt says:

      Rich:

      Gas molecules with three or more atoms like H2O, CO2, and CH4, have bending and stretching oscillatory modes that monatomic molecules like Ar and diatomic molecules like N2 and O2 do not have.

      The natural oscillatory frequencies of these modes correspond to the frequencies of longwave infrared radiation emitted in quantity by substances at typical earth temperatures. So these molecules both absorb and emit radiation well at these frequencies, whereas the non-greenhouse gases do not (so they are very transparent to these frequencies).

  54. Curt says:

    FTOP & mpainter: You say, “LWIR is unable to heat water.”

    Virtually all (>95%) of the LWIR is absorbed by water, and the energy from the LWIR that is absorbed by the water increases the thermal energy of the water above what it would be without the LWIR.

    Now, in most cases, the water is emitting more LWIR than it is absorbing, so you can get into semantic debates when using non-technical language about whether LWIR “heats” water or not. But the energy in the LWIR is definitely absorbed by the water.

    A lot of people make a big deal about the fact that LWIR is absorbed in the very thin “skin” surface layer of the water. All this is saying that the water is opaque to LWIR. But this means also that all of the water’s own LWIR thermal emissions are from this same very thin surface layer. Again, in most cases, the water’s upward LWIR emission is greater than its downward LWIR absorption, but the net energy balance is most certainly affected by the LWIR.

    Rocks are also opaque to LWIR, so the energy from the LWIR is absorbed in the very thin surface layer of the rock. Do we say that LWIR cannot heat rocks?

    In addition, rocks are opaque to visible light. Do we say that visible light cannot heat rocks?

    • SkepticGoneWild says:

      Curt,

      Have you ever parked your car out in the open at night, and come back hours later to find the surfaces of your vehicle warmer than ambient air due to DWLR? And did you notice that the interior of your car is much warmer than the ambient air temperature? After all, DWLR can be in the range of 300 W/m^2 at night. That is a lot of energy striking your vehicle’s surfaces. The interior of your car should be very warm as well. Is it? Why not? That’s the equivalent of parking your car out in the sun at about 9:00 in the morning. 300 W/m^2 of solar insolation will warm the surfaces of your car and heat its interior.

      Do mechanical engineers in calculating roof heat loads, take into account the effect of DWLR? After all, DWLR averages 330 w/m^2 during the day. During the summer, solar insolation can average about 500 W/m^2, sunrise to sunset. So, seems like they should include DWLR in their calcs. But they don’t.

      So, show me where DWLR heats any kind of surface.

      • SkepticGoneWild says:

        DWLR = downward longwave radiation (infrared)

      • Curt says:

        SGW:

        Consider a still clear night where the surface air temperature is just above freezing. Look at the parked cars to see which have frost and which do not. Those cars exposed to the night sky with nothing overhanging them have frost. Those cars under some kind of overhang — tree, carport roof — do not.

        The cars with and without frost can be right next to each other. I have even seen individual cars half and half this way. There is no difference in air temperature between the two cases. (Often, you can notice the pattern on roofs, with the part of the roof under a spreading tree having no frost, but the “open” part of the roof having frost.

        The difference is that the cars under some kind of overhang are receiving DWLR from substances that are about at the same temperature. Those not under an overhang are receiving lesser DWLR from the colder atmosphere at altitude.

        You ask, “Do mechanical engineers in calculating roof heat loads, take into account the effect of DWLR?”

        The answer is yes! When I studied mechanical engineering heat transfer (in the 1970s), I was taught to use an “effective blackbody radiating temperature” for the clear night sky of -20C, at least as a first approximation, and to use -40C to 0C depending on conditions if I wanted to get more precise. Of course, this is very different from the -270C (3K) that would apply with an atmosphere transparent to radiation.

        • SkepticGoneWild says:

          Curt,

          Show me any kind of tables or calculations where DWLR is included in calculations of roof heat loads. You did not provide the information.

          Cars under a carport at night are not receiving the 330 W/m^2 DWLR and should have frost, and the cars out in the open at night and exposed to DWLR should not. So you are telling me that under a carport at night there is well over 300 W/m^2 of IR emittance? More than out in the open? On sunny mornings, do cars exposed to sunshine have frost, and those in the shade do not??

          Walk out from under a carport into the morning sunshine. You can feel the warmth on your skin. Walk out from under a carport at night into the DWLR; do you feel the warmth on your skin? No. The reason you don’t is that DWLR is originating from a colder atmosphere above, and we know from the 2nd Law of Thermodynamics that cold does not transfer heat to warmer surfaces. (That’s what I was taught in my thermodynamics and physics classes)

          • Curt says:

            SGW:

            Let’s take a night at +2C (275K) near-surface temperature. Objects with the common emissivity of 0.95 will emit 308 W/m2 at that temperature. A car under a carport roof will basically exchange 308 back and forth with the underside of the roof for no net exchange, and its surface will be at about the air temperature of +2C. No frost.

            The clear sky on such a winter night will have an “effective blackbody radiating temperature of about -40C (233K). This means its DWLR will have a flux density of only 167 W/m2.

            So a car out in the open under these conditions, if its surface is initially at +2C, it will have a (308-167 = 141) W/m2 radiative imbalance, transferring heat (from hot to cold) by radiative means to the atmosphere. It will cool until its radiative losses decrease enough and the conductive gains from the surface atmosphere increase enough that an energy balance is restored. Under these conditions, the car surface would be below freezing, and frost can form.

            Have you seriously never noticed that frost forms first on surfaces directly exposed to the night sky?

            So in answer to your questions: “So you are telling me that under a carport at night there is well over 300 W/m^2 of IR emittance? More than out in the open?”

            That is exactly what I am telling you! And it is what you should have learned in the first week of the radiative heat transfer section of an introductory course.

            If you are sitting in a temperature controlled room as you read this, you are bathed in ambient thermal radiation of about 400 W/m2 from the walls and other surfaces. Your body surface is emitting about 500 W/m2, so even without clothes you would be transferring heat to the cooler room ambient through radiative exchange (and there is also conductive transfer).

            Because we are so used to being surrounded by this ambient radiation, it is easy to think of it as “nothing”. But it is real, and it is significant!

            My pool pump turns on automatically when the air temperature hits +4C or below for “freeze protection”. The only reason to turn on above 0C is because of radiative losses. But the designers had to know what DWLR there would be to set the threshold. If there were no DWLR, the threshold temperature would have to be much higher.

        • FTOP says:

          That is a creative explanation for the phenomena you describe.

          My understanding is that as the temperature drops the air is unable to hold the moisture of water vapor.
          The earth has gravity and the condensed vapor falls straight down
          If you have an object above it, the moisture hits that object first
          Check the roof of the carport and your magical missing frost would be there
          If your roof had a pinhole and moisture dripped through, the hood of the car would have frost

          Your explanation lacks physical support. How high would the roof have to be before your magical DWLIR wouldn’t warm the car?

          • Curt says:

            FTOP:

            I’m afraid you are missing important parts both the fundamental science and my example.

            You say: “My understanding is that as the temperature drops the air is unable to hold the moisture of water vapor.
            The earth has gravity and the condensed vapor falls straight down.”

            You are describing rain and snow, not dew or frost (which often appear on vertical and downfacing surfaces). Dew and frost can form even in the absence of gravity.

            You ask: “How high would the roof have to be before your magical DWLIR wouldn’t warm the car?”

            First, I am not claiming that the DWLIR from the roof is warming the car. The roof and the car will be at virtually the same temperature, so there is no significant net heat transfer either way.

            But the roof is blocking the optical path between the car and the cold upper atmosphere, preventing the “cool” car from (net) radiating energy away to the “cold” upper atmosphere.

            If the roof were high enough (and broad enough, as this is all diffuse radiation) that it was significantly colder than the car, then the car could radiate net power to the cold roof and end up colder than the local air at the surface.

      • jerry l krause says:

        Hi Guys,

        An interesting discussion of a common observation about carports and their important function—to keep dew or frost from forming on the car’s windows or to keep the car cooler during the daytime. And I believe I have learned something because of it. For I have frequently considered these observations and have been somewhat uncomfortable of my understanding of it.

        It does not matter who said what that caused to form an better understanding with which I am more comfortable. I will attempt to describe what this better understanding. Remember it is only my own understanding which I am judging, not any of your understandings.

        During the day the carport roof absorbs the solar radiation that the upward surface (roof) of the car would absorb if not for the carport roof above it. The carport roof’s temperature increases until it is emitting as much energy as it is absorbing. Which very likely would be the temperature that the car roof would rise to if not for the carport roof.

        Now, the carport roof could be one of two types: a plywood (OSB) planel covered with a roofing material or a thin steel sheet. The first places an insulating layer between the high temperature of its top surface which causes its bottom surface to remain cooler whereas the temperature of bottom of the steel sheet is likely that of its upper surface. It is easy to see why in the first case that the car remains cooler but not so obvious is why the car also remains cooler in the second case. The steel sheet has two surfaces from which radiation can be emitted. Hence the top and bottom temperature of the steel sheet does not increase to as high a temperature as the top surface of the other roof. While it is likely the bottom temperature of the steel sheet will be greater than the bottom temperature of the other roof, it is still appreciably less than the temperature of the car’s roof directly exposed to direct solar radiation would be

        During the nighttime the top surface of both roofs can emit their radiation directly to space through the greenhouse atmosphere. If it doesn’t, the roof cannot cool. Here, I must insert the qualification that the sky must be cloudless for the carport to have the practical utility being considered. In the first case the bottom surface of the roof will become warmer than its top surface temperature because of the insulating factor of the wood and roofing material. And it must be remembered that the temperatures of the car and earth surface beneath the roof are likely above the temperature of the top surfaces of both roofs (because these surfaces cannot directly emit radiation to the cloudless sky) when the top temperature of each roof cools to the temperature of the atmosphere in contact with it as their surfaces radiatively cool.

        Again, it is easy see why the first roof likely keeps the surface temperature of the car above the atmosphere’s dew or frost point temperature. In the case of the steel sheet, its top and bottom surfaces drop to the same temperature and it might seem that dew or frost might form on both bottom and top surfaces. But, if, as I assume, the temperatures of the car and the earth surface beneath the roof are warmer than the top and bottom surface, their emission, absorbed by the bottom surface, prevents the condensation on the bottom surface of the steel sheet. So the previous assumption that temperatures of its top and bottom surfaces are identical is not precisely correct (true).

        So, this is my understanding of the functioning of the carport. Have a good day, Jerry

    • Kristian says:

      Curt, you need to sort out your terminology. You say: “Rocks are also opaque to LWIR, so the energy from the LWIR is absorbed in the very thin surface layer of the rock. Do we say that LWIR cannot heat rocks? In addition, rocks are opaque to visible light. Do we say that visible light cannot heat rocks?”

      The rocks are heated by LW or SW radiation if this radiation comes in AS HEAT, Curt.

      • Curt says:

        OK, I’ll be a little more precise in my terminology:

        Do we say that the energy in LWIR cannot be absorbed by rocks and increase their thermal energy level above what it would be in the absence of this LWIR?

        Do we say that the energy in visible light cannot be absorbed by rocks and increase their thermal energy level above what it would be in the absence of this visible light?

        Kristian, what distinguishes radiation that comes in as heat from radiation that does not come in as heat? What is the physical difference in these two types of radiation?

        • gbaikie says:

          — Curt says:
          June 18, 2015 at 7:23 PM

          OK, I’ll be a little more precise in my terminology:

          Do we say that the energy in LWIR cannot be absorbed by rocks and increase their thermal energy level above what it would be in the absence of this LWIR?

          Do we say that the energy in visible light cannot be absorbed by rocks and increase their thermal energy level above what it would be in the absence of this visible light?

          Kristian, what distinguishes radiation that comes in as heat from radiation that does not come in as heat? What is the physical difference in these two types of radiation?–

          This is interesting question.
          What is the difference of sunlight at 500 watts and a rock at 500 watts.
          One difference is that sunlight is directed light.
          Another aspect is that sunlight comes from an object which
          is 5,778 K. Which means that if you magnify sunlight one can heat something to 5,778 K. Whereas a rock radiating 500 watt per square meter can not have it’s light magnified to equal a rock which is radiating 1000 watts per square meter.

          If you were 20 feet away from a rock radiating 500 watts and say 200 watts per square meter was reaching you, I assume, you could magnify the 200 watt per square meter so as to become 500 watts per square [assuming rock was big enough].

          Or one could have a solar collector getting 500 watts per square meter of it’s reflector and with few of them one could magnify the sunlight so it’s couple thousands of watts per square meter.

          Whereas if had 1 square meter of metal heated so it radiated
          500 watts [so it was about 310 K [36.85 C/98.33 F] one could use hundreds of them and not heat anything over 36.85 C.

          So one can not simply add watts in order to get some temperature.

  55. mpainter says:

    Curt, thanks for response.
    You say that the absorbed LWIR increases the thermal energy above “what it would have been without the LWIR”
    Yes, but this thermal energy is transient, being converted to latent heat within seconds for the absorbed energy is confined to the air/water interface or within a few microns of that. The thermodynamics of the sea surface, as shown by the temperature profile of that, show that the energy is conducted from the subskin to the interface, hence none of the incident LWIR energycan move below the zone where it is absorbed. This is confirmed experimentally. LWIR does not warm water.

  56. FTOP says:

    Curt,

    No argument on your surface description, but you left out the key element “at depth”. Further, the ambient temperatures of earth are too low to support phase change of rocks.

    LWIR can’t make it past a few microns and water experiences its phase changes way before you garner any 33C increase in water temperature. The Gulf of Mexico reaches 30C+ in summer. It is physically impossible for the atmosphere to be the cause of why the Gulf isn’t frozen in summer.

    The forcing mechanism described by GHE fails the water test and water is 70% of the test for the hypothesis.

    • wayne says:

      FTOP, you speak of the 30°C SST. Climate science seems to always wanting to take the TSI, remove 30% and divide by four. I don’t. There is a day and night side. So take that infamous ≈239.5 W/m² and double it back to the actual lit side mean. It is so curious that running that into a Planck integrator program that I have you get that amount insolation is, what, what??, 303.15 K or 30°C. Ever noticed that?

      In fact the SST rarely ever gets over 30°C anywhere but it can regionally if the local albedo is low over a lengthy period. That is just what the real mean insolation, not 240 W/m² or whatever equates to. No ‘enhancement’ necessary and I think you might agree. Does that not rule out l.w. downwelling completely? Just noticed that a few days ago.

      • FTOP says:

        Exactly. It is the tropical water that causes the GHE house of cards to collapse. Tropical water temps far exceed the “average” insolation and the physics of water and DWLIR don’t allow for GHE to be the forcing mechanism for these high temperatures.

        The sun heats the water way beyond the K & T energy budget allows. This heating diffuses through the system by ocean and wind currents for the evaporative latent heat.

        In order for the GHE theory to fail, it only needs to be wrong once. It is wrong at least for 70% of the surface, but I would argue it is wrong completely.

        Some will argue that wind mixes the surface to drive heat to depth, but the surface layer is an infinitesimal portion of the ocean, how hot would a thimble have to be to warm an Olympic swimming pool?

        Watch out. Climate scientists will claim there is no water next.

        • wayne says:

          Yep, I no longer attach the word ‘scientist’ to ‘climate’ except in very rare cases, there are so few actually scientists left and they tend to congregate as being either skeptical or denying it totally, bless them. Maybe should more use the term ‘climatists’, very similar to ‘manicurists’… job being clipping the public tax billions writing little pieces of paper, which is literally illegal as it seems collusion is occurring inside the government itself with these characters. Soon a select group of public prosecutors and attorneys should smart up and have a real hay day. I belive it is termed a high target zone for the prosectors.

      • Tim Folkerts says:

        There is a day and night side. So take that infamous ≈239.5 W/m² and double it back to the actual lit side mean. It is so curious that running that into a Planck integrator program that I have you get that amount insolation is, what, what??, 303.15 K or 30°C. Ever noticed that?”

        Yes, there IS a day and night side! So why does your analysis only include the day side to achieve the 303 K temperature but ignore the night side? If you are balancing the day side radiation and getting 303 K, the only logical extension of this model is to balance the night side and get 3 K!

        More realistically, the night side will not cool down to 3 K but the day side will not warm to 303 K either.

  57. mpainter says:

    Also, Curt,
    Your ” rocks” example does not apply, since they do not cool evaporatively. Different thermodynamics with rocks vs water.

  58. Curt says:

    What about when the relative humidity is 100%?

    More generally, the rate of evaporation (or condensation) is determined by the local conditions at the water/air interface. With downwelling LWIR, the water will end up with a higher thermal energy level than without, in any of these cases.

  59. Michal Garbers says:

    Roy,

    Thanks for the explanation for the theory on how the “Greenhouse Effect” works. It is a shame the “Greenhouse Effect” is not called “Thermal Reradiation” Effect it would reduce confusion.\

    In your explanation it is stated
    “the energy would readily escape to outer space and as a result it has been estimated that the Earth’s average surface temperature would be only about 0 deg. F”

    The moon average surface temperature is ~-65F. It appears the assumption (earth 0F) is based that the earths Albedo would stay the same even if there was no atmosphere. I understand with any modeling assumption need to be made. I just wish to confirm in the model; Earths Albedo is assumed to be unchanged even though the atmosphere is removed.

    If Earth had no or very little atmosphere, its albedo would be close to Mars. And then my primitive calculations do not correlate ”Earth’s average surface temperature would be only about 0 deg F”

    Also,
    I am uncertain why people struggle with the general concept of the Greenhouse Effect (Thermal Reradiation) is valid. A simple comparison of Mars and Earth’s Moon pretty much shows the even the thin Atmosphere Mars is warming the planet.

    Moon Albedo 0.136 Ave temp -65F Solar Energy input 1
    Mars Albedo 0.17 Ave temp -55F Solar Energy input .42
    Earth Albedo 0.367 Ave temp 68F Solar Energy input 1

    Thanks,

  60. mpainter says:

    Curt,
    Thanks for your reply.
    100% humidity? Either it is a) raining, or fixing to rain, b) pre-dawn, c) foggy.
    Would you argue that these conditions mean warming of the sea surface. Rather a difficult stance, it would seem.
    Nighttime means cooling and convection from near depth and a straight temperature profile is the result with the exception of the interface and subskin, and this shows the same profile as daytime: a cooler interface and warming with depth, to approx. one mm.
    To warm the sea with LWIR, you must give the specific process whereby energy of LWIR incident at the interface is from there transferred to depth against this temperature gradient.

  61. Curt says:

    mpainter – Thanks for the points.

    100% humidity is actually quite common. It occurs anytime there is dew, for example.

    And even though there is no evaporation, and may even be condensation, under these conditions, it is very likely that the surface skin layer is emitting more LWIR to the atmosphere than it is receiving back.

    But I still maintain that the fact that it is receiving and absorbing some (a significant fraction) back from the atmosphere leaves the water warmer than if it were not receiving any.

    Any time there is any wind at all, the immediate skin layer is mixed quickly with water below.

    On a separate note, you stated upthread that you were uncomfortable with K&T’s estimate of average evaporative latent heat flux. In their paper, they state how they calculated it, and it is easy to duplicate their calculations.

    They start with assessments that average rainfall on earth is about 1000mm/year, so each square meter of surface receives a cubic meter of precipitation. Consequently, this same amount of water (1000 kg/m^2) must evaporate each year.

    From this, it is easy to calculate, using the latent heat of evaporation for water of 2260 kJ/kg:

    1000 (kg/m^2/yr) * 2260 (kJ/kg) * 1000 (W*sec/kJ) * 1/[3.16×10^7] (yr/sec) = 75 W/m^2

    which is very close to what they estimate.

  62. mpainter says:

    Curt,
    Wind will not warm the ocean for you; it multiplies the rate of evaporation, irregardless of any mixing.
    To show that LWIR warms the ocean, you must give specifics on how this occurs. I have seen nothing yet that will accomplish that.

    • Norman says:

      mpainter,

      Sorry to jump in on your posts.

      Your comment: “To show that LWIR warms the ocean, you must give specifics on how this occurs. I have seen nothing yet that will accomplish that.”

      I still think this is the wrong point on what the above graphic is showing. In the graphic above the Earth’s surface is radiating an average of 396 watts/meter^2 (this is not a regional amount just an average, tropics would radiate more poles less but the overall average has been determined to be roughly that number). The GHG returning radiation (which comes from the surface flux and not created magically in the air, if the surface flux is less then so to is the return flux) is 333 watts/meter^2. The LWIR is not warming the ocean. The ocean is still cooling. The LWIR will not increase the evaporation rate. What it will do is slow the rate the water surface cools. Instead of losing 396 joules/sec-meter^2 it will lose only 66 joules/sec-meter^2. So now what happens? The energy from Shortwave radiation that penetrates until all is absorbed that resulted in warmer water below will also cool slower since conduction would be the main energy transport here. The rate of conduction is related to the temperature difference between two thermal sources. If the surface cooling is slowed so will be the conduction from below leaving the water warmer than it would be without downwelling IR.

      The GHE does not warm the ocean or land surface. It slows down the rate of cooling. The backradiation is always less than the primary surface radiation. The returning radiation does not warm the surface it hits, the surface is still radiating away more energy than it is getting back. The constant flux of solar energy is what warms the surface.

      From nearly all the posts I read most seem to neglect the fact the Earth system is not a closed one. It is receiving and emitting radiation on a continuous basis. Once this fact is well understood the GHE is not so irrational as so many seem to think.

      Questions remain about the extent of the effect and what really happens with an increase of CO2. But I believe the effect itself is based upon actual thermodynamic physics.

      • FTOP says:

        You do realize that the GHE Theory holds that CO2 forcing is not enough, but it’s effect is that it amplifies the forcing from the most powerful GHE — get ready for it — water vapor.

        Sooooo, this means that the water vapor that gains escape energy at the surface leaving a cooler ocean behind is the 90% GHE that turns around and slows the cooling.

        That is the problem with GHE and IPCC fairy tales, they keep tripping over their own inconsistencies.

    • FTOP says:

      I read the wind argument on Skeptical. My analogy is that blowing cold CO2 (exhaled) on your soup can’t cool it because climate scientist suggest it warms the soup in the bottom of the bowl.

      Hey, maybe the missing heat is in the minestrone?

    • Curt says:

      mpainter:

      The absorption of LWIR by the surface skin layer leaves it warmer than it would be without it. Even light winds will mix this layer with those at least millimeters below it.

      The convective overturning that occurs on diurnal cycle will further mix this meters deep.

      • FTOP says:

        Physically impossible.

        The ocean has an average depth of 4,000 meters. IR is fully absorbed at less than 1mm and evaporation leaves the 1mm below slightly cooler because of the energy used in phase change.

        How hot would the IR absorbed 1mm of water have to be to impact an amount of liquid 4,000,000 times larger? The first ten meters of water has 40,000 times the heat capacity of the surface layer. It would need to be over the boiling point of water to budge the temperature of the first meter.

        • Curt says:

          Sunlight only penetrates a few meters, or where it is really clear, a few tens of meters. This is also absolutely trivial compared to the average 4000-meter depth.

          So can we conclude that sunlight cannot heat the oceans either?

          • FTOP says:

            You missed several orders of magnitude, I would characterize 1000 meters as warming to depth. This also represents 25% of the average depth vs. .0000025% for DWLIR.

            http://oceanservice.noaa.gov/facts/light_travel.html

          • Curt says:

            Your own source says that sunlight rarely penetrates beyond 200 meters. And because of the exponential nature of the absorption, virtually all of it, and its energy is absorbed in the top few meters or tens of meters, as I said.

            Just because the occasional photon can be detected at depth does not mean there is any significant energy transferred this far down. It’s like arguing that because you can see the light from stars, they must have a significant heating effect.

          • FTOP says:

            More sunlight forcing reaches 1000 meters than atmospheric IR reaches 4mm. It is clear sunlight warms the ocean not DWLIR. A1mm surface layer does not warm the ocean when a cool wind blows over it,

            Regardless of climate science claims, the sun is hot. The atmosphere is cold and physics demonstrates it is the sun that warms the ocean.

            Because these are facts, every AGW argument using SST as warming proof is false. Dr. Spencer’s data shows no warming and the vilification of Co2 is a political charade.

  63. mpainter says:

    Curt, thanks for the calculations.
    The KT diagram still does not square. Working from 80 W/ sq m, I get 16% as the figure of ocean cooling via evaporation. This figure is wildly off target.
    Using 161W/sq m insolation, I get about 69 W/sq m, using 60% as the ocean heat loss per evaporation. Does this tell you anything?

    • Curt says:

      mpainter:

      The K&T diagram gives 80 W/m^2 evaporative cooling, 17 conductive/convective cooling, and 63 radiative cooling (396-333). So it shows 47% evaporative cooling.

      But that is a global land and sea average. Certainly, the oceans will have a higher percentage. (I don’t think the Sahara Desert has much evaporative cooling…)

      So I don’t see the K&T numbers as being at all inconsistent with the figure of 60% evaporative cooling from the oceans.

  64. mpainter says:

    FTOP,
    House of Cards is the right call, and the ocean is the heap of tumbled AGW cards. In no way do the physics of the ocean square with AGW theory.

  65. mpainter says:

    Drat! Hypothesis, I mean.

  66. mpainter says:

    Norman,
    You seem to have confused two different issues: the KT diagram is an issue separate from the absorbency of water w/r to IR.
    In regard to your claim that oceans cool by conduction, studies have partitioned ocean cooling into its components and determined that: cooling per evaporation : 60%, cooling per radiation: 30%, cooling per sensible heat loss: 10%. These figures are approximations from several different studies.
    Your claim that incidental LWIR will not cause evaporation is incorrect; IR will cause evaporation and this fact is readily shown experimentally. These same experiments show that IR does not warm water because it is converted into latent heat at the surface (water being opaque to IR).

    • Norman says:

      mpainter,

      I need to clarify a point. I agree the ocean surface cools by radiation, evaporation and conduction. The ocean below the surface is what my post was talking about. Below the surface the water does not cool by evaporation. The energy must move from warmer to cooler by conduction. I think my claim may have not been stated correctly. The backradiation from the atmosphere will not speed the evaporation leading to more cooling (things I have read on several posts). The evaporation rate is determined by ocean surface temperature and the ability of the air above to allow more water to evaporate (dry air allows much faster evaporation ratees). The downwelling IR is not heating the ocean surface. The ocean surface is losing more energy than it is gaining by the KT diagram so the backradiation would not act to evaporate more water and cool the surface, that process is already taking place by the temperature of the ocean surface. The backradiation will not heat that surface above what its temperature is currently.

  67. mpainter says:

    Curt
    1. surface of ocean cools primarily via evaporation. Diagram shows 494 W/ sq m incident on surface: 161 W of insolation plus 333 W of backradiation. You seem to be having trouble.Diagram does not add up.
    2. You need to get the temperature profile of the sea surface. This shows thermodynamics and how heat is conducted from warmer subskin to cooler interface day and night. Please do not comment any further until you have studied this. Light winds, high insolation ate the conditions of this profile.
    3. The nighttime convection is a cooling process and you ate trying to make it a warming process. Repeat, diurnal convection removes heat.
    4. And you continue to make wind warm the ocean when wind cools the ocean under all circumstances.

    • Curt says:

      mpainter:

      In the K&T diagram, the convective and evaporative flows shown are NET flows, probably because there are no good measurements of the gross flows.

      That is, the “17” shown for thermals just indicates that the surface to atmosphere transfers average 17 more than the atmosphere to surface transfers.

      The “80” shown for evaporation just indicates that the evaporative transfers average 80 more than the condensation transfers in the reverse direction.

      They do not show the radiative transfers just as net, I think because they have separate measurements for the gross up and down transfers.

      But for a proper apples-to-apples comparisons of NET heat transfers, the (396-333 = 63) net radiative transfer should be used in comparison to the net 17 convective and net 80 evaporative.

      This makes the net evaporative transfer 50% of surface losses averaged globally over land and sea. Since there will obviously be more evaporative losses over water, I don’t see this in any way inconsistent with your figure of 60% evaporative losses from the ocean.

  68. mpainter says:

    In fact, I consider the KT diagram preposterous, as the Ceres satellite shows most of radiation being emitted at the cloud tops. What does this tell you? It tells me that latent heat is the primary means of surface cooling. And studies confirm this: ocean surface cools evaporatively at 60% of total cooling, twice that of radiation (30%).

    • jerry l krause says:

      Hi mpainter,

      You ask: What does this tell you? It tells me that clouds are the earth’s thermostat. For these clouds which are emitting radiation from their tops, are blocking (regardless of the mechanism involved) the upward transmission of LWIR radiation being emitted by the earth surface below.

      Have a good day, Jerry

    • jerry l krause says:

      Hi mpainter,

      Since you questioned my comments above (6/16 at 7:35am if this doesn’t immediately follow these previous comments), I see I can better define what I intended to state. For these clouds which are observed emitting radiation from their tops according to their temperatures, as if they are near black bodies (S-B Law), have to be blocking (regardless of the mechanism involved) the upward transmission of any LWIR radiation, from below, through them.

      Here I am stating a conclusion, which seems forced by the observations you brought to our attentions. For I accept the validity of these observations.

