Help! Back Radiation has Invaded my Backyard!

August 6th, 2010 by Roy W. Spencer, Ph. D.

Measuring The (Nonexistent) Greenhouse Effect in My Backyard with a Handheld IR Thermometer and The Box

Laypersons are no doubt confused by all of our recent esoteric discussions regarding radiative transfer, and whether global warming is even possible from a theoretical standpoint.

So, let’s take a break and return to the real world, and the experiments you can do yourself to see evidence of the “greenhouse effect”.


One of the claims of greenhouse and global warming theory that many people find hard to grasp is that there is a large flow of infrared radiation downward from the sky which keeps the surface warmer than it would otherwise be.

Particularly difficult to grasp is the concept of adding a greenhouse gas to a COLD atmosphere, and that causing a temperature increase at the surface of the Earth, which is already WARM. This, of course, is what is expected to happen from adding more carbon dioixde to the atmosphere: “global warming”.

Well, it is one of the marvels of our electronic age that you can buy a very sensitive handheld IR thermometer for only $50 and observe the effect for yourself.

These devices use a thermopile, which is an electronic component that measures a voltage which is proportional to the temperature difference across the thermopile.

If you point the device at something hot, the higher-intensity IR radiation heats up the hot-viewing side of the thermopile, and the IR thermometer displays the temperature it is radiating at (assuming some emissivity…my inexpensive unit is fixed at e=0.95).

If you instead point it at the cold sky, the sky-viewing side of the thermopile loses IR radiation, cooling it to a lower temperature than the inside of the thermopile.

For instance, last night I drove around pointing this thing straight up though my sunroof at a cloud-free sky. I live in hilly territory, the ambient air temperature was about 81 F, and at my house (an elevation of 1,000 feet), I was reading about 34 deg. F for an effective sky temperature.

If the device was perfectly calibrated, and there was NO greenhouse effect, it would measure an effective sky temperature near absolute zero (-460 deg. F) rather than +34 deg. F, and nighttime cooling of the surface would have been so strong that everything would be frozen by morning. Not very likely in Alabama in August.

What was amazing was that driving down in elevation from my house caused the sky temperature reading to increase by about 3 deg. F for a 300 foot drop in elevation. My car thermometer was showing virtually no change. This pattern was repeated as I went up and down hills.

The IR thermometer was measuring different strengths of the greenhouse effect, by definition the warming of a surface by downward IR emission by greenhouse gases in the sky. This reduces the rate of cooling of the Earth’s surface (and lower atmosphere) to space, and makes the surface warmer than it otherwise would be.

If you have a day where there are patches of blue and clouds, you can point the thermometer at the clouds and pick up a warmer reading than the surrounding blue sky.

I did it this morning (see photo, above). When I moved from a view of the blue sky to the patch of clouds, the sky-viewing side of the thermopile became warmer…even though the thermopile is already at a higher temperature than the sky. The display would read a few degrees warmer than the reading looking at blue sky.

If you perform this experiment yourself, you need to be careful about the elevation angle above the horizon you are pointing being about the same. Even in a clear sky, as you move from the zenith (overhead), down toward the horizon the path length of sky the IR thermometer sees increases, and so you measure radiation from lower altitudes, which are warmer. This makes the effective sky temperature goes up. (This is ALSO evidence of the greenhouse effect, since looking at the sky above the horizon is like adding greenhouse gases to the atmosphere overhead. The (apparent) concentration of greenhouse gases in the lower atmosphere goes up, and so does the intensity of the back radiation.)

Even earlier in the morning, about 5:30, the middle-level clouds were thicker, and I measured a sky temperature in the 50′s F. We will see more evidence of that using air temperatures, below.

This shows that the addition of an IR absorber/emitter, even at a cold temperature (the middle level clouds were probably somewhere around 30 deg. F), causes a warm object (the thermopile) to warm even more! This is the effect that some people claim is impossible.

Remember, the IR thermometer calibrated temperature output is based upon real temperatures, the temperatures on either side of the thermopile.

And if you think this is just an effect of some sunlight reflecting off the cloud….read on.

Evidence from The Box

I have been seeing the same effect in “The Box”, which is my attempt to use the greenhouse effect to warm and cool a thin aluminum plate coated with high-emissivity paint, that is heavily insulated from its surroundings in order to isolate just the radiative transfers of energy between the sky and the plate. This can be considered a clumsy, inefficient version of the IR thermometer. But now, *I* am making actual temperature measurements.

The following plot (click on it for the full-size version) shows data from the last 2 days, up through this morning’s events. The plate gets colder at night than the ambient temperature because it “sees” the cold sky, and is insulated from heat flow from the surrounding air and ground.

In the lower right, I have also circled where thin middle-level clouds came over, emitting more IR radiation downward than the clear sky, and causing a warming of the plate. Since the plate is mostly isolated from heat exchanges with the surrounding air and warm ground, it responds faster than the ambient air temperature to the intensity of “back radiation” downwelling from the sky.

When I woke this morning before sunrise, around 5:30, I saw these mid-level clouds (I used to be a certified aviation weather observer), I measured about 50 deg. F from the handheld IR thermometer.

This supports what people already experience…cloudy nights are, on average, warmer than clear nights. The main reason is that clouds emit more IR downward, change the (im)balance between upwelling and downwelling IR, and if you change the balance between energy flows in and out of an object, its temperature will change. Conservation of Energy, they call it.

(WARNING: a technical detail about the above measurements and their importance to greenhouse theory follows.)
What this Means for the Miskolczi “Aa=Ed” Controversy

Except for relatively rare special cases, the total amount of IR energy downwelling from the sky (Ed) will ALWAYS remain less than the amount upwelling from below and absorbed by the sky (Aa). As long as (1) the atmosphere has some transparency to IR radiation (which it does), and (2) the atmosphere is colder than the surface (which it is), then Ed will be less than Aa…even though they are usually close to one another, since temperatures are always adjusting to minimize IR flux divergences and convergences.

But it is those small differences that continuously “drive” the greenhouse effect.

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146 Responses to “Help! Back Radiation has Invaded my Backyard!”

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  1. bush says:

    I tried this myself years ago and found that on a clear night under high pressure the reading would get low enough to register an error on the instrument. No water vapour little back radiation.

  2. Roy, perhaps you’re omitting something:if the near-surface temperature is 81F and the sky is 34F, which way is heat flowing?

    • If you close the door on a hot oven, which way is the heat flowing? Does that direction determine whether the inside of the oven will be hotter after you close the door?

      • Anonymous says:

        Roy, I’m baffled by your evasiveness. The atmosphere doesn’t act like the closed system of an oven. It is an open system whereby energy is freely transported back to outer space. If it were not then we’d have had irreversible runaway global warming billions of years ago when CO2 levels were 20x times higher, wouldn’t we?

        An increasing number of scientists are now seeing CO2 not as a ‘heat trapping’ gas. To them its nonsensical how a gas can be claimed to ‘trap’ energy selectively only on the outbound journey away from Earth’s surface when our atmosphere is clearly not showing any signal of ‘trapped’ energy inbound at any level in the atmosphere.

        Atmospheric gases indiscriminately absorb then emit IR. But then this whole ‘greenhouse gas’ hypothesis built up to be a climate ‘theory’ premised around IR is a total red herring anyway. It explains almost nothing about our climate being that IR is such a small player in atmospheric heat energy transport. Even the IPPC concede their level of ignorance to be high (see ‘Radiative Forcing agent, Level of Scientific Understanding,’ IPCC TAR WGI SPM (figure 3). John McLean has done an excellent study ‘Climate Science Corrupted’- it can be read in full here:
        http://scienceandpublicpolicy.org/images/stories/papers/originals/climate_science_corrupted.pdf

        Quite probably, in a year or so from now the GHG term will become as redundant as cold fusion and phlogiston and we’ll more likely be referring instead to ‘gases of infra-red emission and absorption’ (or GIREA). I’m just sorry that contrary to the results of your own experiment you seem to cling onto an outmoded failed hypothesis.

        • Yes, I suppose I’m just a sucker for quaint old wives tales. :)

        • Anonymous says:

          Anonymous
          Dr. Spencer’s analogy of an oven is perfectly acceptable. As far as the oven “knows” the whole of the rest of the atmosphere is effectively an infinite heat sink. (The heat capacity of the atmosphere is much greater than the heat capacity of the oven void space).

          The oven radiates its heat to this atmospheric heat sink. Dr. Spencer’s point is that the temperature of the oven will decrease at some rate once the electricity to the oven is turned off until it equilibrates with atmospheric temperature. It is the rate which varies whether the door is opne or closed becasue the door absorbs some the the radiation back into the oven void when the door is closed.

          This does not contradict any thermodynamic principle. With the power turned off to the oven, the oven will eventually reach equilibrium with the atmosphere whether the door is opne or closed. Only the time it takes to do so will vary with door position.

          Now in the situation with the oven power still on, the interior of the oven remains hotter than it otherwise would be with the door open because the door is re-radiating heat (IR) back into the oven void. The final oven temperature is a reflection of the ongoing dynamic equilibrium of heat quanta and oven power setting. Therefore, the cooler oven door is increasing the hotter oven void temperature – which is where all these discussions started, I believe.

          Note in these instances it’s important to realise that this particular oven isn’t controlled with a thermostat, just the “on” switch and heat energy balances.

          Your results are very interesting Dr. Spencer. I’m going to set my own experiment up in my garden too.

          • Anonymous says:

            .

          • Anonymous says:

            Sorry – last post by timheyes. I managed not to identify myself accidentally

          • Anonymous says:

            “It is the rate which varies whether the door is opne or closed becasue the door absorbs some the the radiation back into the oven void when the door is closed.”

            Sorry – editing for clarity; the above sentence should read:

            It is the rate which varies whether the door is open or closed because the door absorbs some the the radiation and re-radiates it back into the oven void when the door is closed.

        • Anonymous says:

          1) Dr. Spencer was not trying to be evasive, but the post just like yours was off point.
          2) Let’s stipulate 20X CO2 billions of years ago. Why would the effect be “runaway”. Would benice if you could remain factual. The proverbial “Disagree but do not…”
          3) If an increasing number of scientists are doing whatever an increasing number of scientists do, then the AGW folks are OK to talk about consensus. Again keep it factual.
          4) Are you anthropomorphizing atmospheric gases? I never knew a gas do things indiscriminately. Each gas absorbs radiation at various frequencies, depending on it possible excited state(s).
          5) Your reference is to a paper full of leaps of faith and innuendo. If an AGW believer did this, you might be yelping. Again keep it factual.
          6) Try not to leave your daytime job, because your prognostication about GHG seems wrong to me. But it does not matter – I said and you said –who cares. Again keep it factual.

      • Anonymous says:

        You prevented convective cooling just like in a real greenhouse. Not much to do with radiation. Just an analogy I suppose.

        • NO. Do not confuse the two. A real greenhouse uses solar, and its glass windows are opaque to IR transfer. I’m talking about nighttime radiative cooling through an IR-transparent aperture.

      • Anonymous says:

        “If you close the door on a hot oven, which way is the heat flowing? Does that direction determine whether the inside of the oven will be hotter after you close the door?”

        I find a problem with the analogy – if the oven represents the earth, then the heating element would have to be the sun – the oven (earth) can only heat up as long as the heating element (sun) is on – on the side of the planet experiencing day. The oven (earth) only cools when the heating element (sun) is off. Ergo the only time the heat is flowing outward is on the side of the planet experiencing night….

        • Anonymous says:

          What I’m trying to say that if the comparison is accurate, then the only place one could measure back radiation is on the dark side:)

  3. Dear Dr. Roy

    I am not a scientist, let alone a climatologist but I study climate or climate change for a hobby and have done so for the last five or six years. And by now I possibly need some help as I am, so to speak, up against it.
    So yes I am a lay person.

    Your thermopile thing seems to me to be measuring temperature -I did not realise that IR radiation = temperature or warmth.-
    Here I am going around thinking that IR radiation was part of the Electromagnetic spectrum equalling energy travelling at or near the speed of light. So how can, I thought, 15° C (average surface temp.) have enough energy to transform that heat back into IR and send it back to where it came from. Distance of IR radiation travel, from the surface to the sky, back again and then all the way out to space seems to be no problem. Yet gamma radiation from the Hiroshima bomb did not reach all of Japan nor did it catch up with the plane that dropped the device.
    What I have so far been thinking is as follows:

    Every square inch of the top of Earth’s surface is in contact with the air, or atmosphere. So why do I get the impression that scientists ignore conduction and the consequent convection of heat from the surface to the atmosphere, all in favour of radiation?

    Conduction between Earth’s surface and the air must surely be happening.

    Adiabatic cooling and warming rates are also never or hardly ever mentioned.
    Even if the heat retention afforded by water vapour (WV) can, even if mistakenly, be named as a Greenhouse Effect it can only have a local, temporary and very variable and therefore a very difficult to trace influence.

    The changing composition of gases (the rise of CO2) in dry or unsaturated air has so far had no influence on the dry air adiabatic lapse rate. It has stayed at 9.8° C /Km. ever since it was first established. (Roughly speaking 3° C pr 1000 feet)

    However when WV is added and the air gets moist the lapse rate becomes one that depends on how saturated the air is. Averages are said to be 6.5° C pr. Kilometre or roughly 2° C pr. 1000 feet.

    While mentioning WV it is worth mentioning clouds too. Dr. Roy, you are one of the few who seems to take clouds into consideration.

    Overall do they cool or do they warm the “close to surface air” (lower atmosphere)?

    I’ll try to answer that one myself and you correct me if I am wrong.

    Well, if and when solar irradiation is a constant then it can only be clouds that determine the force or strength of the sunrays reaching the Earth’s surface and the Earth’s total cloud-cover has to be known at all times in order to work all that out. On a cloudy night there must be more WV in the air generally than there is on a clear night and the lapse rate is slower

    So far, as far as I can see, your box of tricks has only shown that back radiation keeps the air in the cavity cooler than the air outside of it. What does that prove?

    Your box stops convection to the outside air and to a great degree it stops conduction too. – And the temperature inside the box, when and where the sun doesn’t shine, goes down to below that of the outside air temperature. That should tell us all there is more to a warm night air temperature than back radiation. Of course moisture evaporation – dew etc. present in the outside air do not enter the cavity. Conductive heatloss from the surface is also reduced.

