Yes Virginia – Back-Radiation delivers measurable heat – just not very much

Posted: April 14, 2012 by tallbloke in atmosphere, Energy, methodology, weather

Tim C alerts us to this neat bit of empirical work in progress which uses a peltier to measure the warmth generated by cloud cover. Peltiers are those neat little gadgets used to cool computer CPU’s and miniature fridges. You supply power and a heat sink/fan on one side, and they ‘pump’ heat from one surface to the other. They also work in reverse as ‘TEG’s – Thermo-Electric Generators. If you heat one side while keeping the other cool, they generate electricity. This is the ‘Seebeck effect’. I use a couple on my backpacking trips to recharge AA batteries in the winter when my 100g 3W solar panel fails to get the job done. I have two fixed to a heat sink with thermally conductive glue.  The heat sink sits on my 100g woodstove and a pan of cold water sits on top of the peltiers.


Building a Thermoelectric Sensor to Measure
IR Sky Transparency for Cloud, Cirrus and Dust Detection, at Night
Bill Gillespie
Kitt Peak National Observatory
Observing Support Office

This style of cloud detector was invented by Lou Boyd, of Fairborn Observatory in Arizona. He came up with the idea of running a Peltier device, backwards, to measure the sky temperature. The thermolectric (Peltier device) based cloud detector consists of two aluminum plates sandwiching the thermoelectric device inside. The device functions as a night time cloud detector, because it generates a a measurable voltage, when there is a temperature variation across it’s two aluminum plated sides.

To detect clouds, one plate is positioned facing the night sky, while the other plate faces the ground. Infrared radiation from the ground warms the ground facing side of the detector, while the opposite sky facing side either sees a warm, clouded sky, or a cold clear one. On clear nights, a warm ground, and a cold, cloudless sky, produce a voltage on the device because the two plates have different ambient temperatures.

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Under tests, sun shining directly on the cloud detector
produced around 60 millivolts of voltage potential.

Other test results follow:

100% cloudy night = – 0.56 mv

075% cloudy night = – 1.13 mv

050% cloudy night = – 1.14 mv

025% cloudy night = – 1.40 mv

000% cloudy night = – 1.50 mv

Soaking wet detector -1.45 mv

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When clouds roll in, the sky temperature increases due to the clouds emission of infrared radiation, causing the temperature variance between the two sides of the detector to decease. In other words, clouds can be detected at night because their presence causes a voltage decrease in comparison to the detectors voltage output during clear skies.

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In order to monitor these voltage variations, the cloud detector’s voltages are turned into digital pulses, which are sent to a computer through a serial line. A python program will monitor the pulses and generate a graphical report of the current and recent cloud trends. The report will be available on the network for web based cloud monitoring at night.

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Building the circuitry to turn the voltages into pulses, and the microcontroller to turn the pulses into ascii for transmission to the receiving computer is the next phase. The computer will run a Python program to read the ascii from the microcontroller on the cloud detector end of the serial line. Once the values are operated on a log file will be generated and the data set up to be read by the mountain network, so that telescope operators can view the current cloud conditions from the web.

 

Comments
  1. Richard111 says:

    Brilliant! Must try and build one myself. Last year I placed a half square metre of 3mm mild steel sheet outside on a polystyrene sheet supported about a foot above the ground, A thermcouple was bolted to the middle of the steel sheet. At dawn, before the sun rose but the sky above was turning blue, the temperature of the steel sheet rose BEFORE the local air temperature.
    My claim is that was backradiation from WATER VAPOUR in the clear sky above.

  2. tallbloke says:

    Richard: Does your claim exclude CO2 completely? If so, is that because you think the water vapour is absorbing energy from sunlight whereas CO2 isn’t?

  3. Stephen Wilde says:

    I’m inclined to the view that so called backradiation is a consequence of the temperature of the air just about the surface exchanging energy dynamically with the ground below.

    That layer of warm air above the surface insulates the surface from any backradiation from higher up by itself having its own dynamic interaction with the air directly above it and so on upwards That would work on a molecular level.

    If a cloud comes over then the real effect is in raising the temperature of the air just above the surface to a higher level than it otherwise would have been by either reducing the rate of radiative heat loss upwards or by carrying along warmer air in the layer of air beneath the cloud.

    I have a difficulty envisaging that the cloud itsel actually radiates downward UNLESS all the air between surface and cloud and the ground itself is below the temperature of the cloud droplets. And even then it would be the temperature of the air directly above the surface that commands the surface/air energy exchange by interposing itself between cloud and surface.

    Generally the lapse rate ensures that the cloud droplets are cooler than the temperature of the air at the surface and of the surface except where there is a temperature inversion.

    At this point someone usually brings in the non GHG scenario but in that case the mechanics are different and so not helpful in diagnosing our real world situation where the air nearest the surface is packed with GHGs (mainly water vapour) and aerosols.

    The net radiative flow from surface to cloud still changes but only because of the warmth of the air just above the surface and not from direct radiation downward from any higher up.