      Have a good day, Jerry

  69. Günter Heß says:

    Dear Dr. Spencer,

    I do not understand the fuss people make about the so called greenhouse effect.
    It is just the observation that average global surface temperature is higher than the effective radiation temperature oft he earth. This is a result of clouds and greenhouse gases in the atmosphere and the emissivity oft the earth surface. So the main effect is the real atmosphere being a heat resistance towards space. Without greenhouse gases there would be no heat resistance and the average surface temperature would be close by 2 – 10 K to the effective radiation temperature according to the emissivity, which is above 0.9 for the most part.

    Best regards
    Guenter

    • FTOP says:

      Gunter,

      People make a big fuss about GHE because it is the argument used to:

      Deplete coal power production to unsustainable levels
      Promote bird killing monstrosities that produce at best an unreliable intermittent supply while being a blight on the landscape
      Fund moon bat projects like CO2 sequestration
      Vilify a life producing substance that is the building block of all agricultural production
      Justify an international cabal of centralized government promoters to control and ration energy
      Support government policies that literally tax the air we breathe (out)

      All this based on a tenuous GHE Theory that fails to be supported by observational data (the pause) and regardless of the illogical efforts here and elsewhere to prove it depends on CO2 forcing from an atmosphere with 1/1200th the heat capacity of the ocean to some how warm it.

      The generous may chalk this up to herd mentality among scientists, the less generous may attribute this to something more sinister.

      • jerry l krause says:

        Hi FTOP,

        Very well said. It is amazing that so many are not aware of your well stated facts. Or do not care about them.

        I am a scientist and I am embarrassed to be represented by so-called scientists who do not doubt.

        Have a good day, Jerry

  70. mpainter says:

    Norman, thanks for your comment.
    To understand the thermodynamics of the sea surface, search the internet for a profile of sea surface temperature. This gives a profile for day and night conditions, both.
    Daytime: cooling with depth, hence no convection.
    Nighttime is when convection occurs and the subsurface waters overturn.
    LWIR incident on the surface does not cool per se, however, it adds to the evaporation rate and oceans cool per evaporation, as you note.
    My point is that IR cannot contribute to SSTand this point is disputed by the AGW proponents. The reason that they deny this is easy to see; this cuts off the legs of the AGW hypothesis.
    I assure that IR does indeed cause evaporation.
    Convection is the means that subsurface water cool, and this occurs at night, not day. These are all well-established facts.
    There is a certain amount of conduction from the subskin to the interface, involving about one mm of depth, that occurs during both day and night. This conduction is due to cooling the processes (evaporation, etc.).

  71. mpainter says:

    Norman,
    To me it is a very big question as to how much “back radiation” actually is incident on the sea surface. My view is that this is vastly overstated. That was my point in demonstrating the egregious aspects of the KT diagram, which has 333 W/sq m of back radiation shown as being absorbed at the surface.

  72. mpainter says:

    Mike B
    No need to shout.
    See textbooks for standard information.
    I worked it out already, above. Here it is again:
    The Kiehl- Trenberth diagram gives 494 W/sq m absorbed at the surface and oceans is 71% of the surface and cool predominantly by evaporation (there is no dispute concerning this).I use 60% as a representative figure. You can use a lower figure, but this will not change results in the main.
    494 × .71 × .6 = 207 W/ sq m of latent heat via evaporation of oceans being returned to the atmosphere and thence to space. The KT diagram is egregiously wrong: it gives 80W/ sq m.
    Work it backwards and derive ocean cooling evaporation component starting with 80 W/ sq m:
    .71 y × 494=80W/sq m
    (y is figure of evaporative ocean cooling, as a fraction of total cooling).
    Result: .16
    So the KT diagram gives 16% as the fraction of total ocean cooling through evaporation, a figure egregiously incorrect. I repeat, the Kiehl, Trenberth diagram does not reflect reality.
    Go read up.
    Work it out another way

    • Norman says:

      mpainter

      I have to respectfully disagree with your post here. Your 60% evaporative cooling may not be a valid number.

      The KT diagram is very close to what I get when I run the math. They may have used the lower 540 calories/gram for heat of vaporization. I am using the 580 calories/gram for evaporation at room temperature.

      Link for energy loss in perspiration (evaporative cooling):
      http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/sweat.html

      I might have to break up this post because too many links will not post.

      The next step I used was to get an annual rainfall amount (this smears out local and regional or daily effects with their high varability).

      https://en.wikipedia.org/wiki/Earth_rainfall_climatology

      This article claims that 550,000 cubic kilometers of water fall as precipitation in a year. That means this same volume of water must have been evaporated so we can use this value to calculate the amount of water evaporated in a year.

      I think I will break this post up here to include more links. Thanks for your understanding.

      • Norman says:

        mpainter,

        Post two in the series to demonstrate to you that the KT diagram is very valid and accurate and your calculations may have error that you should investigate.

        If my logic serves me correctly. To get 550,000 cubic kilometers of rainfall a year you need to evaporate this same amount.

        https://www.unitjuggler.com/convert-volume-from-km3-to-ml.html?val=550000

        Convert 555000 cubic kilometers to milliliters and you get 5.5 x 10^20 milliliters. 1 milliliter of water = 1 gram of water (converting it over to mass that can be used in the calculation above of 580 calories/gram of water).

        You would need to evaporate 5.5 x 10^20 grams of water a year.
        Multiply this by 580 and then that by 4.184 joules/calorie to get a value in joules (which can then be converted to watts 1 joule/second is a watt).

        Doing the multiplication on a calculator I got 1.335 x 10^24 joules of energy are used a year to evaporate enough water to produce a year’s worth of rain.

        Now to convert all this to watts/meter^2 used in the KT diagram.
        You find the square meters of the Earth’s surface. The Earth has 510 million square kilometers.
        https://www.google.com/search?sourceid=chrome-psyapi2&ion=1&espv=2&ie=UTF-8&q=surface%20area%20of%20earth&oq=surface%20area%20of%20earth&aqs=chrome..69i57j0l5.4655j0j8

        Convert this to square meters at the rate of one million square meters/ square kilometer. You end with 510 x 10^12 meter^2.

        Evaporation rate in square meters per year is
        1.335 x 10^24 joules/ 510 x 10^12 meter^2

        Final calculation to get this to watts/meter^2 is to divide this amount above by the number of seconds in a year (3.1557 x 10^7).

        I ended up with 82.9 watts/meter^2 as the amount of energy that is required to evaporate water. The KT diagram gives this figure at 80 watts/meter^2. I think they are on the money with this one.

        I hope you realize the energy budget is not a real world item. It is a hypothetical to simplify the systems. Some square meters have zero solar energy hitting them, others have much more than 161. Some square meters have a lot more evaporative cooling than others. But the overall average is quite correct.

  73. mpainter says:

    Jerry,
    The radiation comes from the phase change of water vapor to water then to ice. This releases prodigious amounts of radiation, all in accordance with established physics. If you do not understand this, then you lack understanding of the most basic knowledge of meteorology.

    • jerry l krause says:

      Hi mpainter,

      See my comment of 6/16 at 8:27am. I had not read your response of 6/16 at 7:43am when I wrote it. Your statement: “If you do not understand this, then you lack understanding of the most basic knowledge of meteorology.” does not promote any common understanding.

      The phase changes to which you refer occur once and then the cloud particles remain to continuously emit radiation to space based upon their temperature and the S-B Law.

      Have a good day, Jerry

  74. jerry l krause says:

    Hi Gals and Guys,

    Can we take a break from promoting our individual ideas and try to first see what observations we can agree are fact and then explore at length what are the possible consequences of these valid observations?

    For example, mpainter has two times (6/14 at 6:23am and 6/16 at 5:04am) cited the observation that most of the LWIR being emitted to space, as observed from satellite, is from the tops of clouds. I accept this observation as fact and do not worry about what ‘most’ specifically is. So, I ask the question: Does this observed fact have anything to do with the GHE? For it is the validity of the GHE, as defined by Roy, which is the fundamental question. Or, can we agree, regardless of our individual positions on the validity of the GHE, that the validity of the GHE is the fundamental questions of our present dialogue?

    Have a good day, Jerry

  75. mpainter says:

    Jerry,
    The Ceres satellite data has to do with whether the earth’s surface coils predominantly by radiation or by evaporation. The Ceres images show that the earth’s surface cools predominately by evaporation. This observation refutes the AGW hypothesis by diminishing the importance of radiant cooling in the earth’s thermodynamics. There is no question that the KT diagram in the above post exaggerates the role of radiant cooling of the earth’s surface and hence is unrealistic.

  76. mpainter says:

    Above, for coils read cools

  77. mpainter says:

    Jerry, I urge you to read up. The phase change removes most the heat. Read about latent heat required to transform water to water vapor. To vaporize one ml of water requires over 500 times the amount of energy required for increasing the temp of that water by one degree. From water vapor to water releases that same amount of energy. For water to ice, the factor is 80. Please read up on the subject; you will never understand the matter until you are conversant in the basics. You seem to expect others to remedy your deficiencies in this regard. Well meant advice for you.

    • jerry l krause says:

      Hi mpainter,

      Have you read my comments of 6/15 at 7:57am and 6/15 at 10:03am? In these comments I admit my lack of expertise in meteorology and physics, so I consider what recognized authorities such as R. C. Sutcliffe (a meteorologist) and Richard Feynman (a physicist) have written which I can understand apply to the understanding I seek.

      I do not know who mpainter is. That is beyond your belief that your ‘knowledge’ is without limit. I ask you to read what Sutcliffe wrote and what Feynman taught and then comment upon what they stated. For these scientists are known quantities.

      Have a good day, Jerry

      • Mike Flynn says:

        Hi Jerry,

        It’s a mystery to me why anybody claims cloud tops at -40 C emit IR at a greater rate than molten lava at 1500 C, or sea at 25 C, or ice at -20 C.

        Oh well, its obvious I’m not a climatologist. I’m happy with normal physics. I can’t recollect an instance where I had any significant disagreement with Feynman. From your information about Sutcliffe, it’s possible that Feynman disagrees about one or two things. I’m on Feynman’s side, but it doesn’t really make much difference, does it?

        Regards,

        Mike Flynn.

        • jerry l krause says:

          Hi Mike,

          “It’s a mystery to me why anybody claims cloud tops at -40 C emit IR at a greater rate than molten lava at 1500 C, or sea at 25 C, or ice at -20 C.”

          I do not know anybody who claims cloud tops at -40 C emit IR at a greater rate than molten lava at 1500 C, or sea at 25 C, or ice at -20 C. And that is just the point. Cloud tops at -40 C prevent the sea at 25 C (since the earth has such great area of sea at that temperature or more) from radiating to space that which would be radiated to space if not for the clouds at -40 C. Thus, reducing the loss of energy (radiation) to space just as it is claimed that the greenhouse gases do.

          But given the absence of clouds, the greenhouse gases are still in the atmosphere through which the LWIR radiation emitted by the earth’s surface, whether it be molten lava, the sea, or ice, is observed to be readily transmitted to space. For you wrote (6/14 at 6:24pm): “With regard to energy absorbed from the Earth’s surface, you may have overlooked the fact that when the surface loses energy, it cools. GHGs cannot return all the energy to the surface from whence it came. An extreme example of this is noticed in tropical arid deserts at night. Temperatures can drop in excess of 50 C, due to the rapid loss of EMR to the cold sink of outer space. This is reality.”

          Given the context, it is apparent that the 50 C drop is that of the surface and not of the atmosphere. I make this comment because I had to read you statement a couple of times to make it consistent with the most extreme atmospheric cooling I had found in the historic weather records of hot arid deserts. The extreme atmospheric cooling I have found was about 50 F instead of 50 C.

          And you wrote (6/13 at 6:26pm): “A demonstration of the greenhouse effect under laboratory conditions would settle the matter.” This is a wonderful suggestion for I can imagine an actual, not thought, experiment which might do just what you claim. I just observed two days ago that the temperature of the absorbing-emitting surface of my, approximately level, modified Suomi, Staley, Kuhn net radiometer was 212 F (100 C) about solar noon at 45 N. Of course, I had observed similarly high temperatures before and had observed slightly higher temperatures when the radiometer was ‘pointed’ more directly toward the sun.

          The proposed experiment is very similar to the modeled (thought) experiment recently proposed by Roy to support the existence of the greenhouse effect because it does not freeze at most locations during the summer. Without describing the specific details of the experiment I can imagine, it is to simply fill a sandbox with very dry soil, sand, or pea gravel, paint the surface of whatever material with black paint to minimize its albedo, and observe the temperature of its surface with Roy’s Fli7 during a 24 hour period during which the sky is cloudless as possible and the dew point temperature of the atmosphere is low enough that condensation of water vapor on the surface will not occur before the minimum temperature of the surface occurs. The tropical location you refer to is an important factor not only because it maximizes the incident angle of the solar radiation upon the surface, it also maximizes the period during which the surface cools to its minimum temperature after obtaining its maximum temperature. Roy in his thought experiment assumed a 12 hour nighttime during summer that does not exist at the higher latitudes and it period of the nighttime must obviously be a factor of the cooling which will occur.

          Because I tend to write long comments, which I recognize may be an impediment to some even beginning to read my comment, I do not now address your question, which is important.

          Have a good day, Jerry

        • jerry l krause says:

          Need to correct a mistake. Fli7 should be FLIRi7

  78. mpainter says:

    Curt
    Your last touches on my point.
    The diagram shows 333 W/ sq m of back radiation “absorbed at the earth’s surface”.
    You seem to saying this figure should be disregarded. Obviously Kiehl and Trenberth did not mean for it to be disregarded. Now, either that diagram reliably depicts reality or it does not. So let me try and pin you down here :
    Does the 333 W/sq m of back radiation “absorbed at the earth’s surface” represent reality or is it to be dismissed as unreliable data?
    Because if in fact that back radiation is “absorbed at the earth’s surface” then it will have a real effect, as per my calculation.
    You can’t have it both ways; you can’t exaggerate the back radiation for one purpose (GHE) and then dismiss the data as of no consequence when it is shown to be egregious.
    So, come down and declare: is the data reality or not?

  79. mpainter says:

    Curt,
    Also, where you cite “net” atmospheric radiation from the K-T diagram, these figures are contradicted by my latent heat calculations. Clearly latent heat is “net”. Once again, the K-T diagram is egregious and does not depict reality. The reason, of course, is the greatly exaggerated back radiation “absorbed at the earth’s surface”.
    Another simple algebraic calculation:
    .71 × .6 × y = 80 W/ sq m
    Solving for y (radiant energy absorbed by earth’s surface):
    y= 187; hence, back radiation “absorbed at the earth’s surface” = 26W/ sq m (187-161).
    This figure per our understanding of ocean thermodynamics, given latent heat net at 80 W/ sq m.

  80. mpainter says:

    Also, Curt, you are sounding rather dubious with such statements as “condensation transfers”. Was it not you who showed how the 80 W/ sq m was derived? Per precipitation, right? And now you mumble vaguely about “condensation transfer”.
    A red herring. So answer: are we to believe the 333W / sq m of back radiation “absorbed by the earth’s surface” given in the diagram?

  81. Scott Vickery says:

    How is it that carbon dioxide is a supposedly well mixed gas and water vapor is not. Is water vapor measured to the same fidelity as carbon dioxide? I don’t think either is well mixed. It depends on location.

    For example: deserts are arid, cities are higher in carbon dioxide.

    So if carbon dioxide is a well mixed gas, then deserts around the world should show an increasing trend in temperature if water vapor remained equal.

    • dave says:

      “How is it that carbon dioxide is a supposedly well mixed gas and water vapor is not…”

      Because “normal” temperatures allow for the existence of water in a liquid or solid phase, but not for the existence of carbon-dioxide in a liquid or solid phase. Water vapor condenses into cloud at quite a modest height. One mile up, half the water vapor in the atmosphere is already below you.

  82. Curt says:

    mpainter:

    I’m sorry, but have you lost your mind?

    How is my computing a net 63 W/m2 radiative loss from the surface as the difference between 396 upward emitted from the surface minus 333 downward absorbed by the surface “disregarding” the 333 W/m2 absorbed by the surface?

    If I say that I paid $5 for an item, am I disregarding the $15 in change I received (“absorbed”) in return for the $20 I gave the cashier (“emitted”)? Of course not! But it’s exactly the same principle.

    We have very good direct measurements of this downward radiation, including its spectrum (mostly in the absorption/emission bands of H2O and CO2). Of course, some are higher than 333, mostly in the tropics, and others are lower, mostly out of the tropics. But 333 W/m2 is a reasonable estimate of the average. And we know from repeatable laboratory measurements that water and most other natural surface substances absorb the vast majority of radiation of these wavelengths.

    We also have very good direct measurements of the upward radiation from the surface, some greater than 396, and others less, but 396 is a reasonable average.

    One of the first things you learn about radiative heat transfer in any course is that this heat transfer is the net of the two radiative energy streams. So my computing a net 63 upward as the difference between 396 up and 333 down is absolutely textbook, and in no way “disregrads” the downward radiation absorbs by the surface.

    You state, “Clearly latent heat is ‘net'”. But then you mock me for stating that it is “net”! If it is net, it is net of evaporation minus condensation. And no, I am not “mumbling vaguely”. I am, and I was, stating it very clearly.

    But unlike the radiative transfers, we have no good way of measuring the evaporative/condensing heat transfers directly. The net has to be inferred from precipitation measurements worldwide, as K&T (and others) have calculated. (And water vapor that condenses as dew is not available for precipitation.)

    You seem to think that the “60% evaporative losses from oceans” you have seen is some fundamental law of nature. But it is really just the result of calculations very much like K&T have done. K&T calculate 50% of net surface losses from combined land and sea as being evaporative. As I have repeatedly said, I do not see any contradiction at all between this and your figure of 60% evaporative losses from the oceans.

  83. mpainter says:

    Curt,
    I have not lost my mind and you are wriggling like a worm.
    Let me put the question once again:
    In the Kiehl, Trenberth diagram is the 333 W/sq m of back radiation “absorbed at the earth’s surface” bullshit or is it real?

    • Curt says:

      mpainter:

      Since your reading comprehension is somehow worse than your scientific and mathematical literacy, you could not understand what I clearly stated.

      Yes, the 333 W/m2 average “back radiation” absorbed at the earth’s surface is real. As I explained, we have very good direct measurements of its magnitude and even its spectrum.

      The 396 W/m2 average “forward radiation” from the surface is also real.

    • Kristian says:

      mpainter,

      The funny thing about this is that they absolutely refuse to acknowledge how they treat the DWLWIR as a second separate radiant HEAT flux to the surface.

      It’s a straightforward operation:

      The Earth’s global surface on average absorbs 165 W/m2 worth of radiant heat from the Sun (I use the figures from “Stephens et al. 2012”). This could only ever by itself potentially warm the surface to 232K. So how to get from this to 289K? You have to add a second (and much bigger) radiant heat flux (but make sure you don’t call it that) to the surface, this time from the atmosphere. 165+345 W/m2. That’s 510 W/m2 of TOTAL incoming radiant heat to the surface. But, hey, this is too much. It would give a surface temp of 308K. So what to do? Ah, yeah, you subtract the combined conductive/evaporative loss from the surface of 112 W/m2. And voilà! You have 510-112= 398 W/m2. And there’s your 289K.

      In short:

      HEAT IN: 165+345-112 = 398 W/m2
      HEAT OUT: 398 W/m2

      https://okulaer.files.wordpress.com/2014/10/drivhuseffekten.png

      They simply start out from knowing the average global surface temp, postulate the corresponding intensity of its blackbody emission as a real, distinct upward flux. And then work backwards from there to make their total budget add up.

      • Curt says:

        Kristian:

        We have very good direct measurements of the:

        ~165 downwelling shortwave from the sun
        ~345 downwelling longwave from the atmosphere
        ~398 upwelling longwave from the surface

        These measurements include detailed spectral information as well as overall magnitude.

        We have good precipitation measurements that show an average precipitation of ~1000mm/year globally (most estimates are in the 950-1050 range). Using the very simple inference that what goes up must come down, we get a global land/sea average of about 80 W/m2 net evaporative losses.

        None of this uses any assumption of temperatures.

        The only thermal flow that is backed out of the others is the conductive/convective (“thermal” in the K&T diagram). That is derived by assuming approximate overall energy balance of the surface. Even that does not assume any specific temperature.

      • jerry l krause says:

        Hi to whom this concerns,

        Kristian, welcome back. You have focused our attentions to the instruments with which observations are made. However, you also wrote: “The Earth’s global surface on average absorbs 165 W/m2 worth of radiant heat from the Sun (I use the figures from “Stephens et al. 2012″). This could only ever by itself potentially warm the surface to 232K. So how to get from this to 289K?” But first you wrote: “The funny thing about this is that they absolutely refuse to acknowledge how they treat the DWLWIR as a second separate radiant HEAT flux to the surface.” Who are they? Now, back to “The Earth’s etc.” Now I know that 165 W/m2 is a calculated value, not an observed value. I do not have to go the source of your information to confirm this because of the words: “on average.” You can scan back to see my opinion about the common practice of averaging observations that should not be averaged. And maybe somewhere at some time you have referred to where one can access this reference, but not here and now. I would guess that 165 W/m2 is the result of reducing the solar insolation reaching the earth’s surface by the earth’s albedo. But I do not know if this is the case. I do know that 289K is an averaged value for the whole earth and not an observed actual value. Instruments supposedly observe actual values that exist in real time.

        Your comments about instruments has caused me to reflect upon what I am actually observing with my ‘modified’ Suomi, Staley, Kuhn (SSK) net radiometer. I have referred to this radiometer, one-half of SSK net radiometer, on numerous occasions at various times. One consistent observation is that between sunset and sunrise, given clear sky conditions, the temperature of its absorbing-emitting (a-e) surface is always several degree F (often 10 F but sometimes as great as 18 F and as little as 3 F) below the ambient air temperature being measured about 1.5m about the earth’s surface by an instrument which is shielded from direct line to sight to the clear sky by something other than the instrument itself, like a carport or a tree.

        It is commonly argued by some, perhaps you if I understand what you write, that DWLWIR emitted by matter whose temperature is less than that of the surface intercepting this radiation cannot be ‘absorbed’ by this warmer surface. Of course, this is the common 2nd Law argument. What if the argument were valid? As long as the temperature of the radiometer’s a-e surface is above the temperature of any emitting matter above it, it cannot absorb any energy. But the a-e surface still can emit so it cools until its temperature is less than that of the emitting matter above it, then it can absorb the energy of the intercepted radiation from this warmer matter and there becomes a balance between the energy being absorbed by the a-e surface and the energy being emitted by the same a-e surface.

        Before I go further, I would like to read your comments with regards to the previous paragraph.

        Have a good day, Jerry

  84. mpainter says:

    Norman, I put the same question to you: is it bullshit or is it real?

  85. gbaikie says:

    Say, one had a 20 km wall, which was 75 degree latitude, encircle the world. And the wall goes down to the bedrock.
    And had another wall at 60 degree latitude and another at 45 degree latitude. And walls are on both south and north latitudes

    And question is would this change the average temperature of Earth?
    And if so, would it cool or warm and how much temperature difference would result?

    And is it fair to say that according to the Greenhouse Effect theory this should not increase or decrease earth’s average temperature?

    • jerry l krause says:

      Hi gbaikie,

      At first I ignored your comment because I consider thought experiments a waste of time. Then, upon reflection, I saw the possibility that this was not a thought experiment. Was your point only that the greenhouse effect ignores any atmospheric circulation?

      Have a good day, Jerry

      • gbaikie says:

        — jerry l krause says:
        June 17, 2015 at 11:02 AM

        Hi gbaikie,

        At first I ignored your comment because I consider thought experiments a waste of time. Then, upon reflection, I saw the possibility that this was not a thought experiment. Was your point only that the greenhouse effect ignores any atmospheric circulation?–

        Well I would say it was 1/2 a point and 1/2 a question.

        Do I think the greenhouse effect theory ignores circulation?
        Yes.
        Do I think circulation is something that increases average temperature?
        Yes.

        The greenhouse effect theory states that the only warming or only thing which can cause increase in average global temperature is greenhouse gases.

        So my question/point was that I thought the walls would cause a decrease in global average temperature.
        And I wondered whether believers in the greenhouse effect theory would think such walls would cause warming, had no real effect, or if they agree with me that that the walls would lower the global average temperature.
        And of course I would be fascinated to hear what reasons they came up with to explain their answers.

        Btw it’s a similar issue/question of whether Earth’s rotation would cause an increase in the average global temperature.

  86. mpainter says:

    Curt
    See how you reply to my questions.

    Then my calculations are valid and latent heat is 207 W / sq m, not the 80 W/ sq m given in the diagram above. I repeat, the Kiehl, Trenberth diagram is egregious.

    Norman:
    Your argument is with those whose studies partitioned ocean cooling. You don’t like 60% as the evaporation figure? Then get your own, but you will have to justify it. I believe that the actual evaporation figure is higher than 60%, myself.
    80 W/sq m latent heat gives 16% as the evaporative figure. Do you like that?

    • gbaikie says:

      — mpainter says:
      June 16, 2015 at 10:54 PM

      Curt
      See how you reply to my questions.

      Then my calculations are valid and latent heat is 207 W / sq m, not the 80 W/ sq m given in the diagram above. I repeat, the Kiehl, Trenberth diagram is egregious. —

      I would think more than 207 W / sq meter in the tropic.
      But roughly the amount it rains per year globally is the amount which is evaporated.
      And one average one has 1 meter of precipitation:
      http://hypertextbook.com/facts/2008/VernonWu.shtml

      And one cubic meter of water evaporated requires: “Latent heat of evaporation – 2,270 kJ/kg”
      So there 1000 kg in cubic meter of water:
      2,270,000 times 1000 is amount of joules [or watt seconds]
      Which divide by amount of seconds in a year: 31.5 million.
      So I get 71.98 joules per second is needed.
      So 80 watts seems reasonable assuming there is 1 meter on average of rainfall, and assuming there degree of water evaporation which can’t be measured.
      Now one can get water condensing in the air, and evaporating, and condensing, and evaporating and condensing and etc
      But that seems like it could be rather difficult to figure out.

      But seems that the amount it rains globally per year, is where the 80 W for latent heat number is coming from.
      Or I would say if 1 meter of rainfall is global average there needs to be at least 71.98 watts to make that much water to fall from the sky [can’t be less than this- though certainly could be more].

      • gbaikie says:

        “But that seems like it could be rather difficult to figure out.”

        Or example clouds [droplets of water] commonly burn off.
        You can have Marine layer of clouds in the morning and by noon they have disappeared. So they were caused by water evaporating, and formed into clouds, then they evaporated again- with no precipitation involved.
        And of course Earth is covered with oceans, and happens all the time within oceans.

    • Norman says:

      mpainter,

      The KT diagram is for averaged over every square meter of Earth’s surface. There is frozen water at the poles that is not evaporating. If you average out zero with what is happening in the tropics you get a much lower value.

      Where did you come up with the 60% figure? It would not be if I like this value or not, it is more how did you derive it? I could not say if it is a good or bad figure unless I know what it is based upon.

      Evaporation rate is a highly variable process. It is based upon more than just surface temperature of water. The surrounding environment plays a large role in what the rate is (which would determine then the % of energy lost from surface in evaporative cooling). Very humid still air has a very low evaporation rate even if the surface water temperature is high.

      I think it would be much better if you showed how you arrived at this 60% figure and what it means? Is it a global average per each meter squared? Is it the amount of cooling that occurs in low humid windy tropical water so it only applies to certain regions of the ocean and then cannot be used as a global average.

      Help out, thanks!

  87. Curt says:

    mpainter:

    For your 207 W/m2 latent heat of evaporation to be correct, there would have to be average precipitation of 2500 mm/year across the globe. That is patently ridiculous!

    You can argue values between 950 and 1050 mm/year, but 2500 mm/year is so far from reality, it’s not even funny!

  88. mpainter says:

    Curt,
    I use values from the diagram for my calculations. Is this what you mean by ridiculous?

  89. Curt says:

    The diagram says 80 W/m2 for average latent heat transfer. That corresponds to the reality of 1000 mm/year of precipitation.

    Your 207 W/m2 value does not correspond remotely either to the diagram or to reality. That’s what I mean by ridiculous!

  90. Phillip Bratby says:

    I have to say that in all my years working in fluid flow, heat transfer etc, I never came across anything like the K-T diagram. It doesn’t even have the correct units. Averaging non-linear behaviour over 24hours and over the surface of the spherical earth is complete nonsense. And showing anything other than net radiation (i.e. in the direction from hot to cold) is also complete nonsense. How it ever passed peer review and became accepted as mainstream climate “science” must be one of the world’s greatest mysteries.

    • Curt says:

      Philip:

      The units of the K&T diagram are power flux density (W/m2). I find them peppered all over my heat transfer texts.

      If you want to use overall power flows, just multiply by A, the surface area of the earth (m2). They integrated over the area of the earth, and then divided by that area to get a flux density in manageable units. It is trivial for you to undo that last step.