    Dr. Spencer, you may be trying to explain “The Greenhouse Effect” But I have so far not found any evidence of any such an effect and am hard to convince. If there is one, then there is a very good theory as to why the atmosphere, a few miles above the Equator would warm first, and more, than the air just above the surface, I am sure you know the theory (as explained by Prof. Lindzen in Durkin’s swindle film) so I will not elaborate.
    No evidence of “The Hotspot” was ever found for the 0.6 – 0.8 ° C hike in global average temperature that took place during the last century and was said to be man made.
    That must also mean there is no greenhouse warming of +33° C, said to be caused by the “Natural Greenhouse Effect”. – As there is still no HOTSPOT.

    Furthermore if there is IR radiation from the surface to and from “Greenhouse Gases” other than what is being reflected then what is the resulting “Cooling Rate” for radiation?

    Such a cooling rate must be different at all locations and altitudes as radiation is said to be in close relation to temperature If radiation increases exponentially to the power of 4 as temperature increases then the opposite must be happening when temperature decreases.

    I have not found anything that tells me the Troposphere’s temperature is governed by anything other than the energy input: Solar energy converted to heat, geothermal warmth (El Niño etc.) and the air pressure.
    It is colder on top of Mt. Kilimanjaro than in most other places at similar latitudes.

    If it were a fact that winds blow vertically then I doubt your “Greenhouse Effect” would be able to keep us very warm, but they do not. Winds or air movements are mainly horizontal. Therefore air is kept near to the surface at maximum pressure for longer thus keeping the heat in.

    The last thing I am not able to find out of is: If only “Greenhouse Gases” (GHGs) are able to absorb IR radiation emitted by the surface, conduction does not take place, and those GHGs at most make up just a small proportion the Atmosphere, how do the other 95% of gases warm up?

    • Your thermopile thing seems to me to be measuring temperature -I did not realise that IR radiation = temperature or warmth.-
      Here I am going around thinking that IR radiation was part of the Electromagnetic spectrum equalling energy travelling at or near the speed of light.

      The IR thermometer measures the effect that IR radiative transfer has on temperature. When you feel radiant heat coming off a brick wall in the sun, that’s IR, and it DOES impact temperatures, and it IS emitted by everything that has temperature. The global atmosphere receives IR from the global expanse of land and ocean below it. At the land and ocean below receive somewhat less IR from the atmosphere above it.

      Every square inch of the top of Earth’s surface is in contact with the air, or atmosphere. So why do I get the impression that scientists ignore conduction and the consequent convection of heat from the surface to the atmosphere, all in favour of radiation? Conduction between Earth’s surface and the air must surely be happening.

      It is not ignored…in fact, convection transports approximately 4 x as much heat from the surface to the atmosphere as IR radiation does.

      Adiabatic cooling and warming rates are also never or hardly ever mentioned.

      If an air parcel somewhere rises and cools, an equal amount of air somewhere else has to sink. It’s called conservation of mass, or mass continuity. IR radiation causes excess heating at low altitudes, making air layers super-adiabatic, and evaporating water, so that sensible and latent heat CAN be transported aloft. This is all well understood.

      Even if the heat retention afforded by water vapour (WV) can, even if mistakenly, be named as a Greenhouse Effect it can only have a local, temporary and very variable and therefore a very difficult to trace influence.

      Except that it is a “local effect” everywhere on Earth, because atmospheric water vapor exists throughout the atmosphere.

      The changing composition of gases (the rise of CO2) in dry or unsaturated air has so far had no influence on the dry air adiabatic lapse rate. It has stayed at 9.8° C /Km. ever since it was first established. (Roughly speaking 3° C pr 1000 feet). However when WV is added and the air gets moist the lapse rate becomes one that depends on how saturated the air is. Averages are said to be 6.5° C pr. Kilometre or roughly 2° C pr. 1000 feet.

      No one claims that the lapse rate in the troposphere will change significantly. You can warm (or cool) the entire troposphere without changing the lapse rate. Lapse rate tells you nothing about what the temperature of the troposphere is…only how it changes with height.

      While mentioning WV it is worth mentioning clouds too. Dr. Roy, you are one of the few who seems to take clouds into consideration. Overall do they cool or do they warm the “close to surface air” (lower atmosphere)?

      From a radiative standpoint, clouds have a net cooling effect on Earth’s climate. That’s one reason why I think feedbacks in the climate system are negative, and cloud are reducing the (small) warming effect of increasing CO2.

      I’ll try to answer that one myself and you correct me if I am wrong. Well, if and when solar irradiation is a constant then it can only be clouds that determine the force or strength of the sunrays reaching the Earth’s surface and the Earth’s total cloud-cover has to be known at all times in order to work all that out. On a cloudy night there must be more WV in the air generally than there is on a clear night and the lapse rate is slower. So far, as far as I can see, your box of tricks has only shown that back radiation keeps the air in the cavity cooler than the air outside of it. What does that prove?

      Your box stops convection to the outside air and to a great degree it stops conduction too. – And the temperature inside the box, when and where the sun doesn’t shine, goes down to below that of the outside air temperature. That should tell us all there is more to a warm night air temperature than back radiation. Of course moisture evaporation – dew etc. present in the outside air do not enter the cavity. Conductive heatloss from the surface is also reduced.

      The box is meant to reduce all other heat flows except the up-and-down radiative heat flows. You can not explain the large temperature rise I showed in the box when clouds went over this morning WITHOUT infrared radiation coming down from the sky with greater intensity than before of after the clouds. There is no other explanation.

      No evidence of “The Hotspot” was ever found for the 0.6 – 0.8 ° C hike in global average temperature that took place during the last century and was said to be man made. That must also mean there is no greenhouse warming of +33° C, said to be caused by the “Natural Greenhouse Effect”. – As there is still no HOTSPOT.

      The lack of a hotspot has nothing to do with the existence (or non-existence) of the greenhouse effect. It has to do with how we expect warming to occur as a function of height, especially in the tropics (average lapse rates are different in different parts of the world). If real, the lack of a hotspot says there is something wrong in the climate models, which I agree with.

      Furthermore if there is IR radiation from the surface to and from “Greenhouse Gases” other than what is being reflected then what is the resulting “Cooling Rate” for radiation?

      The average cooling rate of the atmosphere in the IR has been estimated to be 2 deg. C per day.

      Such a cooling rate must be different at all locations and altitudes as radiation is said to be in close relation to temperature If radiation increases exponentially to the power of 4 as temperature increases then the opposite must be happening when temperature decreases.

      That’s true.

      The last thing I am not able to find out of is: If only “Greenhouse Gases” (GHGs) are able to absorb IR radiation emitted by the surface, conduction does not take place, and those GHGs at most make up just a small proportion the Atmosphere, how do the other 95% of gases warm up?

      It’s because every molecule of air is colliding with gazillions of other molecules every second. Temperature is a measure of the average velocity of those molecules. If one absorbs extra energy, it very quickly share it with neighbors. These radiatively active molecules are diffused through the entire atmosphere. Look up “Kinetic Theory of Gases”

  4. Pierre Allemand says:

    Well, let me explain what I have just measured :
    I am a consultant in industrial insulating. I own a small IR thermometer (a $50 yellow type like the one you describe) and also a black professional one, sold in a nice black box.
    I have just pointed the yellow thermometer toward the grey sky : found -2°C.
    With the black one pointed toward the same direction : found -20 °C. (It is now just 9:00 pm)
    I am not able to bring any explanation to the difference, apart perhaps differences in components. But, it is the first time I see such a difference between the two readings.
    So, my (provisional) conclusion will be : beware of what you measure, and what you use for it : results may be completely different, and so your conclusions…

    • Pierre:
      It might be the IR frequency response of the two sensors. They usually have a filter to restrict the range of wavelengths they are sensitive to, and the cold one probably more tuned to window frequencies.

    • Anonymous says:

      Hi Pierre,
      have both the thermometer the same spectral bandwidh?
      I own two too, one (the older) has a 7 to 18um spectral response, while the other has just 8 to 14um.
      The thing I can’t explain indeed is that the wider one has the norrower temperature range… I’m very confused on how they work.
      A long time ago, I experienced your issue too. One other culprit could be the different aperture angle of the devices. The wider gets the radiation of a greater area of the sky, even if you aimed the same place in the sky maybe it “seen” areas with different WV concentrations.

      Massimo

      • Actually that makes sense — wider wavelength sensitivity would probably have narrower temperature range — because it includes more water vapor absorption, which will lead to warmer sky temperatures.

        • Anonymous says:

          Dear Dr. Spencer,
          “Actually that makes sense — wider wavelength sensitivity would probably have narrower temperature range — because it includes more water vapor absorption, which will lead to warmer sky temperatures.”
          Yes, you are right. I went to sleep after I wrote that post yesterday night. When at bed, I soon realized that I wrote a silly argument.
          Those thermometers are made to get the temperatures of solid targets, so the design is surely developed to receive the radiation from a Planck’s blackbody not from a more complex gas’ emission spectrum. In theory, they just have to get a narrow sample of the radiation in the middle of the blackbody peak spectrum and correct that value using a temperature based lookup table.
          Good anyways, I learnt one more thing. :)

          Massimo

  5. Dear Dr. Roy

    Thank you very much for your very comprehensive answers to my questions I shall print them all off and learn it well.
    And then, of course I shall have to supplement it with a book about some kind of blunder or something.

  6. kuhnkat says:

    Dr. Spencer,

    I keep asking this question and have not received any answer elsewhere. Maybe you can enlighten me.

    Blackbody radiation indicates a particular temperature of the body emitting it.

    GHG’s emit energy based on the molecular vibrations, torsions etc. and the radiation is limited to narrow bandwidths due to this.

    If my IR thermometer is receiving the molecular bond energy, does it give the correct temperature?? That is, if the frequency of the IR being emitted is based on the properties of the molecular bond why would it be related to temperature in the same way that Blackbody radiation is???

    • Anonymous says:

      No, the temperature of gases measured with an IR thermometer is just a non sense.
      What we are doing here is just an indicative measurement of the back radiation.
      I’m not sure about the clouds temperature, maybe that the IR thermometers could give a good aproximate measurement of their temps, since they are condensed (liquid) particles of water indeed.

      Massimo

  7. kuhnkat says:

    I have one more question.

    The CO2 and water vapor are denser the lower in the atmosphere you go. Just above your head is much more dense than a kilometer up. The temperature of the GHG’s and air just above your head is close to your ambient.

    Why does the IR thermometer show low temps?? If, as you claim, there is backradiation from the CO2 just above your head why is the thermometer not reading this stronger, warmer, signal rather than a COLD signal from kilometers up???

    You are asking me to believe that you are reading temperature of the atmosphere a kilometer or more away from you when the strong signal in your face should be overwhelming any cold signal from up in the troposphere!!!

    WUWT????

    • Anonymous says:

      “Why does the IR thermometer show low temps?? If, as you claim, there is backradiation from the CO2 just above your head why is the thermometer not reading this stronger, warmer, signal rather than a COLD signal from kilometers up???”
      First of all, read my reply to your previous post.

      The IR thermometers read a “cold” temperature because the back radiation is much less than the radiation which had generated it from the ground.
      You can use the following simulator to get an idea of what the IR thermometer shows:

      http://geodoc.uchicago.edu/Projects/modtran.orig.html

      If you set the locality “1976 U.S. standard atmosphere” and go to the sensor altitude of 0km. Looking down you get about 360W/m^2 @ 288K (14.8°C/58.64°F) which is the ground blackbody. The back radiation in clear sky it’s about 259W/m^2(you get it looking up at 0km).
      Then, since the IR thermometer handle the back radiation as it was sourced by a blackbody, you can go to the emulator looking down at 0km and adjust the “Ground T offset” parameter to get the same 259W/m^2 radiation. After a lot of iteration you get the temperature of 263.9K (about -9°C/15.8°F), which it should be what your IR thermometer shows.
      That is no ways the temperature of the sky, but the temperature of the blackbody which has the equivalent radiation of the whole gases in the sky seen by the aperture of the IR thermometer.

      Massimo

      • Anonymous says:

        How does a simple little instrument discriminate between all these signals. Aren’t we told that all photons have the same amount of energy? Isn’t this instrument a “counter” of energy in a range of frequencies?

        Obviously I am ignorant of how the imstrument works.

        You are only talking about infrared from GHG’s reading cold. What about the actual blackbody from the atmospheric components that are NEXT to the instrument??

        Again, I do not understand how the instrument works. If you are getting local blackbody signal of warm, and local and distant signal of cold, why wouldn’t the local signal be the strongest and be what the instrument reads?

        If the instrument determines temp based on the frequencies being received based on Black Body, then won’t the molecular bond energy from GHG’s, which is emitted based on the bond state and not temperature of the molecule, corrupt this black body reading of temperature?

        • Anonymous says:

          “How does a simple little instrument discriminate between all these signals”

          As already said, it doesn’t.
          Just forget it’s a thermometer for this purpose, and immagine it’s just a collector of the IR radiation from different altitudes and places.
          In our case, the fact that it reports a temperature doesn’t mean that it’s the temperature of anything.
          That temperature measured is proportional to the total power flux of the back radiation seen by the thermometer lens, just that.

          “You are only talking about infrared from GHG’s reading cold. What about the actual blackbody from the atmospheric components that are NEXT to the instrument??
          Again, I do not understand how the instrument works. If you are getting local blackbody signal of warm, and local and distant signal of cold, why wouldn’t the local signal be the strongest and be what the instrument reads?”

          It’s a question of IR active gases density. When the thermometer points the sky it receives just few IR photons at the high temperature close to your head because of the very low concentration of the IR active gases (if you want call them GHGs). Most of the photons received by the thermometer come from the upper layers which are at lower temperatures. You can consider the showed “temperature” as the average of the temperatures of all the atmosphere layers (it’s not true of course, because the received radiation is the average of the power flux which is a function of the 4th power of the temperature indeed). If you think this way, you realize that the warm temperature close to the ground has a very little influence on the total amount of radiation received from the whole atmosphere column.

          Massimo

          • I prefer to think of it’s calibrated measurement as a weighted temperature of everything in its field of view…but those weights are so dominated by specific objects that in many circumstances it is, indeed, close to the real temperature of the object it is pointed at.

        • It sounds like you do not understand that greenhouse gases in the atmosphere still leave the atmosphere partly transparent. The IR thermometer DOES see the atmosphere immediately in front of it, as well as most of the rest of the atmosphere along its line of sight. But depending upon the frequency response of the instrument, and how much humidity is in the atmosphere an at what altitude, the relative contributions of those layers change. The final calibrated brightness temperature can be roughly considered to be the weighted average temperature of all of those layers.