    So there is backradiation from just above the surface but from no higher. Anything from higher up gets diverted back upward from interposing air molecues which have warmed via conduction from all around them (non GHGs) or via radiation from all around them (water vapour, other GHGs and aerosols) all of which then send it back up again before it reaches the ground.

    Isn’t the picture of a cloud radiating energy toward the ground as if there were nothing between it and that ground completely misleading in that it focuses the attention on the cloud as the energy source rather than the air molecules just above the ground ?

  4. tallbloke says:

    Stephen:

    Under a clear sky, the top sensor surface gets colder than the bottom. Bigger differential, more voltage. More radiation is hitting the underside of the sensor from the ground than is hitting the top from the sky – yes?

    Cloud can form in situ without “bringing warmer air in with it”, so when that happens, it must be a slowing of the cooling rate of air near the ground which reduces the differential and reduces the generated voltage – yes?

    That slowing of the cooling rate has to be due to downwelling radiation from the cloud. Although the cloud isn’t warmer than the ground, it is warmer than space and the clear air under it and cloud has a greater density of water in it than the air below and space above. So the flux density of downwelling radiation will be greater and this will more vigorously oppose the upward radiation flux from the ground, slowing its escape more than open skies will. Maybe it’s better to think of the radiation as a wave structure rather than as photons to see that, though some say E/M waves are exempt from interfering with each other?

    That seems to be the logic of the situation, but maybe I’m wrong.

    So I agree that the diagram over-simplifies the situation, but the net result is due to downwelling radiation from the cloud nonetheless.

  5. Mydogsgotnonose says:

    You aren’t measuring ‘back radiation’. it doesn’t exist**.

    You are measuring the impedance to IR transmission from the upper plate to space. This increases under clouds because ‘Prevost Exchange Energy’ from the cloud to the plate increases, for a given atmospheric temperature because the emissivity increases. That means less radiative energy
    is lost by the upper plate.

    For thick clouds, emissivity ~0.8, for a clear sky [UK, humid] it’s ~0.2, desert ~-0.1. So, at -10°C, a cloud will produce 217 W/m^2 and a clear desert sky will produce 27 W/m^2. This will be picked up as an increase of temperature difference hence emf [more -ve].

    **Why does ‘back radiation’ not exist? The best demonstration is to explain how real radiation problems are solved. The radiative flux emitted by an isolated body in a vacuum is set by the Stefan-Boltzmann equation, corrected for emissivity. The flux hotter to colder between two bodies at different temperatures is the difference between the two S-B terms corrected for geometrical ‘View Factor’.

    What you have done is to eliminate a great part of the isolated body S-B terms for each body by summing it to net zero, the Prevost Exchange which does not thermodynamic work. When however, you point a radiometer at the colder body, you shield the detector from the energy from the hotter body so magically, you get a signal apparently coming from the colder body to the hotter. Until you put the radiation shielding in the way, this did not exist!

    The fact that ‘climate scientists’, meteorologists and climatologists Worldwide imagine that a pyrgeometer measures a real energy source shows that there are many 1000s of scientific fools so poorly trained in the basics of radiation physics that they have fallen for what is almost certainly in my view, a massive fraud.

    The root climate modelling paper, Manabe and Wetherald 1967 assumed LW up = SW down, a gross exaggeration but OK for a start [you need ~100 °C before radiation > natural convection]. At some time, and I’m trying to find why, someone stated that the radiative flux from the Earth’s surface is the S-B value for a black body in a vacuum, and the missing 333 W/m^2 [Trenberth et. al. 2009] is imaginary ‘back radiation’. No professional scientist should have made that mistake so I suspect it was done deliberately. What it does is to increase energy by a factor of 2.6, a perpetual motion machine. The IR absorbed by the atmosphere by non-existent direct thermalisation increases by a factor of 15.5!

    This in my view is probably the biggest scientific fraud in history.

  6. Ulric Lyons says:

    Stephen Wilde says:
    April 14, 2012 at 10:03 am
    “I have a difficulty envisaging that the cloud itsel actually radiates downward UNLESS all the air between surface and cloud and the ground itself is below the temperature of the cloud droplets.”

    This may help..

    https://courseware.e-education.psu.edu/courses/meteo101/Section4p05.html

  7. Mydogsgotnonose says:

    Ulric: the article to which you refer makes the Big Mistake of confusing impedance to IR transmission with ‘Downwelling IR’.

    There can be no heat transfer from a colder body to a hotter body.! Climate related sciences are unique in claiming there is. It is thermodynamically impossible because it’s all to do with statistical thermodynamics and the proportion of activated states from the interior of a body [kinetic energy] and the outside [radiation, convection/conduction] and Prevost Exchange is what regulates it all.

    This is why every process engineer who sees the Trenberth/IPCC Energy Budget blurts out ‘How could they be so stupid’.