      If you then want overall energy flows in a year, multiply by Y, the number of seconds in a year. They integrated over a year, and then divided this time to get watts. It is trivial for you to undo that last step.

      And, contra to your assertion, these are all valid mathematically linear operations (conservation of energy demands this).

      What would not be valid (and what K&T did not do) is to derive an “average temperature” from these average power flux densities, given the non-linear relationship between (local) power flux densities and (local) temperatures.

      • Phillip Bratby says:

        What I stated remains 100% correct. I don’t need your attempted interpretation as I am fully conversant with the subject.

        • Michal Garbers says:

          Bear with me for a moment,
          Let’s take a room with one light bulb let state the bulb light the room evenly to 100 lumens.

          Let’s add a second bulb of the same intensity the room will be lit to 200 lumens.

          Replace the second bulb with a 50 lumen light the room is lit to 150 lumens

          Replace the second bulb with a 10 lumen light the room is lit to 110 lumens

          This is where the model you understand and are comfortable with comes in conflict with radiated energy.

          In the model you understand and are comfortable with the 100 lumen light would be hot and the 10 lumen light would be cold. And the 10 lumen light could not increase the illumination (Heat) the room. Because it is a cold object (10 lumens) in a hot room (100 Lumens).

          Anything above absolute zero is technically a “hot object” and can radiate heat. This does not change that heat flows from hot to cold it just adds a greater dimension.

    • FTOP says:

      Totally agree, accept I would replace mysteries with tragedies.

  91. dave says:

    “…doesn’t even have the correct units…”

    Indeed. Says on the tin it is about ENERGY – that is measured in Joules, not Watts per Square Meter!

    I am reminded of an anecdote about Field Marshal Montgomery, when he was the commandant of a staff college. A student was given a mark of zero by him for a planning paper. When he queried it, Montgomery explained:

    “The instructions said to put your name on the top-left of the paper and you put it on the top-right. Anybody who goes wrong on the first line…”

  92. mpainter says:

    Norman
    You are a late comer.
    All of your questions I have addressed up thread and excuse for not repeating.

    • Norman says:

      mpainter,

      I looked for an answer to your 60% evaporation cooling you use. I did not see anything except you saying assume 60% based upon some studies but not having any links to these studies. It looks like MikeB called you out on it but you did not answer his post with a link.

      I do not think you did provide support for your 60% evaporative cooling point. You use it in equations but you do not justify its use with links to where you get this. You claim some studies put it as high as 65%. Can you link to the studies so I can read about how they arrived at this.

      I think the 80 watts/meter is correct. I would like support evidence to show this is incorrect. I have done the math and it fits. I have only seen you use 60% as it is an established fact. I did Google searches to see if I caould find this number but I have not yet.

  93. mpainter says:

    Folks
    Try to not be so obtuse. Of course evaporation totals are determined by precipitation totals. My object was to show the egregiousness of the K-T diagram by using data from the diagram to calculate precipitation in an alternate method.Note that I say repeatedly upthread that the diagram is wrong. Is it not obvious that the DWLWIR is exaggerated?
    My calculation shows that the KT diagram is wrong, showing how it does not square with reality. Read the thread. I never claimed my calculation was the correct precipitation amount, but only that the diagram produced egregious results. Those who horn at the middle of the discussion are tiresome.
    The 60% comes from other studies, as I have said repeatedly.
    One more time:
    The KT diagram is egregious, particularly in its depiction of back radiation. My calculations use data from the diagram to show how egregious it is. Never did I claim that my calculations yield true totals of latent heat, only that tthe math was correct. Dense people are so tiresome. Begone.

    • Curt says:

      And as I have repeatedly said, the K&T diagram shows 50% of surface losses over total land and sea are evaporative (80 out of 160). That is in no way inconsistent with 60% of sea losses being evaporative.

      This is not difficult…

    • gbaikie says:

      “Folks
      Try to not be so obtuse. Of course evaporation totals are determined by precipitation totals. “”

      I would say precipitation totals are roughly equal to evaporation at the surface. X amount leaves the surface and roughly same returns to the surface per year.
      So somewhere around 80 joules of water evaporates from the surface [mostly ocean] and such amount of water falls to surface as rain. So this one tends to rule that there is 80 times 2 amount of water being evaporated from the surface as there would a corresponding increase in rain, or it would force there to be more rainfall than there is.

      So this means the “back radiation” of 333 joules can not be mostly evaporating the water at the surface of the oceans.
      And as we know this “back radiation” can not directly heat the water beneath the surface. So indicates that the “back radiation” can not warm the surface.
      Or if “back radiation” is not causing an increase surface evaporation [the top 1 mm of water should absorb all of this energy] them it’s also not warming the land surface on Earth, either.

      This also means the any CO2 increase would not increase surface evaporation and cause more rainfall. But what would cause more rainfall is more warmed tropical warm moving poleward. Or it’s warmer water which causes an increase rainfall, and back radiation does not warm water.

      But many believer think that back radiation “represents” the lack of energy radiated to to space, than any actual warming effect, so if warm surface water is prevented from cooling as much as it would otherwise, than warmer tropical water could travel further poleward. And if it actually did this, than higher CO2 levels could cause more rainfall and cause more warm water going poleward and warming these regions.

    • MikeB says:

      “My object was ……to calculate precipitation in an alternate method”

      What alternative method was that, Mr. Painter? Just making it up?

  94. mpainter says:

    Curt,
    I agree that the 207 W per sq m is ridiculous; that is the point.
    It is ridiculous because the KT diagram is ridiculous, namely in 333 W of back radiation.
    You insist that 80 W / sq m is the correct figure for latent heat? I do not dispute that nor is my math meant to dispute that. To get 80 W/ sq m you must use 27 W/ sq m in my formula:
    .71 × .6 × 188 W/ sq m = 80 W/ sq m.
    You have to show where my math is wrong.

  95. mpainter says:

    To clarify,
    Back radiation = 27 W/ sq m; add that to insolation for 188 W/ sq m.

    Norman,
    I don’t have a link. Richard Perlwitz gave one on an earlier thread a few weeks ago. Go look and please STOL suggesting that I make things up. And derive your own figure if you don’t like mine

    • gbaikie says:

      -Norman,
      I don’t have a link. Richard Perlwitz gave one on an earlier thread a few weeks ago.-

      You should provide the link or withdraw your claim.
      But I believe I found it:

      — Jan P Perlwitz says:
      May 2, 2015 at 10:15 PM

      “Yes, radiation and conduction are but minor compared to evaporation, which accounts for over 70% of the energy loss at the interface.”

      According to a very recent estimate by Wild et al., Clim. Dyn., 2014, http://link.springer.com/article/10.1007%2Fs00382-014-2430-z

      the partitioning between latent heat flux, net longwave radiation, and sensible heat flux is 100, 53, and 16 W/m^2, respectively, at the ocean surface, i.e., latent heat flux accounts for about 60% of the total heat flux between ocean and atmosphere. And even if it was 70%, the remaining 30% wouldn’t be a negligible fraction.

      “You have utterly missed the significance of the temperature profile: this shows that incident IR does not reside in the skin but is entirely transient and cannot be conducted to depth.”

      That the infrared radiation “resided” in the ocean skin or that it was “conducted to depth” is not what is said.

      Please try at least to understand the argument.

      Absorption of longwave radiation doesn’t mean that the radiation “resided” in the absorbing medium. Upon absorption it is transformed into heat energy, i.e., the kinetic energy distribution of the water molecules at the top of the ocean skin is incrementally shifted to higher values. Without this, there couldn’t be any incremental increase in the latent heat flux from the top of the ocean skin to the atmosphere, either. This incremental heat increase at the top of the skin isn’t conducted downward, but it decreases the temperature differential between the cooler top of the ocean skin and the bottom of the ocean skin. Heat conduction from the bottom of the skin to the top is reduced, since that is a function of the temperature gradient. More heat is retained in the subsurface ocean layer, until a temperature gradient in the skin is reached at which incoming radiation and outgoing heat are in equilibrium again. The new equilibrium requires a slightly increased subsurface ocean temperature.

      “In the final analysis, it will be determined that water vapor feedback is negative, as well as clouds and ECS is too low to have any effect on the global temperature index.”

      More wishful thinking I would say. What you believe will be the outcome of the final analysis is in contradiction to the empirical evidence. For instance, the water vapor content in the atmosphere has been increasing with global warming like it has been predicted. More water vapor means a stronger greenhouse gas effect. This is a positive feedback.—
      http://www.drroyspencer.com/2015/05/uah-v6-0-global-temperature-update-for-april-2015-0-07-deg-c/#comments

  96. mpainter says:

    And if it comes down to whether .6 is a viable figure for representing ocean cooling via evaporation, I will win that, Norman. Go research.

  97. mpainter says:

    Your comment says it well. If the back radiation does not cause evaporation nor warming of the ocean, then what the hell can it do? Or CO2?
    My point in the tangle above was to demonstrate that diagrams like the KT diagram above are portraying the GHE egregiously.

  98. Norman says:

    mpainter,

    Your comment I think directed at me “If the back radiation does not cause evaporation nor warming of the ocean, then what the hell can it do?”

    I will repeat again. It slows down cooling. It does not add energy to the surface since it is always less than the energy being given off by the surface. The evaporation rate is already determined by the surface temperature of the water and other factors. The backradiation will not increase this temperature and add more energy to cause greater evaporation. It will slow the rate of cooling which will allow the sun to add more energy into the system. Not a real hard concept.

    • jerry l krause says:

      Hi Norman,

      You are correct. It is amazing how some twist slowed cooling rate into warming.

      Have a good day, Jerry

  99. m says:

    Norman,
    My comment was not directed at you specifically.
    I note what you claim regarding back radiation,that ” it does not add energy to the surface”.
    Note the KT diagram where it says of the backradiation “333 absorbed by surface”.
    Will you please explain how the surface can absorb such energy without any energy being added, and you can’t, of course. Net energy flow is the makes sense, the KT diagram falls to the ground when you examine the claims built into it.

  100. Curt says:

    mpainter:

    You are misinterpreting what that “60% evaporative losses” means. Let’s take a step back.

    The earth’s surface absorbs 161 W/m2 of solar radiation, averaged over the surface and over a year. You seem to agree that this is at least a reasonable number.

    So to stay in at least approximate thermal balance, the surface must lose very close to this same 161 W/m2, again averaged over the surface and over a year. Not even the worst alarmist thinks the imbalance is much over 1.0 W/m2.

    This 161 W/m2 of total losses represents 100% in all of these analyses. You are using the wrong denominator in your equation.

    There are 3 significant modes for the surface to lose energy: conductive/convective, evaporative, and radiative. These three sum up to 100%, which is ~161 W/m2 globally.

    The net contribution of each of these three modes is a “net” contribution, because all three of these have transfers in both directions:

    Conductive/convective: Surface to atmosphere (dominant in daytime), atmosphere to surface (dominant at night)

    Evaporative: Surface to atmosphere (evaporation dominant especially in daytime), atmosphere to surface (condensation, common at night)

    Radiative: Surface to atmosphere (upward LWIR), atmosphere to surface (downward LWIR).

    K&T’s analysis produces the following net values for globally averaged land and sea:

    Conductive/convective: 17 W/m2 (11%)

    Radiative: 63 W/m2 (39%)

    Evaporative: 80 W/m2 (50%)

    The analyses you cite for ocean alone that say ~60% of losses from the ocean are evaporative went through a very similar type of analysis as K&T. But again, “100%” of these losses matches the solar power absorbed.

    You tell Norman that you will “win” that 60% of ocean losses are evaporative. Fine.

    But what you will not win is your assertion that the many, many direct measurements of downward LWIR are off by over an order of magnitude, as you assert, especially when these very measurements were key in producing the 60% figure you keep citing.

    • gbaikie says:

      — Curt says:
      June 17, 2015 at 2:18 PM

      mpainter:

      You are misinterpreting what that “60% evaporative losses” means. Let’s take a step back.

      The earth’s surface absorbs 161 W/m2 of solar radiation, averaged over the surface and over a year. You seem to agree that this is at least a reasonable number.–

      I think it could be somewhere near close.
      But also would say about +80% is absorbed by the ocean
      and 90% absorbed by the ocean is absorbed below the skin surface. Or the ocean absorbing sunlight is much difference
      than non transparent surfaces- a sidewalk- or land surfaces in general.

      Another element is than about 1/2 of earth surface absorbs most of the sunlight. And this half of Earth is warming the other part of the earth.
      Or tropics is about 40% of surface of earth, or 80% of the half of the earth which warms the other half of Earth.

      –So to stay in at least approximate thermal balance, the surface must lose very close to this same 161 W/m2, again averaged over the surface and over a year. Not even the worst alarmist thinks the imbalance is much over 1.0 W/m2.–

      I wonder if I am worse than the worst alarmist. I never figure it out. I would guess that Earth has added more than 1.0 w/m2 per year. I have done some numbers on number years it take if all the sunlight was absorb by the ocean, to increase it’s average temperature by 1 C- centuries, and so in terms of tens of thousands of years, I think the ocean has warmed by around 1 C, so I would guess that would require more than one 1.0 W/m2 imbalance.

      And generally I think we are still recovering from the Little Ice Age, which same as saying we have had imbalance
      for more than a century [the oceans are warming and sea level is rising. Or the sea level fell during the Little Ice Age, and that would also would required imbalance in terms of loss of heat from the Ocean.
      But don’t how many watts it is- but needs to add up to a lot of heat which gained or lost per year.

      –This 161 W/m2 of total losses represents 100% in all of these analyses. You are using the wrong denominator in your equation.

      There are 3 significant modes for the surface to lose energy: conductive/convective, evaporative, and radiative. These three sum up to 100%, which is ~161 W/m2 globally.–

      [Not saying much as this is the only way to transfer heat.]

      –The net contribution of each of these three modes is a “net” contribution, because all three of these have transfers in both directions:

      Conductive/convective: Surface to atmosphere (dominant in daytime), atmosphere to surface (dominant at night)

      Evaporative: Surface to atmosphere (evaporation dominant especially in daytime), atmosphere to surface (condensation, common at night)

      Radiative: Surface to atmosphere (upward LWIR), atmosphere to surface (downward LWIR).

      K&T’s analysis produces the following net values for globally averaged land and sea:

      Conductive/convective: 17 W/m2 (11%)

      Radiative: 63 W/m2 (39%)

      Evaporative: 80 W/m2 (50%)–

      So massive amounts of water flows from the poles to the Equator, and equal amount flows from tropics to poleward.
      Of course atmosphere itself also flows cold air from poles which replaced with warmer air drawn poleward. But the ocean is much biggest player.
      So the Gulf stream for instance, probably more or less unrelated to radiate transfers, so the leaves Conductive/convective or evaporation. But sure how it’s scored. Is mostly evaporation, is it counted as Conductive/convective.
      But it seems there seems that there is agreement than Europe would be about 10 C cooler without the warmed waters of Gulf Stream.

  101. mpainter says:

    Curt,
    Once again you ignore the 333 W of back radiation “absorbed by the surface” shown in the KT diagram. Something tells me that you will always ignore it, even one put a big red X on it, you would ignore it.
    Tell me Curt, how much energy does this back radiation add to the surface? Any idea?

    • Curt says:

      mpainter:

      For the umpteenth time, I have NOT ignored it. Each time, I explicitly include it in my explanation, and you miss it every time. That is why I have to question your reading comprehension!

      I said in my last comment of the various heat transfers:

      “Radiative: Surface to atmosphere (upward LWIR), atmosphere to surface (downward LWIR).”

      I explicitly included the “back radiation absorbed by the surface” there. In other comments I have used the numbers (396-333)=63 and you have still accused me of ignoring it!

      As I have explicitly said before, this 333 W/m2 is real energy absorbed by the surface, but of course less than the equally real 396 W/m2 emitted by the surface.

      I repeat once again that you are misinterpreting what those saying “60% of ocean heat losses are evaporative” mean. The key is: what is 100%. It is (or should be) very clear that 100% losses are those that match the solar input of 161 W/m2 alone, not of solar plus DWLWIR. So they are saying that evaporative losses are (0.6 * 161) or (0.6 * whatever average solar is absorbed by the oceans).

      You are tying yourself in knots by using the wrong denominator!

      And I ask you once again, do you really believe that all of the detailed measurements we have of the downwelling longwave infrared radiation, often with detailed spectral content, are off by over an order of magnitude? (They say 333 on average, you say 27.)

  102. mpainter says:

    Curt, the KT diagram does not add up, period. The diagram shows the surface as absorbing back radiation and then emitting it as…more radiation. The implicit claim of the diagram is that none of the absorbed back radiation is converted to latent energy and the ocean covers 71% of the earth.
    How now, defender of the faith?

    • Curt says:

      mpainter:

      The 2nd LoT requires that the radiative transfer from a hotter body to a colder body be greater than the radiative transfer from the colder body to the hotter body. Clausius understood this and stated it very clearly 150 years ago. You really should try to keep up!

      And when there are multiple modes of inputs and outputs of heat transfer, as there are here, it is not at all valid to try, as you are doing, to match one particular input with one particular output. You can’t say, for example, that the DWLWIR input should go into evaporative output, whereas the solar input should go into the radiative output. Where on earth did you get that kind of idea?

  103. Norman says:

    mpainter,

    Is this the article link you had mentioned?

    http://link.springer.com/article/10.1007/s00382-014-2430-z/fulltext.html

    I am starting to understand what you are seeing but I do not think you are absorbing it. The ocean graph shows evaporation removing 100 watts/meter^2. The radiation is still removing a lot more energy from the surface but because of backradiation the evaporation dominates as the energy remover since it does not come back.

    Radiation flux from the surface of the oceans is given as 409 watts/meter^2. But the loss by radiation is only 53 watts/meter^2 because of backradiation. If you removed GHG from the atmosphere then radiation would be 4 times as great an energy remover.

    Notice the backradiation will not heat the water surface as it is adding less energy back than is lost.

    I am thinking your are backwards in your criticism of the GHE. You think the atmosphere is generating this IR that goes to the surface and heats it. I look at it as the surface is losing this 396 watts/meter^2 (total surface land and sea). Some of this outgoing energy is absorbed by GHG in the atmosphere and then redirected back to the surface (not warming it or adding energy, just lowering the rate of loss).

    With GHG you lose 63 watts/meter^2 from the surface by radiation (the sun adds 161 watts and this added energy which would be 161-63 leaving forever to get a surface imbalance of positive 98 watts/meter^2 which 97 watts/meter are then removed by evaporation and thermals).

    If you have no GHG in the atmosphere you lose 396 watts/meter^2 so the cooling rate is much faster. I think that is all that is being said. Sometimes I believe people over complicate the budget.

    mpainter you should listen to Curt, he seems one of the most knowledgeable posters on this blog. His posts seems solidly grounded in actual textbook physics and I generally learn a lot by reading them.

  104. mpainter says:

    Curt, its quite simple, it really is. If 333W of back radiation is absorbed by the ocean, it will in crease evaporation by…?
    How much, Curt?

    • Curt says:

      mpainter:

      You first! You have ignored my repeated requests to answer the following question:

      “do you really believe that all of the detailed measurements we have of the downwelling longwave infrared radiation, often with detailed spectral content, are off by over an order of magnitude? (They say 333 on average, you say 27.)”

      It isn’t a difficult question…

      • dave says:

        “…333…27…It isn’t a difficult question.”

        It isn’t a difficult answer.

        27 upwards (or 23, according to the diagram), is the net of 356 upwards and 333 downwards. Whether these numbers really should be used as independent “building blocks” in this way, is the difficult question.

        • Curt says:

          But mpainter is arguing that the gross downward radiative power flux density from infrared cannot be greater than 27 W/m2. I have been asking him why the direct measurements of this downward flux density are around 333 W/m2.

          Does he really believe these measurements are off by over an order of magnitude? He has been dodging the question for days now.

    • wayne says:

      mainter, you are good talker and I really don’t have the time but there is one question no one ever seems to even ask. If there is 333 Wm-2 downwelling ir and all atmospheres radiate isotropically, in every direction, where is the equally large red arrow pointing upward? Care to carry that question with you and ask others as you probe? I am curious of the answers you get back.

      (I’ll give the only real answer much later, probably on some later thread but those who push that KT graphic as true and correct will not like the implications.)

      • wayne says:

        Oops, missed the ‘p’ key I see and please just ignore any other typos, didn’t even re-read that one.

      • Curt says:

        wayne:

        Finally someone has identified the most important weakness of the K&T diagram!! Now we can have some real discussions!

        When the K&T diagram models the atmosphere as a “single shell” (the dark blue band on the right), this “shell” cannot have the required isotropic radiation.

        This limitation has long been known. A better model (but still imperfect, of course) is a “two shell” model for the atmosphere. Willis Eschenbach showed it nicely at WUWT six years ago:

        http://wattsupwiththat.com/2009/11/17/the-steel-greenhouse/

        Of course, like any finite-element modeling of a continuous system, it is only a crude approximation.

  105. sky says:

    By claiming that “temperature is an energy balance issue,” Roy gets the discussion off on the wrong foot. CHANGE in temperature is what energy balance controls. The actual temperature LEVEL is quite a different matter. Alas, fundamental physics is seldom the strong suit of meteorologists.

  106. Norman says:

    mpainter,

    I was reading your post above where you stated:
    “The Kiehl, Trenberth diagram does not properly add up. It gives 333W/sq. m of DWLWIR “absorbed by the surface”. Add the 161W/sq. m of insolation, and the earth’s surface absorbs a total of 494 W/sq. m radiation.
    Some simple math:
    Assume sea surface cooling is 60% via evaporation. Ocean=71%of earth’s surface, hence:
    .60×.71×494= 207 W/sq. m. Compare this to the figure of 80W/sq. m of evaporation in the Kiehl, Trenberth diagram.
    These sort of diagrams show how badly distorted is the GHE by the present body of “consensus” science.”

    I can see the error in your thinking. You are only taking energy that is moving down and using this in your equation to come up with the absurd value. You have to subract 396 watts from the downward energy to have a valid amount of energy hitting the surface from the combination of backradiation and solar insolation. The energy at the surface is not just the downward flux, it is losing energy at the rate of 396 watts. If you subract the 396 from the 494 you get a correct value of actual energy absorbed by the surface of 98 watts which is removed by thermals and evaporation (or the surface would be really hot!).

    I think that is where you confusion comes from in your counter posts to Curt. You are forgetting the surface is a furious radiator as well as absorber.

    Have a nice day.

  107. mpainter says:

    Norman, you are full of it, as in up to the eyebrows.
    You say: “I think you are backwards in your criticism of the GHE”
    BS button on you, Norman. All of my criticism has been directed toward the Kiehl- Trenberth diagram of this post. I have not used the term GHE that I can recall.
    Regarding the KT diagram of the post, it shows the surface absorbing 333 W/ sq m of back radiation and emitting that as more radiation. However, the surface is 71% ocean and so this absorbed energy must be converted to latent energy. The question that I put is how much? ThevKT diagram etrs in that it omits this conversion of absorbed back radiation to latent energy.

    • dave says:

      “…conversion of absorbed back radiation to latent energy…”

      It may well be that the energy of incident IR, absorbed – as it is – in the topmost skin of the water of the oceans causes local boiling in which case energy is transferred back to the atmosphere – by evaporation and then condensation into liquid droplets – in a forced, feedback, manner. Anybody who has followed the discussion over the last thirty years knows that this possibility never crosses the minds of establishment scientists*. Paradigms of “trapping” are ever in their heads.

      The KT diagram, of course, confuses the different processes over land and ocean, resulting in a hopeless muddle. And the drawer does not seem to know whether he is supposed to illustrate a “one slab atmosphere” or a “many slab atmosphere.”

      It is still the case, however, that the oceans emit thermal IR radiation in large amounts, because that is the way water works. In other words, it would be quite wrong to say that the oceans cool only by evaporation.

      *It is unbelievably difficult to find investigations of this precise issue. The question has fallen into the cracks between scientific specialisms.

      • mpainter says:

        Dave,
        Thanks for your comment. So climate scientists do not investigate such issues? That means to me that they do not consider such issues.I regard this as a reflection on the state of the science and especially of the invetigators in this field.

        However, for me there is no question whether or not LWIR causes evaporation.

  108. mpainter says:

    Correction, last sentence: The KT diagram errs in that…

  109. jerry l krause says:

    Hi Guys and Gals,

    Here’s something you might consider. You might think it has nothing to do with your present efforts but I know it has something to do with mine,

    In an article titled—Mathematics and Science Learning: A New Conception (Science, Volume 220, pp 477, 478)—Lauren B. Resnick wrote: “Several studies show that successful problem-solving requires a substantial amount of qualitative reasoning. Good problem-solvers do not rush in to apply a formula or an equation. Instead, they try to understand the problem situation; they consider alternative representations and relations among the variables. Only when they are satisfied that they understand the situation and all the variables in it in a qualitative way do they start to apply the quantification that we often mistakenly identify as the essence of “real” science or mathematics.”

    Have a good day, Jerry

  110. jerry l krause says:

    Hi Guys and Gals,

    I have a question of which I really hope more than one of you would answer. In fact I wish Roy and each of you would all give your answer to this question. The question is: Why do you deal with average values for the earth as if they are real when an average earth clearly never has existed, does not exist, and never will exist?

    Have a good day, Jerry

    • gbaikie says:

      — jerry l krause says:
      June 17, 2015 at 8:27 PM

      Hi Guys and Gals,

      I have a question of which I really hope more than one of you would answer. In fact I wish Roy and each of you would all give your answer to this question. The question is: Why do you deal with average values for the earth as if they are real when an average earth clearly never has existed, does not exist, and never will exist? —

      I blame it on the Greenhouse Effect theory.
      So it starts with the hopeless stupid model.
      Or some Swede or Englishman asks why aren’t we freezing to death. Benjamin Franklin gave them a clue, when noted the effect of the Gulf Stream:
      “Although first observed in 1513 by Ponce de Leon, the Gulf Stream was not charted until the early 1770s by Benjamin Franklin, with the help of a Nantucket sea captain.”
      http://oceanservice.noaa.gov/facts/bfranklin.html

      • Noram says:

        gbaikie

        I may be wrong in my response to your question: “The question is: Why do you deal with average values for the earth as if they are real when an average earth clearly never has existed, does not exist, and never will exist? –”

        I believe the purpose to generate an average is to look for trends which would be very difficult with just a bunch of data points.

        A real world example. A teacher wants to see how effective her teaching methods are. She has 100 students and gives out tests on the material at intervals to see what her students are learning.

        Individual test scores are all over the place (things like 100,87,94,76,…) and she cannot see an overall trend. She adds all the scores up and divides by the total number to get an average test score. As the year goes on with averaging she is able to see if the overall test scores are going up, down or staying the same.

        With the Earth system you can learn nothing of a trend on energy balance by taking the temperature at one specific location to try and see if things are changing. It is a daunting task and takes lots of effort. Bias can damage the outcome (like in my test score example if the teacher fudges a few numbers to make her look better).

        I think the main goal of a global average is to try and get some way to determine a trend line.

        • jerry l krause says:

          Hi Noram.

          I’m sorry I did not thank you for responding to my question even though I disagree with what you wrote. Maybe I should not have even disagreed with you. But I did pose the question as an attempt to begin a dialogue about the issue of averaging.

          Have a good day, Jerry

      • jerry l krause says:

        Hi gbaikie,

        Thank you for your response. Think you for your very good response.

        Noram, you did not even read carefully what gbaikie wrote. It was my question to which he was responding and that was clearly written in his response. So, I doubt if you thoughtfully read what he wrote further. He referred to a specific factor, the Gulf Stream, which we have long known affects the climate of Northern Europe. So, if we are to understand trends in their climate maybe we should begin by studying the ‘trend’ of the Gulf Stream and not some average trend of the whole planet. And I know there are scientists who have and are studying the Gulf Stream. The problem is their efforts are lost in the focus upon the average trend of the whole planet.

        I do not know if you have noticed, but weather always occurs locally. Even very large, unpredictable, weather systems such as hurricanes directly affect only a relatively small portion of the earth’s surface.

        Now, I read that an El Nino event, a generally unpredictable event, in the equatorial Pacific is occurring. This we have now observed is related to abnormal local weather at very distant locations from the Pacific Ocean. We know this because we know the average high diurnal temperatures and the average low diurnal temperatures during specific times of the year at these specific locations. But while these specific departures from local average values of temperature and precipitation etc. are observed, the average values of earth are not observed to change significantly. This is because the high temperature of a day and the low temperature of a day is clearly influenced by the presence or absence of clouds. At almost any specific location, the high temperature of the day will be greater than the average high temperature of that location and season and the low temperature of the day will be less than the average low temperature when there are no clouds. But the average temperature of this specific day will be very close to the average temperature of the day at the that specific location and season.

        Hence, there can be no doubt that averaging the temperature of the day destroys information about a specific location and averaging temperatures of the world destroys even more information which needs to used to begin to understand the meteorology and climatology of the earth.

        Have a good day, Jerry

    • Steve Milesworthy says:

      Average values are used for illustrative purposes, but are derived from real measurements and calculations taken from all over the Earth.

      Considering your previous post, if you “understand the problem situation” and apply them to “alternative representations and relations among the variables” then you can do the same calculation for different locations on the planet at different times of day and then start to refine each calculation.