          • Anonymous says:

            Dear Dr.Spencer,
            Excuse me, maybe due to my bad English, but I’m not sure. What did you mean with “I prefer to think of it’s calibrated measurement as a weighted temperature of everything in its field of view…but those weights are so dominated by specific objects that in many circumstances it is, indeed, close to the real temperature of the object it is pointed at.”?
            Are you supposing that the IR thermometers give the average of the objects’ temperature in its field of view or what else?
            If that is what you mean, then I agree of course.
            But if we are arguing about the sky temperature I’m not sure about the real meaning of the temperature read by the IR thermometers.

            Massimo

      • Massimo,
        I tried your suggestion of using the Modtran model to calculate what an IR pyrometer should see and get a much different result:

        Model Input: CO2 (ppm> 375
        CH4 (ppm) 1.7
        Trop. Ozone (ppb) 28
        Strat. Ozone scale 1
        Ground T offset, C -24.2
        hold water vapor 1
        Water Vapor Scale 1976 US Std Atm
        Locality no clouds or rain

        Sensor Altitude km 0
        Looking down

        I get -24.2 C for “what the pyrometer should see” to match 259 W/m2, 264 K instead of your -9 C. Why?

        • Anonymous says:

          Hi Hockey Schtick,
          sorry, I understand that I could have confused you using K and °C together, but that’s what Dr.Archer’s Modtran on-line simulator does indeed.

          Of course you read 264K, which is the same of my -9°C.

          (0°C = 273.15K so -9°C = 264.15K)

          Have fun.

          Massimo

          • Massimo,
            The Modtran model indicates “back radiation” is 259 W/m2, but Kiehl-Trenberth say it is 324 W/m2. Why such a large difference?

          • Anonymous says:

            Hi Hockey Schtick,
            <>
            AFIK that Modtran application simulates the radiation flux for the condictions you set on its left panel. I don’t know which atmospheric condictions K-T used for their paper. I never read that paper indeed.
            Anyways, as first point K-T paper was issued in 1997, and maybe they used a “Tropical Atmosphere” instead of the “1976 U.S. standard” one. Muna Loa 1997 CO2 concentrations shows 360-365ppm, but the paper maybe refers to some years before. Using this new setup the “back radiation” computed by modtran rises to 347.912W/m^2.

            Secondly, K-T could have used a global average atmospheric condictions.

            I’m not a scientist, I’m just an electronic engineer and to be honest I’ve many doubts about the Modtran reliabilty for the simulation of the radiative budget.
            For example. Dr. Archer’s Modtran application give us the “Intensity” as W/(m^2 * cm-1), but if you look inside its “Whole output file” the used units for radiance is w/(m^2 * cm-1 * sr) (where sr is a steradian which is about 1/12th of the whole sphere surface). This could be good for the incoming radiation from the Sun which is made of almost parallel rays, but what can we tell about the outgoing radiation?
            Dr. Archer’s Modtran simulates only the outgoing radiation on the zenith, (which is what the radiometers on the satellites do indeed). This could be good enough to the evaluate the radiative budget of a planet without IR active gases (such the supposed GHGs are). But the outoging radiation of an atmosphere made up also of IR active gases should exit the atmophere with many different angles because of the scattering. Thus, in this case the whole outgoing radiation is not accounted for the budget.
            But as already said, I never read the K-T paper indeed, maybe they measured that non-zenithal outgoing radiation too.

            Massimo

  8. Let me make it a little simpler than Massimo did…

    When looking up from the ground, the total amount of IR radiation being measured is from all different altitudes, at all different temperatures. This is because the atmosphere is anywhere from very opaque, to very transparent, depending on the IR wavelength in the band the IR thermometer measures.

    For an IR thermometer sensitive to wavelengths from, say, 8 to 14 microns, you could plot a weighting function profile that shows the proportions of IR energy being received from different altitudes. It could be that Modtran that Massimo mentioned allows you to do that…I don’t know.

    • Anonymous says:

      No, maybe I’ve bad explained my intention.
      The Dr.Archer’s Modtran application has not been thought to do what I suggested of course. I just wanted to show how very different is a 259W/m^2 blackbody shape from the equivalent simulated back radiation spectrum.
      The first it is the effective radiation of a solid which is proprotional to its body temperature, while the second is a spectrum with the same integral of the blackbody but which is the result of the scattered IR from different places and altitudes. Looking to that spectrum it’s obvious that in no ways it is the temperature of anything.
      Anyways, the IR thermometers give the temperature of that equivalent blackbody of course.

      Massimo

  9. Bryan says:

    Roy when I read your blog I almost agree with you but there is a gray area where I’m not sure.
    The term the “backradiation warms the atmosphere” can have two meanings.
    The insulation meaning I agree with(normal blanket), the electric blanket meaning I don’t.
    Another thought experiment should clarify whether we are both on the same hymn sheet.
    At night well away from sundown and dawn we have a magic switch that can make the atmosphere disappear and appear.
    For eight consecutive quarter hours we switch on and off the atmosphere.
    We then draw an Earth surface temperature against time graph for the two hours.
    The graph would show (I think) in the periods where the atmosphere is off, a fall of temperature against time.
    When the atmosphere is on a smaller drop of temperature.
    At no time, would there be an increase in the Earth surface temperature.

    • Excellent question and example, Bryan. Everything you say is absolutely true.

      But when those who claim that adding greenhouse gases cannot warm the atmosphere, they are missing the same part you left out: the sun is continuously adding energy. I have repeatedly said this is a requirement for an actual temperature RISE to happen. You have nailed it.

      Sometimes, though, I do leave that part out just to yank their chain….Because THEY left it out when they claimed it couldn’t happen. :)

  10. Kevin says:

    Dr. Spencer,

    Before you get too carried away with your thermopile you might want to investigate an optical phenomena called Narcissus. This is a common problem with Infrared imaging and radiometry systems (Yes Virginia it is possible to image infrared light, see the Spitzer Space Telescope). When extreme Narcissus occurs in an infrared optical system the system literally cannot see the forest since it is blinded by the infrared light from its own trees.

    More than likely your device has a focusing lens which allows it to select a small angular cone of the incoming infrared light. Without this lens the thermopile will most likely respond as a Lambertian receiver and accept light from a very broad cone. The problem is that lens is at “room” temperature and is emitting quite a bit of infrared light itself. Normally this is not a problem for these devices since they are intended to measure temperatures that are quite a bit higher than the temperature present at the device.

    In higher performance infrared imaging or radiometry systems lenses are replaced with mirrors to reduce the effects of Narcissus. To achieve extreme levels of performance the measuring instrument is placed into the cold vacuum of space, i.e. the future James Webb Space Telescope.

    I am not inferring that you are not measuring something; my concern is that you may actually be measuring the temperature of your instrument along with the backradiation. Further, without detailed knowledge of the construction of your instrument it may be impossible to properly determine what portion of the signal is backradiation.

    I still enjoy reading your posts, keep up the good work.

    Cheers, Kevin.

    • Anonymous says:

      Hi Kevin,

      “The problem is that lens is at “room” temperature and is emitting quite a bit of infrared light itself. Normally this is not a problem for these devices since they are intended to measure temperatures that are quite a bit higher than the temperature present at the device.”

      I evidenced this issue about the polyethylene film of the Dr.Spencer’s “BOX”.
      Just for your information. I opened my own IR thermometers and both have a thermocouple inside to compensate the effects of the lens self emission.

      About “Narcissus”, I guess you are talking about the subject of this nice paper from the University of Brandenburg Germany.

      http://www.fh-brandenburg.de/~piweb/mitarbeiter/papers/Inframation_Polarization.pdf

      Massimo

    • Kevin, the people who design these things are not that stupid. They account for the thermal emission by the lens.

      • Anonymous says:

        Roy, yes you can measure the temperature of the lens and use that information to more accurately calculate the temperature of the surface imaged by the instrument.

        However the presence of the warm lens limits the sensitivity of the instrument at lower temperatures. If this was not the case they could just put the James Webb Space Telescope in a low Earth orbit, measure the temperature of the mirrors and get the same sensitivity as they will get by placing it where they plan to.

        These low cost infrared imaging radiometers are intended and designed for measuring temperatures that are higher than room temperature. Yes, you can measure the temperature of gases with them, but this measurement is full of potential errors and should in no way be considered an absolute measurement of infrared radiance arriving at the surface of the Earth.

        I am not assigning stupidity to anyone, just pointing out that using an instrument to make a measurement that is not designed for is ripe for errors and misinterpretation.

        Cheers, Kevin.

        • well, mine was designed to read to -76 F. I suspect they have an algorithm to convert the small temperature drop at the detector to the huge temperature drop it is looking at. It works pretty well looking in my freezer, -5 deg. F

          • Anonymous says:

            Roy, fair enough, but please remember that the instrument you have is designed to INFER the temperature at a remote location by INTERPRETING some mostly understood rules of how infrared radiation behaves. It is not designed to (and MOST CERTAINLY DOES NOT) measure the infrared radiance (i.e. the backradiation) arriving on the surface of the Earth. Therefore the signal you receive PROBABLY varies from what you intended to measure.

            Again, I am not saying that you are not measuring something, just suggesting that you exercise caution when interpreting what you have measured.

            As an aside you might want to investigate just how difficult it is to make absolute radiation measurements when light (visible or infrared) is involved. You can in fact purchase a “standard” light bulb from NIST, and then, with lots of expensive spectroradiometery equipment establish an absolute measurement that approaches an uncertainty of plus or minus 2%. This takes extreme attention to detail including distances, reflectances of materials, etc. etc. Having participated in this exercise I will state that making absolute radiation measurements in the visible wavelengths is difficult at best. Making the same measurements with the same accuracy in the infrared wavelengths is damn near impossible. I would much rather establish a measurement of voltage or distance or any other number of measurements to accuracies of less than 1%.

            Cheers, Kevin.

          • Oh, I’m fully aware that the IR measurement of the sky is only partly related to the back radiation. For one thig, the handheld IR thermometers are only sensitivite to a restricted range of wavelengths that are chosen to be centered on the IR “window”, to actually minimize atmospheric absorption…which is what I am measuring. But your comments are true of any kind of remote sensing…it’s never *exactly* what you want to measure, so you calibrate it to be related — weakly or strongly — to what you want to measure.

            I would wager that the IR thermometer’s measurement of the sky is indeed proportional to the broadband IR effective temperature (brightness temperature) of the back radiation. It might have an offset of 30 or 40 or 50 deg., but it is still mostly measuring downwelling IR radiation from the sky.

          • Anonymous says:

            I fully agree.

            Massimo

          • Jan Pompe says:

            Hi Dr Roy,

            It might be worthwhile checking before making statements like this

            “For one thig, the handheld IR thermometers are only sensitivite to a restricted range of wavelengths that are chosen to be centered on the IR “window”, to actually minimize atmospheric absorption”

            I agree it would be nice and in an ideal world would be so, but the real world is rarely so obliging. It’s pretty unlikely that the cheap device is so finely “tuned” here is a thermopile I’ve used in my designs for some remote temperature sensing and am planning to use in a differential pyrgeometer.

            http://datasheet.octopart.com/ZTP-135S-GE-datasheet-5414032.pdf

            the transmission spectrum is on page 2. It is a very similar spectrum as pyrgeometers with silicon lens that are actually designed to measure downwelling IR.

            http://miskolczi.webs.com/AO_Vol32_No18_1993.pdf

            The problem for the engineer is finding a filter suitable and robust, against the elements, enough to measure a select band. For example see figures 3 & 4 for what UV does to polyethylene over time. We take what we can get then rely on our ability to calibrate it to give a measure of the parameter we want to measure.

          • I did check. The manufacturers almost always list the wavelength range as 8 to 14 microns. I realize this is not a perfect filter…they never are. But my understanding is that this has been one of the advances in recent years, the availability of thermopiles with some selectivity in the frequency range of interest.

          • I am only repeating what I have read. More recent thermopile designs are more selective to the 8-14 micron band, to avoid atmospheric attenuation. If you Google this range of wavelengths for IR thermometers you will find most of the designs now use try to restrict sensitivity mostly to this range.

          • Anonymous says:

            Pointing it in your freezer minimizes the molecular bond IR it receives. You get mostly black body at such a short range!!

            Pointing at the night sky you get a lot of molecular bond IR which only APPEARS to be a temperature due to its frequency.

        • Anonymous says:

          I partially agree with Kevin when he says that our IR thermometers have not sensitivity for low temperature, but I guess they should be good at the specified lower measurement limit (my own have -10°C/0F and -30°C/-22°F). At least for the limited accuracy they have at those limits (+/-3°C or +/-5.5°F) and keeping the thermometer at the temperature its lens is compensated (0-50°C the first and 23-28°C the second, note how the second having larger temperature range is more restrictive about its operating temperature range).

          Anyways, I would emphasize one other aspect about the temperature readings made by the IR thermometers.
          If the target has not an isothermal surface then the thermometer fails the measurement because it averages the incoming radiation power not the temperature.
          That’s was well explained in the user guide of those devices a long time ago. Now no one mention it, but I’m sure that the issue has not been solved.
          Probably, the reason this information is missed is because for little temperature differences the error is well under the instrument accuracy.
          But measuring the sky temperature I believe it’s another story.
          AFIK most of the backradiation should come from the troposphere and the stratosphere. The temperature there falls down almost linearily from 25°C/298K at ground to about -80°C/193K at abt 18km. Now we must decide which is best between averaging the temperatures or the radiation powers. Being a linear fall down the average temperature should be (193+298)/2=245.5K, while using the radiation we get ((193^4+298^4)/2)^-4=260K (which is what the IR thermometers should measure).
          Not really the same value.
          Of course, the above computation is oversimplified, there are many other things to consider such as the different air density and composition as function of altitude and displacement in the viweing angle of the instrument, but I guess they shouldn’t adjust that discrepancy.

          Massimo

  11. Bill Hunter says:

    its puzzling what caused the 3 degree warming of the plate then cooling just before sunrise. The difference between the plate temperature and outside temperature plus any reasonable number for the sky radiation doesn’t add up to sufficient radiation to override the radiative cooling.