  8. tallbloke says:

    From Ulric’s Penn State link:
    “Don’t be too concerned with specific values of downwelling IR and the units. Most of the time, I simply use this plot to qualitatively assess the radiative effects of nighttime clouds.”

    Lol. :)

  9. steveta_uk says:

    MDGNN, I’ve been very curious about your ideas some some months now but cannot locate any references to support what you are saying.

    For example, a google search for “Prevost Exchange Energy” has only a few 100 hits, and as far as I can tell they are all from you, or a few other blog commenters (SpartasucIsFree, for example) who’s writing manner is so similar I assume it is also you.

    All I can find about Prevost’s real ideas are just a very primitive description of radiative heat transfer that preceeds any real descrptions by nearly 100 years, which means Prevost was well ahread of his time, but that dones’t make him correct – he talks of heat as a fluid moving between objects to bring them into thermal equilibirum, but I can’t find the part about “cannot do any work” that you always quote.

    And why bother with Prevost anyway? Surely Maxwell, Plank, Boltzman, etc., would be better sources? If not, surely this casts a very large amount of doubt on Prevosts view of the world.

    So please provide some references that will allow me to work out why you are so sure that back-radiation is a myth.

  10. tallbloke says:

    Mydogsgotnonose says:
    April 14, 2012 at 10:51 am

    You aren’t measuring ‘back radiation’. it doesn’t exist**.

    You are measuring the impedance to IR transmission from the upper plate to space. This increases under clouds because ‘Prevost Exchange Energy’ from the cloud to the plate increases, for a given atmospheric temperature because the emissivity increases. That means less radiative energy
    is lost by the upper plate.

    Thanks Mydog, that sounds like a clearer explanation of what I was trying to say at 10.41

  11. Mydogsgotnonose says:

    Hi Steveta: http://en.wikipedia.org/wiki/Radiative_equilibrium

    The importance of Prevost’s law is that it started out on the road to moderrn physics. The article refers to its modern nomenclature as ‘photon gas’ or ‘electromagnetic radiation.

    In short, ALL the S-B flux from a colder body in the view factor of a hotter body is exactly nulled by a proportion of the opposite flux and what’s left over is the net flux. What you do when you point a pyrometer at the colder body is to reveal the S–B emission from the colder body to the hotter body. <b<Until you put the shield in place, that Prevost radiative flux did not exist.

    This is so elementary a concept that I cannot for the life of me understand why so many people imagine it can do real thermodynamic work [which is to be converted to kinetic energy at the hotter body].

    Now I could be being stupid and dogmatic, but can you fault my argument? Sorry if I am going on but this Emperor has no clothes and with the totally wrong claim in climate science that IR from the Earth’s surface is the S-B black body level, it has made climate science’s predictions totally useless. And you must realise that the 3-5 times greater predicted heating than reality even if you assume modern warming is GHG-AGW, which it ain’t, is offset by imaginary cloud aerosol optical physics, except for thin clouds.

    Over to you!

  12. Richard111 says:

    TB: I don’t have any qualifications to make a precise claim. What I look at is the IR signature bands of H2O and CO2. Next I make an assumption that as sunlight reaches the upper atmosphere and Rayleigh scattering shows up as blue sky then near IR will be affecting H2O. CO2 might be seeing something at 2.7 microns but that is only one band. H2O will be affected by five bands.

    The point of my experiment being that before sunlight reaches the surface, ‘something’ is warming the plate. Again I make the assumption this is ‘back radiation’. What else could it be? Next time we gat a period of cloudless days and nights I will run the experiment again.

  13. tallbloke says:

    Mydog says:
    When however, you point a radiometer at the colder body, you shield the detector from the energy from the hotter body so magically, you get a signal apparently coming from the colder body to the hotter. Until you put the radiation shielding in the way, this did not exist!

    And conversely when you point the device at the ocean surface you’ll see 390W/m^2 rather than the ~65W/m^2 ‘net flux’ from ocean to air.

    I think this is why NASA’s latest energy budget diagram doesn’t show the mythical ‘upward’ and ‘downward’ components separately like the old Keihl-Trenberth cartoon.

  14. Stephen Wilde says:

    TB said:

    “it must be a slowing of the cooling rate of air near the ground which reduces the differential and reduces the generated voltage ”

    Yes indeed. Slowing the cooling rate of AIR NEAR the ground. Not the ground itself.

    And as we have seen from N & K (and all the prior science about the standard atmosphere) the temperature of the air near the ground itself is determined by pressure plus insolation.

    So if the air near the ground tries to get warmer from any cause other than insolation or pressure the consequent expansion will reduce the temperature proportionately (lower density at the surface) and the effect of a greater energy input to the AIR NEAR the ground from clouds will be a net zero and so the ground surface temperature change caused by the cloud must be net zero too.

    And it isn’t a matter of the cloud radiating directly to the surface as if they were two distinct items separated by a vacuum, it is a matter of the cloud dropets both radiating and conducting to each lower molecule in sequence downward ultimately to the air molecules nearest the surface.