      Such calculations could be the basis for a simple weather/climate model.

      • jerry l krause says:

        Hi Steve,

        “Considering your previous post, if you “understand the problem situation” and apply them to “alternative representations and relations among the variables” then you can do the same calculation for different locations on the planet at different times of day and then start to refine each calculation.”

        Steve, the problem is we do not yet “understand the problem situation”. If we did, I would not read the debate and argumentation that I read on this blogsite and others. At one time the understanding of Copernicus and Galileo was strongly refuted by the ‘establishment’. I no longer see any debate and argumentation about this issue. This is what to “understand the problem situation” means to me. Roy titled his post—What Causes the Greenhouse Effect—as if it has been established there is the Greenhouse Effect as he defined it in his post.

        Have a good day, Jerry

        • Steve Milesworthy says:

          It’s an engineering problem. A complex engineering problem. There is no magic.

  111. mpainter says:

    gbaikie,
    Yes, Jan (not Richard)
    Perlwitz and I went round and round. He admitted that the referred sea surface temperature profile was correct in its essentials, or he did not contest the validity of it but seemed to accept it, I should say, a tacit admission.
    I have seen other such partitions referred to on other blogs, giving slightly different figures, these predating this link.
    As long as my memory serves me, I will be relying on it. Thanks.

  112. MikeB says:

    mPainter

    The oceans do not cool predominantly by evaporation. This is in your imagination. Look at the diagram again.
    The oceans cool predominantly by RADIATION: 356 watts per square metre compared to 80.

    In the original Kiehl Trenberth paper, KT97, the surface radiation was given as 390 W/sq.m. This figure can be calculated directly from the Stefan-Boltzmann equation. The Earth’s surface, and especially the Ocean, can be regarded as a near blackbody in the infrared (Wilber et al. (1999) estimate the broadband water emissivity as 0.9907). Thus, Stefan-Boltzmann gives 390 W/sq.m for a surface at an average temperature of 15 degC.

    The 356 W/sq.m is a refinement made in the 2009 update by Kiehl and Trenberth which takes account of the fact that the planetary surface is not all at the same temperature and so emissions from warmer latitudes and seasons is disproportionately higher.

    What you must NOT do, is subtract radiation coming in from radiation going out and pretend you have anything meaningful. If you do that then you have nonsense figures which do not make sense from the Stefan-Boltzmann perspective. This would indicate that the average surface temperature was -90 deg.C, clearly wrong. Throwing information away is generally a bad idea and in this case there is no justification for it, except obfuscation.

    • jerry l krause says:

      Hi MikeB,

      “The oceans do not cool predominantly by evaporation. This is in your imagination. Look at the diagram again.
      The oceans cool predominantly by RADIATION: 356 watts per square metre compared to 80.”

      Mike, do not be critical of mpainter. Notice Roy’s big red X on the diagram. Roy is the one who claims that oceans and all the earth surface does not emit radiation upward. And he said he would correct any errors that were pointed out. But if he considers this red X a mistake, he has not corrected it.

      Have a good day, Jerry

      • Tim Folkerts says:

        Jerry, you need to re-read what Roy said, because you completely misinterpreted a key idea in this whole post.
        “the energy flows I have marked with an “X” would not exist without GHGs

        And note, the big red x upward from the surface is ONLY on the “356 W/m^2 from surface to atmosphere” not the “40 W/m^ from surface to space”. Indeed the 40 W/m^2 to space would INCREASE (and all the other numbers would rearrange as well) if we suddenly magically made the atmosphere transparent to IR.

        • gbaikie says:

          — Indeed the 40 W/m^2 to space would INCREASE (and all the other numbers would rearrange as well) if we suddenly magically made the atmosphere transparent to IR.–

          It would seem to be magic.
          How can make any gas completely transparent to large spectrum of IR?
          Of course we loosely say your atmosphere is transparent to sunlight, and it’s fairly transparent if only passing thru say 10 meter of it- or about 10 kg per square meter of it.
          But when sunlight goes thru 10 km of it [or 10,000 kg per square meter] of it, loses about 300 watts of the 1300 watts enter it at the top of the atmosphere.
          And it seems if you heated a square meter of something on the Earth surface so it radiated at 1300 watts per square meter that the 1300 watts of IR would loss 300 watt before leaving earth atmosphere.
          As in take one square meter of iron and heated it to 100 C
          or a 10 meter by 10 meter sheet of iron, it seems one would lose about 1/3rd of it. If made it 10 by 10 meters, one should be able to easily detect it from orbit particularly if it’s during the night.

          And if one had less atmosphere say something around 1/10th of Earth atmosphere [1 ton per square meter] it should have less loss going thru the atmosphere.
          So it seems generally regardless of gases involved if there is 10 tons per square square there will be losses going thru it.

          You could compared Earth atmosphere to Mars atmosphere.
          So Mars is very thin atmosphere with total of about 25 trillion tons of mostly CO2 [with about 210 ppm of water vapor] and it has less than 1 ton per square meter.

          [Mars surface area is 144.8 million square km, or 144.8
          trillion square meters, so far less than 1 ton per square meter.]

          And if heat a square meter of iron so it emitted 1300 watts per square meter, one should not loss anywhere near 300 watt before it leaves the atmosphere.
          Or one could say Mars has less greenhouse effect than Earth, but Mars has more greenhouse gases than earth has per square meter.

          Another aspect of Mars is it should cost less energy to heat one square meter of iron to 100 C on Mars as compared to doing it on Earth. Or on Earth one has much more convection loss because of it’s denser air and higher gravity. And this factor should more significant as compared to the difference of air temperature.
          So for example Mars air temperature could be -100 C at nite, whereas Earth’s air temperature could be more than 100 C warmer, say 10 C.

          Now, CO2 is very transparent to Sunlight and if not for the dust in atmosphere one almost gets as much sunlight on Mars surface as compared to top of the Mars atmosphere.
          Or might say dust in atmosphere of Mars is it’s major greenhouse effect. So if on Mars during on if major global dust storm, you might loss as much as 300 of the 1300 watts
          from the heated square meter of iron as passed thru atmosphere into space. But it seems that such dust in the atmosphere would have little effect upon the energy cost involved with heating the square meter of iron- or it’s related mostly to the lack of air convection losses.

        • jerry l krause says:

          Hi Tim,

          The big red X denies that the earth surface emits any radiation. It does not matter what happens to this radiation after it is emitted and this emission from the surface is clearLy not dependent upon there being an atmosphere or what an atmosphere might be.

          And immediately after your quote of Roy’s he stated: “Thus, those energy flow arrows marked with an “X” in the above diagram represent huge flows of energy which can affect temperature, if they really exist.” I just reread all that Roy wrote following the diagram and this statement and I found nothing which quantitatively addressed the numbers he had X’d out.

          Plus, I made in comment in the context of the debate that was occurring relative to the question whether the oceans primarily cooled by evaporation or radiation.

          And how do we know it would take magic to make the atmosphere nearly ‘transparent’ to IR?

          Have a good day, Jerry

    • sky says:

      MikeB:

      When incoming LWIR intensity is subtracted from outgoing, the NET radiative cooling rate (i.e. effective heat transfer) is obtained! Until this basic distinction between heat transfer and directional intensity is thoroughly grasped, egregious confusion will grip your mind.

    • wayne says:

      “The oceans cool predominantly by RADIATION: 356 watts per square metre compared to 80.”

      If MikeB truly believes that statement scratch him from knowing anything of thermal radiation.

      Or, perhaps he just forgot to perform the subtraction such as 396 minus 356 of that which is opposing in direction and simultaneous and that the two streams can never be addressed without a stated knowledge of the other. Besides, he chose the wrong figure accoring to that TFK graphic.

      I’ll take the liberty to fix his statement (corrections bolded):

      “The oceans do not cool predominantly by RADIATION: Radiative cooling being 396-333 or 63 watts per square metre compared to 80 plus 17 sensible cooling.”

      There, a little of Mike’s confusion removed from this thread.

  113. mpainter says:

    Wayne,
    Your point is taken. Note the 396 W/ sq m of radiation from the surface, which includes the emission of the absorbed 333 W. So there is now, in an isotropic radiative flux
    396 + 333 = 629 W of outgoing IR flux. There are all sorts of absurdities that can be concocted from this diagram. It does no good to point it out to Curt, Norman, MikeB, and such defenders of the faith, who will never allow themselves to be pinned down on such discrepancies. They are not capable of taking note of them.
    But, as Philip Bratby pointed out up thread ” showing anything other than net radiation is complete nonsense”.

  114. mpainter says:

    MikeB
    In fact, I misstated. I meant evaporation, not precipitation.
    Curt refuses to answer my questions. Perhaps you will:
    How much evaporation can be expected when the oceans absorb 333 W/sq m of LWIR?

  115. mpainter says:

    MikeB
    ” The oceans do not cool predominantly by evaporation.”
    ” The oceans cool predominantly by radiation”
    ###
    Wrong, MikeB. No one else would make such a claim but you.

    • MikeB says:

      You’re not arguing in the school playground now. You must support your assertions by reference to some scientific paper or other. You can’t just make things up.

      There are many peer-reviewed papers supporting what I say, Kiehl and Trenberth, 1997, being just one. So far you have offered none!

      • jerry l krause says:

        Hi MikeB,

        “You must support your assertions by reference to some scientific paper or other.”

        MikeB, don’t you know the deck is stacked. Of course you can find peer-reviewed papers supporting what you say. But, can you find peer-reviewed papers that challenge what you say? Of course not.

        Have a good day, Jerry

      • jerry l krause says:

        Hi MikeB,

        As I read further from the bottom, maybe I support your point of view. But you still cannot make the argument that you have made. Just because something is peer-reviewed does not mean what is stated in the article is valid.

        Yes, I quote authorities to support my viewpoint, but usually it is to inform others of something that seems not well known. In other words to demonstrate that the coin has two sides and both cannot be correct.

        Have a good day, Jerry

  116. mpainter says:

    MikeBLike the others, you ignore the evaporation from the surface.

  117. mpainter says:

    Curt,
    Oh yes, let’s assume the figures are inviolable.
    So we have 333 W/sq m of LWIR being absorbed by the oceans.
    Sooooo.. how much evaporation, Curt?

  118. Steve Milesworthy says:

    In the second paragraph “miniscule” is spelt wrongly.

    The correct spelling is “significant”.

  119. mpainter says:

    MikeB
    Otherwise your comment is Stefan-Boltzmann gobbledegook which has no application to the oceans.

  120. mpainter says:

    MikeB
    And here we go again;
    161 W/ sq m insolation plus 333 W/ sq m LWIR
    equals 494 W/ sq m total radiation on the ocean. For that we get 80 W/ sq m of latent heat. Does not square.
    I repeat, the KT diagram is egregious.

  121. mpainter says:

    MikeB
    Your perspective seems to be that any thoughts unsanctified by approved doctrine is impermissible.
    And we the skeptics are supposed to genuflect before such a diktat.

    • MikeB says:

      Quite right! That is how science works. I expect you to be able to substantiate the ridiculous made-up facts that you assert.

      It is not acceptable to just make up non-facts or pluck them out of thin air. If you can’t substantiate it, and I have asked several times, then I simply dismiss it as the rambling of a fool.

      However, you’re not yet quite as bad as Doug Cotton, he made up a whole new branch of non-physics, on the basis that people like you wouldn’t be able to tell.

      But you are in 2nd place.

  122. mpainter says:

    Norman
    I’m sorry, but you seem incapable of following the train of my reasoning.
    I will try once more, for the last time.
    The LWIR is absorbed at the surface is 333 W/ sq m as per the diagram.
    This energy, incident upon the ocean, by the laws of radiative physics, is converted to latent energy, at least in part. You have it all emitted as radiation. So does the diagram and this is incorrect.

    • Curt says:

      mpainter:

      You seem to think that this DWLWIR must be specifically “earmarked” for latent heat generation. It’s as if you were arguing that one of several money flows into a common bank account could be earmarked for a certain class of expenses from that account.

      It is clear to me now that you have never done any real energy balance problems, especially any that have several subsystems within the overall system. If you had, you would have learned long ago that flows to and from a system or subsystem cannot be partitioned that way.

      From the K&T diagram, considering the solid/liquid earth as the subsystem we have:

      161 W/m2 input from the sun (gross and net)

      63 W/m2 net output in longwave radiation
      17 W/m2 net output in thermals (conductive/convective)
      80 W/m2 net output in evapotranspiration

      The outputs almost balance the inputs – so far so good. (By the way, I don’t agree that we know the flows with this precision.)

      And note that 50% of the output power flux is evaporative. (This is computed in the same way as the figure you keep citing of 60% evaporative output power flux from oceans is evaporative.

      All three net output figures are really the differences between lesser inputs and greater outputs. We only have direct measurements of the gross flows for radiation, so K&T show these alone as opposing gross flows. But let’s represent all three this way, using possible values (the exact values are not important:

      (396-333)=63 W/m2 net output in longwave radiation
      (20-3)=17 W/m2 net output in thermals (conductive/convective)
      (82-2)=80 W/m2 net output in evapotranspiration

      So looking at this more finely, we have four gross inputs:

      161+333+3+2=469 W/m2

      and we have three gross outputs:

      396+20+82=468 W/m2

      Still close to balance in the same way.

      But for you claim, as you have, that the specific 333 gross flow of DWLWIR must be specifically allocated “at least in part” to latent energy, is meaningless. There is no physical mechanism to earmark it in this way.

      • gbaikie says:

        –But for you claim, as you have, that the specific 333 gross flow of DWLWIR must be specifically allocated “at least in part” to latent energy, is meaningless. There is no physical mechanism to earmark it in this way.–

        Well suppose that sunlight was not transparent to ocean water.
        If ocean were muddy or completely filled with algae then the ocean would be far less transparent.
        So were the ocean not as transparent as it is, then the ocean would unable to absorb as much energy and therefore most of the 161 watts the world absorbs would go towards evaporating the ocean.
        So rather than sunlight passing thru the ocean to depths of of 100 meter, if ocean were murky, the sunlight could only go say to only an 1/10th of meter. And this would dramatically effect earth climate.

        The dispute is about the 333 watt being absorbed by the surface. Most of sunlight reaching the Earth surface is not absorb at the surface but rather, most is absorbs more meters under the ocean surface.
        So imagining that earth surface absorbs 333 watts per square would have a more dramatic effect than if the ocean were not transparent to sunlight.

  123. mpainter says:

    Who is this Noram person?

  124. Norman says:

    mpainter,

    You seem also incapable of following my reasoning so we are equal in this aspect.

    Just so you can understand this. For the moment remove incoming solar flux from the KT diagram above and what are you left with?

    You have a downwelling IR of 333 watts/meter^2 but you have an upwelling IR of 396 watts/meter^2. Now I ask you with your physics knowledge and understanding of radiative physics, how can 333 watts/meter^2 heat the surface when it is losing energy at the rate of 396 watts/meter^2?

    Your logic does defy basic function here. I guess in your world if you are filling an outdoor pool at the rate of 333 gallons a minute your pool will fill up even though you have a hole that is pouring out at the rate of 396 gallons a minute. What physics did you study when you attended college?

  125. JDAM says:

    Is it true the surface of the moon cools 6 times slower than the surface of the earth?
    The Moon’s surface cools at a rate of 0.8°C/hour.
    The Earth’s surface cools at a rate of 5°C/hour.

    • gbaikie says:

      — JDAM says:
      June 18, 2015 at 10:05 AM

      Is it true the surface of the moon cools 6 times slower than the surface of the earth?
      The Moon’s surface cools at a rate of 0.8°C/hour.
      The Earth’s surface cools at a rate of 5°C/hour.–

      What is true is that we tend measure the surface of the Moon
      and measure the air surface of Earth [5 foot above the ground in the shade in a white box].

      A large factor of warming the ground is the the angle of the sun.
      When the sun is higher than 45 degree above the horizon- which related to time of day and latitude and season- then
      on a level surface, one has most intensity sunlight falling on a square meter of the level ground.

      So if at equator and at equinox, and it’s noon, the sun will be 90 degree above the horizon. Or directly over head.
      Each hour that sun will change by 15 degree. Or 3 hours before noon the sun was at 45 degree above the horizon and 3 hours after noon, the sun will be at 45 degree above the horizon.
      And so 4 hours before or after the noon, the sun would be 30 degrees above the horizon.
      When the sun is at about 30 degree above the horizon, the level ground will receive about 1/2 as much sunlight per square meter. And receives less sunlight at 5 to 6 o’clock.

      Or on clear day, when the sun is higher than 45 degrees, the level ground will receive about 1000 watts per square meter. And when it’s at 30 degrees above horizon, the level ground will receive about 500 watts.

      Whereas with the Moon, when sun is high in the sky the ground receives about 1360 watts per square meter and half of that when sun is 30 degree above the horizon.
      On the Moon if you point a square meter at the sun, that square meter will receive 1360 watts. And if sun is at 30 degree, the shadow coming from the pointed square meter will be about 2 square meters.

      The length of shadow of something pointed at the sun in indicates the amount sunlight hitting a level surface.

      So on Earth as on the Moon, the early morning’s and late afternoon’s has long shadows. And winter is the time of longer shadows.

      So when sun has about 1000 watts it can warm the surface to about 70 to 80 C [160 F]- whether on the moon or Earth.
      And on Moon when sun is high above horizon and it’s getting about 1360 watt [the Moon and earth varies it’s distance from the Sun and this has variation of 1414 to 1321:
      “Earth distance: 1,413 to 1,321”
      https://en.wikipedia.org/?title=Sunlight ]
      So on Moon when about 1360 watt per square meter it can heat surface to 120 C.
      And finally to perhaps the point, when the Moon has the sun blocked by earth [eclipse] that surface temperature can lower by 100 C [100 K] in a period of 2 hours.

      Now if took frying pan heated to 120 C and put it outside, it should cool to air temperature within about 1 hour.

      So very roughly speaking, Moon and Earth cools at around the same rate. The slow rate you think is related to Moon is related it’s long night and the lower surface temperatures cooling slower [and beneath lunar surface the ground is warmer than the cold temperature the surface of the ground can cool to].

    • MikeB says:

      Is it true the surface of the moon cools 6 times slower than the surface of the earth?

      NO!

      Where did you get these non-facts from? mpainter by any chance?

      If the Earth cooled at 5 Deg.C per hour, imagine what the temperature would be in the morning!

      “At lunar sunset, the moon’s surface temperature shown in Figure 1 is about -60°C. Over the next 30 hours, it drops steadily at a rate of about 4°C per hour”
      http://wattsupwiththat.com/2012/01/08/the-moon-is-a-cold-mistress/

  126. mpainter says:

    Norman
    I understand you. You think that radiation incident on the surface is instantly emitted 24 hours a day on every square meter whether night or day, land, sea, ice or what. Latent energy does not enter your thinking.

    • Norman says:

      mpainter,

      I am glad you understand me because I do not understand your comment. Radiation is being emitted continuously 24 hours a day on every square meter at the rate determined by the Steffan-Boltzmann relationship (temperature and emissitivity for the particular material involved). Radiation will be continuously emitted regardless if the surface is exposed to incoming radiation or not. It really does not matter the amount of radiation incoming on the outgoing (except for reflection). The incoming radiation will be absorbed and turn into kinetic energy of molecules that will emit the radiation. I am not sure of the actual speed at which this process takes place. Latent energy does enter the thinking.

      Maybe I am not so clear. I will try again. As you show patience with me I will be considerate of you.

      You have 161 watts/m^2 solar incoming hitting the surface, you have 333 watts/m^2 backradiation hitting the surface. As you demonstrated this adds to 494 watts/m^2 hitting the surface. I point out the surface is continuously losing energy at the rate of 396 watts/m^2 by radiation. The difference between the Energy gained by the surface and energy lost is 98 watts/m^2 (or if you want joules/second). These joules will add energy to the surface. Now what happens, the surface heats and you have convection and conduction removing 17 watts/m^2 from the surface and moving it upward. Then of that remaining energy 80 watts/m^2 is used up in breaking the hydrogen bonds holding an H2O molecule in liquid phase and allowing it to break free to become vapor.

      Based upon what rational logic do you assume “Latent heat does not enter your thinking”??

      I think you are making things far more complex than they are, it really is simple to understand the KT energy budget and it is not flawed as you are trying to argue.

  127. Norman says:

    JDAM,

    Is this the source of your post?

    http://www.principia-scientific.org/why-lunar-heating-cooling-disproves-the-greenhouse-effect-theory-on-earth.html

    That is such a bad article scientifically.
    One they are using the rate of cooling of a boiling pot of water which loses heat in all three ways (where the moon only is losing energy via radiation).

    There calculation of the moon’s cooling rate is horrible science. I read some of this group’s material for a period of time but they are so poor in science it is more a propaganda sight.

    Here is what the measured rate of heat loss is for the moon’s surface.

    http://diviner.ucla.edu/science.html

    The temperature drops from 390 K to around 110 K in 6 lunar hours then drops the other 10 K in the remaining night hours.

    The moon day cycle is 27 days (give or take hours) so 6 lunar hours would be around 4.5 Earth days. You have 108 hours in 4.5 days. Now divide the 390-110=280 K in 108 hours instead of falsely stretching the time and you have the moon cooling at a much more realistic rate of 2.6C/hour.

    Now how much of the boiling water cooling they used to get the 5 C/hour was lost by convection and conduction so you can compare the two properly.

    • jerry l krause says:

      Hi Norman,

      You do not cite another observed fact about the moon which is found at http://diviner.ucla.edu/science.html. Which is that the moon’s surface was observed to be generally covered by a 1-2cm thick layer which is an extremely good thermal insulator. Might this insulating layer be the principal explanation of the lunar surface temperature oscillation which you cite?

      Have a good day, Jerry

      • Norman says:

        jerry l krause,

        You are probably correct. The insulating property of the moon’s surface dust may slow down the rate of cooling as the energy gained by the solar day moves up slowly to be radiated away. I would think if the surface was less insulating the temperature drop and gain would be much more rapid. The insulating part seems to work on both sides, the daytime heating curve is similar to the cooling curve.

        My main point on the post was to discredit the horrible science found at PSI.

        • jerry l krause says:

          Hi Norman,

          “I would think if the surface was less insulating the temperature drop and gain would be much more rapid.”

          My understanding is quite the opposite. There is little discernible lag in the ‘daytime’ temperature plot because there is so little thermal inertia involved. And there is so little thermal inertia involved because of the insulating layer. So, the temperature of the surface nearly keeps pace with the solar radiation being absorbed with the result most of the absorbed radiation is immediately being emitted back to space. Hence, very little energy is conducted through the insulating to be stored as sensible heat in the lunar soil below the insulating layer. Then at sunset, the surface has already cooled very near its lowest temperature of the lunar diurnal cycle. So during the long night the sensible heat that has been stored in the soil below the insulating layer during the daytime is slowly emitted emitted to space because of the extremely cold temperature of its surface.

          Thank you for responding to my previous comments and I hope you will give me some feedback about these comments.

          Have a good day, Jerry

    • gbaikie says:

      “http://diviner.ucla.edu/science.html

      The temperature drops from 390 K to around 110 K in 6 lunar hours then drops the other 10 K in the remaining night hours.”

      Yes but that is lunar hours which are the time it takes for the sun to appear to move 15 degrees times 6 [60 degrees].
      So Lunar day is about 27 earth days [655 earth hours]
      so 655 divided by 360 degrees is 1.8 earth hours per degree, times 15 is 27.3 earth hours per 1 earth hour. And 6 lunar hours equals about 164 earth hours.

  128. mpainter says:

    Jerry,
    You got the very devil in you and I love you for it.
    XXX
    mpainter

  129. mpainter says:

    Norman,
    It is no good attempting a dialogue unless you agree on basics. My understanding is that the back radiation is LWIR and that LWIR is absorbed in the first few microns of the sea surface. For example, the 15 micron wavelength is absorbed within three microns of the interface.
    I intend this as preface to my answer to you. Do you agree on those points?

  130. mpainter says:

    Be patient Curt

  131. mpainter says:

    Feeling a little distempered? If you have faith in your science the ramblings of a fool should not upset you.
    What non-facts, pray tell.

  132. mpainter says:

    The above intended for MikeB

  133. Norman says:

    mpainter,

    You have more knowledge than myself on the distance traveled by IR in water. I have looked at some sights and most would agree with your conclusion. I will accept this as factual informtion for you to be able to project a position of your understanding into my own view on the issue.

    I would still like to point out that the same water molecules that will absorb the downwelling IR will also be the ones emitting it (warm water below will not emit IR through the water to the surface, radiation is a surface phenomena). So you have these surface molecules losing energy at the rate of 396 joules/sec-meter^2. The surface will quickly cool unless there is a source of energy to make up this loss, some from warmer water below some from the downwelling IR. Your radiating surface water molecules a few microns thick are losing 396 joules/sec and you have this downwelling source of energy at 333 joules/sec-meter^2 to replace the loss, these surface molecules will still be moving slower after each second and will have less energy to break the hydrogen bonds necessary to enter a vapor phase.

    Now you have the solar incoming energy that is not stopped at the surface. It penetrates as deeply as the turbidity will allow and is converted to thermal energy. This source will keep the thin surface molecules vibrating and radiating and evaporating at the equilibrium rate.

  134. mpainter says:

    Norman, you do not have to depend entirely on me; there is plenty of info on the web concerning the radiative and absorbency properties of water.
    But I am glad you accept my claim so now we can proceed.
    My response depends on the partition of ocean cooling into its components, that is, evaporative, radiative and sensible heat loss.
    Several studies have examined this question and have derived slightly different figures. I propose to use evaporative cooling: 60%, radiative cooling, 30%, sensible heat loss, 10 % which fractions are representative of all studies, in the main.I will use the 60% figure for some calculations. If you disagree with that 60% figure, you can use your own.
    Now, I wish to illustrate something with a simple calculation:
    Insolation from the KT diagram is put at 161 W/sq m; oceans cover 71% of earth’s surface. So:
    161 × .71 × .6 = 69
    or 69 W/sq m is latent heat through oceans. Add latent heat from land (evapo- transpiration) and the sum is close to 80 W/sq m, perhaps a little less, say 77 W/sq m.
    And so we see by the above calculation that the latent heat on the Kiehl-Trenberth diagram, that is, the 80 W/ sq m, is well accounted for by the energy from insolation.
    Now, what do we do?
    Because we still have the back radiation of 333W/sq m “absorbed at the surface” in terms of the diagram. If this back radiation is real, then the ocean surface is absorbing more energy via back radiation than via insolation.
    Remember also that it is absorbed within a few microns of the interface.
    Soooo… how much latent heat results from the 333 W/sq m absorbed at the ocean’s surface?
    This is one reason why the KT diagram is egregious. My calculations up thread, which you, Curt, and MikeB ridiculed, were intended to reveal this flaw, not to derive some new figure for latent heat. This point missed you utterly. Let’s see if this comment has any effect.

  135. mpainter says:

    I should caution you.
    I know that you will have a great temptation to attribute to radiative heat loss all or most of the absorbed energy of the 333 W.
    You need to know that numerous studies have examined, empirically, the effect of LWIR directed at the surfaces of water, experimentally, with measured results. There is no question LWIR will evaporate water very effectively- from the top down.
    So be forewarned and do some homework before you start throwing figures of radiative heat loss on the wall.

  136. mpainter says:

    Kristian
    You say that the figure for back radiation is pulled out of their hat?
    This is contrary to what Curt says who repeatedly claims multi-measurements of back radiation.

    • Kristian says:

      Of course he does. That’s because he doesn’t understand (or, more precisely, doesn’t want to understand) what’s actually being MEASURED (as in ‘physically detected’) and what’s simply being COMPUTED (as in ‘mathematically calculated’). It is ALWAYS the heat and nothing but the heat (what he calls the ‘net flux’ or ‘net transfer’), the actual thermal transfer of energy, from warmer to cooler, that’s being measured, NEVER any of the two conceptual halfs of a radiant heat transfer. How Curt ever managed to convince himself that you can somehow physically isolate, extract and directly observe only one ‘side’ of an integrated heat transfer as if the other one, occupying the exact same space, weren’t there at all, is beyond me.

      • Curt says:

        Kristian:

        You wonder “How Curt ever managed to convince himself that you can somehow physically isolate, extract and directly observe only one ‘side’ of an integrated heat transfer as if the other one, occupying the exact same space, weren’t there at all.”

        Well let’s see… First you use a sensor that’s only open to radiation in one direction. Then point it in different directions. That’s a start.

        Second, if you want, you can supercool the sensor, so even if the actual signal you are sensing is a net exchange, it is not significantly different from the gross input. (A lot of the early IR sensors were supercooled, mostly to reduce measurement noise, but it had the advantage of essentially eliminating the difference between gross and net.)

        Third, isolate small wavelength bands on the sensor, usually through refraction angles (like a prism). With the outgoing radiation from the sensor at near blackbody (and trivial if it is supercooled), as you scan through the wavelengths, you can get dramatically different readings depending on the varying emissivities as a function of wavelength of the source.