  12. John Millett says:

    Dr Spencer,

    Couldn’t higher emissivity inside the box than outside account for the lower night-time temperature inside than out?

    Couldn’t reflected surface radiation from the mid-level cloud account for the temperatures inside and outside the box becoming equal temporarily?

    • liquid clouds only emit and absorb IR…very little reflection.

      If the emissivity in the box is higher, its absorptivity is correspondingly higher. This does not lead to a lower or higher temperature…it instead increases the rate at which the inside radiatively equilibrates to the outside. So, High emissivity is chosen for the material on the inside of the box to maximize this rate of equilibration between the plate’s emission upward, and the sky’s emission downward.

      • Anonymous says:

        So, why do I get sunburnt on my yacht though sheltering under a canopy?

        Thanks for the explanation re emissivity.

        You say: “The plate gets colder at night than the ambient temperature because it “sees” the cold sky, and is insulated from heat flow from the surrounding air and ground”.

        Given that the thermometer outside the box also “sees” the cold sky but, unlike the one in the box, is not insulated from convection effects, doesn’t it follow that it’s the insulation that makes the difference, not the cold sky?

        • yes, in a sense, the “insulation makes the difference”. But do you not understand why temperatures fall at night? For temperature to fall, there needs to be an energy sink, and I am trying to isolate the plate from other energy sources in the environment that slow the energy sink’s effect on temperature. The temperature doesn’t fall just because the sun isn’t shining.

  13. Bill Hunter says:

    It obviously overrode the cooling because the plate warmed up 3 degrees.

    It reacted like somebody opened the insulated box. Was that the source of energy inflow?

  14. Fulco Bohle says:

    Roy,

    I studied the text of Ference Miskolczi a year ago and although I stil have problems with it at some points, I think it needs te be rewritten with more details.

    You need to keep in mind that this new theory is about an athmosfere at equilibrium, which is on earth never the case.

    The fact that Aa is not equal to Ed means there is no equilibrium. At night the greenhouse effect wil force a delayed cooling and during the day a delayed warming.

    I think, than the greenhouse effect is about the decay time of vibrational modes and the mean time between collisions in a gas. A molecule of water or carbondioxide must transfere it’s vibrational energy to kinetic energy before it falls back by sending a IR photon (1 microsecond). This can only be done by colliding with an other molecule, which on earth is almost always N2. N2 as we know is not a greenhouse gas and therefore acts like a battery. At knight things work the other way around, N2 collides with greenhouse gas molecules and activates vibration modes which lose energy by sending a photon (IR). And of course things are a little more complicated.

  15. JAE says:

    Roy: IMHO, all you have shown so far is that objects (including air) radiate at an intensity that is related to their temperature. That has been well-known for over 100 years. Back-radiation, per se, is not a “greenhouse effect.” It is only radiation toward the surface.

    • backradiation from the atmosphere IS the greenhouse effect.

      • Anonymous says:

        OK, then, does the limited path-length of the radiation downward “beat” the convection upward? Show your work.

      • Anonymous says:

        Dammit, the “anon” here should be JAE, PhD (but just a chemist)

      • Anonymous says:

        Well, you still need to show how an increase in the amount of GHGs causes an increase in temperature with empirical evidence (you have not done that, yet, and nobody else has, either. You are just demonstrating what everyone has known for 100 years!). Does convection and the lapse rate “negate” the warming efects of backradiation? If the amount of water vapor doubles in the air all-of-a-sudden, like after a thunderstorm, does the air get warmer (no)?Show your work.

        • Actually, in science you can’t “prove” things to be true…only falsify hypotheses. Back radiation from the sky is the only way to balance the energy budget of the atmosphere and Earth’s surface, and weather forecast models, climate models, and 1-D radiative concetive-model all need to use it to produce realistic temperatures.

          So, if you want to falsify the hypothesis that the greenhouse effect exists and that it is necessary to explain surface temperatures on Earth, then do so. Show your work.

          • Anonymous says:

            The heat capacity, Cp, of air is about 1 joule/gram/K. It is trivial to show that the total amount of energy received each day in, say, July is just enough to heat the air in the first 5 km of the column an average of about 1-2 degrees C. Of course, that average represents something like 10 C at the surface and 0 C at 5 km. This amount of heating can be explained without resorting to any greenhouse effect, except for thermalization reactions. It can be explained as simply the absorption and storage of energy by the air.

            I don’t think that the earth is warmer than the BB equations (SB) “predict,” only because of radiation; heat storage also plays a large role. Blackbody calculations don’t work well for a substance which is “soaking up” heat, as well as radiating thermal energy. Which is all real substances, I guess.

  16. Harold Pierce Jr says:

    RE:
    Temperature Data from Death Valley Falsifies the Enhanced AGW Hypothesis.

    Hello Roy!

    The following a little off topic but revalent to the down-welling IR discussion.

    Please go to the late John Daly’s website “Still Waitng for Greenhouse” at http://www.John-Daly.com.

    On the home page scroll down and click on “Station Temperature Data” On the world map click on the USA. In the section “Pacific” click on “Death Valley”.

    The geaph is the annual mean temperature at the Furnance Creek weather station and is empirical field data that falsifies the enhanced AGW hypothesis.

    A desert is a region of low relative humidity, a low biomass of plants and animals, little or no free running or standing water, and mostly cloudless skies.

    After sunrise the land heats rapidly because there is little vegetation to block the sun rays and air also heats rapidly by conduction and convecton.

    After sunset, the air and land cools rapidly and mostly by conduction and convection. However, some heat will be lost as out-going long wave length IR. If CO2 has any effect on warming the air around the weather station, then we would anticipate a slight but discernible increase in temperature over time that should correlate with inceasing concentration atmospheric CO2.

    As you can readily see, there is no increase of annual mean
    temperature over time for the four seasons. Since Death Valley is below sea level, we don’t know the amount CO2 in the air. We only know that the concentation will increase with time.

    Roger Pielke Sr et al have pointed out that Tmean is not a good metric for analysis of temperature records. The three month sample period is a little long and will have some weather noise which a function of changing sunlight and possibly the ENSO. We can eliminate the effect of sunlight by analyzing one day or a number of different days of the record for Tmax and Tmin. In particular, we can start with equinoxes and soltices for the four seasons.

    I would really like to do this, but I’m organic and pheromone chemist and don’t know how to use data base and spread sheet programs.

    You should also check out temperature plots for other desert location such as Tombstone and Dodge City.

    • even a tiny change in atmospheric humidity will swamp the CO2 signal they are looking for in the Death Valley data. The drier the atmosphere is, the more sensitive temperatures will be to a small change in humidity.

      • Anonymous says:

        I don’t think this is true. If it were, then certain days in Death Valley would be far, far hotter than others within a given season, due to some humid air drifting in. That doesn’t happen. If it’s 100 F in, say, Phoenix any you venture downwind from an irrigated field, you will see an increase in humidity. But it remains at 100 F. I have seen no empirical evidence anywhere that an increase in ghgs anywhere produces an increase in temperature. Hence, my reluctance to completely buy into the “greenhouse effect.” Without that empirical evidence, we are still dealing with a nice hypothesis only.

      • Anonymous says:

        Good Morning Roy!

        What CO2 signal? There isn’t any! The trend lines for the four seasons look flat to me. If there is a so-called CO2 signal, then it is obviously quite small.

        Suppose we plot the annual humidity for the four season and compare these to temperature plots. If you are right, then temperature plots should correlate quite closely to the humidity plots.

        Notice that plots for the summer and winter are parallel. In winter the air is colder and more dense than in summer and would have a higher concentration of CO2 and OH2. If air pressure is about the same for the two seasons, then we would expect the winter trend line to have a greater slope than the summer trend line, but that is not evident in the plots.

        You might say that this would not be the case because there would be about the same amount of CO2 and OH2 in the vertical column air above the weather station. In real air there is no such stable vertical column of air due to wind, convection and the rotation of the earth.

        BTW, what is evaluation of John Daly’s website?

        • Anonymous says:

          Yup you are so right. But I cannot figure our why would you take four seasons. Is that not three seasons too many?

  17. Bernd Felsche says:

    The emissivity of most gases is well below 0.95. I don’t have any idea (near-enough) what is being measured when the instrument is pointed at the sky. Other than that it’s probably not the “temperature” of gas molecules.

    Even water vapour has low emissivity at altitude because its concentration is still low.

    Any heat transfer by back-radiation is likely to be swamped by negative, convective feedback. Higher surface temperatures provide more energy to drive convection. Horizontal and vertical.

    German meteorologist Dr Wolfgang Thüne explains this (in German) at http://www.derwettermann.de/allgemein/zum-marchen-von-der-erde-als-treibhaus/ Sorry, I don’t have enough time to translate accurately.

    Those who think that back-radiation provides significant nett warming are invited to personally bask under the back-radiation of a clear, summer sky in their back yard.

    • Actually, they are basking in it! Without back radiation, that clear summer day would be too cold to go outside without a heavy coat.

      • Anonymous says:

        So, you are saying that there would be no thermalization/heat storage of the air and surface and water? I could reflect all the backradiation with a foil suit, and I would still be hot on a hot summer day, no? Like many in this field, I think you are overly-focused on radiation. The planet STORES heat. It is not just a radiative world!

        • Anonymous says:

          Could you point me to the silos, where the heat is stored. I think I have an energy conversion business idea.

  18. Anonymous says:

    Dr Spencer said: “This supports what people already experience…cloudy nights are, on average, warmer than clear nights. The main reason is that clouds emit more IR downward, change the (im)balance between upwelling and downwelling IR, and if you change the balance between energy flows in and out of an object, its temperature will change. Conservation of Energy, they call it”.

    The temperature of the cloud was 50 F, that at the surface, over 70F. That is, the down-welling flux was weaker than the up-welling one. The energy imbalance and the rate of temperature change would have been reduced; the cooling would have slowed; but the temperature would not have risen.

  19. Kevin says:

    Ok, I was going to take the ten steps and give up on the whole “greenhouse effect” thing, but I can’t stop. Maybe I need some medication.

    Just a couple of observations;

    Yes indeed some gases in the atmosphere selectively absorb infrared radiation, then warm, then emit less than 50% of this energy back towards the Earth (as defined by the geometry of the sky .vs. the surface), then in a final important step cool down by an amount equal to the total radiation emitted. There I admitted it; “back” radiation a.k.a. “downwelling” radiation exists! So the next question is SO WHAT ?

    I again stand by my professional experience regarding the difficulty of measuring absolute radiation in the form of visible or infrared light to much better than a couple of percent. Check for yourselves, if you go the the NIST (formerly the USA Bureau of Standards) website you will see that you cannot purchase a “standard” radiation source that is measured by them to much better than a few percent. If you check out the internationally agreed to standard of optical radiation it is based on measuring the total radiation from a pool of molten Platinum held at its melting (or freezing depending on your perspective) temperature. NIST actually does this expensive measurement every few years and then compares a few “master” tungsten filament light bulbs to the measurements. These light bulbs are then locked in a safe, after which when you purchase a “calibrated” light bulb they transfer the measurements from the “master” standard to your secondary standard. Every step in this process adds about ½ percent to the overall uncertainty of the total radiation. In contrast you can have a resistance calibrated by NIST to parts per million if you wish to spend the money for that service. You can easily get a distance standard based on the accuracy of the dimensions of the Helium Neon molecules (accurate to small fractions of nanometers). Optical Radiation is easily the most difficult physical phenomena to measure absolutely. I stand behind this statement.

    Roy, are you sure your freezer is at -5 degrees F ? That seems pretty low for most consumer freezers. Most of my freezers are at about 10-25 degrees F ? Perhaps you should crosscheck your IR thermometer with a simple bimetallic or HORRORS a mercury thermometer ? It may be at -5 F, but it may actually be a bit higher and the assumptions made in designing the IR non-contact thermometer have some flaws.

    As a final observation, it is not sufficient to postulate only the effects of additional “greenhouse” gases, it is critical to also understand the effects of the lower levels of “non-greenhouse” gases. I am still (while fully admitting to the existence of “backradiation”) of the opinion that increases in “greenhouse” gases actually work to increase the speed of heat/energy through the atmosphere since slightly more heat/energy travels through the atmosphere at the speed of light (quite speedy) versus at the speed of heat (comparatively sluggish). I also believe that this effect is so small we probably could not spend enough money to measure it.

    Cheers, Kevin.

  20. anonymous2 says:

    What about this post ?
    http://www.john-daly.com/artifact.htm
    (I hope not having put it into the wrong psot!)

  21. Ric Werme says:

    I like your box experiment. One thing I’ve been tempted to explore is a larger structure, perhaps an old satellite receiver dish, and set it up to radiate as best it can. Then use it to chill water over summer nights to use for cooling the next day. Sort of solar heating in reverse.

    Of course, it would be tempting to experiment with it during the day, but hassles with blocking the sun and dealing with convective winds would be a substantial challenge.

    I haven’t worked through any math, but I suspect the satellite dish will be nowhere near large enough for anything but measuring how big a radiator I need. I do have property on a mountainside – perhaps I can use/enhance the cool nighttime draft for the same effect. :-)

    • you can’t magnify an extended source of thermal radiation.

      • Ric Werme says:

        Sorry – I should’ve been more clear. I didn’t want to use the satellite dish to focus anything – I wanted something with a managable area for the experiment that would left chilled air accumulate where it could take up heat from water or some other storage medium. A big bucket would do, or maybe an old bathtub (those are heavy suckers!).

        I figure an old satellite dish is fairly light, readily findable (“Hey bud, I’ll take that old dish out of your backyard for only 10 bucks, for free if you help!”) and easy to coat with foam, paint, cut holes in, etc.

  22. Kevin says:

    Dr. Spencer,

    I do not enjoy repeating these quotes, but I feel that I should;

    “It might have an offset of 30 or 40 or 50 deg., but it is still mostly measuring down welling IR radiation from the sky.” Wow!!, the whole calculation of the alleged ”greenhouse” effect is based on the transfer of radiation between two surfaces based on Kirchhoff’s equations and an offset of ”30 or 40 or 50 deg” is acceptable? We do need to note that the energy is related to temperature by the FOURTH power, i.e. energy = temperature^4. I would expect and accept that I would be fired from my job if I glossed over this difference.

    “the IR measurement of the sky is only partly related to the back radiation”, with respect, what kind of dodge is this? When I make measurements I need to show how they relate to measurements that other people have made. We call this “verification”, a critical step in the process of demonstrating how any system performs against what the theory predicts it will do.