    So the cloud doesn’t warm the surface at all. The additional warmth in the the air between surface and cloud causes expansion to negate the warming effect on the surface that would otherwise have occurred.

    Cloud over water is a different scenario. In that case there are complications caused by increased humidity developing under cloud cover (each cloud floats on its own bubble of humidity) so as to reduce the rate of evaporation from the water surface which gives a warmer water surface and that gets dealt with by changes in windspeed and direction as per Hans Jelbring’s hypothesis.

    The point being that one doesn’t need to propose downward radiation from any point above the molecules that are directly above the surface whether it be land or sea. The rate of surface cooling is simply a consequence of the presence of an atmosphere of a given mass and density plus the level of insolation.

    Radiation up, down or laterally within that atmosphere is a consequence of the temperature achieved by pressure plus insolation and not a cause of that temperature.

    Note that I’m just playing with the concepts in this post so there may be a flaw and I await hearing as to what it is from the more capable physicists here.

  15. tallbloke says:

    Richard, read what Mydog is saying carefully. I think he’s right. The water vapour absorbs energy from sunlight. This increases the emissivity of the sky, and that ups the ‘impedance’ against energy leaving the ground.

    The reason why the rate of cooling of the ground is slowed by a colder object (the cloud) makes clearer sense to me in the terms Mydog casts it.

    The way I see it now is that there is radiation, it is buzzing about in all directions, and there’s more of it when there is an object with higher emissivity up in the sky than when it’s clear. This increases ‘impedance’ and slows the rate at which the upward component percolates to space via a myriad number of interaction with air molecules and water vapour (and 0.039% CO2).

    Right Mydog?

  16. Mydogsgotnonose says:

    Thanks Tallbloke for the ocean upwelling idea. But remember that the Trenberth cartoon has in it the apparent fixed point in the climate models that the IR from the earth’s surface is the S-B level in a vacuum rather than the the measured value of 23 W/m^2, also this was not in Manabe and Wetherald.

    The Ocean gives 390 W/m^2 temperature signal because it is effectively a black body. However, the real energy transfer to the air will be mostly evaporation plus some convection and a much small amount of radiation than the S-B level.Also, it will be a torla of 160 W/m^2.

    Go to engineering texts, e.g. McAdams, and you’ll see that because convection and radiation are coupled [meaning the same activated sites can transfer either way] you have to use an empirical heat transfer coefficient that is a function of the lateral air velocity, hence the convective boundary layer thickness.

    I can prove this with the windbreak story. I’m on the beach and the air temperature is 25 °C, sand temperature 30 °C, not too comfortable. So, I put up a windbreak and the sand temperature rises to a nice 45 °C as radiation increases to make the sum of that and convection equal to the Sun’s SW energy coming in.

    Assuming 0.85 emissivity of the sand, 0.2 absorptivity/emissivity for the air [it’s sunny, so it’s got no clouds] also that half the increased radiative flux comes back as ‘back radiation, really increased impedance to IR flux from the earth to space, and you have just increased ‘back radiation by ~5 times net GHG-AGW! On that basis the UHI is responsible for global warming!

    Another trick question i have developed is to ask why, if ‘back radiation’ at 333 W/m^2 is real, and it’s twice solar SW heating of 160 W/m^2, don’t passive solar panels work at night?

    Yesterday a warmist argued that it’s because you need the energy from the SW to make the back radiated LW heat the panel contrary to the 2nd law of thermodynamics. I then told him that he was claiming the S-B constant is a variable dependent on local atmosphere composition! He went away to think……

  17. tallbloke says:

    Stephen: I think what you are saying is pretty near when you look at the situation in the large. Here we are dealing with ‘local effects’ where radiation is playing a role, until it gets mitigated by the processes you are describing.

    So, clouds form at night, there is an effect due to radiation, but:
    1) It is a pretty small effect as the test results show, and
    2) The other processes in the atmosphere start working to mitigate it anyway.
    3) Nonetheless, the whole night will be warmer under cloud cover than a clear sky.

  18. tallbloke says:

    Mydog: where does this 23W/m^2 measurement come from? and what the heck is a Torla? :)

    “Also, it will be a torla of 160 W/m^2.”

  19. Mydogsgotnonose says:

    Tallbloke; you are correct about the impedance issue. The physicists think of it as the ‘photon gas’ being reflected at filled sites on the hotter body which, because it is hotter transfers more energy from the interior to those states than are filled by the photons from the cooler body.

    The other key concept is that climate science makes the elementary mistake of imagining that an IR activated GHG molecule [the activated state for the atmospheric radiative body] remembers its history. It doesn’t and as soon as another thermally activated GHG molecule emits the same energy photon in a random direction, Local Thermodynamic Equilibrium is restored.

    This is why there cannot be any direct thermalisation. I suspect all the experiments involve finite vessels with thermalisation of pseudo-scattered IR at the walls. Nahle has reportedly shown that using a Mylar balloon, so walls a few thou’ thick, there is no warming. So, in the atmosphere, the GHGs act as a heat transfer fluid to heterogeneous interfaces or space. The inference is that all the atmosphere even over the horizon is in the view angle of the earth’s surface.