        Now lets say you want to understand the radiative exchange between a surface at 288K and an upper atmosphere at 255K or so. You put your sensor supercooled by liquid nitrogen to 77K in between. First you point it down and scan through the wavelengths. The radiative output of the sensor is so low that you wouldn’t have significant errors even if you ignored the difference between gross and net.

        Now point it up and repeat the process. You now have good measurements of the magnitude and spectral distribution separately of the upward and downward gross radiative flows.

        You don’t want to pay for a liquid nitrogen cooling system for your sensor? Then just use a sensor at ambient temperature that has been calibrated against one of these, so the proper adjustments can be made.

        Your argument is equivalent to asserting that you can’t use a tire pressure gauge to calculate the absolute pressure inside a tire because it only measures the relative pressure between the tire and ambient atmospheric. Of course, just adding the atmospheric pressure, say from a barometer, does it.

  137. Norman says:

    mpainter,

    It seems Roy Spencer has already devoted an entire thread to the topic of your interest.

    http://www.drroyspencer.com/2014/04/can-infrared-radiation-warm-a-water-body/

    I did not see you posting in this thread but many views are entertained.

  138. mpainter says:

    Oh. I forgot my reference:
    The Kiehl, Trenberth study from which the above diagram was taken. Now MikeB can’t complain about lack of references.

  139. mpainter says:

    Thanks Norman,
    I am aware of the thread.I have very definite views on that topic.

  140. jerry l krause says:

    Hi gbaikie,

    Almost missed your comments of 6/17 at 8:01pm. For that reason I submit these comments at the end of things so you, and others, will be less likely to miss your comments and my comments. For the information of others, gbaikie divided the troposphere with high walls every 15 degrees latitude or so to limit the atmosphere’s circulation from the equator to the poles and back.

    And in reply to my comments he replied: “Do I think circulation is something that increases average temperature? Yes.”

    First I must commend you that your comments were entirely qualitative. No numbers. I have had to ponder your reply for I had never considered the question. I know that during the arctic winter, the surface temperatures at the North Pole do not cool to the extremely low temperatures that can be observed during the South Pole’s winter season. But this difference I understand is due to the difference of elevations of the two pole’s surfaces. I know that the North Pole temperatures do not cool to near the low temperature that one would expect for an ice surface that does not intercept any solar radiation for half a year. And I understand this is because its subsiding atmosphere is warmed adiabatically as it subsides. And I understand this subsiding atmosphere is due to atmospheric circulation. So if I did not ponder further, I might conclude the circulation does increase the average temperature. And the katabatic winds rushes down the slopes of the Antarctic Continent is evidence that the atmosphere, during the southern winter, subsides over the continent.

    But I also know that the atmosphere could not subside over the North Pole or South Pole during their respective winters if cold air masses did not regularly drift toward lower latitudes. Which would obviously cool the usual seasonal temperatures of these lower latitudes. Hence, at one general region circulation warms and at another it cools. So based on this I would conclude that the coin stands on edge and that circulation causes neither warming nor cooling. Or maybe a better answer would be: I don’t know.

    However, I have noticed something that I wonder if you have considered. I have just focused on the general time of the solstices. Which, even if the elevations of the two polar surfaces are quite different, have somewhat similar atmospheric circulations. What if we focus upon the atmospheric circulation that occurs near the equinoxes?

    Now, I propose the elevation difference makes a great difference. The Arctic region has an approximately 10,000ft thick surface, atmospheric, layer that the Antarctic region does not have. So, toward the northern spring this not insignificant volume of atmosphere begins to warm and as it warms it must expand. This expansion must drive (force) a circulation of cooler, drier, atmosphere toward the lower latitudes whose surface, being mostly land, is also warming relatively rapidly. There is only one specific region of these lower latitudes, with is a pool of warm sea water, which is so accessible to this forced circulation of the cooler, drier, arctic atmosphere. It is the Gulf of Mexico. And thunderstorms and tornados are a feature of the spring time climate of the lower Mississippi drainage region. And these thunderstorms and tornados strongly lift the surface atmosphere to high altitudes further promoting the flow of the surface atmosphere from the north and south to where they collide with each other. Producing more thunderstorms and tornados.

    Then, near the northern fall this not insignificant volume of atmosphere begins to cool and as it cools it must contract. Commonly it is not said that the atmosphere is pulled anywhere, but that is what I imagine to happen as this volume of atmosphere contracts as it cools. The contraction should pull surface atmosphere from the lower latitudes toward the higher latitudes. Now, much of this lower latitude atmosphere is warm and humid. But now the colder, drier atmosphere of the higher northern latitudes is also being pulled northward and there is nothing to lift the warm, humid atmosphere as it pulled northward. So there are few thunderstorms and more fog and low clouds, because of the cooling land surface it is being pulled over.

    As I said, this is what I imagine and I cannot give any reference to any peer-reviewed article which directly support what I imagine. This does not mean that they don’t exist, just I do not know about any. But I certainly would be interested in your comments, if any.

    Have a good day, Jerry

    • gbaikie says:

      –I know that the North Pole temperatures do not cool to near the low temperature that one would expect for an ice surface that does not intercept any solar radiation for half a year. And I understand this is because its subsiding atmosphere is warmed adiabatically as it subsides. And
      I understand this subsiding atmosphere is due to atmospheric circulation.
      So if I did not ponder further, I might conclude the circulation does increase the average temperature. And the katabatic winds rushes down the slopes of the Antarctic Continent is evidence that the atmosphere, during the
      southern winter, subsides over the continent.–

      I would say that the Antarctic is big gorilla down south, and as you suggest with the katabatic
      winds- it’s drawing in high elevation air [which not very warm- though would have more moisture] whereas with north polar regions has a vast plain of low elevation frozen sea ice, and Greenland a bit player in the entire “eco system”.
      Of course, duh, I am also forgetting about the Gulf Stream which is the biggest gorilla of arctic.

      — But I also know that the atmosphere could not subside over the North Pole or South Pole during
      their respective winters if cold air masses did not regularly drift toward lower latitudes. Which
      would obviously cool the usual seasonal temperatures of these lower latitudes. Hence, at one
      general region circulation warms and at another it cools. So based on this I would conclude
      that the coin stands on edge and that circulation causes neither warming nor cooling. Or maybe a
      better answer would be: I don’t know. —

      As said elsewhere, if one were to thoroughly mix the ocean, you would instantly cool the world [for humans] but it also warms the ocean and therefore warms the world [eventually]. And mixing the entire oceans would inhibit cloud formation in tropics, and Oceans would warm very fast [in terms of
      centuries].
      But anyhow, we are in ice box climate- have been for millions of years. Ice box climate is cold ocean
      polar ice, and glacial and interglacial periods.
      What has cause this climate is the arrangement of land areas in relationship to ocean areas.
      The position of Antarctic at south pole [it been moving there for over last 50 million year] is one
      of major elements causing us to be in Ice box climate.

      In general terms, I would say the land ocean arrangements has caused cooling by a net result of the
      inhibiting global circulation.
      Another element was the joining of south and north America.
      Or imagine an El Nino when there is open stretch of ocean from Africa to Asia??
      There also Bering straits, and India crashing into Asia. Also imagine vast polar ice sheet without
      it being stuck to a continent- tens of millions square km of floating ice dumping cold water into the tropics [and tropical surface water flowing back]. All such polar ice would do in warm the world- until such time that polar regions were too warm to form ice in the winter [but even just cooler water would fall].

      Anyhow it’s all very interesting and etc.
      But idea of the walls was mainly about making it simpler
      and based on the fact that most of sunlight shines in the tropics.
      And everyone should know that tropics is the heat engine of earth. It’s geometry.
      And btw, it’s why you should not try to make Germany the solar capital of the world- even if Germany wasn’t cloudy in the summer.

      • gbaikie says:

        oh mis pasted it. This should been above’s beginning:

        — I have had to ponder your reply for I had never considered the question.
        I know that during the arctic winter, the surface temperatures at the North Pole do not cool
        to the extremely low temperatures that can be observed during the South Pole’s winter season.
        But this difference I understand is due to the difference of elevations of the two pole’s surfaces.–

        Yes. Or average elevation of Antarctic is:
        “Antarctica is the highest continent on Earth: average elevation is 8,200ft (2500m). The elevation
        at the South Pole is 9,300ft (2835m). ”

        So due to dry air lapse rate is closer to 8 C per 1000 meter, and 2500 meters being 24 C cooler
        vs sea level. And we could say that higher elevation is significant factor.
        Related to topic, there is a polar wind circling Antarctic. So this wind would act vaguely
        like a human made wall. And so I would say that is another factor which makes the south pole cold.
        But it seems that a 20 km high wall would work better. And the circular wind is also mixing so it’s
        both a cooling and warming effect.
        Now want to check Greenland’s average evelation:
        “The mean altitude of the ice is 2,135 metres (7,005 ft).[1] The thickness is generally more than
        2 km (1.2 mi) and over 3 km (1.9 mi) at its thickest point.”
        https://en.wikipedia.org/wiki/Greenland_ice_sheet
        So it seems Greenland has a high average elevation also. As recall average temperature is
        around -11 C or it’s about 40 K warmer than Antarctic average temperaqture.
        Of course Greenland is smaller and not in at middle of Pole, and isn’t surrounded by sea ice.
        And Greenland is more like the southern pensulia of Antarctica- which pokes furthest towards and is
        the nearest to the 60 degree latitude and has milder temperatures compared to most of Antarctica.
        Anyways not going to go into it all, but I would say the polar wind around Antartica may be causing
        it to to be about 10 C cooler.

  141. mpainter says:

    Kristian
    An IR astronomer on a blog talked about measuring back radiation
    It was interesting because he insisted that the contribution of CO2 to total back radiation was only 3%, and water the other 97%. I was impressed. But he seemed to measure this by some spectral method. Is this what Curt means by measuring?

    • Kristian says:

      The point is, mpainter, under no circumstance are you measuring “back radiation” (that is, the smaller conceptual half of a radiant heat transfer). You are always measuring (physically detecting) the actual thermal (thermodynamic) energy transfer, the heat (the ‘net flux’). You cannot possibly isolate and extract one perceived half of a full, integrated process.

      The spectral instruments (spectroradiometers), like the AERI, allegedly ‘measuring’ DWLWIR from the atmosphere to the surface use cryogenically cooled detectors (at 67-77K). In other words, the spectral flux measured and analysed is “forward” radiant heat transferred from cool air layers (mostly directly above the instrument) to a much colder detector, NOT to the warmer surface beneath.

      No one in their right mind would seriously believe that you could simply open up a hatch to the sky and let all its ‘cold’ radiation flow freely down to a warmer sensor at, say, surface or room temperature and then everything would be fine, full DWLWIR spectrum revealed.

      If you read about these instruments, you will soon discover to what lengths one has to go in order to capture the spectrum (hence, the photons) one wants to analyse. Either you need a very hot IR source shining at and through your sample and then onto the detector behind, or you need a detector that is cooled down to cryogenic temperatures and then try the best you can to shield it from the instrument’s own radiation.

      The point is:

      There’s a potential for radiative transfer all the time and everywhere, simply from objects and regions having temperatures (a certain amount of ‘internal energy’ and a certain ‘heat capacity’) and possessing various radiative properties depending on their composition and configuration.

      But these potentials are only ever realised when two such objects or regions at different temperatures and/or radiative properties come into thermal contact of some kind.

      In other words, the cool atmosphere has the potential to transfer energy by radiation to the ground, but it can never realise this potential unless the ground becomes even cooler. But cool a detector just above the warm surface to a temperature way below that of the atmosphere and the radiative potential of the air is spontaneously realised as a radiant heat transfer along just that path, not between cool atm and warm sfc, that is, but between cool atm and super-cold detector.

    • wayne says:

      Kristian seems to be describing that correct.

      I have such an infrared spectroradiometer spectrum, very high resolution, taken by the Raytheon Corp. Not taken pointing upward as many assume but horizontally to not add the complexity of varying density and temperature with altitude that would bring into play other unnecessary assumptions. The path length able to detect is no more that 1000 feet. Horizontal is upon the assumption that all radiation from the atmosphere is isotropic in nature (all directions) so direction matters not as long as the surface is not in the view cone. This is such, that I gather, from which that 333 Wm-2 (“back radiation”) comes from, not taken pointing up into the atmosphere but instead right here at the surface just meters above, and horizontally, and that explains why that figure in KT’s graphic is so high as you would expect just a few degrees cooler than the surface. As Kristian says, it is really normal forward radiation from the gases. Very misleading on that graphic. Might view that as taking the radiative temperature of the air just above the surface (the 356 Wm-2) that cannot read the window frequencies.

      • Curt says:

        Climate science has many, many problems, but not knowing which way to point a sensor is not one of them.

  142. Norman says:

    mpainter

    On a post above you state: “You need to know that numerous studies have examined, empirically, the effect of LWIR directed at the surfaces of water, experimentally, with measured results. There is no question LWIR will evaporate water very effectively- from the top down.”

    I looked up some equations on evaporation.
    https://en.wikipedia.org/wiki/Penman_equation

    and
    http://www.engineeringtoolbox.com/evaporation-water-surface-d_690.html

    Note in the first link the first term in the equation is net radiation. If this term is negative it will decrease the evaporation amount by some amount. The net radiation of outgoing and incoming backradiation (without solar) is negative and will reduce the rate of evaporation from the surface.

    I found no equation where it makes any claim that IR will directly cause evaporation. The process I have seen are that the water temperature is what determines the evaporation rate regardless of the heat source. The IR is absorbed by the surface and is converted to heat, this heat is then what warms the fluid and causes the water to evaporate not a direct conversion process.

    I do have a question to ask you. What does it mean to you that the KT diagram shows 396 watts/meter^2 being emitted from the surface? What is going on when you are losing this energy? What does this do to the surface that is emitting it? Does the surface warm as it emits? If you remove the 161 solar input from the KT diagram (Sun is removed from the system) what happens to the surface based upon the KT diagram. Finally how can 333 watts/meter^2 warm the water surface when simultaneously as it is absorbing this energy it is losing 396 watts/meter^2. I have yet to see you consider this radiation loss in any of your posts or equations. The elephant in the room that is ignored. Why?

    Would you also mind taking the time (using the Penman equation putting in realistic variables for it) and show me how a net negative radiation flux can increase evaporation rate? You have a minus 63 watts leaving the surface, how does this produce increased evaporation?

    Thanks and have a good day.

  143. mpainter says:

    Kristian
    You seem to be saying that the measurement is obtained only with respect to the conditions pertaing to the measuring device and has only a limited validity when applied to generalizations. This makes sense.

  144. mpainter says:

    Norman
    I put on you the obligation of accounting for the 333 W/ sq m, or so i thought. Kristian is right when he claims that the AGW proponents will never admit to that as being an independent flux of energy. I hope that you will realize that this 333 W/sq m needs to be accounted for in terms of its evaporative effect on the oceans. I am about to give up.

    • mpainter says:

      Oh, Norman, if you are going to argue that 333 W/sq m of LWIR incident on the surface of the ocean does not cause evaporation, lets just call off the discussion. I dont think it would do any good to continue.

  145. mpainter says:

    Curt,
    About that 27 W/sq m.That was added to the insolation figure (161 W/sq m) to achieve a sum of 188 W/sq m which calculates to the 80 W/ sq m given as latent heat loss. This calculation has been superceded by my post at June 18, 4:02 pm. This latest calculation needs no DWLWIR to account for the 80 W of latent energy lost. In other words, insolation alone accounts for 80 W of latent energy in the KT diagram.

    so here we are again, and still you refuse to answer the question I have been putting:

    How much evaporation from the 333 W/ sq m of back radiation?

    I hope you realize the need to address this question.

    • mpainter says:

      MikeB, you are invited to this party, too.

      • MikeB says:

        mpainter, thank you for the invitation.

        Let me tell you my position. I am a physicist, although climate science is not my field. I am sceptical that the greenhouse effect threatens any catastrophic consequences for our planet. The greenhouse effect is certainly real; all scientists will concur with that. The only question is ‘what will be the effect of adding even more CO2 to the atmosphere?’ I believe the answer is not much, the warming effect from CO2 is something we already have (33 deg.C worth) but adding any more will have only a marginal effect. There is science to support that view.
        On the other hand, I think some of the sceptical arguments from amateurs who deny the greenhouse effect altogether, do more harm than good. They give the impression that ALL sceptics are uninformed, anti-science, or both. They merely provide ammunition to those who say ignore the crazy sceptics.

        As Dr. Robert Brown from Duke university said to another commenter on another current blog
        “As a physics professor and with the greatest of good will, you should never, ever post again on the subject of radiation or physics in general, certainly not unless you are going to take the time to learn some of both”

        So, likewise, with the best will in the world, why not try to learn? Start here…..
        http://scienceofdoom.com/roadmap/confusion-over-the-basics/
        …or anywhere else. But start to learn. I would like to see ‘informed’ sceptics. They add weight to our view, science-deniers do not.

        • Mark Stoval says:

          “The greenhouse effect is certainly real; all scientists will concur with that.”

          That is demonstrably false.

    • Curt says:

      mpainter:

      I have already told you repeatedly that you can’t partition inputs and outputs that way.

      Let’s say you and your spouse both deposit your paychecks into a joint checking account. You cannot validly say that “60% of my paycheck goes to the mortgage payment and none of my spouse’s” and that “25% of my spouse’s paycheck goes to food, and none of mine”. All the funds are co-mingled, so it makes no sense.

      Similarly, all of the energy inputs are co-mingled at the earth’s surface, so it is not valid to ascribe a particular output like evaporation, to a particular input like DWLWIR.

      And since you have not caught it before, I will state it one more time. You are misinterpreting the sense in which analyses state things like “60% of oceans losses are evaporative”.

      These all start from the premise that the only significant gain is the solar input — 161 W/m2 in K&T over land and sea, possibly a little different over sea alone.

      So the total losses must total this same 161 W/m2. In other words, 161 W/m2 is always 100% in their analyses.

      There are three significant loss mechanisms from the surface, but all are just NET losses, with two-way exchanges. In these analyses, the NET from each of these exchanges is calculated before a percentage is calculated. So in the K&T analysis, the 333 downward is subtracted from the 396 upward to get 63 NET upward, which is ~39% of the total 161 that must be lost to match the input from the sun.

      The 17 NET upward thermal is ~11% of this and the 80 NET upward latent is ~50% of this.

      But remember, this “50% of global surface losses are latent heat” in the K&T diagram are part of a system where there is already 333 DWLWIR. You are acting as if it were not.

      Now, the closest I can come to answering your question, which is actually meaniningless for the reasons shown above, is this:

      If the existing 333 W/m2 of existing DWLWIR were somehow removed (the atmosphere made transparent to LWIR), the earth’s surface would cool significantly. This would significantly reduce all three modes of loss, including latent heat loss.

    • Curt says:

      By the way, I have answered your question multiple times now. It’s time for you to step up and answer mine.

      Do you really think that the measurements of the downwelling longwave infrared radiation are off by over an order of magnitude? (333 typically reported compared to your 27)

  146. Norman says:

    mpainter,

    Maybe instead of giving up you should explain what is going on with the loss of 396 watts. You constantly ignore this. Maybe Kristian can explain it to you.

    What is difficult for you to grasp here. All the effects going on are happening in the skin layer of water (radiation out, radiation in, evaporation, convection and conduction with air).

    Please think about it rather than being so convinced of your own reasoning is correct (which it is not you have convinced yourself you are right and your ego self worth will not allow any violation of your own inner thought process to be challenged so you just ignore what you don’t want to hear).

    The same water molecules of the surface that are breaking free of the skin and evaporating are also part of the molecules that are involved in radiating away 396 watts and are the same that are absorbing 333 watts. All the same molecules all involved in the same continous process.

    Did you even look at the link for the evaporative rate??? If you did not then you are need to give up on yourself. What do you not grasp about the concept of NET RADIATION????

    NET RADIATION IS WHAT DETERMINES EVAPORATION RATE according to the equation I linked to!!! LOOK AT IT PLEASE! Net radiation without solar input is negative. The water molecules are radiating away 396 watts and only receiving 333 make up. The net energy is negative. Wow if you can’t understand that take a physics course and quit pretending you know something when you know less than nothing becasuse you cannot learn anything!

  147. RF says:

    To my mind there are (or should be) less convoluted ways to describe how the earth’s climate functions, and therefore probably more representative of reality. Tim Casey’s “The Shattered Greenhouse” ( http://greenhouse.geologist-1011.net/ ), for example, does this quite well, using physics that (most) everyone knows, understands, and can agree on; the best explication you will find in texts or blogs, skeptic or otherwise. In it you will also find a thorough history of the “greenhouse effect(s)”, from its muddled beginnings to how it has been misunderstood and misrepresented down to the present day.

    • jerry l krause says:

      Hi RF,

      Thank you. Thank you. Everyone who has a thought about the greenhouse effect should read what Tim Casey has written. His writing (scholarship) should be an example to which we should all aspire.

      Go and read everyone.

      Have a good day, Jerry

      • Tim Folkerts says:

        I disagree.

        Tim Casey starts by expounding on various historical understandings, along with various modern introductions to the GHE. Yes, the historical understanding were not all correct. Yes, many modern sources (especially introductory ones) are incomplete (or even well off the mark). But the same could be said about “weight” or “how airplanes fly” or “electron orbitals”. This does not mean that we can’t explain weight or make planes fly or predict the spectrum of H2 gas. And it doesn’t mean that there are not good, modern explanations. You just have to be willing to dig past high school level introductions.

        When he gets to his own analysis, he
        * focuses on conduction (which is mostly immaterial) and ignores convection (which is critically important)
        * misses that fact that emissivity (and hence albedo) for LWIR can be (and is) different from the emissivity (and albedo) for visible light. Rather than neither emissivity nor albedo mattering, BOTH matter independently.

        • gbaikie says:

          “And it doesn’t mean that there are not good, modern explanations. ”

          I have not seen them.
          And you continue to say that without greenhouse gases earth would have average temperature of -18 C.
          Which is adequate to demonstrate to me that you have not seen them.

          One thing which is causing the problem is ideal blackbody.
          Now what the ideal blackbody actually would be a very good passive refrigerator in space.
          One can do better job at cooling by first blocking the sunlight for reaching planetary surface.
          And with active refrigeration one also do better.
          But to have a surface be 5 C in sunlight when the moon is 120 C in sunlight, is a pretty good passive refrigerator.

          So to get -18 C one needs reflectors and a pretty good passive refrigeration and that is “unnatural”.
          Or not going to get a planet with ice at it’s equator and Earth distance from the Sun. Or the frostline of this solar system is beyond Mars.

          Though we do have some glacial ice at the tropics on Earth- at high elevation. And one could call that proof that Earth has a refrigerator. And there is aspect of having water at the tropics in it’s deep depths which is close to freezing is also proof of some kind of refrigeration.

          Now when you say Earth would have average temperature of about -18 C, one could say Earth already has average temperature of -18 C. But it depends upon what you regard as proper way to measure Earth’s average temperature. If it’s 5 feet above the ground in shaded white box near sea level. Then it’s roughly 15 C.
          And it’s average is accordingly 15 C average largely because the tropics is much warmer than 15 C on average. Or one has large areas of ocean which average over 20 C.
          So if thought tropics was only thing important, one could say Earth’s average temperature was more than 15 C. And if you though only below 1000 meter of ocean surface was important, than Earth average temperature is about 3 C.

          It seems if measure air temperature at sea level in white shade box and the planet had average temperature of -18 C
          than that planet could have warm temperatures- room temperature type temperature, but obviously a large part of what you were measuring would need to be much colder than this.
          For instance, one could have a slow rotating planet, so if walk at 1 mph you could keep up with the sun. Or half the planet was very cold and the part in sunlight was warmer.
          Btw, no amount of greenhouse gases would cause that world to have a warmer night.

          So my question to you, is if imagine a planet without greenhouse gases at earth distance from a sun like star which had average temperature of -18 C, what would the range of temperatures measured in the white shaded box.
          How cold would the pole be during it’s winter? How warm would an average day be in the tropics?

          So what probably unwittingly doing is putting the planet at distance between Earth and Mars and having be an ideal blackbody.
          Or ideal blackbody at earth distance is 5 C [regardless of where one is on the planet [it’s a refrigerator]].

        • RF says:

          “Yes, the historical understanding were not all correct. Yes, many modern sources (especially introductory ones) are incomplete (or even well off the mark). But the same could be said about…“

          No, the same cannot be said, not anywhere near to this degree.

          NASA
          There are many greenhouse gases, but the most abundant greenhouse gases are water vapor and carbon
          dioxide. Shortwave radiation from the sun passes through greenhouse gases, but longwave radiation is
          absorbed by them.
          Greenhouse gases absorb longwave radiation that is emitted by the surface of the earth.
          Subsequently, they re-emit the energy as longwave radiation in all directions. About half of the re-emitted
          longwave radiation does escape into space, and contributes to the planet’s radiative equilibrium.
          About half of the longwave radiation emitted by the gases is directed back toward the surface of the earth. As
          a result, a continual exchange of longwave radiation takes place between the surface of the earth and the
          atmosphere above it.
          The longwave radiation contained in this exchange causes the warming effect known as the greenhouse effect.
          This phenomenon is known as the greenhouse effect because, like the glass on a greenhouse, the atmosphere
          traps some of the energy beneath it.

          National Oceanic and Atmospheric Administration (NOAA)
          What is the Greenhouse Effect?
          The glass walls and roof of a greenhouse allow most of the sun’s light in, but do not allow most of the heat to
          escape. This causes the temperature inside the greenhouse to be warmer than outside. The
          earth’s atmosphere, and in particular carbon dioxide (CO2) and water vapor (H2O), acts like a greenhouse,
          trapping heat and making the earth warmer.

          National Oceanic and Atmospheric Administration (NOAA)
          The greenhouse effect occurs when the atmosphere of a planet acts much like the glass in a greenhouse. Like
          the greenhouse glass, the atmosphere allows visible solar energy to pass through, but it also prevents some
          energy from radiating back out into space. The greenhouse effect insures that the surface of a planet is much
          warmer than interplanetary space because the atmosphere traps heat in the same way a greenhouse traps
          heat. Certain gases in our atmosphere, called greenhouse gases, tend to reflect radiant energy from the
          Earth’s atmosphere back to the Earth’s surface, improving the atmosphere’s ability to trap heat.

          National Oceanic and Atmospheric Administration (NOAA)
          The greenhouse effect occurs when the atmosphere of a planet acts much like the glass in a greenhouse. Like
          the greenhouse glass, the atmosphere allows visible solar energy to pass through, but it also prevents some
          energy from radiating back out into space.

          Draft of IPCC Fourth Assessment Report
          The name “greenhouse effect” comes from the analogy with a greenhouse made of glass which allows sunlight
          to enter but restricts infrared energy from leaving, thus warming the interior. […] The natural greenhouse
          effect is neither harmful nor mysterious. Its basic principles are well-understood and are firmly based on
          fundamental physics.

          University Corporation for Atmospheric Research (UCAR)
          Have you ever been inside a greenhouse on a cold winter day? It might be cold outside, but inside the
          greenhouse lush green plants flourish in the warmth and sunshine. Greenhouses are made of glass and are
          designed to hold heat inside. The atmospheres of some planets are able to trap energy just like a greenhouse.
          Energy from the Sun can enter the atmosphere, but not all of it can easily find its way out again.
          What blocks the Sun’s energy from escaping a planet’s atmosphere? Unlike a greenhouse, planets do not
          have a layer of glass over them! Instead, molecules in the atmosphere called greenhouse gases absorb the
          heat.

          University of Michigan
          The “greenhouse effect” is the heating of the Earth due to the presence of greenhouse gases. It is named this
          way because of a similar effect produced by the glass panes of a greenhouse. Shorter-wavelength solar
          radiation from the sun passes through Earth’s atmosphere, then is absorbed by the surface of the Earth,
          causing it to warm. Part of the absorbed energy is then reradiated back to the atmosphere as long wave
          infared radiation. Little of this long wave radiation escapes back into space; the radiation cannot pass
          through the greenhouse gases in the atmosphere. The greenhouse gases selectively transmit the infared
          waves, trapping some and allowing some to pass through into space. The greenhouse gases absorb these
          waves and reemits the waves downward, causing the lower atmosphere to warm.

          Cary Academy, North Carolina
          The term “greenhouse effect” describes how the gasses in Earth’s atmosphere retain the radiant energy from
          the sun instead of letting it fly back out into space. The glass of the greenhouse works the same way.
          In short, the gasses in the atmosphere let energy in and don’t let all the energy back out.

          The University of Winnipeg
          As glass in a greenhouse traps heat inside, gases in the upper atmosphere trap some of the heat escaping the
          Earth, creating a greenhouse effect.

          The University of the Western Cape, South Africa
          A greenhouse is made entirely of glass. When sunlight (shortwave radiation) strikes the glass, most of it
          passes through and warms up the plants, soil and air inside the greenhouse. As these objects warm up they
          give off heat, but these heat waves have a much longer wavelength than the incoming rays from the sun. This
          longwave radiation cannot easily pass through glass, it is re-radiated into the greenhouse, causing everything
          in it to heat up.