    Cheers, Kevin.

    • Anonymous says:

      Hi Kevin,
      I believe that Dr.Spencer means that if you know the bias you can weight and remove it to get a meningful measurement.
      My only doubt is how much that measurement is meaninful or meaningless because we don’t really know the origin of that bias. Thus how it changes as a function of the GHGs concentration is unknown.

      Massimo

    • The point I was making was a qualitative one…that the atmosphere emits IR downward. Since my handheld IR thermometer does not cover the entire IR spectrum, but is more tuned the the window region, there can be expected to be an average offset between what I measure and the true effective emitting temperature of the atmosphere as seen from the surface.

      Of course we need numbers as accurate as possible for most applications.

      • Anonymous says:

        Dr. Spencer,

        With respect, what the H—L is a “qualitative” point? As an engineer I am not afforded the luxury of “qualitative” points!! My designs either work or they don’t. They do not “sort of” work better than other designs, either they perform better or they do not perform better.

        As an alternative thought experiment I suggest the following;

        1 – Allow a cast iron kettle to reach equilibrium with
        room temperature

        2 – Heat a thin sheet of aluminum foil to 200 degrees F

        3 – Place the cast iron kettle on the aluminum foil

        4 – Please explain how the “back-conduction” from the aluminum foil causes the temperature of the cast iron kettle to permanently rise ?

        Again, I do respect the quality of your work and I appreciate your role in stopping us from running over the “cap and trade “cliff, but we still have vastly different conceptions of how “heat” works in the real world.

        Cheers, Kevin.

        • Kevin, either you have not read what I have written, or your mind is totally closed. I suspect that you and I actually believe the same things about thermodynamics, but you have not thought about the system that has a continuous outside source of energy, and how that allows things to happen which cannot happen in a closed system.

          • Anonymous says:

            Dr. Spencer, we are clearly experiencing some crosstalk regarding this subject. My initial training was as a “sparky” (Electrical Engineer), with additional training and experience in “bending light” (aka optics) including absolute radiometry measurements among other things.

            Strangely enough both of these disciplines can be traced to the Laws of Thermodynamics and Maxwell’s Laws. Hard as I try I have not violated any of these laws yet. I bet I could be very rich as soon as I figure out how.

            I have read and understood most of your writings. And we probably have a closer agreement about the laws of thermodynamics than what comes through during some of these posts.

            So as a “sparky” let me summarize my position as such:

            Climate Science treats the atmosphere of the Earth as a “DC” (i.e. direct current) system. If “X” power comes in, and “Y” percentage of this power leaves, the energy leftover (i.e. X times (1-Y)) causes warming.

            An electrical engineer will treat this as an ”AC” (i.e. alternating current) system. We view all of the non-active (i.e. those elements that are not creating energy by breaking molecular bonds) as the equivalent of capacitors (or inductors). Capacitors slow the changes in current/heat and the only important question is how much do the capacitors slow the flow of current/heat. We in the “sparky” community would refer to the weather as a 1/(60 (secs) * 60 (mins) * 24 (hrs)) = 0.0000115 Hertz system. So our question becomes what are the equivalent values of the capacitors that the climate science community call ”greenhouse” gases ?

            Note that electrical engineers are quite adept at alternating between “DC” and “AC” systems. In fact most real electrical circuits incorporate both “regimes” to perform the function they are designed to accomplish. We call this “bias” and effectively establish different DC voltages (i.e. temperatures) within portions of our designs to make the available semiconductor devices perform the desired AC functions (usually a gain/amplification). I have yet to see how gases in the atmosphere can in any way accomplish the “bias” function that we accomplish in our systems.

            You wrote: “but you have not thought about the system that has a continuous outside source of energy”, in fact I postulate that what you consider a “continuous outside source of energy” is in fact a 0.0000115 Hertz source of energy. This is why I have on several occasions referred to the “speed of heat” in my posts (probably to your annoyance). What you consider a “continuous” source of energy, we consider just another “AC” source.

            Note that all of the laws of thermodynamics are directly convertible to equivalent laws in the electrical domain. If my “DC” power supply gives me 1 kilowatt-hour I can charge a small capacitor to a maximum voltage quickly, or I can charge a large capacitor to the same voltage slowly, but I can’t do both.

            I’m sure that my postings may appear to be “close minded”, but perhaps we are both somewhat equally guilty of this all too human tendency? Just for reference, I am not employed in the “Energy” Industry, and I don’t have a dog in this hunt. I would just like everybody to reach the correct conclusions based on the laws of physics that we in the engineering field have used to improve many aspects of the human condition (yes, some mistakes have happened, but does everybody really want to ride a mule to work everyday?).

            Cheers, Kevin.

          • Kevin, I have to smile when I read your post. You are like me…you believe that others think in the same terms as you. In your case, you use metaphors from electrical engineering. Unfortunately, they mean nothing to me. :) Not your fault.

          • Anonymous says:

            Dr. Spencer, yes I do think in terms most familiar to me. But I also have “cross” training. I have in fact attempted to design thermal control systems using infrared temperature sensors, as well as direct sensors (RTDs, thermocouples, etc.) In my experience absolutely measuring temperature remotely by measuring infrared radiation is good to perhaps +/- 1 degree C (on a really good day when you control all of the parameters). Alternatively, if you exercise enough attention to details you can accurately measure the temperature of a small volume to +/- 0.01 degree C using direct sensors. That’s a difference of 2 orders of magnitude.

            I have enjoyed exchanging thoughts with you. I do hope you prosper after the shakeup that is coming to the climate science community.

            Cheers, Kevin.

          • I do not dispute any of that…you are clearly experienced.
            My point is that even if my cheap, hand-held IR thermometer is only good to 5 or 10 degrees, and I don’t really know its exact spectral response, it can still be used to demonstrate the existence of downwelling IR from the sky, and that its intensity increases with increasing humidity of the air mass.

          • Anonymous says:

            Dr. Spencer, I agree, down welling radiation exists and varies with the composition of the atmosphere (specifically the percentage of water vapor, etc.). You have presented empirical evidence that increases in water vapor (with its inherent increase in thermal capacity) causes an increase in down welling IR radiation. No dispute here.

            Still enjoying your efforts to overcome the language barriers between climate science and the rest of us.

            Cheers, Kevin.

  23. bill fish says:

    I am impressed by the responsiveness of Dr Spencer. I wonder if he gets any sleep.

    I recall in some previous entry where someone in Utah built a conical reflector which can boil water during the day and cool things at night. For the cooling effect to take place, the cone has to be directed to the sky with clouds or massive objects absent in its line of view.

    Suppose the source of the sky’s IR radiation which reaches the focal point of the conic reflector – long and narrow – is contained in a (virtual) cylinder of radius R – R being comparable to the average radius of the conic reflector – and extending upwards to, say, 1000 km. Let the mass placed at the focal point of the reflector be M. Now, all this is being done at night. Let the mass be gas in a small spherical bottle whose walls are transparent to IR. The mass will radiate IR radiation from the focal point. If the mass of the gas is doubled, the number of emitters is doubled. It seems that there is a critical mass Mc where radiation emitted is equal to radiation absorbed. If mass is less than Mc, then the gas temperature in the bottle increases since more energy is absorbed than emitted. If mass is greater than Mc, then the gas temperature in the bottle decreases.

    Question: has anyone done such an experiment?

    • Anonymous says:

      Hi bill fish,
      what I think is that it’s not a question of mass but of radiation ratio.
      I mean, when you put under the open sky a solid object it emits and receives radiation via a lambertian path.
      See wikipedia here:

      http://en.wikipedia.org/wiki/Lambert's_cosine_law

      In this case the ratio of its emitted/received radiation is the one for its current temperature.
      When you put the same object in the focus of a parabolic mirror, the outgoing radiation of the object is fully emitted into a narrower column of the amosphere. While only the incoming “back radiation” from that column which is perpendicular to the diretrix of the parabolic mirror is focused on the object. This means that the emitted/received ratio is increased. Thus, since the object loses more radiation than the it receives from the atmosphere, its temperature falls until it reach a new radiation equilibrium.

      Massimo

      Obviously, the object in the focus must be upward thermally insulated to avoid its direct radiation exchange with the open sky.

    • you cannot magnify the thermal radiation from an extended source. You can, however, restrict its field of view. Since the downwelling IR is the weakest at zenith (straight overhead) you could focus on that part of the sky, but then you have reached the limit of what you can do using the sky as an energy sink.

  24. Bill Illis says:

    A telling experiment would be “is there any back-radiation coming from the atmospheric windows?” I’m assuming the Greenhouse theory assumes there is none but this part is not exactly spelled out anywhere.

    Well the Modtran calculator says there is still radiation coming down (or radiation levels if you are looking up from 50 metres height) which is equivalent to a blackbody curve radiation intensity of 260K.

    http://geoflop.uchicago.edu/forecast/tmp/rad.12200636.gif

    So it is a small amount of back-radiation, but it is higher than the back-radiation that CO2 exerts at 7 kms high.

    http://geoflop.uchicago.edu/forecast/tmp/rad.12200915.gif

  25. Macha says:

    Off topic….slightly Roy, but my first time here..whats your view on this report in relation to how you ( or anyone else ethat has done) uses the data ?

    http://www.climatechangefraud.com/climate-reports/7491-official-satellite-failure-means-decade-of-global-warming-data-doubtful

    macha (West Australia).

    We have similar issues with data from Darwin and Ciarns(faily remote)

    • much ado about nothing. John O’Sullivan doesn’t know whether the NOAA-16 AVHRR data make in into a climate dataset, anyway. Even if they do, it is not done blindly, without quality control checks.

  26. Mike Edwards says:

    Dr Spencer,

    At the end of your main piece you said:

    “Except for relatively rare special cases, the total amount of IR energy downwelling from the sky (Ed) will ALWAYS remain less than the amount upwelling from below and absorbed by the sky (Aa).”

    Strangely enough, I think that Miskolczi’s own paper shows one example where the downwelling from the sky IS greater than the amount upwelling from the Earth’s surface. This related to the winter in Antartica IR measurements, if I’ve interpreted them correctly.

    I view this as the extreme case where the Earth’s surface is in semi-permanent “night” and all its heat energy is actually being supplied by the atmosphere above – ie the air above the Antartic polar region *is* warmer than the surface during this period, so that heat energy is flowing downwards.

  27. Harp;d Pierce Jr says:

    ATTN: Anonymous

    The reason for analyzing the Tmax and Tmin data for the four seasons is to check for any seasonal effects.

    In “Climate Change and Global Warming”, A. Masterman reports the results of his analysis of the CET on a month-by-month basis at 30 year intervals. He found that Tmeans of months of the fall season were significally warmer than that of the other seasons by t-tests. For several months the Tmeans have remained unchanged for about 300 years.

    His study of the CET didn’t determine if the cause of the warmer fall months is due to Tmax or Tmin or both. His study confirms Roger Pielke Sr et al claim that the annual Tmean is not useful for metric for climate studies.

    The big question is: What is the cause of the seasonal effect?

    The URl for his paper is:

    http://www.usefulinfo.co.uk/climate_change_global_warming.php

    I can’t imagine that Phil Jones et al and most climate scientists are unaware of this article. There are lots of hobby and amature scientists, and the internet allows them to publish their works which would be readily rejected by
    scientific journals.

  28. I have been trying to find out the amount of heat added to the surface temperature by the Earth’s hot interior; in particular the ocean heating from the molten interior. Clearly it has SOME effect. Moreover by heating the bottom of the sea, the effect will be circulation.

    I have no idea how one would measure the amount of heat conducted to the seas by the magma below them, but surely there is some. We talk about the average temperature of the Earth (to a tenth or even a hundredth of a degree) but we don’t mean by average an average from the center to the outer edge of the atmosphere. We’re interested in the biosphere. All the climate models I have seen seem to ignore the heat coming up from below.

    There are 50 – 70 surface eruptions and lava flows each year on and. That implies about 200 per year under the seas. I have no idea how much heat that adds to the system, but it must be some. Moreover, it’s unpredictable and surely changes from year to year. The usual effect of a land eruption is atmospheric cooling (an observation first made by Benjamen Franklin) but the effect of an underwater lava flow would be heating and the heat circulated from the bottom to the surface. Is this all a trivial amount of heat or is it significant compared to, say, the re-radiation from the CO2 layers back to the surface? Or somewhere between trivial and significant? I have no idea, but I don’t find it in any of the expositions on our climate models.

    Jerry Pournelle
    Chaos Manor

  29. Ric Werme says:

    The notes I’ve heard about it are that the energy is minor compared to many other factors.

    Aha – here’s a 2010 paper that looks really good. About .06 w/m^2, more at tectonic spreading zones, and (perversely?) at Greenland and Antarctica.

    http://www.solid-earth.net/1/5/2010/se-1-5-2010.pdf

  30. JAE says:

    Well just what is the bottom line, here? Does the backradiation heat the world or is the backradiation simply a measure of the world’s temperature. Still absolutely no demonstration of any “greenhouse effect” here, as far as I can discern.

  31. chrisd says:

    Dr. Spencer, are you aware that the title of this post is being used by some as “proof” that the greenhouse effect does not exist?

    They’re simply quoting the title and leaving it at that, with the obvious implication that “Here’s a real climate scientist who proves that the greenhouse effect is completely nonexistent.”

  32. Noblesse Oblige says:

    I live most of the year in SW COlorado at an altitude of 7100 ft. We have near desert conditions; humidity is usually very low (10-25% during the day). With little water vapor and 24% less atmosphere to begin with, the temperature trajectory during the day follows the sun much more closely than lower altitudes with humid climates. In June before the late summer rains come, it is not uncommon to see temps in the mid 80s during the day, dropping to the 30s at night. But let a wisp of a cloud pass overhead at night and boom: we are up to the low 50s. The effect is striking. I understand this as increased back-radiation from a cloud that is cooler than the ground, like the second rod in the jar in Roy’s earlier gedanken experiment.

    • Anonymous says:

      The “cloud effect” you describe might be due, at least partly, to the fact that the clouds slow convection. Also the air is more humid and water vapor stores twice as much heat as does all other air molecules.

      • I’ve spent a lifetime studying and publishing on atmospheric convection. At night, under a clear sky and light winds, convection can be assumed to stop. There is none at night. That’s why a strong temperature inversion develops at night! Convection would destroy that inversion (for instance, mechanically-induced, rather that buoyancy-induced, as during windy conditions).