    Basically climate science has completely cocked it up IMHO!

  20. Mydogsgotnonose says:

    23 W/m^2 is from Trenberth et al 2009.Energy Budget. It’s the 63 W/m^2 measured net IR from the Earth’s surface minus the 40 W/m^2 that goes thrrough the atmospheric window.

    Torla is my version of total.

    Since Trenberth/IPCC claims to the 23 W/m^2 you add 333 W/m^2 ‘back radiation’, the IR absorbed in the first 30 m of the atmosphere is increased 15.5 times reality. This is an appalling fraud IMHO.

  21. JuergenK says:

    Richard111 says:
    April 14, 2012 at 8:50 am
    … At dawn, before the sun rose but the sky above was turning blue, the temperature of the steel sheet rose BEFORE the local air temperature….

    Just before sun rises air reaches it’s dew point. Water vapour releases it’s energy and condenses. Maybe that’s why temperatures often rise before sunrise, but …

    I did a similar experiment. During day I measured the temperature with a pinpoint thermometer directly on ground. My measuring place was shielded from direct sunlight.

    I recorded the differences between open sky and shielded sky at different times and at different cloudy conditions.

    The temperatures rose when the themometer saw the sky and fell immediately when shielded.

    Temperatures rose only a tiny 0.1 degree when the sky was overcast and grey. Temperatures rose up to 0.6 degrees, when the sky was blue.

    I suppose that’s caused by scattered down dwelling blue light , not by infrared light.

  22. tallbloke says:

    Kitt Peak is 6,875 feet above sea level, high above the Sonoran desert in Arizona in nice clear air. It must get a pretty big dose of solar radiation during the day. It looks like it was a nice sunny day when the photo of the test does in sunlight was taken too.

    For a typical cloudless atmosphere in summer and for zero zenith angle, the 1367 W m-2 reaching the outer atmosphere is reduced to ca. 1050 W m-2 direct beam radiation, and ca. 1120 W m-2 global radiation on a horizontal surface at ground level.

    So lets assume the 1120 W/m^2 and compare the 60mv generated in the sunshine test result with, say the 1.15mv 50% cloud cover result to mimic the global cloud albedo. We need to add on say 33% (half is radiated to space and it’s a clear day not at average cloud cover) of the 16% of incoming 1367W/m^2 TOA solar shortwave for the daytime absorption in the atmosphere too, so maybe somewhere around 1193W/m^2 total.

    Wattage due to ‘back radiation’=1.14*(1193/60) =22.7W/m^2

    This is satisfyingly close to Mydog’s 23W/m^2 derived from the Keihl Trenberth cartoon.

    I declare this a provisional win. :)

  23. Mydogsgotnonose says:

    Thanks Tallbloke.

    Of course, you must realise that because of this scientific lunacy, which appears to come from Meteorology [Trenberth did not do Physics – Maths and Meteorology], climate science exaggerates the warming of the first ~30 m pf the atmosphere [the IR absorption depth] by [333+23]/23 = 15.5!

    I repeat, scientific lunacy coupled with arrogance coming from being funded by lots of money from politicians who have wanted to create the new World currency, the Euro/Amero and perhaps the ‘Ausso’, on the back of carbon taxation and trading of the new commodity, also evil neo-colonalism, appropriate for the Belgians, of the Eucalyptus plantations in Africa for ‘carbon offsets’.

  24. tallbloke says:

    Mydog: Small thanks to me; Big thanks to you and everyone who contributes here for invigorating my crash damaged brain. I just generate conjecture and do easy sums on calculators. :)

  25. davidmhoffer says:

    Confronted with an actual valid experiment showing back radiation exists, the nay sayers assemble arguments of a complexity bordering on absurdity to explain away the actual data. IR can, and has been in many papers, measured directly. Arguing that there is no back radiation and coming up with explanations as to why that are so complicated that they would make Occum LOL is the very hijacking of science seen every day from the CAGW crowd.

    Sad that there are skeptics so devoted to defeating the CAGW fraud that they have turned to the same tactics of misrepresentation and determined confirmation bias as the fraudsters themselves.

  26. Mydogsgotnonose says:

    davidmhoffer; can you provide references showing how the ‘back radiation’ was measured?

    remember, as a process engineer I have measured ad predicted coupled conduction and convection and helped develop a number of pyrometer designs for on-line measurements in metallurgical plants so i am hard-wired with the physics in a way few others alive can be.

    No process engineer believes the claim, and i have communicated around the World with these experts.

    Please tell me whether you understand that the very act of shielding one half of the energy exchange in the colder direction automatically reveals what climate science claims is ‘back radiation’?

    If you do understand this, please join us in decrying this appalling error which has had very serious consequences.