          University of Cincinnati College of Engineering, Ohio
          Greenhouses are much warmer inside than the air is outside because the glass is transparent to light and
          allows short-wavelength light to pass through and heat the contents of the greenhouse. It also reflects back
          the longer wavelength heat radiating within the greenhouse, thus preventing if from passing back out.
          In a glass greenhouse, heat builds up and gets trapped due to presence of carbon dioxide and other heat
          trapping gases in the upper atmosphere.
          CO2 is analogous to glass.

          Miami-Dade Environmental Resources Department, Florida
          The phenomenon gets its name from the similarity to a garden greenhouse. Visible light passes through the
          glass ceiling and walls of a greenhouse. Some of the light is absorbed, some is reflected back, and a portion of
          it is radiated out as heat. Glass, like carbon dioxide (CO2) and other “greenhouse gases” doesn’t allow heat to
          escape back out.

          Science Encyclopedia
          The greenhouse effect is the retention by the Earth’s atmosphere in the form of heat some of the energy that
          arrives from the Sun as light. Certain gases, including carbon dioxide (CO2) and methane (CH4), are
          transparent to most of the wavelengths of light arriving from the Sun but are relatively opaque to infrared or
          heat radiation; thus, energy passes through the Earth’s atmosphere on arrival, is converted to heat by
          absorption at the surface and in the atmosphere, and is not easily re-radiated into space. The same process is
          used to heat a solar greenhouse, only with glass, rather than gas, as the heat-trapping material.

          California Environmental Protection Agency
          Simply put, the greenhouse effect compares the earth and the atmosphere surrounding it to a greenhouse
          with glass panes. Plants in a greenhouse thrive because the glass panes keep the air inside at a fairly even
          temperature day and night, and throughout the four seasons of the year. Just as the glass lets heat from
          sunlight in and reduces the heat escaping, greenhouse gasesand some particles in the atmosphere keep the
          Earth at a relatively even temperature.

          The Atmosphere, Climate & Environment (ACE) Information Programme (UK)
          The greenhouse gases in the atmosphere act in a similar way to panes of glass in a greenhouse (see Figure 2
          below). Radiation from the Sun (consisting mainly of visible and ultraviolet (UV) radiation) can travel through
          glass into the greenhouse. When this radiation is absorbed by objects in the greenhouse, it is re-radiated as
          infrared (IR) radiation, or heat. This heat cannot escape through the glass, so the greenhouse warms.
          up.

          The U.S. government’s Environmental Protection Agency
          The energy that is absorbed is converted in part to heat energy that is re-radiated back into the
          atmosphere. Heat energy waves are not visible, and are generally in the infrared (long-wavelength)
          portion of the spectrum compared to visible light. Physical laws show that atmospheric constituents—
          notably water vapor and carbon dioxide gas—that are transparent to visible light are not transparent
          to heat waves. Hence, re-radiated energy in the infrared portion of the spectrum is trapped within the
          atmosphere, keeping the surface temperature warm. This phenomenon is called the “greenhouse
          effect” because it is exactly the same principle that heats a greenhouse

          (These citations were culled from http://www.ilovemycarbondioxide.com/pdf/Greenhouse_Effect_Poppycock.pdf , http://www.ilovemycarbondioxide.com/pdf/Greenhouse_Effect_Poppycock_updated.pdf — where more can be found along with links to sources.)

    • wayne says:

      Yes, thank you very much RF for that link. It adds so much credence to what I have working on and off for the last couple of years that it is not even funny. The light on Arrhenius’s flaws are priceless and now thoroughly referenced, just what I needed and I will give him credit in the references.

  148. RW says:

    The GHE is not really driven by so-called ‘back radiation’, i.e. downward LW emitted from the atmosphere to the surface, but instead is driven by radiative resistance to outer space cooling by radiation from the atmosphere into space. That is, the surface is ‘instantaneously’ largely opaque to its emitted IR, i.e. a significant amount of surface IR power is absorbed or captured by the atmosphere (‘blocked’ from passing through the atmosphere into space), but the system must be making the push towards radiative balance at the TOA by radiation emitted up in the atmosphere; but in order to do that it must also push back the other way at the same time, because absorbed IR is re-radiated by the atmosphere both up and down.

    Downward LW at the surface has little to do with the underlying mechanism driving the GHE, and more involves how the surface energy balance, i.e. the net of about 390 W/m^2 gained, is physically manifested at the surface/atmosphere boundary.

    • RW says:

      The key is when the atmosphere absorbs upwelling surface IR, the atmosphere re-radiates this absorbed energy both up and down. As stated before, the system must above all be making the push toward radiative balance at the TOA via absorbed IR re-radiated up in the atmosphere, but in order to make that push, it has to push back the other way, because absorbed IR is re-radiated both up and down. The push back the other way towards the surface ultimately requires the surface and lower atmosphere to be emitting at a significantly higher rates in order for the surface and the whole of the atmosphere to be pushing through the required 240 W/m^2 needed to be passing out the TOA, thus elevating the temperature of the lower layers and ultimately the surface above what they would otherwise be in the process.”

      I would say this is really the underlying physics of the GHE, and not the manifesting downward LW passed to the surface. Also, while all of the LW incident on the surface acts to warm the surface (assuming none is reflected), not all of it is actually added to the surface or not all of it is acting to increase the surface temperature, as much of the absorbed LW flux at the surface is replacing non-radiant flux leaving the surface but not entering the surface as non-radiant flux, making it a net zero flux across the surface/atmosphere boundary.

      Another part of the problem is the use of the term ‘back radation’, as it implies surface emitted IR absorbed by the atmosphere and re-radiated back to the surface. In reality, there is really no such thing or at least no way to discern or quantify that. It’s just downward LW emitted from the atmosphere to the surface, where there are multiple energy inputs to the atmosphere besides surface emitted IR absorbed by the atmosphere. Post albedo solar power absorbed by the atmosphere and emitted down the surface as IR would not be ‘back radiation’ as the term implies, but instead ‘forward radiation’ from the Sun yet to reach the surface (a key distinction). Then of course there is latent heat of H20 flux from the surface, the water and energy of which eventually forms clouds, which radiate substantial IR from the atmosphere to the surface. Of course, the system is far too complex and chaotic to do any accurate accounting of how much downward LW at the surface is sourced from each. All the flux inputs, radiant and non-radiant, to the atmosphere get mixed together and are no longer traceable in any way.

      • Tim Folkerts says:

        RW, I like your first paragraph. It seems a good explanation of the basic physics.

        Then I wonder why you change your tone. You had just said “re-radiates this absorbed energy both up and down</b" and "it [IR radiation] has to push back the other way [down]”. But in the 2nd paragraph, you now seem to distance yourself from the downward part — the the GHE is “not the manifesting downward LW”.

        The two are inseparable. Why treat them that way in the first paragraph, but not in the 2nd?

        • RW says:

          Because of the amount of IR ultimately passed from the atmosphere to the surface is itself is not the underlying driving mechanism of the GHE, even though it greatly influences the surface energy balance. The vast majority of upwelling IR absorbed and re-radiated back downward does not pass to the surface, but instead is absorbed by layers below and subsequently re-radiated again, both up and down, and the cycle repeats until the deficit IR flux needed to be passed out the TOA for balance with the Sun is achieved.

          Roy puts far too much emphasis on downwelling IR emitted from the atmosphere directly to the surface, and not what needs to be happening at the TOA.

      • Kristian says:

        RW says, June 19, 2015 at 10:43 AM:

        “I would say this is really the underlying physics of the GHE, and not the manifesting downward LW passed to the surface.”

        No, you get that exactly wrong, RW. The hypothetical rGHE (that is, the actual ‘extra’ surface warming) is ultimately ONLY about the increase in the downward LW radiation from the atmosphere to the surface.

        Even the concept of the ‘raised effective emission height’, purporting to be the more sophisticated and sort of advanced-level version of the rGHE hypothesis, really boils down to a matter of “extra surface heating by back radiation”. It’s the same story, only told in a bit more convoluted way.

        The thing is, you cannot get the surface to warm unless you feed it with more energy directly. According to the rGHE hypothesis. And that’s what they do.

        It’s after all impossible to warm the air layer (or any specified air layer) just above the surface on a permanent basis if the surface itself didn’t warm first. Natural buoyancy sees to that. Try warming air that is free to float up. It doesn’t matter what kind of method you use; a torch or a radiative imbalance at the ToA. It won’t work. It will spontaneously rise to restore hydrostatic equilibrium.

        That is why you will HAVE TO start the warming to ‘push back’ on your perceived radiative ToA imbalance AT THE SURFACE. The surface only warms the rGHE way when the “back radiation” increases.*

        BTW, since 2000, atmospheric DWLWIR to the global surface of Earth hasn’t increased at all as expected (after all, both tropospheric column CO2, WV and clouds have gone up). In fact it has decreased instead. This is partly the reason why the global surface, during this same period, has greatly strengthened its ability to cool by radiation. This straight from CERES …

        *You should read Guy Callendar’s original paper on this whole issue. The rGH EFFECT comes as a direct result of the lowering of the ‘average depth of atmospheric downward radiation’ or “sky radiation” as he called it, and hence from an increase in downward LW to the surface that the surface can only compensate for by warming:
        http://onlinelibrary.wiley.com/doi/10.1002/qj.49706427503/pdf

    • RW says:

      I think the basic problem here in a nutshell is Roy is mixing the how and why of the manifesting atmospheric physics with the underlying mechanism driving the GHE, which is just radiative resistance to outer space cooling by radiation from the atmosphere into space. The underlying driving mechanism itself is really independent of the manifesting physics that ultimately determine the surface energy balance. Though of course, those manifesting physics are initially driven by the underlying mechanism.

      The how and why regarding the amount of IR being passed from the atmosphere to the surface is the result of all the physics mixed together, and is not really the underlying driving mechanism. I really don’t think Roy has this quite right and/or is not characterizing it correctly. Hence why so many people are perpetually confused and/or can’t understand or grasp it.

      • gbaikie says:

        It seems it may be true that doubling of CO2 would cause
        an increase of 1 C to global average temperature. Though the the recent increase in CO2 levels which may have increased global temperatures has not been measured.
        Why the added warmth from increased global CO2 has mot been measured is probably due the lack of understanding warming and cooling mechanism of natural variation and the lack of precise enough measurement that could measure it.

        CO2 is not the control knob of global temperatures, and it appears that rising or falling global CO2 level is more of result of warming and cooling than something which causes warming and cooling.
        Or rising levels of CO2 will not cause much warming and high level of CO2 should not seen as way to prevent global cooling.

        Or humans at the moment have no known way to predictably affect global temperature- we can’t prevent an Little Ice Age from occurring, nor create a warming period such as Medieval warm period [or our recent warming period which began around 1850].

        It appears to me that CO2 levels do not affect global evaporation rates, and therefore CO2 is not agent that forces rise or lowering of global temperature. Likewise it doesn’t appear to me that the major greenhouse gas, water vapor is forcing agent, which causes increase or decrease
        in levels of other greenhouse gases [or water vapor itself].

  149. mpainter says:

    Norman,
    It has been a fruitful thread for me, I have learned some things and it has helped clarify some issues. Thank you for your participation, likewise all others.
    The notion that LWIR absorbed by the ocean which does not cause evaporation or any other discernable effect, but only is part of a hypothetical radiative exchange, is for me another untestable hypothesis from the AGW inventory of such.
    I pass, thank you.

    • RW says:

      Evaporation is mostly temperature dependent, not incident flux dependent. In general, energy flux out is temperature dependent and energy flux is not.

    • RW says:

      You would benefit from understanding that the balance at the surface is the sum of the fluxes in and out where the additive superposition principle applies to the effects of energy (and power) to the temperature of a system:

      https://en.wikipedia.org/wiki/Superposition_principle

      If the surface with an emissivity of about 1 is at a temperature where it radiates about 390 W/m^2 as a consequence of its temperature (dictated by the S-B law), the universal requirement is all power in excess of 390 W/m^2 entering the surface must be exactly offset by power in excess of 390 W/m^2 leaving the surface. In short, all power in excess of 390 W/m^2 must be net zero across the surface/atmosphere boundary, and only a net of 390 W/m^2 is actually being added to the surface to replace the 390 W/m^2 radiated away as a consequence of its temperature (about 288k).

  150. mpainter says:

    MikeB
    Did you get the impression that I deny the GHE? I assure that I do not.

  151. mpainter says:

    Curt,
    I think that the whole KT diagram is egregious. Kristian has explained convincingly why the “measurements” are not to be relied upon, so I simply do not agree that the numbers on the diagram have any validity, insofar as the LW radiant flux is concerned. Now, please don’t go around saying that I am a “slayer” or some such type.
    I simply believe that the GHE has been mischaracterized, as per the KT diagram.

    • Curt says:

      mpainter:

      Kristian’s conclusions are just bizarre. He goes off the rails in the middle of his analysis. Let’s look at it in a simple example.

      We have a surface at about 289K, and a somewhat colder atmosphere above it that can absorb and emit in the infrared.

      We will use Kristian’s cryogenically cooled sensor, held at or below the boiling temperature of liquid nitrogen of 77K. We accept that what this sensor really detects is the net heat transfer to the sensor from what it is pointing to. (We even accept, for the sake of argument, that it can only do this for objects hotter than it…)

      Now, a body at 77K can thermally emit at most 10 W/m2 (what a perfect blackbody would emit). I should also point out that more than 98% of this would be at wavelengths longer than 20 um, useful if we are doing wavelength-specific measurements.

      So we point this sensor downward at the surface (which is warmer than the sensor, of course) and measure a net heat transfer corresponding to 390 W/m2 from the surface to the sensor. Even if we assume this net transfer is a gross upward radiative power transfer from the surface, we will not be more than 10 W/m2 off.

      Next we point this sensor upward toward the sky (which also is warmer than the sensor) and measure a net heat transfer corresponding to 327 W/m2 from the atmosphere above to the sensor. Again, even if we assume this net transfer is a gross downward radiative power transfer, we will not be more than 10 W/m2 off.

      (By the way, these would be typical measurements.)

      So we can conclude from these measurements that the ground is emitting upward a thermal radiation power density of 390 – 400 W/m2, and that the atmosphere is emitting downward a thermal radiation power density of 327 – 337 W/m2.

      Now, the only reasonable inference from these measurements is that the surface and the atmosphere would still be emitting this same radiation even if we remove the supercooled sensor.

      So now the upward radiation from the surface, no longer blocked by the sensor, would travel higher in the atmosphere, where most of it would be absorbed, and the downward radiation would reach the surface, where virtually all of it would be absorbed.

      If you use Kristian’s logic, the surface and the atmsophere would have to change how they radiate based on whether or not there is a supercooled sensor in the path, which has ridiculous causality problems.

      Kristian seems to think that heat is an actual physical substance, rather than just a way we account for net energy transfers in exchanges. This old “caloric” theory of heat was discredited well over 150 years ago.

      • Kristian says:

        Curt says, June 19, 2015 at 7:57 PM:

        “Now, the only reasonable inference from these measurements is that the surface and the atmosphere would still be emitting this same radiation even if we remove the supercooled sensor.”

        Curt, Curt, Curt. No, the only reasonable inference from these measurements is that what you have measured on two separate occasions is a radiant heat flux from a relatively warm place to a much colder introduced object. Heat transfers happen like that. They spontaneously move energy as soon as you bring two objects or regions of different temperature into thermal contact. If no such temperature difference, then the energy is rather kept inside the objects as internal energy, maintaining their separate temperatures. However, they always have a potential to transfer some of their internal energy out and away, if they’re only faced with a colder object or region.

        Have you ever had a look at the surface energy budgets of Venus and Mars, Curt?

        If you did, and if you applied your UWLWIR-DWLWIR approach to them, what would you find?

        • Curt says:

          Kristian:

          Why do you insist on treating “heat” as a real physical substance, rather than a convenient accounting mechanism, a metaphor really? The caloric theory of heat was thoroughly discredited 150 years ago!

          On the other hand, we understand energy-carrying electromagnetic radiation as a real physical entity, with the energy understood down to the photon level. (And there is no “heaton”.)

          We also know that objects emit EMR as a function of their own state and properties, and NOT as a function of what they are radiating to, as you effectively believe.

          Here’s how one of my introductory texts in engineering thermodynamics, written in the 1960s, introduced the subject of radiative transfer:

          “All surfaces radiate energy, but there will be a net energy exchange only if the bodies are at different temperatures. Two facing parallel plates will radiate to each other, but if their temperatures are equal there is no net energy transfer.”

          Here’s how Clausius explained it in the 19th Century:

          “The principle may be more briefly expressed thus: Heat cannot by itself pass from a colder to a warmer body; the words “by itself”, however, here requires explanation. Their meaning will, it is true, be rendered sufficiently clear by the exposition contained in the present memoir, nevertheless it appears desirable to add a few word here in order to leave no doubt as to the signification and comprehensiveness of the principle.

          In the first place, the principle implies that in the immediate interchange of heat between two bodies by conduction and radiation, the warmer body never receives more heat from the colder one than it imparts to it. The principle holds, however, not only for process of this kind, but for all others by which a transmission of heat can be brought about between two bodies of different temperature, amongst which process must be particularly noticed those wherein the interchange of heat is produced by means of one or more bodies which, on changing their condition, either receive heat from a body, or impart heat to other bodies.”

          He believed that radiative heat transfer was a two-way exchange of energy way back then!

        • Tim Folkerts says:

          Kristian. A simple thought experiment. I create a 1m x 1m surface on the earth that is at the temperature of the sun (~ 5800 K). According to you (if I follow your arguments), the sun somehow ‘knows’ that is should not radiate to this 1 m^2 because is at the same temperature as the sun. But the sun happily radiates to the REST of the surface, since the rest of the surface is much cooler.

          But now suppose I move the hot patch 100 m away. What happens now — remembering that the sun cannot know this movement has occurred for over 8 minutes. Does the sun continue avoiding radiating to the original location for 8 minutes and “keep the energy inside”? Or perhaps the photons/heat/EM energy that HAD been heading 100 M away shift and suddenly go to the now-vacated spot? Maybe the sun anticipates the move and 8 minutes before I move the patch the sun rearranges what energy to ‘keep inside’ and what energy to send so as to work for the future configuration.

          You have a serious relativity problem in your worldview. You speak of “simultaneous” changes, but that isn’t consistent with relativity. The radiation would need ‘prior knowledge’ or ‘instantaneous information exchange’ in order for your model to work. The problem is more extreme here than in Curt’s example, but the principle is the same.

          • gbaikie says:

            — Tim Folkerts says:
            June 20, 2015 at 3:03 PM
            Kristian. A simple thought experiment. I create a 1m x 1m surface on the earth that is at the temperature of the sun (~ 5800 K). According to you (if I follow your arguments), the sun somehow ‘knows’ that is should not radiate to this 1 m^2 because is at the same temperature as the sun. But the sun happily radiates to the REST of the surface, since the rest of the surface is much cooler.

            But now suppose I move the hot patch 100 m away. What happens now — remembering that the sun cannot know this movement has occurred for over 8 minutes. Does the sun continue avoiding radiating to the original location for 8 minutes and “keep the energy inside”?–

            The photon knows cause it’s traveling at speed of light- in photon time is doesn’t have an 8 min delay- in fact photon doesn’t have a delay if travels across the universe due to time dilation.

            But we should not assume a photon or the sun knows anything,
            instead energy is transferred or heats something which is cooler than it is.
            Or the heat of the sun does not heat itself, instead it’s energy comes from the fusion of hydrogen and other light elements. Or:
            “In the modern microscopic interpretation of entropy in statistical mechanics, entropy is the amount of additional information needed to specify the exact physical state of a system, given its thermodynamic specification. Understanding the role of thermodynamic entropy in various processes requires an understanding of how and why that information changes as the system evolves from its initial to its final condition. It is often said that entropy is an expression of the disorder, or randomness of a system, or of our lack of information about it. The second law is now often seen as an expression of the fundamental postulate of statistical mechanics through the modern definition of entropy.”
            https://en.wikipedia.org/wiki/Entropy
            So in this way of seeing it, the sunlight could not add any information to the 1 meter square area which was ~ 5800 K.
            Nor can sunlight add any information to a 1 meter square which was about 120 C at earth distance. Or at Earth distance the sun temperature is about 120 C.

          • Kristian says:

            Tim Folkerts says, June 20, 2015 at 3:03 PM:

            “Kristian. A simple thought experiment.”

            Tim. Have you ever contemplated the surface energy budgets of Venus and Mars?

            If you did, and if you applied your UWLWIR-DWLWIR approach to them, what would you find? Consistency?

            Venus:

            The global surface of Venus only absorbs ~17 W/m2 of radiant heat from the Sun. Let’s say about 7 of these are lost again via conduction. That leaves 10 W/m2 of radiant heat (net LW) from the surface up (could be more, could be less, but at least somewhere between 0 and 17 W/m2).

            Now, Venus’s global surface is at an average of 737K, so should, according to the Stefan-Boltzmann equation radiate a steady 16,730 W/m2 (assuming emissivity=1). This is your UWLWIR ‘flux’. So how large must the DWLWIR ‘flux’ be to make the net LW a mere 10 W/m2? Simple answer: 16,720 W/m2. So the slightly cooler air above the surface manages to return 99.94% of the upward radiation from the ground. As if it were a perfect blackbody surface at 736.9K, facing the surface across a vacuum say 1mm wide, emitting 100% according to its temperature.

            That is quite astounding, isn’t it? Considering this ‘air’ is made up of 96.5% CO2 and 3.5% N2. That’s some serious ‘pressure broadening’ going on there, wouldn’t you say?

            Yes, I know, I know, the CO2 at the surface of Venus and about 4 km up is above its critical point, so can be considered a gaslike supercritical fluid rather than a proper gas.

            So let’s assume this layer of supercritical fluid, then, acts like a 100% IR opaque (but SW transparent) blackbody that can only conduct and convect the surface heat through. If it’s a perfect absorber of IR from the surface, and a perfect (re)emitter of 99.94% back down (from a temperature 0.1K lower), then at the top of this supercritical layer, ~4 km above the ground (at a bit less than 74 bar), it would emit an UWLWIR flux of say 14,200 W/m2. To the now proper CO2 gas above.

            The question remains: How many % of the 4km supercritical “surface” IR emission would this CO2 atmosphere let through and how much would it absorb and radiate back? 99.9% 90% 75% 50% 33% And remember, from this level there are still tens of km (and of lapse rate) through ‘free air’ up to the haze and cloud levels.

            Mars:

            A funny circumstance persists on the Red Planet. Its mean global surface temperature (according to the ‘Thermal Emission Spectrometer’ (TES) instrument aboard the ‘Mars Global Surveyor’ (MGS), and so now generally accepted by NASA, based on multiple-year measurements with no real trend observed, albeit slight interannual differences, its central estimate corroborated by the overlapping and succeeding investigations of the ‘Mars Climate Sounder’ (MCS) spectrometer aboard the ‘Mars Reconnaissance Orbiter’ (MRO), plus in situ surface measurements from various landers like Viking 1 and 2 and Pathfinder) is 210-211K.

            Which happens to match EXACTLY its apparent planetary emitting temperature in space: (587 * [1 – 0.235])/4 = 112.3 W/m2 => 210-211K. (Global albedo of Mars: TES gives 0.232, MCS 0.239.)

            So where Venus has a T_sfc of 737K and a T_erl of 232K, so a 505K rGHE as defined, and Earth has a T_sfc of 288K and a T_erl of 255K, and thus a 33K rGHE as defined, Mars has a T_sfc of 210K and a T_erl of 210K, and accordingly, a zero Kelvin rGHE.

            But that’s not the funniest part.

            We all know about the Moon, how it absorbs ~298 W/m2 from the Sun on average and how it readily emits it all back to space, and how it manages to do this from a mean global temp of 197K rather than the ‘required’ or ‘expected’ 269K. How does it manage? Through its huge temperature swings of course. The larger its surface temperature amplitudes, the colder a body can be on average and still maintain a radiative equilibrium between input and output. Courtesy of the ^4 exponential relationship between radiative output and temperature. Straying from the isothermal condition allows the surface of the body to put out more radiation than what its physical temperature average would suggest. The larger the deviations, the more it puts out relative to its mean temp. The Moon, for instance, puts out 3.5 times as much LW to space from its global surface over a year as what its actual average global temperature of 197K would imply.

            This physical phenomenon occurs on Venus and Earth as well, although hardly at all on Venus, and to a fairly moderate extent on Earth (the quoted 390-400 W/m2 is probably in reality ~430-440 W/m2).

            However, on Mars, the effect would be considerable. Not at all as large as on the Moon, but still much larger than on Earth. The normal temperature range is about 133-293K, meaning, even as the global average temp of the Martian surface is really 210-211K, implying an emitted flux of ~112 W/m2 (if isothermal), its actual annual global output would be much bigger, most likely upwards of 200 W/m2.

            So where lies the problem?

            The problem is that the Martian planetary system as a whole, in space, only absorbs and therefore also only emits back a total radiant flux of 112.3 W/m2.

            200+ W/m2 out from the sfc (real UWLWIR), 112 W/m2 out through the ToA (OLR). And its worse than that. The atmosphere also absorbs some of the incoming from the Sun directly (maybe ~15 W/m2) and some energy is conducted>convected from the surface and into the atmosphere (maybe ~10 W/m2).

            How does all this add up?

            On Earth the atmospheric “back radiation” (your DWLWIR ‘flux’) would of course cover for it. On Mars (as on Venus) it’s not that straightforward:
            https://okulaer.files.wordpress.com/2015/06/two-mars-spectra.png

            80-90% of the surface radiation appears to go unimpeded through the wide ‘atmospheric IR window’ and straight to space. It might ultimately be a bit less due to atmospheric dust, but clearly still 75-80% (there’s not much to disrupt that nice blackbody Planck-curve outline, is there?). The 96% CO2 concentration can do nothing outside its specific spectral absorption bands.

            75-80% of 200 W/m2, how much is that? 150-160 W/m2. That’s already ~40% more than Mars’s entire absorbed flux from the Sun. And then we still haven’t included the atmospheric output.

            Looking at those spectras, it seems pretty clear the Martian atmosphere would be utterly incapable of emitting a DWLWIR ‘flux’ to the surface to counter the 200+ W/m2 of outgoing (UWLWIR).

            Something’s gotta give …

          • jerry l krause says:

            Hi Kristian,

            Yup, its me again.

            It is amazing how you and nearly everybody else can overlook (ignore) the fact that the Venus atmosphere has an approximately 10km thick overcast (the well-known sulfuric acid droplets) above about 50km altitude at which its atmospheric pressure is earth-like.

            Have a good day, Jerry

          • jerry l krause says:

            Hi Kristian,

            I forgot to mention the observed fact that the temperature of the top of the Venus cloud deck is earth-like and to review that it is an generally accepted observation that earth-based cloud tops emit as if they are near-perfect blackbodies according to their temperatures.

            Have a good day, Jerry

          • Kristian says:

            jerry l krause says, June 21, 2015 at 7:54 AM:

            “It is amazing how you and nearly everybody else can overlook (ignore) the fact that the Venus atmosphere has an approximately 10km thick overcast (the well-known sulfuric acid droplets) above about 50km altitude at which its atmospheric pressure is earth-like.”

            Jerry, I am not ignoring it:
            https://okulaer.wordpress.com/2015/06/20/why-atmospheric-mass-not-radiation-p2/

          • Tim Folkerts says:

            Kristian, I agree that the theory must able to deal with the atmosphere of Venus. I just don’t agree it is such an insurmountable challenge as you seem to think it is.

            Considering this ‘air’ is made up of 96.5% CO2 and 3.5% N2.
            Plus …
            Sulfur dioxide 150 ppm
            Argon 70 ppm
            Water vapour 20 ppm
            Carbon monoxide 17 ppm
            Helium 12 ppm
            Neon 7 ppm
            Hydrogen chloride 0.1–0.6 ppm

            So we have a whole whole soup of GHGs.

            That’s some serious ‘pressure broadening’ going on there, wouldn’t you say?
            Plus doppler broadening (due to temperature). Plus the atmosphere is ~ 40 times thicker, meaning all the values should be multiplied by 40 to be comparable to earth’s atmosphere. And CO2 is ~ 100,000 times thicker than on earth, meaning even the ‘atmospheric windows” are not really clear windows on Venus.

            So we have a thick soup of many GHGs with considerable broadening of the lines, blocking IR over pretty much every wavelength within short distances.

            “As if it were a perfect blackbody surface at 736.9K, facing the surface across a vacuum say 1mm wide, emitting 100% according to its temperature.”
            No. 0.1 K at a lapse rate of ~ 10 K/km is 10 m, not 0.001 m. Given the common understanding that GHGs absorb completely in their absorption bands over lengths on the order of 10 meter, this suddenly sounds PERFECTLY reasonable.

            For more details, you might want to read http://journals.ametsoc.org/doi/pdf/10.1175/2011JAS3703.1

            PS. I notice you completely sidestepped the issue I brought up — that your “potential” model has serious shortcomings visa vis relativity and time delays as objects move.

          • Curt says:

            Tim:

            You say that ‘[Kristian’s] “potential” model has serious shortcomings visa vis relativity and time delays as objects move.’