  33. Reader says:

    Just looked at the temps on the Java tool at the discover site. Not really looking like the temperature is dropping yet.

    We might be going for the record after all it seems.

  34. Massimo or Dr. Spencer,

    I tried my own IR thermometer experiment and here’s my result:

    sky temp at noon on a clear day: -9C
    sun temp at noon on a clear day: 3.2C

    (no box, just an instantaneous measurement)

    for my location, date, and time, an online insolation calculator tells me the solar insolation was 827 Wm-2. Since about 40% or so of incoming solar radiation is IR, that leaves 331 Wm-2 to be picked up by the IR thermometer pointed at the sun and sure enough, 3.2C corresponds to 331 Wm-2.

    the -9C from the clear sky corresponds to 276 Wm-2 “back-radiation,” close to what you predicted.

    however, when the IR thermometer is pointed at the sun, shouldn’t it be seeing both 331 from the sun and 276 from “back-radiation” for a total of 607 Wm-2 or 48.51C!

    • Anonymous says:

      Hi Hockey Schtick,
      as already said I’m not a climate scientist, just an electronic engineer.
      Despite that, because of my job (I’m ab industrial control boards designer) I had some experiences about measurements and in particular measurement errors.
      I don’t believe you can use the Dr. Archer’s Modtran simulator to get reliable measurements from your IR thermometer for your actual situation.
      Dr. Spencer surely knows better, I’m just saying how I would do it, but I’m not sure about what follows.
      To give to your thermometer’s readings a meaning, you should know EXACTLY the air composition at that moment and the EXACT ditribution and concentration of the gases along the altitude at least. Then you should adjust the incoming radiation in the range of your IR thermometer for the absorbance, trasmittance and reflectivity of the medium.
      It’s a very tricky job, I’ve many doubts you could ever have a meaninful reading from it.
      Into another post here, Dr.Spancer stated that he does “qualitative” measurement with the IR thermometer and Kevin (which is an engineer like me) opposed his perplexity about the usefulness of “qualitative” measurements.
      I agree with him, but I don’t believe that Dr.Spencer meant that he used or use any “qualitative” measurement for his statistic job.

      Massimo

      • Massimo,

        I realize that an accurate determination of ‘back-radiation’ or solar IR radiation is not possible with an IR thermometer.
        However, you will find many descriptions of the ‘greenhouse effect’ which show that the atmosphere somehow doubles the surface IR radiation to emit 240 W/m2 to space and 240 W/m2 back to earth, such as shown in figure 1 of Richard Lindzen’s paper to produce a total of 480 W/m2 at the Earth surface:

        www-eaps.mit . edu/faculty/lindzen/198_greenhouse. pdf

        I am simply trying to show that if this was anywhere near correct or possible that I wouldn’t be obtaining the readings I have from the IR thermometer in comparing the RELATIVE readings for a clear sky and clear sky + Sun.

        Do you agree?

        • Anonymous says:

          Hi Hockey Schtick,
          the Sun blackbody doesn’t radiate in the LW band which your IR thermometer uses to work.
          Your “clear sky + Sun” and “clear sky only” relative measurement give you just an idea of the amount of the Sun SW radiation directly absorbed and converted to SW by the atmosphere which reach the ground.
          Given your measurement, the only thing you can state for sure is that the emitted down radiation of the atmosphere due to the Sun SW to LW conversion which reach the ground is about 54W/m^2, while the “back radiation” is about 276W/m^2.

          Massimo

          • Anonymous says:

            Massimo,

            if the sun doesn’t emit in the far IR band then why are there innumerable graphs on the net by reputable organisations that show it does?? Having little power at earth orbit and not emitting are two different situations.

          • Massimo,
            Anonymous is correct – about 40% of solar radiation is IR
            e.g. look at the area under the curve of observed solar radiation in the IR at the top of the atmosphere in this post:

            hockeyschtick.blogspot . com/2010/08/is-greenhouse-effect-is-based-on-cool . html

          • Anonymous says:

            Hi Hockey Schtick & Anonymous.
            There is “IR” and “IR”. Near infrareds (NIR) are in the so called short wave band (SW), while the far infrareds (FIR) are into the long wave band (LW). Some reports the LW/SW border at 4um.
            The ones we are talking about, which your IR thermometer uses to work, are placed in the spectrum between 4-18um (many have narrower band having the lower wavelength limit at 8um). If you look at the graph it stops at 2um. Over that wavelenght the Sun power density is negligible. Your thermometer works there.
            There are two more issues aiming the Sun with the IR thermometer to take into account. The first is the emissivity of the fresnel lens embedded in the thermometer viewing path, which is stimulated by the direct incoming Sun radiation in the UV, visible and near infrared bands. The second is the view angle of that lens. I already discussed this issue with Dr. Spencer in another post. The IR thermometer averages the radiation on its whole view angle. This is not a problem when the target object is greater than the footprint of viewing angle and the target is almost isothermal. If the Sun dish is smaller than the projected view angle onto the immaginary sky background, your thermometer reports a much lower temperature than the one reported if the Sun dish fully fill the view angle.

            I did a check this morning with my 1:12 focused thermometer, when aiming the Sun it read 79°C (I’m in North Italy just 48km West from Milan). If I put a 2mm thick glass layer between the Sun and the thermometer aperture (which is a discrete LW IR blocker), I read just 75°C. This doesn’t mean that much to me, because we should consider the glass self emission due its temperature, and its UV filtering coefficient (which I don’t know), but it states that most of the measured radiation passed throught the glass layer so it was not in the LW IR band, but in the UV-Vis-NIR bands.

            Massimo

        • Anonymous says:

          Hi Hockey Schtick,
          I’ve a doubt, maybe I missed your point: what did you mean with “clear sky + Sun”?
          I explain. If you meant that you are measuring the radiation aiming the Sun dish then the previous message I wrote is what I think about.
          If your “clear sky + Sun” meant “clear sky by day” instead and your “clear sky only” meant “clear sky by night”, then you should take account of the thermal capacitance of the ground surface which is facing your “clear sky” to extrapolate the Sun contribution to the atmospheric “back radiation”. There are many other factors which could change day and night such as the humidity which should be accounted too.
          I don’t believe it’s so easy to get it with an IR thermometer.

          Massimo

      • Despite the fact that we do not know the exact frequency response of any given IR thermometer, we can still make some interpretations from the readings which are physically meaningful. For instance, I have found the clear-sky temperature at zenith to vary from 40 deg. F during very humid conditions to 0 deg. F during moderately humid conditions. In fact, since as a meteorologist I am keeping track of how humid the atmosphere is on any given day, I have gotten pretty good at predicting what the IR thermometer will read when I point it at the sky.

        When REAL fall air arrives in Alabama, I’m sure I will measure well below 0 deg. F.

    • don’t trust IR thermometer measurements when pointed at the sun. The sensor is designed to screen out shortwave solar energy as much as possible, which is where most of the Earth’s absorbed energy comes from.

  35. wayne says:

    There are so many commenters above who seem to really want to know exactly what this is all about. I have some notes when I was decyphering K&T’s energy budget so I’ll just post these notes as is, sorry, it’s text.

    It is bound to run off the right edge so just select all of the text and paste back into your text editor such as notepad so you can study it.

    You will notice this spread doesn’t agree with some of K&T’s arrows (such as the 324 back radiation, my spreadsheet computed 222 W/m2 downward radiation), so put in a spreadsheet and you might have a better solution. Forgive if you should find a mistake but I think it’s solid in it’s own right (no “windowing”).

    My notes on Kiehl/Trenberth/et al Global Energy Budget.

    If this runs of right edge,
    copy and paste in a text editor.

    This exactly matches K&T initial parameters but unable to get
    a viable balanced results when trying to include “window” parameters
    used in their budget. When included it wants to set the mean
    altitude to zero and send more radiation down than up and that
    seems impossible by geometry of rays (SW & LW) above a sphere
    and this seems to show that.

    A 1368 Sun – at TOA ( specified )
    B 342 Sun – TSI ( A / 4 )
    C 77 Out – Reflected by the atmosphere ( specified )
    D 30 Out – Reflected by the surface ( specified )
    E 235 In – Total SW radiation from the sun ( B – C – D )
    F 67 Up – Absorbed by the atmosphere ( specified )

    G 168.0 Dn – Absorbed by surface and converted to LW ( E – F )
    H 222.0 Dn – Absorbed LW from the atmosphere by the surface ( N * (1-M) )
    ( This is the “back-radiation” and some call it the
    “greenhouse effect” though it is merely due to
    the fact that ALL MATTER ALWAYS RADIATE in random directions
    and in a blackbody manner and some gases also strongly interact
    in bands per the molecule type being addressed.
    (zero K case ignored here) )
    —–> subtotal 390.0 ( 14.8 ºC )

    I 78.0 Up – Evapo-transperation to atmosphere ( specified )
    J 24.0 Up – Thermals to atmosphere ( specified )
    K 288.0 Up – Remainder of total LW flux to atmosphere ( G + H – I – J )
    —–> subtotal 390.0

    L 6368 Altitude of mean atmosphere (m) (geometric adjusted)
    ( This is the fitted mean atmosphere altitude of F&T budget )
    M 0.5142 Ratio emitted by the atmosphere toward space (geometric adjust)
    ( ( acos( rEarth / (rEarth + L) ) + pi/2 ) / pi )

    N 457.0 In – Total flux absorbed by the atmosphere from surface ( F + I + J + K )
    Out – see h & o.

    O 235.0 LW emitted by the atmosphere to space ( N * M ) ( -19.4 ºC )

    P 390 Expected radiation at the surface ( specified )
    Q 235 Expected radiation at TOA ( specified )
    R 0.0000 Diffs squared ( used to best fit, then = zero )

    rEarth = 6371008 m

    Geometric adjust — Earth is a sphere not a flat plane so altitude causes more
    LW photons to intersect the space hemisphere than the surface hemisphere,
    e.i. the two hemispheres are not equal except at exactly sea level when the
    radiation occurs.
    ( See: dip of horizon effect
    or exsecant where altitude = rEarth * exsecant(dip) opposite eq above )

    -wayne

    • Anonymous says:

      Darn. “pre” tag didn’t engage so you might have to do some “text” work if you really want to know what is hidden within. Cheers.
      -wayne

    • Anonymous says:

      More simply, consider that the net effect of radiation adsorption is a random re-direction of the absorbed radiation by the almost instantaneous re-emission of the absorbed radiation. Since the atmosphere is thin compared to the radius of the earth, the effect is to re-direct half of the radiation back to earth and half to outer space. The earth will heat up until the 1/2 emitted to outer space plus what passes through the atmosphere w/o absorption is equal to what is absorbed. The key in using this well known formulation to get real values is to notice that night sky irradiance shows about 20% of outgoing IR in the “IR window” is backscattered toward the surface. Hence, only about 80% of un-absorbed radiation gets out.
      SO: outgoing radiation must satisfy (in an equilibrium time)

      ~238 W/m2 = I(earth)*{a/2+b(1-a}}
      where a=well known approximation of atmospheric absorption~.65 and b~.8.(my eyeball value). (238=(albedo*solar constant)/4)

      This gives I(earth)=393W/m^2 => 288.6K;

      I(earth)*a/2= radiation from “top” of atm. into space = 127.7W/m^2 => T=217.85 K =
      -55.3C

      a=.64 and b=.805 gives 1961 standard atmosphere values for surface and top of troposphere temps.

      Simple analysis from a simple mind!

      Mogman

      • Anonymous says:

        Hi all, I just have to state some of my observations about the calculations that predict how “much” warmer the surface of the Earth is because of the ”greenhouse” effect.

        First, please note that I do not dispute the existence of “back-radiation” from gases in the atmosphere. Also I acknowledge the existence of “back-conduction”. Now you might think me crazy, but think of this scenario; two molecules meet in a bar, a warm molecule and a cold molecule. When they introduce each other and shake hands the cold molecule “back-conducts” some energy/heat to the warmer molecule. Then, of course the warm molecule “back-conducts” EVEN MORE energy/heat to the colder molecule. This is the essence of the Second Law of Thermodynamics. The whole “backwards” versus “forwards” direction of the energy/heat transfer is relative only to which molecule you are. There is nothing different in this relationship if the heat/energy is transferred by conduction or radiation. There is nothing special about the “kinetic” nature of “infrared active” gases, all materials demonstrate the energy retained within by vibration of the molecules, solids, liquids, gases…. Temperature is the measurement of this vibrational (a.k.a. kinetic) energy.

        Now some specific observations about the “greenhouse” effect’s alleged ability to cause the temperature of the surface of the Earth to be higher than in the absence of these gases;

        1 – Why is an alleged “Energy Budget” of the Earth (including its Atmosphere) calculated in units of POWER (i.e. watts/unit area)? Watts are units of POWER, Joules/Photons/BTU’s are units of ENERGY. While this seems a slight difference, in fact with a cyclical system (i.e. Energy is input only about 50% of the time) this is a FATAL omission. I am of the opinion that the proposed “Energy Budget” of the Earth should be ignored until such time as the proper units can be employed (using more of my tax dollars I expect).

        2 – Why are not the proper coordinates being applied, (i.e. spherical coordinates versus 2 dimensional surface coordinates)? Surely spherical coordinates are well understood and available in many computer tools. I do vaguely remember spending lots of late hours in Graduate School struggling with converting Maxwell’s Equations into spherical coordinates. I hate to say it but it seems to me that the climate science community are the ones insisting the “Earth is Flat”.

        3 – I rarely see any discussion of the thermal capacity, thermal conductivity, or thermal diffusivity of the materials involved in this massive heat transfer problem. If only we could ignore them in the real world.

        4 – Lest you think I have no experience with the marvels of computer modeling you would be mistaken, we have a saying: “If your hardware does not perform as predicted by your model you need to improve your model”. Anybody care to fly on a plane I design after I declare “This Airplane Is Very Likely To Remain Airborne?”

        I do hate to be hyper-critical, but perhaps some in the Climate Science Community will be open to this summary;

        1 – Your Energy Budget is using the WRONG units

        2 – The Geometry of your calculations/models is using the WRONG coordinates

        3- You have ignored OTHER REAL properties of the actual materials in the System

        4 – You are overly dependent on your computers

        Again, I respect Dr. Spencer’s work, but I still remain a denier of the “greenhouse effect” hypothesis, not a denier of the existence of ”back-radiation”. I believe the science has made a mistake and is incorrectly attributing the results (i.e. a permanent rise in the temperature of the Earth) to the “greenhouse effect”.