  27. davidmhoffer says:

    Mydogsgotnonose says:
    April 14, 2012 at 5:35 pm
    davidmhoffer; can you provide references showing how the ‘back radiation’ was measured?
    >>>>>>>>>>>

    Search for downwelling IR and you will find plenty of papers using a variety of techniques.

    Mydogsgotnonose says:
    Remeber, as a process engineer>>>>

    Appeal to authority. A favoured and much discredited tactic of the CAGW crowm.

    Mydogsgotnonose says:
    I have measured ad predicted coupled conduction and convection and helped develop a number of pyrometer designs for on-line measuremnts in metallurgical plants>>>>>>>>>>

    Appeal to authority.

    Mydogsgotnonose says:
    so i am hard-wired with the physics in a way few others alive can be.

    Appeal to authority.

    Mydogsgotnonose says:
    No process engineer believes the claim, and i have communicated around the World with these experts.>>>>>>>>>>>>

    Appeal to authority coupled with an unverifiable claim of support from anonymous sources. By the way, I used to do pre-sales technical design in for electronic components for several semi-conductor companies and my customers ranged from aerospace to smelters. In order to do my job, I had to understand the objectives of the design engineers, the physics involved, and the characteristics if the discrete components (analog and digital) involved. I dealt directly with PhD’s in a vast variety of fields from geology to physics to chemistry and plenty of process engineers in the mix.

    Mydogsgotnonose says:
    Please tell me whether you understand that the very act of shielding one half of the energy exchange in the colder direction automatically reveals what climate science claims is ‘back radiation’?>>>>

    I’ll tell you that SB Law states that a body at 300 K radiates at 459.3 w/m2.
    If it is exposed to deep space which has an effective temperature of 2.7 K, it radiates…. 459.3 w/m2.
    If it is exposed to another body which is also at 300 K, it radiates at….459.3 w/m2.
    If it is exposed to another body which is at 10,000 K, it radiates at…. 459.3 w/m2.
    If it is exposed to the sun, at millions of degrees K, it radiates at…. 459.3 w/m2.

    Wave your arms all you want, these are the facts. A body at a given temperature and a given emissivity radiates at a given w/m2 NO MATTER WHAT ELSE IS GOING ON AROUND IT.

    Clouds are composed of water droplets. They have a temperature and an emissivity. They radiate on that basis. Where the radiation goes, they couldn’t care less, and it cannot just disappear. It must travel in a straight line until it strikes something that can either absorb or reflect it, and it doesn’t give a damn what the temperature of that something is.

  28. Stephen Fox says:

    A couple of questions.

    TB, you put ‘The other processes in the atmosphere start working to mitigate [the radiation effect] anyway.’
    If the upper air were warmer, would that not likewise inhibit the other processes, by which I assume you mean conduction/convection?

    Stephen, you wrote ‘So the cloud doesn’t warm the surface at all. The additional warmth in the the air between surface and cloud causes expansion to negate the warming effect on the surface that would otherwise have occurred.’ Did you mean ‘the cooling effect on the surface’?

    Thanks for another interesting thread all.

  29. tallbloke says:

    459.3

    Not if more than half the energy is leaving the surface via other means it isn’t.

  30. davidmhoffer says:

    tallbloke says:
    April 14, 2012 at 6:36 pm
    459.3
    Not if more than half the energy is leaving the surface via other means it isn’t.>>>>

    Wrong. If energy is leaving the surface by other means, then the amount of energy required to MAINTAIN the body at 300 K changes. Provided that the body is maintained at a temperature of 300K, the amount of energy it radiates is 459.3. How much is leaving by conductance or other processes is immaterial.

  31. Stephen Wilde says:

    Stephen Fox asked:

    “Stephen, you wrote ‘So the cloud doesn’t warm the surface at all. The additional warmth in the the air between surface and cloud causes expansion to negate the warming effect on the surface that would otherwise have occurred.’ Did you mean ‘the cooling effect on the surface’?”

    I see the ambiguity.

    I mean that the expansion fails to result in cooling that would otherwise have been caused by that expansion because there is extra warmth in the air AND

    that the expansion prevents the warming of the surface that would otherwise have been caused by the extra warmth in the air

    Thus a net zero outturn.

  32. Stephen Wilde says:

    Does the cloud detector record downward radiation from the cloud itself or downward radiation from the temperature of the air molecules directly above or in contact with the sensor ?

  33. Bryan says:

    Perhaps I’m missing something here, but all I can find is heat being transferred from the source(warmer ground) to a sink(colder sky).

    “Infrared radiation from the ground warms the ground facing side of the detector, while the opposite sky facing side either sees a warm, clouded sky, or a cold clear one. On clear nights, a warm ground, and a cold, cloudless sky, produce a voltage on the device because the two plates have different ambient temperatures.”

    This is similar “proof” to the satellite spectrogram showing bites around 15um.
    All that shows is thermalisation of upward radiation from the warmer Earth surface.

    Neither prove or disprove backradiation.