            His “potential” model is also completely unable to deal with non-thermally generated electromagnetic radiation (e.g. fluorescence, LEDs, lasers) and/or non-thermal absorption of EMR (e.g. photosynthesis, photodiodes, photovoltaics).

          • Kristian says:

            Tim,

            You might be right. That the super-dense (66.6 kg/m^3) CO2 ‘air’ directly on top of the 737K Venusian surface somehow manages to emit a DWLWIR ‘flux’ equal to 99.94% of the UWLWIR from the surface, or as from a perfect blackbody (100%) at 736.9K.

            Or you might be wrong. To me it seems like you’re mostly just trying to casually pass private conjectures and speculations off as facts in the hope that this whole issue might just go away.

            Check out spectralcalc.com or any similar site and have a look for yourself.

            https://okulaer.files.wordpress.com/2015/06/venus-sfc-air-emission.png
            (In-band (2-30 microns) radiant emittance: 12,198 W/m^2)

            The point is this:
            Even if the atmosphere managed to return 90% of the UWLWIR, you would be 1663 W/m^2 short. If it returned 95%, still 827 W/m^2 missing. What about 99%? 157 W/m^2 to go. You see where I’m going with this? It needs to return it all. Across the full IR spectrum. No dents or divots of any kind.

            And if the atmospheric layer emitting its perfect blackbody flux down to the ground happens to be say 2K or 1K or 0.5K cooler than it, then even this wouldn’t be enough. Even with a 100% spectral output, it couldn’t emit 16,720 W/m^2. Because it would simply be too cool.

            You know of course, Tim, that on Earth, you wouldn’t normally find the air at ten metres above the ground to be only 0.1K cooler. The tropospheric temperature gradient doesn’t progress in such perfectly linear fashion already from the sfc/air interface. It’s not like the air 1 m up is 0.01K cooler than the surface and the air 10 cm above 0.001K cooler. There’s a drop (in the mean state) of 1K at least from the actual surface to the air just above. Being much less turbulent at the Venusian sfc/air interface, the difference might be much smaller there, but a mere 0.1K decline in temperature from the sfc to the ‘air’ 10 m up is still rather unlikely, I would say. I don’t think we have any in situ measurements determining such an average difference, though, to settle the matter. What we do have are parameterised radiative transfer models and a bunch of assumptions to go along with them. One such assumption is that there’s a continuous two-way transfer of radiant energy up and down between sfc and air, providing a net value between them: UWLWIR minus DWLWIR equals the radiant heat. And assuming this, you must also assume that the atmosphere above the Venusian surface sends back 99.94% of the upward surface flux.

            BTW, Tim, there’s also Mars …

          • Kristian says:

            Tim Folkerts says, June 20, 2015 at 3:03 PM:

            “According to you (if I follow your arguments), the sun somehow ‘knows’ that is should not radiate to this 1 m^2 because is at the same temperature as the sun. But the sun happily radiates to the REST of the surface, since the rest of the surface is much cooler.”

            No, then you are not following my argument, Tim.

            It is simply about a different way of viewing basically the same phenomenon, how to attain and maintain a system’s (dynamic) thermal equilibrium.

            The imagined “amplified internal radiative transfer loop” of the rGHE surface/atmosphere system is actually just one mental model of a real-world situation, one way of viewing how a steady state (and its equilibrated temperature) is achieved and made to last. A radiative view.

            This is how the build-up phase towards a (dynamic) thermal equilibrium is pictured in the rGHE hypothesis: A continuous cycle of steadily intensifying radiative transfers of energy going on inside the atmosphere, between its different parts and between it and the surface, looping up and down and back and forth, warming in all directions, until the steady state is reached.

            Another way of picturing such a process is this: A simple and gradual storing up of internal energy (microscopic/molecular PE and KE). That’s how thermodynamic systems are normally seen warming. A certain transfer of energy to the system adds energy to it, the energy accumulates inside, thereby raising its temperature. It accumulates because the system is not yet warm enough to release as much energy as comes in. A this is simply because it does not yet contain enough internal energy. The more energy is stored up, however, the warmer the system itself gets, and the more energy it manages to shed. And the more energy the system itself manages to shed, the slower its warming rate, until it has reached a temperature that allows it to get rid of as much energy per unit time as it absorbs from its energy source. At this point, the system is filled up of internal energy providing it with a temperature just high enough for the balance between input and output to be attained.

            There’s a distinct ‘philosophical’ difference between energy being statically held inside a warming thermodynamic system, an internal energy storage manifesting itself in – and causing – the system temperature, and energy being constantly dynamically transferred in and out of the system as a result of this temperature (and the surrounding temperature).

            Or is there?

            What the rGHE hypothesis in effect seems to be saying is that, in the new steady state, the solar flux is finally once again given a free pass all through the system: Conceptually, 240 W/m^2 come in to the surface, warming it; the 240 W/m^2 then move out from the surface, cooling it back, and is absorbed by the atmosphere, warming it; the 240 W/m^2 then finally move out from the atmosphere, cooling it back, and ends up in space, where it originally came from. The overall temperature at this stage will thus not change. It warms as the Sun shines, and cools back down when the Sun’s out of sight.

            The point is that, in addition to this lossless ride of the solar heat through the Earth system, one now imagines a continuously (and constant) cycling loop of radiative energy transfer up and down between the surface and the atmosphere, absolutely necessary to maintain the steady-state temperatures enabling this equilibrated situation to last. This transfer loop, in the steady state, is really a zero-sum game*, but that doesn’t mean it can be removed in any way. If it’s removed, then the entire Earth system will freeze cold.

            *480 W/m^2 up from the sfc, of which 240 go straight to space (the ‘solar flux’ part) and 240 to the atm, after which the atm returns those 240 back to the sfc, adding it to its 240 W/m^2 in from the Sun and enabling it to emit 480 W/m^2 up to space and the atmosphere in combination. In this way 240 W/m^2 always moves up and down between the sfc and the atm, while at the same time an equal flux always escapes to space:
            https://okulaer.files.wordpress.com/2015/06/rghe-loop.png

            The thing is, in the opposing ‘world view’, the orange internal atmospheric radiation loop in the diagram linked to just above doesn’t really exist. It’s imaginary. A mental construct. What exists is rather the statically stored ‘internal energy’ held within the thermal mass of the atmosphere system, giving it its steady-state temperature (and temperature distribution). There is no continuous thermodynamic zero-sum transfer game of radiative energy fluxing back and forth between the surface and atmosphere, heating all the way, keeping the temperatures up. The temperatures are already there. In the energy storage, the total molecular KE of the atmospheric bulk. As long as the heat (from the Sun) keeps coming it, the storage is fine, because it sheds to space only what it needs to to balance the incoming.

            What the ‘radiationers’ perceive as somehow the cause of the steady-state temperatures (that necessary, eternally cycling radiative transfer loop), is really just an apparent effect of those steady-state temperatures (caused simply by the storing up of ‘internal energy’). ‘Thermal radiation’ is called thermal radiation because it is the result of warmth, of temperature, not because it causes that warmth or temperature.

            And, yes, all objects radiate. In all directions. But they don’t transfer energy in all directions. They have the potential to, but this potential is only realised towards objects or surroundings cooler than themselves.

            It’s in the transition zone between the microscopic (the individual photons, the (hypothetical) quantum world) and the macroscopic (the probabilistic averages of photons, the (observed) world of thermodynamics) that interesting ideas quite often emerge. There is (as of yet) no point in trying to tie the two realms together from totally opposite ends. That’s when you end up believing (perhaps even without knowing or wanting to) that a conceptual DWLWIR (or UWLWIR) ‘flux’ can actually itself have a thermodynamic effect.

  152. mpainter says:

    Correction, not the whole diagram, but those parts marked by Roy with a red X… chuckle chuckle

  153. mpainter says:

    RW
    Water absorbs LWIR within a few microns of the surface (air/water interface). This of course increases the kinetic energy of the molecules of that minute interval accordingly.

  154. mpainter says:

    Curt:
    ” ..you cannot partition inputs and outputs that way”
    ####
    Are you saying that the KT diagram does not partition inputs and outputs?
    It certainly appears to do just that.

    • RW says:

      Yes, I think he’s saying just that.

    • Curt says:

      mpainater:

      If that sentence had to stand alone, it would say:

      “You cannot partition the relationship of particular inputs to particular outputs that way.”

      That should have been clear from the rest of the post. K&T does not “partition the relationship of particular inputs to particular outputs” that way.

  155. mpainter says:

    Okay.
    Let me see if I understand this KT diagram. It can partition the radiative flux and label it and assign values. That’s the right way. But when I wish to discuss one of those partitions and ask certain questions and point out certain aspects, then that can’t be done. Its just wrong to do it _that_ way.
    I think that I’m beginning to understand.
    Now, lets see if I understand this back radiation correctly. It is 333 W/sq m, says right there on the diagram. It is absorbed by the surface, says right there on the diagram. Of course, this means that if the surface is water, like the ocean, maybe, then that back radiation is absorbed by water molecules, as per the laws of physics. And those water molecules acquire more kinetic energy, as per those same laws.
    Am I right so far? Good.
    Now I am beginning to understand where I goofed, because I consult the KT diagram and it shows that the kinetic energy thus acquired is given up as radiation and radiation _only_. Yes, I see now. I thought that the increase in kinetic energy would lead to evaporation, but no, not one molecule is evaporated, but all acquired kinetic energy is released as radiation.
    My goof, but I can explain my thinking so that maybe folks won’t think that I am too awfully stupid.
    See, I was thinking that since the back radiation is LWIR that it would be absorbed in the first few microns of the interface, according to the rafidiative absorbtive properties of water and that minute interval being exactly where the evaporation takes place, well, I just thought that the water would evaporate, you see. But now I see my mistake, that I partitioned in a way that I was not supposed to. Looking at the diagram, I thought that I could, because the diagram partitions this stuff, but obviously K&T know this partitioning business and I don’t. Well, fellas, gotta go.

    • Curt says:

      mpainter, you say: “I consult the KT diagram and it shows that the kinetic energy thus acquired is given up as radiation and radiation _only_.”

      IT DOES NO SUCH THING !!!

      (Yes, I’m shouting!)

      Only because of your complete confusion about the most basic principles of thermodynamics could you conclude that!

      If you took an actual thermodynamics course, say in engineering, you would understand in the first few weeks that you cannot match up particular inputs to particular outputs in this way.

  156. mpainter says:

    Curt
    PleSe excuse me if I decline to talk about “heat”, whose definition in the field of thermodynamics has undergone a transmutation in the last one hundred years or so and transmutation has led to interminable semantic arguments on the igosphere

    • Curt says:

      I am very happy to dispense with the use of the term “heat” in these discussions.

      (Should I say its use creates more heat than light…?)

  157. mpainter says:

    Drat, read blogosphere, please.

  158. mpainter says:

    Curt says
    IT DOES NO SUCH THING!!!
    ###

    Ah, have I been partitioning again? Humbly beg yer pardon.
    Let me de-partition here:
    161 W/sq m + 333 W/sq m = 494 W/sq m
    Using .6 as the ocean evaporation figure, I compute:

    .71 × .6 × 494 W/ sq m =
    210 W/sq m for latent energy.
    How’s that Curt? Any better?

    • Curt says:

      No it’s not!

      I’ve told you repeatedly that the 100% reference value for those studies you cite that say “60% of oceans losses are evaporative” is the solar input value alone.

      You haven’t even acknowledged that, let alone started to come to grips with it.

      You are simply showing that you don’t have the conceptual background to process the most basic concepts in thermodynamics. (I see now why so many science and engineering schools consider thermo a washout course.)

  159. Kristian says:

    Haha, I find this whole ‘discussion’ only cute.

    A massive atmosphere on top of a solar-heated planetary surface surely necessitates that surface to become warmer than if that atmosphere weren’t there. But it’s the atmosphere’s MASS that necessitates it, not its radiative properties. In fact, the atmosphere being massive and thus supplied with (‘non-radiant’) energy from the heated surface necessitates it to be IR-active also. You simply cannot have a massive atmosphere that isn’t IR-active.

    • FTOP says:

      You can lead a horse to water, but it will still drink the Kool-aid.

      Occam’s Razor demands that the entire IPCC story be abandoned.

      The physics of water invalidate Ocean warming by anything other than the sun
      The lapse rate explains the temperature profile
      The sun is the only heat source
      Thermodynamics does not allow heat to move from atmosphere to surface

      Yet with all this information, you have this AGW-CO2 story that just won’t die. Belief trumps science every time

      Instead we see a rigged temp anomaly process
      Mann’s trick
      Arguments that wind warms the ocean

      The insanity never ends

  160. mpainter says:

    Curt
    Its true that I have had no instruction in thermodynamics.
    If I took a course in that, would the instructor teach that 333 W/sq m of LWIR incident on the ocean would produce no evaporation?
    I fail to see the value of such instruction, if that is taught.
    Insolation, at 161 W/sq m, accounts for the latent energy of the diagram (using the .6 figure for calculation) and this accords with all of our understanding.
    Is it your position that the back radiation of the KT diagram causes no evaporation?
    So far, you have refused to specifically address this issue. Time to do so.
    So, does the BR cause evaporation it not?

    • Curt says:

      mpainter: You say, “Its true that I have had no instruction in thermodynamics.”

      Unfortunately, that’s painfully obvious, because you demonstrate again and again that you don’t have the conceptual framework to start addressing these issues.

      There’s a reason it takes multiple years of hard work to get even a first degree in subjects like this. It takes solving hundreds and hundreds of problems of escalating complexity, figuring out why you got some of them wrong, etc. etc. etc.

      You have not even started this process.

      You ask, “would the instructor teach that 333 W/sq m of LWIR incident on the ocean would produce no evaporation?”

      No, and I have never claimed that either. But if you asked your instructor whether this 333 W/m2 would go exclusively into latent heat loss or radiative loss, he would have told you (as I have, repeatedly) that you cannot match particular inputs to particular outputs that way.

      And because you don’t have the necessary background, you don’t realize that I’ve answered your questions multiple times already, most recently here:

      http://www.drroyspencer.com/2015/06/what-causes-the-greenhouse-effect/#comment-193510

      So to repeat and simplify, yes there would be more evaporation with DWLWIR than without, because the surface would be hotter than without. But there would also be more radiative losses and conductive/convective losses.

  161. Norman says:

    gbaikie says:

    “So Lunar day is about 27 earth days [655 earth hours]
    so 655 divided by 360 degrees is 1.8 earth hours per degree, times 15 is 27.3 earth hours per 1 earth hour. And 6 lunar hours equals about 164 earth hours.”

    I just saw your correction to my error about lunar time. Thank you spotting the mistake and correcting it!

    Even so the PSI lunar cooling rate is still a little phony! Since they can easily see lunar cooling is not a linear phenomena yet they treat it as such in their calculation of lunar cooling. This seems as if it is intentional misleading to confuse those unlike yourself that are sharp and trying to keep accurate information the key in a discussion.

  162. mpainter says:

    Curt says:
    “that’s painfully obvious”
    ####
    You seem to be saying that there is no accounting for the 333 W of BR, at least not in the manner of my approach.
    The diagram says that this “absorbed by surface”. You can only hold your views if you ignore that portion of the diagram, imo.

    • Curt says:

      It’s now completely clear to me that you don’t even have the sophistication to understand the arguments. This is pointless.

  163. mpainter says:

    To continue:
    1. 333 W/sq m “absorbed by surface” all according to the KT diagram.
    2. This absorbed within a few microns of the surface of the ocean, according to the absorbency of water.
    3. Kinetic energy of this minute interval increased accordingly.
    3. Resultant evaporation? Well, none, according to the KT diagram and those like Curt, who declares that you are untutored insophisticate if you expected any evaporation.

    Well, Curt, here is the truth: I never expected any evaporation because the diagram is egregious, specifically those components marked by Roy with a big red X.
    This has always been my point.
    The 80 W/sq m of latent heat is due to insolation, and not back radiation.
    The 333 W of DWLWIR is the dog that didn’t bark, conspicuous by the lack of evaporation which certain would issue from such energy.

    • Kristian says:

      The 333 and 396 W/m2 ‘fluxes’ are not radiant fluxes at all. They cannot separately do anything. Because they are not real, distinct entities in any way. They are temperature/emissivity potentials, mathematical terms in the Stefan-Boltzmann equation. The only real radiant flux here, the sole transfer of energy by radiation, the only thing that is ever actually detected physically (as in any heat transfer situation), is the 63 W/m2 of average upward radiant heat from the surface. And that’s it. The rest is just mental constructs and temperature effects.

      If the global surface of Earth were at an average temperature of 255K rather than 288, then if we pointed our cryogenically cooled detector downwards, it would ‘read’ a radiant heat flux of ~230-240 W/m2. If the combined conductive+evaporative loss stayed the same as in the current state, 95-100 W/m2, then the radiant loss would still have to be 60-65 W/m2 (after all, ~160 W/m2 is still coming in from the Sun). In other words, if we then pointed our cold detector upwards to the sky, we would expect 170-180 W/m2 worth of radiant heat to well down to it.

      The budget would still be balanced, the ‘net’ would still be the same, only now with a colder surface. The only problem with this being that we know the surface to be much warmer. And then we start out by taking for granted that the reason for this is somehow those mathematically derived 333 W/m2 down from the sky. But they are not the cause of anything. They are a simple result of temperature.

    • cwilson says:

      mpainter:

      “Resultant evaporation? Well, none, according to the KT diagram and those like Curt, who declares that you are untutored insophisticate if you expected any evaporation.”

      I have never said that there is no resultant evaporation, and I have explained to you repeatedly that I have never said that.

      The fact that you fail to understand both basic science and clear English is what really makes you an “untutored insophisticate”.

  164. A Good presentation but then we would not expect anything less.

    Dave

  165. mpainter says:

    cwilson? Is that you Curt?
    Curt: ” I have never said that there was no resultant evaporation…”
    Given: 333 W/sq m of LWIR absorbed at the surface of the ocean, calculate the latent energy resulting.

    Curt: “You cannot partition the realationship of particular inputs to particular outputs that way”

    • Tim Folkerts says:

      You STILL don’t get it. Your statements belie two errors in your thinking. Curt has patiently tried to explain; maybe hearing it from a second sourse will help.

      1) You say: “The 80 W/sq m of latent heat is due to insolation, and not back radiation.” No, the rate of evaporation is due to the energy actually *IN* the water, ie the mean KE of the molecules ie the temperature of the surface of the water. Whether that energy might have come from insolation or LWIR or an electric heater or a nuclear reactor is immaterial. The whole idea of thermal energy is that it is randomize, with any knowledge of the source having been erased. (And of course, external factors like wind and humidity matter that matter no matter what the water temperature/source of heating for the water.)

      2) The climate is a highly non-linear system. Changing any one variable changes the others in non-linear ways. You can’t use a “simple algebraic calculation” like evaporation = 0.6 sunlight” because there is no simple algebraic relationship! Even a quadratic function would probably not some close to work.

      So to steal a phrase from Fermi, your whole “0.6 of solar radiation” thread is “not even wrong”. It is a non-starter for two completely valid and completely different reasons!

  166. mpainter says:

    Tim Folkerts,
    Thanks for your comment, Tim.
    You say: “The rate of evaporation is due to the energy *IN* the water..”
    I agree, but nowhere in the above discussion was the “rate” of evaporation discussed, so I can’t see how your first point is relevant.

    Concerning the absorption of LWIR by water, this takes place within a few microns of the air/water interface. For example, the 15 micron band is absorbed within three microns. Thus the kinetic energy gained by the ocean from LWIR is confined to a minute interval of a few microns at the surface. So yes, evaporation depends on the temperature of the water, among other factors, but the absorption by water of LWIR is a special case. The absorbed energy is transient; it is converted to latent energy within a few seconds, or perhaps radiated.

    Concerning your second point:
    You come in at the middle of the thread and raise issues that have been addressed and settled, such as the partition of the ocean cooling. This was discussed in detail up thread. I will repeat for the benefit of newcomers some of the facts presented.
    The partition of the cooling of the ocean into its components (evaporation, radiation, and sensible heat/conduction) has been addressed by several studies. These give slightly different % for each component, but a representative partition would be : evaporation-60%; radiation-30%;sensible heat/conduction-10%. Now you come and asert that such partitioning is not validly applied by way of my simple computations. The only support that you give for this assertion is to cite the complexity of climate processes and the need to apply quadratic functions.
    You also seem a bit confused when you say “your .6 of solar radiation” which term I never employed. That figure I always used as the figure representing ocean ocean cooling via evaporation, as per the studies. I agree that insolation is the means whereby the ocean gains its heat.
    So your first being irrelevant and your second based on lack of understanding of the assumptions behind my calculation, I do not see how you can justify your closing remark. I will not object if you choose to withdraw it.

    • Tim Folkerts says:

      “I agree, but nowhere in the above discussion was the “rate” of evaporation discussed, so I can’t see how your first point is relevant. “

      [scratching my head] You were discussing 80 W/m^2 of latent heat, ie the rate at which thermal energy of liquid water is changed to latent energy in water vapor (removing 80 W/m^2 from the oceans and delivering 80 W/m^2 to the atmosphere).

      The 8o W/m^2 latent energy *is* evaporation. The 80 W/m^2 is evaporating water at some rate.

      “The absorbed energy is transient; it is converted to latent energy within a few seconds, or perhaps radiated.”
      Or perhaps it warms the water. I agree that IR is in some ways special because it interacts with the very top layer only. But there is no reason this energy must only go into evaporation. This was Curt’s point. If the energy gets even 0.1 um into the water, that water molecule would undergo 100’s of collisions (maybe millions — I haven’t done any calculations) before it might get to the surface to evaporate. I don’t see any reasonable conclusion other than the energy getting thermalized long before that molecule thinks of evaporating.

      Adding 333 W/m^2 from BELOW the surface would certainly create a different profile WITHIN the water than adding 333 W/m^2 of IR from the top. But the topmost surface would still be the same, to lose the same total energy.

      • gbaikie says:

        — Tim Folkerts says:
        June 21, 2015 at 1:15 PM

        “I agree, but nowhere in the above discussion was the “rate” of evaporation discussed, so I can’t see how your first point is relevant. “

        [scratching my head] You were discussing 80 W/m^2 of latent heat, ie the rate at which thermal energy of liquid water is changed to latent energy in water vapor (removing 80 W/m^2 from the oceans and delivering 80 W/m^2 to the atmosphere).

        The 8o W/m^2 latent energy *is* evaporation. The 80 W/m^2 is evaporating water at some rate. —

        The 80 W/m^2 of latent energy may be close enough in terms of evaporation from the surface as it’s about the same amount that falls from the sky as rain.
        But Earth is covered in clouds and sunlight would warm the cloud and evaporate the water droplets- so that evaporation would be included in the 80 W/m^2 of latent energy- as it doesn’t alter the amount of rain fall [it only delays it].
        So you draw $1000 from the bank, you buy something, and money may exchanged a number times before it returns to a bank- which similar to counting the exchanges of evaporation
        with rainfall- the $1000 returning to a bank is rain, or taking the money from the bank is rain- either way.

        • Tim Folkerts says:

          “But Earth is covered in clouds and sunlight would warm the cloud and evaporate the water droplets- so that evaporation would be included in the 80 W/m^2 of latent energy- as it doesn’t alter the amount of rain fall [it only delays it].”

          No, the 80 W/m^2 would specifically be energy associated with liquid on the surface turning into vapor (and then that same vapor re-condensing to for the clouds). The KT diagram is about energy transfers from one region to another. So water that evaporates from a cloud within the atmosphere and re-condenses in the atmosphere is only about energy within the atmosphere and isn’t in the diagram. (Just like IR from one part of the atmosphere to another part is not included).

  167. mpainter says:

    Tim Folkerts:
    As for “errors in my thinking” I see none yet. I do see plenty of error in the KT diagram, for reasons stated by me in this thread.

  168. mpainter says:

    Tim,
    Your point is taken, but I do not regard using units of power as a discussion of evaporation rates per se, which rates depend on water temperature, wind, and atmospheric humidities as well, and these were not discussed, not in the sense that the _process_ of evaporation was dicussed. Please let’s not get into one of those dumb semantical tussles.
    I will refute your first point thus:
    My point was that SST is due to insolation, quite simply. I wonder that you missed such an obvious point. Please, no more of these kind of tedious quibbles. To reiterate: Ocean heat, SST, sea water temperature, by whatever name or metric is the result of insolation. In that sense the 80W/sq m is owing to insolation.
    I consider your point #1 a pointless quibble and a non-sequitor to boot; an irrelevancy, as I stated.

    LWIR is absorbed by water as kinetic energy _understood_ or stated explicitly in all my previous comments.
    I mentioned radiation in my last comment as one way that the energy of absorbed LWIR was released.
    My ultimate point is this:
    LWIR absorbed at the water’s surface is transient and may be accounted for as 1) evaporation : 60%; 2) radiation: 30%; 3) conduction/ sensible heat: 10%.
    333 W/sq m of LWIR absorbed in water will result in 200 W/sq m of latent heat, according to this partitioning.
    But, as we see, insolation, at 161W/sq m, well accounts for the 80 W/sq m of latent heat.
    That 200 W is not to be found because it does not exist. Hence the KT diagram is egregious.
    Go read the thread.
    Another absurdity of the KT diagram is that it utterly omits the UWLWIR component of the atmospheric radiant flux (the complement to the back radiation in an isotropic radiative flux) as if the authors decided that it simply did not exist.
    Repeat, the KT diagram is egregious.

    • Curt says:

      mpainter:

      No how many times it is explained to you, you will never understand the difference between net and gross transfers.

      Don’t ever become an accountant — you would quickly end up in jail.

  169. Norman says:

    mpainter,

    Why do you still completely ignore the 396 watts the surface of the water is losing at the same time it is receiving 333 watts?

    The error in your thinking is very obvious and simple to explain but you want to ignore this issue. Why are you doing this? What motivates you to not be able to see this energy leaving the surface of the water?

    I would maybe have respect for your posts if you even considered what does losing 396 watts do. But you don’t you ignore it and act like you know something. If you actually took a class in physics you probably could not understand it and get frustrated when it was explained to you in simple basic terms from many different views.

    Explain why you ignore the energy that is constantly leaving the ocean surface?

  170. mpainter says:

    Norman
    With me it is a matter of science, not faith. I have no faith in the KT diagram because I know that 333 W/sq m of LWIR absorbed by the ocean will produce latent energy and there is none in the diagram attributable to that energy flux. The diagram shows that this is absorbed and returned 100% as radiation. This I have explained repeatedly. You are in the position of having to explain why that BR does not produce latent energy, yet you seem utterly bewildered as to my views.

    • Norman says:

      mpainter

      Huh? The KT diagram shows clearly the surface losing energy at the rate of 396 watts/meter^2

      You state: “because I know that 333 W/sq m of LWIR absorbed by the ocean will produce latent energy and there is none in the diagram attributable to that energy flux.”

      How can 333 returning energy produce latent heat when the same surface is losing 396 watts/meter^2? If your surface is losing energy continuously it will not create higher rate of evaporation. Why do you think it will?

      I think I have explained to you numerous times in any way I think you might possibly comprehend your point “You are in the position of having to explain why that BR does not produce latent energy, yet you seem utterly bewildered as to my views.”

      Why is because the BR is LESS than the energy that the surface is losing continuously. I am bewildered by your views as why you cannot comprehend what and NET energy flux is and how it behaves. If the NET energy is negative (as is with surface loss vs BR gain) the surface will cool and latent heat drops. You ignore the 396 watts of loss and ONLY add energy gained. Why do you continue in this view when it makes no sense? How does this position make you a more competent individual? What is driving you to keep pursuing this course? Read what Curt is saying to you and think about it. Take a piece of paper and draw it out, stop being so pigheaded and reconsider, think about the loss of energy, please. I think you will benefit to correct your error. Learn from you mistakes and grow in intellect.

  171. mpainter says:

    Curt,
    No one has explained to me satisfactorily why the back radiation has gained an exemption from the laws of physics.
    The KT diagram is so bad that it does not even show the radiative flux correctly, but leaves out half of it, the UWLWIR complement to the DWLWIR.

    • Curt says:

      My god, mpainter, you are just spectacularly incompetent!

      The “396” that is staring you in the face is the UWLWIR complement to the DWLWIR.

      And if, as I have repeatedly explained, you just treated the LWIR as a 63 upward flux (as the 80 evaporative and 17 thermal are net fluxes), all your objections would vanish!

  172. mpainter says:

    Curt
    My understanding of the hypothetical atmospheric IR flux is that it is isometric and that UWLWIR=DWLWIR.
    The KT diagram shows the DWLWIR component of this flux, but the authors pretend that the upwelling component of this flux does not exist. More egregious science.

    • Tim Folkerts says:

      Your understanding is wrong.

      UWLWIR depends on the temperature & emissivity of the surface.
      DWLWIR depends on the temperature & emissivity of the atmosphere.

      Since the temperatures and the emissivities are different, the IR in the two directions can and will be different.

      I can see how some people misunderstand Kirchhoff’s Radiation Law and think it says the power in and out must be the same, when it instead says the emissivity must equal the absorptivity. Perhaps that is what you were thinking of.