        Cheers, Kevin.

        • Steve says:

          Kevin,

          Well put, though energy is energy, Watt (Joule/sec/m^2)or Newton. What led me to asking many questions about AGW was imagining energy transfers in terms of Newtons or Force.

          Immediately that led me to gravity, which makes Convection possible. Gravity is sort of ‘Free’ energy that powers convective currents.

          From there once you start calculating how fast objects cool via radiation alone, then add convection, and then add Evaporation (Latent Heat of Vaporization), it becomes obvious fairly quickly that cooling via radiation is quite minimal compared to convection and especially evaporation. Evaporation towers are not there for their looks, it is very efficient.

          Another consideration is that a cubic meter of water at 15 C (average Sea Surface Temp) contains about 1.2 Billion joules. Trapping 2 Watts (from CO2) at the surface only serves to slow the cooling rate by ~87,000 joules over 12 hours, it takes 4.1 Million joules to cool the cubic meter of water 1 C. Trapping 2 Watts via radiation isn’t heating jack crap, especially when convection is free to continue unabated and in fact increases along with surface temp.

          N2 and O2 are not without temp and most certainly emit and absorb radiation, ALL substances do. One property of most matter (I say most because there are likely exceptions) is that if it emits in a certain wavelength, it also absorbs in that wavelength. Does this make N2 and O2 GHG’s? The Troposphere has a mass of 4,000 Trillion metric tons. It takes a great deal of energy to heat it, day and night temps vary by as much as 30 C in arid areas. 99% of the energy is within N2 and O2. CO2 only accounts for .7 grams per cubic meter where N2 and O2 account for 1300 grams at sea level.

          Lastly, I agree with you that the GHE is poorly understood, and at best poorly explained. Is it just ‘trapped’ radiation? Is it just H2O, CO2, CH4, etc? Greenhouses work by preventing or slowing convection. Trapping radiation does not do this. Clouds slow convection, but are more apt for cooling, etc.

          And the repeated statement that the Earth would be 33 C cooler without GHG’s is bunk. The surface of the Earth would get just as hot and transfer energy to N2 and O2 via conduction, just as it does today, and that air mass would rise and cool do to convection, just as it does today. So what that CO2 would not ‘trap’ an additional 87,000 joules over 12 hours, on average.

          • Anonymous says:

            Steve, I fully agree with your initial calculations!

            Kirchoff’s equations for radiative heat transfer assume materials that have an infinitesimal thickness and an infinite thermal capacity, which is quite a bit different than any actual known REAL materials.

            You wrote: “And the repeated statement that the Earth would be 33 C cooler without GHG’s is bunk.” I fully agree, ironically the actual effect of increasing “greenhouse” gases in the atmosphere is that “slightly“ more heat/energy flows through the the atmosphere at the speed of light (quite speedy) versus the “speed of heat” (comparatively sluggish).

            This is in fact THE EXACT OPPOSITE of the proposed “greenhouse gas hypothesis”. I also postulate that this effect is so small that we probably cannot spend enough money to measure it. I postulate that increases in “greenhouse” gases actually cause the temperature of the gases in the atmosphere to more quickly approach their “theoretical” temperature. I.E. the gases in the atmosphere “warm” every slightly faster after sunrise and “cool” down every slightly faster after sunset. Due to the overwhelming thermal capacity of the water and the rocks on the surface of the Earth these changes are much less than our ability to measure them accurately.

            Cheers, Kevin.

    • Wayne, the geometric source of error you are talking about is small compared to other uncertainties in the K&T energy budget. They show their best estimate (for back then, 1997), obviously dependent on various assumptions thy have made that others may or may not agree with. I would say any of the numbers could easily be off by 10 Watts per sq. m. A couple could be off by 40 or 50 Watts. But we know that the top-of-atmosphere imbalance must less than 2 Watts or so, otherwise we would see more long-term warming or cooling…they have merely forced those two numbers to be equal by adjusting other numbers.

      • Anonymous says:

        I totally agree but that geometric source of error was no error, it is the core of what that spreadsheet was showing. I know K&T just forced the equality. That is one of the thing in physics we all know must equate overall, out equals in. I gave those figures to maybe have some interested commenter who hadn’t yet delved into the energy balances so far published to take a more realistic view if we are going to have a single shell atmosphere and form a “budget” as K&T did.

        K&T’s budget I viewed as a single, one meter thick shell at some altitude above the surface which is then called “our atmosphere”. Nothing happens between the ground and the bottom of that layer, and above that one meter layer is space (TOA). Everything that happens, everything, happens in that compressed one meter of “atmosphere”. The 6368 meter altitude is where that layer must reside to get K&T’s numbers to balance and it would have a geometric adjustment, take it from a sailor who knows celestial navigation and what altitude does to light rays on our sphere, LW outgoing photons included. And all that showed was there is no 324 coming down from the sky, the real number is much lower or you would turn your radiative thermometer upward toward space and would read on the average 2ºC everywhere, not the approx. minus 22ºC which is closer to the reality, coming from the ~222 W/m^2 downwelling.

        In other words, I was just toying around when I built that spreadsheet but it did surprise me what it came up from K&T’s figures from that special one layer aspect! Plan later to morph that spreadsheet using Miskolzci’s figures just to see what are the differences.

        • Anonymous says:

          May be that what I write below is just a silly thing, but I would like to know how and if the TOA outgoing radiation has ever been measured indeed.
          The measurement of the incoming radiation is easily performed because of the very long distance which separates the Earth from the Sun, so the radiation flow can be considered fully perpendicular to the radiometer detector.
          But the measurement of the whole Earth outgoing radiation is a real enigma for me.
          I can’t figure out how a radiometer (no matter its monochromator is a grating based or a Michelson interferometer based one), which needs a coherent light at its input slit could measure the whole outgoing energy since it measures only the one which is perpendicular to the input slit having spatial coherence.
          I’ve no doubts that a satellite which measures the outgoing radiation spectrum can be used to estimate the ground temperature doing some transmission related adjustments (if these adjustments are accurate of course). What I can’t really figure out is how can satellites estimate the absorbtion of a GHG just using this kind of measurements, since we should take account of the GHG spreaded radiation which exits the TOA with different angles and for that is not “seen” by the monochromators.

          Does anyone knows?

          Massimo

  36. Jan Pompe says:

    Dr Spencer,

    Except for relatively rare special cases, the total amount of IR energy downwelling from the sky (Ed) will ALWAYS remain less than the amount upwelling from below and absorbed by the sky (Aa). As long as (1) the atmosphere has some transparency to IR radiation (which it does), and (2) the atmosphere is colder than the surface (which it is), then Ed will be less than Aa…

    you seem to be saying that the atmosphere can keep absorbing more IR than it emits ad infinitum without warming how is this?

    BTW There will be times and places (not all that rare) when the atmosphere is warmer than the surface for instance in the evening when the surface cools fast than the air. Then there the frosty nights and the arctic winters as in figure 5 of his latest paper (i’m not sure I agree with his use of “rare” there).

    Then there is his measurements which he does plot in an earlier paper:
    http://met.hu/idojaras/IDOJARAS_vol111_No1_01.pdf

    Fig 1 you will see some above the (x=y) regression line and some below where the air is warming or absorbing more than it emits (warming) and some above where it is emitting more than it absorbs (i.e) cooling. They are more evenly divided than one might suppose if temperature inversion was such a rare event.

    • you seem to be saying that the atmosphere can keep absorbing more IR than it emits ad infinitum without warming how is this?

      Jan, it’s because the lower atmosphere loses heat convectively. IR is NOT the only mechanism of heat transfer in the atmosphere, you know.

      They are more evenly divided than one might suppose if temperature inversion was such a rare event.

      Sighhhh. For average conditions in any geographic local, persistent inversion conditions only occur in polar airmasses over land. I am NOT talking about the fact that inversion conditions occur almost everywhere in the world over land every night, which is what is what is observed when one uses individual radiosonde profiles. I’m talking climatological averages here.

      • Anonymous says:

        Christopher Game replying to the response of Dr Spencer on 2010 Aug 27 at 6:53 AM to Jan Pompe’s comment of 2010 Aug 16 at 3:20 PM.

        Dear Dr Spencer,

        You write:
        “Jan, it’s because the lower atmosphere loses heat convectively. IR is NOT the only mechanism of heat transfer in the atmosphere, you know.

        They are more evenly divided than one might suppose if temperature inversion was such a rare event.

        Sighhhh. For average conditions in any geographic local, persistent inversion conditions only occur in polar airmasses over land. I am NOT talking about the fact that inversion conditions occur almost everywhere in the world over land every night, which is what is what is observed when one uses individual radiosonde profiles. I’m talking climatological averages here.”

        Dr Spencer, you make an important and valuable distinction. We may be getting towards an agreement. I think the following argument may be useful.

        A single climatological average profile is I think made by averaging the temperature at each altitude for a large sample of individual sounding profiles. In general, the inversions will be completely removed by the averaging process, I think. This is because the altitudes and shapes of the inversions are scattered. The 61-year dataset of averaged soundings analysed in Miskolczi 2010 has no inversions, so far as I recall (I do not have access to the pictures right now.) One can calculate the single set of radiative fluxes reaching the boundaries for such a single climatological average profile.

        A different calculation is to calculate the many respective sets of radiative fluxes reaching the boundaries for each of the many respective individual radiosonde profiles that go into the constitution of the single climatological average. Then one can average the many respective sets of boundary fluxes. The many respective sets of boundary fluxes take into account the presence of inversions in many of the respective individual radiosonde profiles.

        How to relate these two calculations? Which is the one to use to test the claim that Aa = Ed?

        The single calculation of the set of radiative fluxes for the single climatological average profile does the t^4 calculation after the averaging operation. The many calculations of sets of fluxes for the many respective individual soundings do the many t^4 calculations before the averaging operation. The averaging operations are different. They cannot be expected to commute with the t^4 calculations.

        On the one hand, if the order of procedure of calculating first the average and then the t^4 were the one of interest, one would get the result that you emphasize, that since there is no inversion, one must have Aa > Ed. But one would then have artificially removed the inversions that are there in nature.

        On the other hand, the claim that Aa = Ed is intended to refer to each respective individual respective radiosonde profile, as a good approximation. The proper order of procedure is then to do the t^4 calculations first, and then the averaging. The result is different from the climatological-averaging-first result.

        You make it clear in your “Sighhhh” comment that you are referring to the climatological-averaging-first result: “I am NOT talking about the fact that inversion conditions occur almost everywhere in the world over land every night, which is what is what is observed when one uses individual radiosonde profiles. I’m talking climatological averages here.”

        There is a well-known mathematical difference here.

        (2^4 + 3^4) / 2 = (16 + 81) / 2 = 97 / 2 = 48.5
        [(2 + 3) / 2]^4 = 2.5^4 = 39.0625

        This kind of difference will affect both Aa and Ed.

        It seems that we have been talking past each other, thinking about different things. Miskolczi 2010 tests the Aa = Ed claim by use of the TIGR2 dataset which consists only of individual radiosonde profiles, not by use of the 61-year dataset, which consists entirely of pre-averaged climatological profiles, that was used to track the time course of the global average Planck-weighted greenhouse-gas optical thickness.

        I have no problem agreeing that the order of procedure to which you refer, that of doing the climatological averaging first and then the calculation of t^4 for the boundary fluxes, must yield Aa > Ed; there are no inversions in the climatological average profile. But I note that it is an entirely artificial calculation, quite unsuited to a very precise test of the Aa = Ed claim for individual soundings, precisely because it artificially removes the inversions.

        One cannot, in one’s head, or a priori, do the calculation for t^4-then-averaging case. To test the Aa = Ed claim, one must actually do full calculations for the t^4-then-averaging case, individual profile by individual profile first, then average. It is also required by the laws of physics to take into account the reflectivity of the interface between atmosphere and the land-sea body, and the emissivity of the materials of that body. Each of these two factors respectively and cumulatively calls for a reduction of the unadjusted value of Aa, as required by the Stokes-Helmholtz reversion-reciprocity principle for reflection at an interface, and by Kirchhoff’s law for emissivity, as explained by Planck 1914. The emissivity is already built into Ed, which accordingly does not require adjustment.

        One will find that for some individual radiosonde profiles Aa Ed. The scatter is small because the individual quantities in any one individual radiosonde profile are mostly governed by the fairly small range of temperatures in the lowest 300 meters of the atmospere. The scatter is remarkably small, and the averaging gets one closer to Aa = Ed than the extremes of the scatter; just how close is an empirical matter.

        Yours sincerely,

        Christopher Game

  37. Reader says:

    thinking about K&T and wayne’s post above. if we were to measure the energy in the atmosphere at a specific instance in time, can we estimate what it would be ?

    i.e. would it be roughly 390Wm-2 for example ?

  38. Shooshmon says:

    Hey Roy,

    Most of these posts and comments are going way over my head, I would like to ask a few questions not related to this post. How can the statement be made that we are adding co2 at an unprecendented rate? I have personally concluded that this is false and that surely co2 was being added at a much faster rate during the time of the dinosaurs. Also, since we had 7,000ppm of co2 in the atmosphere and are now down to 390ppm, isn’t this proof that the earth filters out extra co2? This is why I think global warming is a hoax. I think this turns the argument into “the earth cannot and will not do something it has already done, even though it is on a smaller scale. Nobody has adequately explained this to me.

  39. Alan S. Blue says:

    Is there another post with the mechanics of “The Box?”

    Specifically: What are the rough dimensions and composition of the shield? (How deep is the well?)

  40. PJP says:

    Interesting … tried pointing my thermometer at the sky tonight: -13F (10pm), air temp ~ 64F.

  41. Jeff says:

    a cooler object/gas/vapor cannot ever heat up a hotter object … 2nd law …

  42. JAE says:

    “The Troposphere has a mass of 4,000 Trillion metric tons. It takes a great deal of energy to heat it, day and night temps vary by as much as 30 C in arid areas. 99% of the energy is within N2 and O2. CO2 only accounts for .7 grams per cubic meter where N2 and O2 account for 1300 grams at sea level.”

    Folks, please give this statement some thought!

  43. KuhnKat says:

    So, how much IR comes from the moon when it is full?? How much SW comes from the moon during night and day time?