    Backradiation exists but as Mydogsgotnonose says above since it cannot do thermodynamic work it cannot be called heat.

  34. Mydogsgotnonose says:

    Let’s take the hypothetical case of two plates with perfectly insulated backs at the same temperature, 300 K ,with a view factor of unity, unit emissivity, and absorptivity,.What you are saying is that they each emits electromagnetic radiation at a rate of 459.3 W.m^2. The implication [shown in the Trenbetrth ‘cartoon’] is that that flux is absorbed by the opposite plate and re-emitted. Thus the new emitted flux is 918.6 W/m^2. This then adds to the energy being emitted from the other plate making 1,337.9 W/m^2………………………..

    Congratulations, you have made your first supernova. Did you learn that from Blue Peter [a UK children’s programme.]?

    The reality is far more subtle. What you must understnd is that before EM energy can be converted to extra kinetic energy [=heat], and vice versa, you must create an activated state of the right energy level either on or near the surface [dense matter] or withing the volume [gas]. The key issue is that you then impose the Principle of Indistinguishability devised by J Willard Gibbs in the 1890s. This states that statistical thermodynamic entities have no memory so behave identically to all other quantum entities.

    So that activated state can either re-emit a quantum of EM energy [called a photon by Planck] or it can be added to internal kinetic energy with equal probability. The net energy flow, in or out, depends on whether new empty states are filled from internal or external energy..

    So, we have the two plates at 300 K: they are losing no heat, neither can they gain any. It’s easy to understand. No net energy transfer from inside to outside and vice versa means an equal number of the activated states is being filled per unit time from within as without and all the states are filled. Thus 50% of the energy in those states comes from ‘photons’ emitted by the other plate but that energy is immediately replaced by identical photons emitted in random directions from those energy states. This is why physicists think of a photon gas. Thus there is no net transfer of EM energy to kinetic energy at either plate, no thermodynamic work. The net energy transfer from each plate is zero, not 459.3 W/m^2.

    Now reduce the temperature of Plate 2 to 290 K. Because of Wien’s Displacement Law, the statistics change. A number of the sites at Plate 1 cannot be filled by ‘photons’ from Plate 2 because its upper energy cut-off is lower. Therefore those states can only be filled from kinetic energy.. There are second order effects but you see the principle I hope: there is now net heat transfer from Plate 1 and it will transfer energy to plate 2 until they are of equal temperature again. So, net energy transfer starts at SB1-SB2 = 58.25 W/m^2 then falls to zero.

    At that new equilibrium there is no net energy transfer because half the ‘photons’ emitted are replaced by the same number coming from the other plate. Now put a radiometer facing Plate 2’s internal face. Because it shields the radiation from Plate 1 and by definition has to act as a black body,in terms of counting every photon from Plate 2, it measures a flux of 401.1 W/m^2, the S-B level for that temperature for a black body. However, without the radiometer in place, net energy transferred to plate 1 is ZERO.

    Now pay attention to the Trenberth cartoon. It claims 396 W/m^2 is emitted from the Earth, the black body S-B level for 16 °C when the real level is 68 W/m^2. the difference, 333 W/m^2 is claimed to come from the atmosphere as ‘back radiation’, the S-B level for a black body at ~3.7 °C. That is lunatic physics because emissivity of the atmosphere will be <~1.

    So the claim that 356 W/m^2 is absorbed by the GHGs in the ~30 m of the atmosphere near the earth’s surface is a factor of 15.5 too high, sheer scientific lunacy. I do hope you read this and understand it because your claims are only true for a black body in a vacuum with no other emitter in line of sight. For all other cases, the flux is less, and at equal temperatures, net zero with the Prevost Exchange energy communicating between the two emitting/absorbing density of states to control emissivity relative to absorptivity. .

    There are a few more complicating issues in that S-B actually gives the total emission from the activated states, half in and half out, but for a real body there is only one surface so the inward energy cancels out. In the case of a perfectly insulated back, the complete s-B energy flux adds 100% to the internal kinetic energy being transferred to the single emitting face! Also, in transient heat transfer, Kirchhoff’s Law of radiation does not apply.

  35. Tenuc says:

    tallbloke says:
    April 14, 2012 at 6:36 pm
    459.3

    Not if more than half the energy is leaving the surface via other means it isn’t.

    Yes, and because a body radiates in all directions, let’s not forget those photons being radiated back into the surface of the land, the energy of which is moved rapidly down by conduction.
    Confounding issues here are what happens at the frozen poles, ground moisture level, type/level of vegetation cover and seasonal changes.

    The deep ocean responds differently as the wrongly named ‘back radiation’ is thermalised in the nano scale air/water boundary layer which cools it via the enthalpy of evaporaton. Over shallow seas/lakes, coastal boundaries and marshland thing get even more complicated, but it is what is certain is that the simplistic radiation models used by many CAGW psudo-climatologist is completely wrong. Energy can only do work efficiently once, after this systems are only left with waste energy. The law of MEP ensures that in complex dynamic non-linear systems, like Earth’s weather / ‘climate’, the amount work is always maximised and the low level waste energy is ejected into space.