      Even more egregious, Curt just explained that the 396 W/m^2 is the UWLWIR, yet you REPEAT your incorrect claim that it is not in the diagram. That I simply DON’T understand.

  173. mpainter says:

    Then the atmospheric IR flux is not isometric? Fine, I can live with that notion.
    The 63 W/sq m NET comes from insolation, it is plain. The rest of the radiation chases its tail up and down.. or is it down and up?
    Allow me to REPEAT my query about latent energy. How about it Tim, ball’s in your court; where’s the 200 W/ sq m of latent heat, or does not the absorbed back radiation add kinetic energy to those water molecules? Or perhaps there is no 333 W/sq m being absorbed by the ocean. The diagram is egregious.
    Come, Tim, let’s hear you say that the back radiation gets absorbed without adding kinetic energy.

    • Curt says:

      mpainter:

      The 396 upward on the K&T diagram is from the surface, not the atmosphere, so the fact that it is not the same as the downward from the atmosphere at the surface does not violate the isotropic property you talk about (but which you have no idea what it signifies).

      Until you get these basic, basic concepts down, you have no business claiming others are making “egregious mistakes”.

      I have a “Where’s Waldo” problem for you. Look all over the K&T diagram and find the UWLWIR component from the atmosphere that you have missed.

      And for the umpteenth time, the people talking about “60% of oceans losses being evaporative” DEFINE the 100% level as only being equal to the solar input.

      Oh, for extra credit, on top of your mixing up of gross versus net flows, explain why your use of the 0.71 factor in your equation, when the other units are already normalized to density values, is completely invalid…

  174. Norman says:

    mpainter,

    A question for you to answer. What do you believe the source of the atmosphere back radiation is?

  175. wayne says:

    Tim, bad play. You know enough of physics to be particularly dangerous to unsuspecting minds. Here you are, once again, leaving out some of the correct factors because you know them well and they marginalize what you are currently selling and I know that you know them well just to prop up your mistaken mindset that you are feeding to someone (mostly misleading cr.p).

    You said:
    UWLWIR depends on the temperature & emissivity of the surface.
    DWLWIR depends on the temperature & emissivity of the atmosphere.

    But that is not even complete on the DWLWIR is it Tim, just T and ε? What of the window frequencies that even TFK isolated out in their paper and graphic as from portions of non-interacting frequencies? Do you imagine downwelling thermal radiation from the general atmosphere is going to radiate what is does not even have the capability to absorb? Of course not (Kirchhoff). You know that yet you leave it out.

    What of the the overall path length? Do you imagine that an atmosphere 1/100th or 1/1000th as thick as ours is going to radiate to the surface just as much since you say it only depends on temperature and emissivity? You know that yet you leave it out. Strike two. Come on with some more fantasies, I just need one more strike. 😉

    Also, re-read mpainter’s comment. You didn’t even understand what he was asking. *(that is, if there is 333 downwelling from the entire general atmosphere (the 396 is upwelling from just the surface) where is the 333 arrow pointing upward, for atmosphere gases radiate equally in all directions as you have said many times, if you have a down you must also have an up, where is it?) Please answer his question, I am also very curious of your answer.

    • Curt says:

      wayne:

      The K&T diagram presents a simplified analysis with 4 subsystems: sun, earth, atmosphere, and space. Only the transfers between subsystems are shown, not any transfers within a subsystem.

      This is completely standard practice in thermodynamic analysis, by the way, because transfers within a subsystem do change the energy of that subsystem.

      So if there is 333 downward IR originating in the atmosphere at the surface, there is also 333 upward at that altitude. The K&T diagram is not in error for not showing it.

      The K&T diagram does show upward IR originating from the atmosphere. Can you find it?

    • Tim Folkerts says:

      Wayne, I could point out several MORE factors that you left out (and that I suspect you don’t even know). But you OUGHT to know that none of the factors you named, nor the ones you don’t even realize, really pertain to the discussion at hand.

      The one truly valuable thing you say is “if there is 333 downwelling from the entire general atmosphere … where is the 333 arrow pointing upward, for atmosphere gases radiate equally in all directions.”
      If that truly IS his point (which it may well be), then I did indeed miss it. (And that makes your other comments even FURTHER off on a tangent, rather being helpful).

      “Please answer his question, I am also very curious of your answer.”
      OK, since neither you or mpainter seem to know …

      The KT diagram shows only 4 subsystems: the sun, outer space, the surface as a whole and the atmosphere as a whole. As such it ONLY shows transfers from one subsystem to another subsystem. This is a very common approach — dividing a system into a small, simple set of subsystems for ease of conceptual understanding.

      What you and mpainter seem to want is a more detailed diagram, one that divides the atmosphere into multiple subsystems — eg layers 100 m thick. We could consider say 200 such layers up to 20 kn high, each with their own temperature, composition, clouds, and density. This is laudable, but it is a vastly more complicated problem that what KT are presenting.

      In such a system, consider the 396 W/m^2 from the surface. A sizable part of it would be absorbed by the first layer by the GHGs. Another sizable part — in the “atmospheric window” — would make it through that first layer. Most of that radiation in the atmospheric window would get absorbed in higher layers by clouds (and a little by GHGs since the ‘edges’ of the window are not perfectly clear). And the remaining 40 W/m^2 would still escape to space.

      So now we have a diagram with 201 different arrows going up from the surface to the 200 various layers (and one to space).

      Similarly, the 333 W/m^2 coming down as “back radiation” is actually the sum of 200 different arrows from 200 individual layers. The bottom layer would contribute a big part of that (call it 150 W/m^2); most of the rest would be from clouds higher up radiating through the lowest layer through the atmospheric window (call it 150 W/m^2 too); the remainder would be from higher layers radiating at the edges of the ‘window’ (which would have to be 33 in this case to make the numbers work).

      So … what we REALLY would want in this much more detailed sort of analysis is an arrow of 150 W/m^2 DWLWIR from the bottom layer to the surface, and a series of arrows from the bottom layer to higher layers (and space) that add to 150 W/m^2.

      Yes, the total DWLWIR from the bottom layer is equal to the UWLWIR from the bottom layer and would be clear in this diagram (if you can say anything is “clear” in a diagram with 1000+ arrows connecting 203 systems). The total UWLWIR from the 2nd layer is equal to the total UWLWIR from the 2nd layer. Etc.

      If this is really what mpainter was after, I hope it helps.

      KT combined those 200 layers into one for their diagram. While each layer individually radiates the same up and down, the TOTAL radiated up (239 W/m^2) does not need to equal the total radiated down.

      **********************************
      Similar analysis would have to me done for all the OTHER
      modes of energy transfer.
      * The 80 W/^2 of latent heat would be 200 arrows to 200 layers that all ADD to 80 W/m^2. \
      *The 78 W/m^s of incoming sunlight absorbed by the atmosphere would be 200 arrows to 200 layers that all ADD to 78 W/m^2. * The 239 W/m^2 outgoing IR would be 201 arrows from 200 layers & the surface that all ADD to 239 W/m^2.
      Etc.

  176. mpainter says:

    Norman, Curt, Tim,
    Is the back radiation “absorbed by the surface” without adding kinetic energy to the molecules of the surface?

    • Curt says:

      How has anything we said possibly led you to believe that?

      The back radiation “absorbed the the surface” does add kinetic energy to the molecules of the surface. By the same token, the forward radiation “emitted by the surface” reduces the kinetic energy of the molecules of the surface.

      Since the magnitude of the forward radiation is greater than that of the back radiation, the NET effect of the infrared radiation exchange is to reduce the kinetic energy of the molecules of the surface.

  177. mpainter says:

    Curt, thanks for your reply. I am happy to have it. Can’t finish now. Nite.

  178. mpainter says:

    No one has yet addressed the question of how the ocean could absorb 333 W/sq m of LWIR without yielding a consequent component of latent energy, say 60% of the absorbed energy thus converted. The 80 W/sq m of latent energy is confirmed by precipitation totals and corroborated by partition of ocean cooling, assuming 161 W/sq m insolation. The egregious KT diagram shows all back radiation as absorbed at the surface and emitted as IR.

    Curt, have you ever tried to calculate how much latent energy results from the back radiation?
    Curt: “You cannot partition the relationship of particular inputs to particular outputs that way”
    Sure you can, Curt.

    • Curt says:

      mpainter: You say, “No one has yet addressed the question…”

      Bull! I’ve directly addressed it multiple times, and you have always ignored my responses. I don’t think you even comprehend it!

    • gbaikie says:

      — mpainter says:
      June 22, 2015 at 6:07 AM

      No one has yet addressed the question of how the ocean could absorb 333 W/sq m of LWIR without yielding a consequent component of latent energy, say 60% of the absorbed energy thus converted. —

      An average of 333 watts per square meter of longwave IR would significantly increase the amount of latent heat connected to evaporation.
      Therefore you can rule out the idea that 333 watts is absorbed
      by the surface [most of the surface is ocean and the 333 watts could only heat the top mm of the ocean. It would also prevent dew on the surface during night.

      Instead the more knowledgeable believer/worshiper believes that without trace gases of CO2 and H2O, the average surface would radiate 396 watts. So serious believers, think the greenhouse gases [and/or things like droplets of water in clouds] reduce 396 watts which is “suppose” to be radiated by 333 watts.
      So the 333 watts would not so any work.
      They might justify the use of watts [energy which can do work] by imagining that were the earth surface absolute zero
      it would do work [but it actually wouldn’t].

      • gbaikie says:

        Btw, the goes back to the mis-named greenhouse effect [which btw the is like the mis-named Lefties which call themselves “progressive” as they are against progress, want everyone else to live in mud huts [and don’t have babies- what going to do in mud huts for entertainment is a mystery].
        Anyways, in an actual greenhouse it is the warm air which keeps things warm, not the radiant effect of trace gases.

      • Curt says:

        gbaikie:

        You say, “the more knowledgeable believer/worshiper believes that without trace gases of CO2 and H2O, the average surface would radiate 396 watts.”

        You get the argument exactly backwards, which just demonstrates that you don’t even understand the issues in play!

        More knowledgeable people understand that without these traces gases, and the (measurable) radiation they emit downward, there is no way the average surface could radiate 396 watts we actually measure, because it receives far less than that from the sun.

        Oh, and you absolutely do not understand the concept of thermodynamic work here. Watts are units of energy per unit time, period, whether work is involved or not.

        • gbaikie says:

          “In physics, power is the rate of doing work. It is equivalent to an amount of energy consumed per unit time. In the SI system, the unit of power is the joule per second (J/s), known as the watt in honour of James Watt, the eighteenth-century developer of the steam engine.”
          https://en.wikipedia.org/?title=Power_%28physics%29

          The human body is about 100 watts, and it it has about
          1.5 square meter of surface, and has mouth temperature
          of about 98.6 F [37 C, 310 K].
          How many watts per square meter do you radiate?

          • Curt says:

            gbaikie:

            My kids weren’t allowed to use wikipedia as a resource even in middle school! The person who wrote that is clueless about the difference between work and heat transfers, a fundamental thermodynamic distinction.

            As for me, my torso skin temperature is typically 35C (308K) with 0.95 emissivity. So it has a radiative flux density of

            0.95 * 5.67E-8 * 308^4 = 485 W/m2

            My limbs have slightly lower temperature, but it’s a good estimate that I am radiating about 485 * 1.4 = 675 watts.

            But I am bathed in ambient thermal radiation right now of about 425 W/m2 from the walls and other surfaces, for a total I am receiving of about 595 watts. So the net would be about 80 watts from me to the room by radiation. None of this power transfer does any thermodynamic work.

          • gbaikie says:

            Thanks, Curt.
            So you say, you are radiating a net amount of 80 Watts. Or
            average of about 50 watts per square meter.

            This seems strangely similar to the earth surface, which according chart, has an average net loss of 40 watts per square meters.

          • Curt says:

            Why should that be strange, when there are similar temperature levels and temperature differences?

          • gbaikie says:

            I don’t think it’s strangely similar because 37 C is
            near 15 C.

            And I think a house is kept warm just like a greenhouse is kept warm- which works by reducing convection losses.
            Also people normally warm or cool a house by making air warmer or cooler. Rather than heating or cooling walls.

            People wear clothes to reduce convection losses rather radiant losses.

            Also Earth evaporates a lot water, and people perspire.

            Wiki:
            “In humans, sweating is primarily a means of thermoregulation which is achieved by the water-rich secretion of the eccrine glands. Maximum sweat rates of an adult can be up to 2–4 liters per hour or 10–14 liters per day (10–15 g/min•m²), but is less in children prior to puberty.”
            https://en.wikipedia.org/wiki/Perspiration

            And finally while on the topic it’s somewhat strange that tropical creature did evolve to perspire. Yeah I know, about the theory that humans may have been mostly scavengers and therefore traveled long distances, but still it seems somewhat strange.

  179. Norman says:

    mpainter,

    Why do you think you can partition incoming radiation ia particular way?

    The total radiatnt energy reaching the surface is 494 watts/meter^2. The total radiant energy leaving the surface is 396 watts/meter^2. You have an excess of 98 watts/meter^2 reaching the surface. Of that 98 watts/meter^2, 80 watts/meter^2 are removed by evaporation. Why do you think calculating the % or energy of the backradiation is used for evaporation is so important? The total amount of radiation absorbed – the radiation lost will be all that is needed to see how much energy is removed by evaporation and the graph shows this result.

    I think you are mentally stuck on something but I am not sure what it is.

  180. mpainter says:

    Kristian
    Thanks for this. Interesting contrast of two views of atmospheric thermodynamics.
    The radiating IR loop produces adsurdities which twists the proponents into pretzels when they try to defend their views. As seen on this thread.

  181. mpainter says:

    Tim,
    I think it would be well to review the beginnings of this discussion:

    My comment above:

    Curt, my understanding if the hypothetical atmospheric IR flux is that it is isometric and that UWLWIR=DWLWIR.
    The diagram shows the down welling component of this flux, but the authors pretend that the upwelling component of this flux does not exist. More egregious science.

    To which you responded:

    Your understanding is wrong.

    A most unequivocal statement from you.

    I have seen nothing from except a refutation of the notion of an isometric atmospheric IR flux. More convolutions in AGW science, it seems.

    • Tim Folkerts says:

      mpainter,

      See this comment. It may address some misunderstandings. There seem to be some confusion as to specifically which ‘corresponding’ UWLWIR radiation you are referring to.

      http://www.drroyspencer.com/2015/06/what-causes-the-greenhouse-effect/#comment-193691

      • Tim Folkerts says:

        To clarify.

        Typically people discuss DWLWIR from atmosphere to surface and UWLWIR from surface to atmosphere. That is what Curt and I have been addressing. This UWLWIR *IS* in the diagram.

        You seem to be discussing DWLWIR from the bottom layer of the atmosphere downward and UWLWIR from from the bottom layer upward. This UWLWIR is *NOT* in the diagram (and should not be). The closest equivalent is the UWLWIR from the TOP layer upward to space. This *IS* in the diagram, but, due to the difference in temperature at the top and the bottom, this is NOT the same magnitude as the DWLWIR downward at the bottom of the atmosphere.

  182. mpainter says:

    Tim,
    Where is the “confusion”?
    Yes it was that comment of yours which prompted my last. I see nothing in your referred comment which permits the inclusion in the KT diagram of the down welling portion of the atmospheric IR flux and the exclusion of the upwelling portion, assuming an isometric flux.

    • dave says:

      “…inclusion in the KT diagram…”

      As already mentioned, the artist did not know whether he was meant to illustrate a single-slab atmosphere or a multi-slab atmosphere. Probably, he did not know the difference.

      Somebody called Toronto Anne did a PROPER reworking (IMO) of this diagram, years ago, on this blog. Later, she improved it by distinguishing the different processes above ocean and land.

      If I remember rightly, she advised not to waste time looking further into these issues, unless and until UAH and RSS showed three months of clear anomalies outside the range of the 1990’s.

      • Tim Folkerts says:

        As already mentioned, the artist did not know whether he was meant to illustrate a single-slab atmosphere or a multi-slab atmosphere. Probably, he did not know the difference.”

        Yeah. In the same way that Da Vinci didn’t know whether Mons Lisa was meant to illustrate a man or a woman, and probably didn’t know the difference. Or when Bill Belichick was drawing up plays for the Super Bowl, he didn’t know if they were meant for the offense or defense, and probably didn’t know the difference.

  183. mpainter says:

    Tim, then in fact my understanding was not wrong and the atmospheric radiative flux is isometric, after all.

  184. Curt says:

    Where is the “confusion”?

    Fundamentally, you confuse the idea that emission of radiation originating from a location within the atmosphere is isotropic (true) with the (erroneous) conclusion that the total radiative flux at that location, including that originating from elsewhere.

    Why are you not also objecting to the description of the shortwave flux in the K&T diagram as well? It shows a large imbalance downward. It’s not isotropic at all!

  185. mpainter says:

    Curt
    Hunh?

  186. mpainter says:

    Curt,
    Do I understand you to say that my understanding was correct, notwithstanding the understandably confused lack of understanding of the radiative atmosphere?

  187. Curt says:

    mpainter:

    You just continue to demonstrate that you don’t have the conceptual framework even to start to grapple with these issues, let alone come to the correct conclusion.

    I repeat my earlier question. The K&T diagram shows a shortwave flux in the atmosphere that is strongly net downward. Do you believe that to be in error?

  188. mpainter says:

    Fine Curt, let us end the discussion. I am tired of your snark and personal sniping.

  189. gbaikie says:

    I tend to think CO2 could cause some amount of warming, so something like, if or when global CO2 reaches 800 ppm, there could be as much as 1 C added to global average temperature.
    Ten years ago, I thought doubling CO2 level might cause 2 C or more of warming.
    Currently, I of the opinion that over last hundred years or so
    Global CO2 has risen about 100 ppm, though we have only had somewhat accurate measurement for less than 60 years and it seems likely to me that over next 100 years CO2 levels may rise another 100 to 200 ppm. So by year 2115 we could have a global CO2 level as high as 600 ppm though it’s as possible that could rise to 500 ppm or less.

    So I think CO2 could cause warming, though the increase of 100 ppm of CO2 over last 100 year has not appeared to cause
    an increase in global temperatures.

    Nor does it appear that any human activity has caused any increase in global temperature. Though this lack of evidence
    of human caused increase in global temperature could be due
    to not having a system to accurately measure global temperatures. And I will note that there is other human activity which may have caused an increase on global temperature, other than a rise in global CO2 levels- and likewise such human causes has also not be measurable.
    The declaration of “human fingerprints” upon global temperature is quite simply, false.
    The build up of urban areas over the last hundred years, does cause the local region to warm, this is referred as Urban Heat Island Effect [UHI effect]. And most people in the world live within to these urban effected area.
    There has been far less attention on UHI effect as compared to trying to determine if CO2 levels have increased global temperature. In terms gross dollar spent by the public the difference may be about 1000 to 1. Or for each billion spent related to effects of CO2, a million dollars has been spent on UHI effect. And this public policy [like most] has been hopeless idiotic.
    So amount warming which occurred in urban area is basically unknown but it’s somewhere around a 5 C increase in average temperature for some urban area over the last 100 years.
    And due to small percentage of land area which is urban area, such levels of increase in temperature has been immeasurable in terms of global temperature increase.
    Unlike other human causes which may result in an increase in global temperature, UHI effect is easily measurable in terms of it’s local effect upon temperature.
    Anyone can [and many have] easily measured the warming effects of UHI, this is not mere fingerprints, this is murder weapon, hundreds of witnesses, fingerprint, DNA, video coverage, motive and etc. And this UHI effect is cause of urban area having “heatwaves” and has affected the most amount of people. If one can claim that a tropical creature has died from warming, the murderers have been UHI effect. This has more evidence than OJ Simpson murdering his wife- and only doubt in the matter would be due to biaes
    and motivations of those judging the matter.

    Though it’s also well known the tropical human is killed far more from cold conditions than warm conditions, though it should said that global warming has never been about the welfare of human beings, in fact those people with “most concern” [or cry about it the most] would rather have far less human living than the current 7 billion souls. So such people as Al Gore would be very pleased if we could “get” less than 1 billion people living on Earth [or anywhere else]. So global warming has been about saving the environment, and largely about saving the environment from what humans could do it, rather saving the environment for the sake of human’s well being living in this environment.

    So idea that CO2 levels, as originally suggested as a possible theory to explain cause of glacial period, at this point in time is not considered very likely [or impossible].
    And idea that CO2 levels could cause some kind of insane amount of global warming, was never serious believed by anyone- it was seen as necessary exaggeration, to somehow cause fear and therefore political action to do something about human CO2 emission.
    Due to lack of information some people have had a somewhat rational fear than a doubling of CO2 could cause more than 5 C of global warming. But it’s not particularly rational to worry about something due to lack of information, and cooling my 2 C is more likely [has happened in past] and would have far more bad consequences. So to be clear I doubt we will have cooling of more than 1/2 of degree nor warming of more than 1 C in global temperature, or at least there is no known or explainable cause of this happening within 100 years. Though in regions [rather than global] there have been and will be greater variation of average temperature, nor is there any evident that global changes in temperature causes an increase in regional changes.

    So, I tend to think that increases on global CO2 levels would cause a small increase in average global temperature.
    Though there seems to be more evidence that increasing global temperature would slightly increase global CO2 levels. Or perhaps said differently, were we in a 100 year period of cooling, human CO2 emission could appear to have less effect in terms of rising global CO2 levels.
    A simple way to understand current “climate science” is that models projecting future warming have over estimated the effect of increase levels of CO2.
    So instead of CO2 and any “forcing” increasing temperature by 2 C, it’s 1 C, plus throw in the “lack of attention” on “natural climate variability”.
    An analogy of this, would be idea that increasing electrical production in a nation will cause increase in economic growth and reduction in poverty [people become more wealthy]. So there was some agreement with past increases in electrical power capacity of nation, which increased economic but there was “natural variability” which also caused economic growth. So mostly ignore the “natural variability” and come up with X amount of increase in electrical power results in 2% economic growth, so then electrical power is increase by 150% greater than the X amount which suppose to cause 2% growth, and result is about 0% economic growth.

    So that would be simple, and that’s basically what has happened with “climate science”. But it’s is obviously a poor understanding of economics and climate. Though both may be more or less a practical way of understanding either. But the view that Earth was the center of the universe was also a practical way of understanding the universe.

    But point I was trying to get to was regarding Mars and Mars having 28 times more CO2 per square meter as compared to Earth. And I was lately searching for some estimate of how much warming this CO2 should be causing on Mars. And it appears to have become unfashionable recently to say this CO2 causes much warming on Mars. Though idiots are still saying if increase CO2 on Mars by thousands of times it’s current level this would warm Mars.
    Also I would guess, that attempts at understanding Mars climate will have a significant future effect upon understanding Earth Climate. And this is roughly due to sheer number of spacecraft currently at Mars.

    • wayne says:

      gbaikie, I noticed you are now addressing Mars.

      Do not you not gather from all that has been said in all of these sites and threads along with what you have said that if enough nitrogen was added to the Mars atmosphere to bring its unit columnar mass up to match that of Earth’s unit columnar mass that there would then definitely be a greenhouse effect raising the Mars surface above the now near theoretical black body radiative temperature of ≈210 K? So in that light what causes this ‘greenhouse effect’? The co2 at 28 times Earth’s or the mass that was added? Come on, its not that hard to correctly think through. I am surprised you are still questioning this.

      Yes, Mars would then show the same effect as seen on the Earth but diminished due to Mars’ greater semi-major axis and less solar influx. It is the mass, sorry, not the quantity of co2 per unit column. So how does that new realization play into the current ‘discussion’ about Earth and co2?

      (I used http://nssdc.gsfc.nasa.gov/planetary/factsheet/marsfact.html for the 210.1 K blackbody temperature and the 210K mean surface temperature.)

      • gbaikie says:

        — wayne says:
        June 28, 2015 at 9:33 PM

        gbaikie, I noticed you are now addressing Mars.

        Do not you not gather from all that has been said in all of these sites and threads along with what you have said that if enough nitrogen was added to the Mars atmosphere to bring its unit columnar mass up to match that of Earth’s unit columnar mass that there would then definitely be a greenhouse effect raising the Mars surface above the now near theoretical black body radiative temperature of ≈210 K? So in that light what causes this ‘greenhouse effect’? The co2 at 28 times Earth’s or the mass that was added? Come on, its not that hard to correctly think through. I am surprised you are still questioning this.–

        I believe that Martian nites would much warmer than they are right now.
        But it would reduce Mars noon and sun zenith direct sunlight
        by about 100 watts. So right now Mars would get [depending how close the sun was in the orbit] about 500 to 700 watts.

        And added earth like atmosphere would decrease that to about 400 to 600 watts per square meter. Plus tropics of mars would be much warmer than rest of Mars.
        Or tropics with earth like thick atmosphere would mostly get sunlight when sun was 45 degree or higher than horizon.
        But for say region higher than 40 degree latitude, they get less sunlight at 45 degrees above the horizon. So when sun is at 30 degrees above horizon, the sunlight goes thru twice as much atmosphere as compare when sun is 45 degrees or more above the horizon.
        So midlatitude on Mars [25 to 65 degrees] would have much lower average temperature- as midlatitude are also much cooler than tropics on Earth.

        So tropics on Mars would be similar average temperature as deserts on Earth in the tropics. And so average temperature in tropics on Mars with earth type atmosphere could near 0 C. It might have highest daytime temperature of 20 C, and lowest night time around -20 C [roughly].
        And since average temperature of US is about 12 K cooler than our tropic. And area in same geographical location on Mars may likewise around -12 C average temperature.
        Or Mars global average temperature might be around -10 to 15
        C. Or it could colder, but I am counting on Higher CO2 levels.
        Or such a Mars would make Earth coldest glacier period seem quite warm. And for human beings, it would feel or for practical purpose of human, be colder than present Mars. Despite increasing mars average global average temperature by 40 to 50 C.

        • gbaikie says:

          Continuing:
          –Yes, Mars would then show the same effect as seen on the Earth but diminished due to Mars’ greater semi-major axis and less solar influx. It is the mass, sorry, not the quantity of co2 per unit column. So how does that new realization play into the current ‘discussion’ about Earth and co2?

          (I used http://nssdc.gsfc.nasa.gov/planetary/factsheet/marsfact.html for the 210.1 K blackbody temperature and the 210K mean surface temperature.)–

          Adding more atmosphere would lower Mars Blackbody temp, as with:
          Moon vs Earth
          Black-body temperature (K) 270.7 254.3
          http://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html

          Or adding significantly more atmosphere would always increase the Bond albedo. The moon does have darker surface, but if put atmosphere on the Moon, the Moon would become much brighter in our sky.

          In above I didn’t add a global ocean for Mars. And probably make Mars tropics warmer by add bodies of water.

          I skip adding earth like atmosphere on Mars and instead just add a tropical ocean- that would significantly increase Mars average temperature. And just adding ocean [100 meter deep on average] would also increase Mars average temperature by 40 to 50 C. And it would be cold for humans- though they need spacesuit/pressure suit to wander about the surface. But they can swim under water without pressure suits. So they have house with white picket fence 10 meter below the water. Or other than being underwater it could be similar to Earth.
          And with tropical ocean on Mars, Mars could appear from space to be wearing a wide black belt.

          • gbaikie says:

            –And it would be cold for humans- though they need spacesuit/pressure suit to wander about the surface.–
            Correction:
            And it would not feel or for human purpose be cold for humans- though they need spacesuit/pressure suit to wander about the surface.

            What I mean for practical purposes. Is Mars unchanged, would lower heating bills for homes as compared to US or Europe.
            Also in spacesuit on Mars, you don’t have problem with being cold. Spacesuits would instead need refrigeration to avoid getting to warm. On moon with Apollo evaporated ice to keep the astronauts cool. So if in Mars night in spacesuit, one would need less cooling- though warmed boots might be useful.

            And if add atmosphere, it’s going to cost more energy to heat homes and you could easily freeze to death in a spacesuit.

    • wayne says:

      My, quite honestly I didn’t mean for that simple thought to evidently shake you so. Very sorry, carry on.

      • gbaikie says:

        🙂

        I not really much of a Mars fan.
        I think the Moon is more important.
        And if to up to me, I might prefer Mercury or Venus over Mars.
        But I do think NASA should explore the Moon [to find if and where there could be commercially minable water] and then explore Mars to see if human settlements are possible.
        NASA should also focus on finding abundant underground liquid water on Mars as that would be important discovery related potential future Mars settlements.

        But for me main purpose of Moon and Mars, is actually Earth. Need to mine the moon to get to the point of having commercially viable SPS [space power satellites] so Earthlings can get cheap and abundant electrical power, globally. It can not be done within 50 years, but if we do lunar exploration then could lead it to be possible in less than 100 years.
        Which is related of course to the whole global warming issue and fossil fuel use, and perceive shortage within 50 to 100 years.
        Or one could say it’s a pathway to US not needing it’s vast amount of coal reserves. And one bet that if there is a global shortage of fossil fuel, US will change it’s mind about using this coal.

        But I do tend to go on and on.