    The amounts we are talking about for backradiation and overall CVO2 involvement aren’t particularly huge. How does the moon’s reflection of Solar radiation and emission of IR to the earth compare?

    Is it even in the models?

  44. Derek says:

    Hi Roy,
    I just logged on to your great blog and found this article. I am a little late to join the conversation, but I found the whole thing so interesting that I thought I would post this just in case you picked it up.

    I find the whole concept of heat very interesting. I am particularly interested in how gases such as O2 and N2 get hot, since if I understand the article correctly they are completely unaffected by radiation. Say we have a flask of N2 and we heat the flask with a Bunsen burner. The flask will receive radiation from the burner and get hot, but what is the mechanism for the radiation from the flask from affecting the N2 molecules? Is it simply by conduction when each molecule touches the flask wall, and then by contact collisions between molecules?

    Another question – when air expands as it rises in the atmosphere it cools. What form is the energy in which is lost and where does it go?

    I presume that the thermopile, although it gives readings in degrees C, is only able to detect radiation and so cannot detect the temperature of O2 and N2. It would therefore be no use in measuring the temperature of those gases unless they had some H2O or CO2 in them? Is that right?

    I will re-visit the blog in the hope you can help.

  45. mircea says:

    Dear Dr. Spencer

    Regarding the Ed=Aa it looks to me that the logic is as follows: It starts from observing that Sd=Aa-Qa (Sd – down radiation, Aa – accepted radiation and Qa – heat transferred from GHG to the rest of gas particles).

    Ed=Aa-Qa is valid because any radiation received from Aa that goes toward Eu in infrared form is actually Et (Infrared transmitted to space).

    If the rest of gas particles are colder than GHG particles receiving the Aa (which increase their temperature) then Qa is positive and the result is that all the gas particles increase their temperature.
    If the rest of gas particles are hotter than GHG receiving the Aa then Qa is negative and the result is that all particles decrease their temperature.

    When the system is at equilibrium then Qa = 0 and results Ed=Aa.

    Looking at the chart it results: Qa=26 for Trenberth values and Qa=12 for Miskolczi values.

    Also, in this case Eu (up radiated energy) depends only of the heat entering the system: Eu=Q+K+Qa (where Q is heat absorbed by atmosphere fron solar radiation, K is heat from earth to air through non infrared means).

    Does this make sense?

    Thank you!
    Mircea Dochia

  46. mircea says:

    Just to expand my above post:

    This means that if Qa is linear dependent of T (where T is transparency), e.g Qa is Qa=a-b*T (where a and b are constant) then Er (infrared energy radiated from ground will be independent of T. Which means that Tground (ground temperature, Tground=sigma*Er^0.25) is independent of T too.

    However, while it looks plausible I cannot say if the above are true or not.

    Mircea

  47. Christopher Game says:

    Christopher Game replying to Dr Spencer’s article Help! Back Radiation has Invaded my Backyard! posted August 6th, 2010, with a “technical detail” end note:

    What this Means for the Miskolczi “Aa=Ed” Controversy

    Except for relatively rare special cases, the total amount of IR energy downwelling from the sky (Ed) will ALWAYS remain less than the amount upwelling from below and absorbed by the sky (Aa). As long as (1) the atmosphere has some transparency to IR radiation (which it does), and (2) the atmosphere is colder than the surface (which it is), then Ed will be less than Aa…even though they are usually close to one another, since temperatures are always adjusting to minimize IR flux divergences and convergences.

    But it is those small differences that continuously “drive” the greenhouse effect.”

    Dear Dr Spencer,

    Mike Edwards commented about this at 2010 August 13 at 9:35 AM, but I did not see a reply from you to him about this. Mike wrote:

    “Strangely enough, I think that Miskolczi’s own paper shows one example where the downwelling from the sky IS greater than the amount upwelling from the Earth’s surface. This related to the winter in Antartica IR measurements, if I’ve interpreted them correctly.”

    Your “technical detail” note above seems to invite further comment on Miskolczi’s 2010 paper.

    You have also, in a previous blog article, put the challenge to answer your question:

    “You could help clarify things by answering the following question:
    If atmospheric layers A and B each contain greenhouse gases, under what conditions will we find that the rate of absorption by layer B of layer A’s thermal emission equal the rate of absorption by layer A of layer B’s emission? Your answer to that question could potentially remove all my objections to this key issue.

    You asked me to wait for a reply to this question, but one does not seem to be forthcoming.

    First may I offer my own reply to this general theoretical question and then come back to your technical detail?

    The concept of radiative exchange equilibrium is used by Planck 1914 (page 40, section 46) in preference to the currently common textbook concept of (pointwise) radiative equilibrium (Chandrasekhar 1950 page 290, Mihalas and Mihalas page 338, Goody and Yung 2nd edition page 20) for very good reasons.

    The physics of our present problem is expressed in the Stokes-Helmholtz reversion-reciprocity principle, in Kirchhoff’s law of radiative absorptivity and emissivity, in the second law of thermodynamics, and in the radiative definition of local thermodynamic equilibrium (Chandrasekhar 1950 pages 7 and 288).

    My answer to your challenge question is:

    Let any two spatial regions, A and B, of the atmosphere each be small enough to be characterized by its respective single temperature, tA and tB, local thermodynamic equilibrium conditions prevailing inside each separately of A and B. Let there be any other moderate transport processes including convection affecting A and B directly, and let there occur other forms of transport including departures from local thermodynamic equilibrium in the other regions of the atmosphere.

    Then A and B are in radiative exchange equilibrium if and only if tA = tB .

    Moreover, if two small regions A and B of the atmosphere are each separately in radiative exchange equilibrium with a third small region C, then, regardless of other forms of transport that may be occurring, A and B will be in radiative exchange equilibrium with each other.

    When two small regions A and B, each in respective radiatively defined local thermodynamic equilibrium, regardless of other forms of transport that may be occurring, are such that tA > tB, then net radiative transfer will transport heat from A to B. This is due to a slight generalization of the classical form of the second law of thermodynamics.

    I think this answer is the one you are asking for. Please correct me about this.

    Coming back to your technical detail.

    May I refer you to Figure 3 of Miskolczi 2010? There it is shown that most of Aa and Ed are respectively absorbed and emitted in the lowest 300 meters of altitude. This is a key fact for the present problem. I have not seen elsewhere precisely this kind of calculation of the altitude profiles of the contribution densities of Aa and Ed, though I think the general finding is well known.

    May I also refer you to Roland B. Stull, Meteorology for Scientists and Engineers, 2nd edition, 2000, ISBN 9780534372149 ? Stull describes temperature inversions. The relevant ones are mostly those that affect the lowest 300 meters of altitude, the nocturnal inversions. Also frontal inversions are described. They are not exceptional and not rare. You are of course long familiar with these facts, though as a beginner I am still learning about them. The point here is that we are not concerned with whole atmosphere inversions, for what matters here is the optical-thickness weighted effect. The nocturnal inversions, mostly in the lowest 300 meters of altitude, are very heavily optical-thickness weighted.

    Combining the altitude of most of the absorption and emission of Aa and Ed demonstrated by Miskolczi’s Figure 3, with the frequent occurrence of strongly optically weighted inversions as described by Stull, one sees that it is quite natural to consider a priori that cases of Aa < Ed can occur often enough, and Aa = Ed will not be too rare. Miskolczi's dataset has many soundings with inversions, an extreme case being shown in Figure 5.

    The above is not a full answer to the question of whether Aa = Ed is a good general statement. It is evident from Miskolczi's (2010) Figure 5 that Aa = Ed is not exactly true in every sounding; there is chaotic scatter on either side of equality. What is true that Aa = Ed is a rather good approximation in general and it is true much more closely on average than in individual soundings, with chaotic scatter on either side of equality. The average is the questioned case indicated in diagrams like the KT97 one that you show in your blog article on Miskolczi 2010.

    That blog article of yours also dismisses Miskolczi's (2010) emissivity factor 0.967 as a "fudge", and, since you offer no alternative suggestion, this seems to imply that you think he should use a unity emissivity and zero reflectivity. KT97 do the latter because the closeness of Aa to Ed is not a big issue for them. Perhaps you will offer an alternative suggestion now?

    But I submit to you that it is physically correct to adjust the Aa values of Table 2 for emissivity and reflectivity. The values of the factors are not so easy to know exactly. A guide may be in Briegleb 1992 cited by Kiehl and Trenberth 1997. Briegleb 1992 writes at page 11,482: "Given the computed reference values for water, ice, nondesert land, and sandy desert, the values of e that yielded best fits were chosen to be 0.97, 0.95, 0.95, and 0.88, respectively." These values bracket Miskolczi's 0.967. Briegleb apparently did not have time to read Helmholtz 1856 or Kirchhoff 1860 or Rayleigh 1894 or Planck 1914, and his values may perhaps suffer accordingly. Thus one may have some doubts about the exact values of the emissivity and reflectivity adjustments that should be applied to Aa. Perhaps you will offer your opinion about suitable values.

    I think it important to use the Stokes-Helmholtz reversion-reciprocity principle for reflection at an interface in this situation, as recommended by Rayleigh and Planck, but ignored by most modern experimentalists except Clarke and Parry (1985) Lighting Research & Technology 17 (1): 1-11, who find it accurately obeyed when read strictly in the special circumstances in which they tested it precisely.

    Thus I submit to you that it is not a priori apparent that Aa must exceed Ed always, or even on average. The closeness in the turbulent conditions of the atmospheric boundary layer, I submit to you, is a matter for empirical investigation.

    As I have just indicated, the weakest point of the empirical case for the equality holding very closely indeed on average must be the weakness in our empirical knowledge of accurate values for the emissivity of the land-sea body and for the reflectivity of the interface between the atmosphere and the land-sea body.

    You write: “But it is those small differences that continuously “drive” the greenhouse effect.”

    As I read you, perhaps misunderstanding you, you mean to say that the occurrence of the greenhouse effect requires by definition that Aa > Ed because this will ensure that there is upwards ‘radiative diffusion’, so to speak, of heat through the atmosphere, which is, as I perhaps mistakenly read you, your definition of the greenhouse effect. Perhaps I misread you here. Please correct me. My understanding is that the dry adiabatic lapse rate is approximately accounted for not by the presence of greenhouse gases but simply by hydrostatic equilibrium and the ideal gas law, as stated in section 2.7.2 on page 51 of Holton, 4th edition, 2004, ISBN 9780123540157.

    The occurrence of Aa = Ed depends on inversions at low altitudes where there is a large optical density, and above these in the troposphere the temperature lapse rate is predominantly normal while the optical density is lower. Thus Aa = Ed is quite compatible with upwards ‘radiative diffusion’, so to speak, at higher altitudes in the troposphere.

    Other definitions of the greenhouse effect are by Inamdar and Ramanathan 1997 ( G = Sg – OLR ) and by Stephens Slingo and Webb 1993 ( G = Sg / OLR ). These definitions do not require that Aa > Ed , but require only that Sg > OLR .

    Yours sincerely,

    Christopher Game

  48. Christopher Game says:

    Christopher Game responding to Dr Spencer’s article Help! Back Radiation has Invaded my Backyard! posted at 2010 Aug 6.

    Dear Dr Spencer,

    You write: “The IR thermometer was measuring different strengths of the greenhouse effect, by definition the warming of a surface by downward IR emission by greenhouse gases in the sky.”

    This looks like a considered definition of the greenhouse effect. Is this your most considered defintion of the greenhouse effect?

    It is not the only definition of the greenhouse effect. For example Inamdar and Ramanathan 1997 define the greenhouse effect quantitatively by G = Sg – OLR, and Stephens Slingo and Webb define it quantitatively by G = Sg / OLR.

    Does your above definition have a quantitative formulation? If so, please very kindly tell us what it is. Do you use a quantitative definition of the greenhouse effect? If so, please very kindly tell us what it is.

    Yours sincerely,

    Christopher Game

  49. Ned Nikolov says:

    Dear Dr. Spencer

    I’ve enjoyed reading your article about ‘greenhouse radiation’ as well your other postings. I have talked to you on the phone a few months about the ‘greenhouse effect’. If you remember I mentioned that I’m working on a comprehensive research paper on this subject. My draft manuscript is almost finished now (about 40 pages single space). I’d be happy to send it to you for review towards the end of this month, if interested. I believe it will answer fully all your questions regarding the ‘greenhouse effect’. But here is the bottom line:

    1. Yes, the atmosphere provides an extra warmth to the Earth surface, but this atmospheric thermal effect has nothing to do with trapping or absorption of outgoing infrared radiation. It’s a totally different mechanism that’s responsible for the extra heat.

    2. The atmospheric thermal effect is not 33C as claimed by the current theory, but actually 133C .. Yes, the way these 33C have been traditionally calculated is mathematically wrong!

    3. The current ‘greenhouse’ theory violates the First Law of thermodynamic big time, because observations show that we at the surface receive 44% more radiation from the troposphere than from the Sun. Specifically, the average down-welling LW flux from the troposphere is 343 W m-2 while the total shortwave flux absorbed the ENTIRE Earth-atmosphere system is 239 W m-2 (343/239 = 1.44). In other words, the lower troposphere contains much more energy than provided by the Sun. How can this be explained in the context of the current theory given that the heat storage capacity of air is negligible?

    4. Thermal radiation cannot be trapped by substances of high absorptivity. This fact is well known in the insulation industry, where radiant heat loss from buildings is reduced by using ultra-low absorbing materials such as aluminum. These material reflect LW radiation. There are no such reflective materials in the atmosphere, and because IR travels with the speed of light, trapping of radiant heat in the free atmosphere is physically impossible.

    5. The term ‘greenhouse gas’ is a complete misnomer. There are no gases that can cause a NET warming. Changing the chemical composition of a gas mixture cannot increase the gas’ internal energy and lead to higher temperature. This is a high-school level physics.

    4. It it physically insane to think that altering the chemical composition of 0.04% of the atmospheric mass could increase the Earth’s equilibrium surface temperature by 3.2C as the latest IPCC fantasy claims.

    Overall, the current ‘greenhouse theory’ is a vivid example of a physical absurdity taken to the extreme. But it also reveals a lot about human nature and the so called ‘scientific method’ as implemented by humans, for it is mind boggling how such an unphysical theory could survive 180 years, and even become the basis for policy decisions that will dramatically impact the world’s economy and the well-being of humanity.

    Cheers,
    - Ned