    The only sensible place to measure radiation is at the boundary between atmosphere and space – everything else in just guesswork or deliberate fudges to support ones own agenda.

  36. Mydogsgotnonose says:

    PS Forgot to add that when you have combined convection and radiation,. it’s coupled because the activated states can transfer energy to adsorbed gas molecules or emit photons.

    The sum of convection and radiation is the input energy at steady state. Radiation in an atmosphere is alwayys well less than S-B predicts in a vacuum.

  37. tallbloke says:

    David Hoffer says:

    Wrong.

    You’re sounding like Mosher. Stop it.

    Provided that the body is maintained at a temperature of 300K, the amount of energy it radiates is
    459.3.

    This would be true if the body was a theoretical black body in a perfect vacuum. But since we’re talking about the planet we live on, it isn’t.

    On Earth we have x joules of energy leaving the surface as latent heat of evaporation, y joules leaving as long wave radiation, and z joules leaving by thermal convection plus any others I forgot.

    By the time it leaves local space all this energy is converted to longwave radiation, and a proportion of it has been re-radiated back towards the surface, battling convection all the way.

    Carry on.

  38. davidmhoffer says:

    Provided that the body is maintained at a temperature of 300K, the amount of energy it radiates is
    459.3.

    This would be true if the body was a theoretical black body in a perfect vacuum. But since we’re talking about the planet we live on, it isn’t>>>>

    OK, to be more accurate, a theoretical black body emitts 459.3 w/m2 and an actual body emitts something less than that based on itz emissivity. Provided that the body is maintained at a temperature of 300K, it radiates exclusively based on itz temperature and emissivity. If it is losing energy via conductance or any other mechanism, the energy required to maintain an equilibrium temperature of 300K rises, but the amount of energy radiated changes not at all.

  39. davidmhoffer says:

    guy with deformed dog;
    Let’s take the hypothetical case of two plates with perfectly insulated backs at the same temperature, 300 K ,with a view factor of unity, unit emissivity, and absorptivity,.What you are saying is that they each emits electromagnetic radiation at a rate of 459.3 W.m^2. The implication [shown in the Trenbetrth ‘cartoon’]>>>>

    1. I said not a thing about Trenberth’s cartoon.
    2. The implication you claim is entirely your own. I said nothing of the sort.

    You example results in a net energy xfer of 0. SB Law satisfied. Laws of thermodynamics satisfied. Occums razor satisfied. No supernova postulated. If you want to have a discussion then please refrain from putting words in my mouth and then criticizing me for them.

    SB Law:

    P = 5.67 * 10^-8 * T^4

    There are no terms in that equation (even if you include emissivity) that pertain to how much energy is being transferred from the surface by other means, nor are there terms in that equation that pertain to how much energy is being received by any means, nor are there terms in that equation that pertain to how much radiance the body is exposed to by other bodies. For a given T and a given emissibity, you get a given P in w/m2.

    Clouds have a T and an emissivity. They radiate on that basis.

  40. davidmhoffer says:

    tallbloke;
    You’re sounding like Mosher. Stop it
    >>>>>>>>

    If I was sounding like Mosher I would have just said “wrong” and there would have been no words after that.

  41. Mydogsgotnonose says:

    David; you fail to understand that you cannot use the S-B equation on its own except for an isolated body in a vacuum, and in Sapce you have the microwave background. You ALWAYS have to use the difference between two S-B terms because emissivity and absorptivity are operational parameters and only equal at equilibrium

    Aarhenius made this mistake and was called out for it by Angstrom and Bohr.

    In 1967 Manabe and Wetherald made a slightly less bad mistake [assumed all radiation, no convection but some time afterwards someone in climate science decided to go the whole hog on phantasy physics.

    I have measured these things. Have you?

  42. tallbloke says:

    David Hoffer says:
    “If it is losing energy via conductance or any other mechanism,”

    Then it isn’t a black body.

    The only way to consider Earth as a black body is to consider a ‘shell’ at a point high enough that all other means of energy exchange other than radiation have ceased to be operative. In Earth’s case, that means it’s radiating the same ~240W/m^2 into space it receives form the sun averaged over the diurnal cycle. Since re-emitted radiation goes in all directions, that means half of anything which didn’t get through the window (40W/m^2) is radiating back towards the surface, about 200W/m^2.

    That gets re-absorbed and 440W/m^2 leaves the surface and atmosphere in the form of x,y and z to account for the TOA situation. Around 107W/m^2 of that is latent heat, thermal convection and conduction, so that leaves 333W/m^2, which isn’t far off what is measured to be buzzing around in the atmosphere.

    How are we doing?

  43. davidmhoffer says:

    Mydogsgotnonose says:
    April 14, 2012 at 9:35 pm
    David; you fail to understand that you cannot use the S-B equation on its own except for an isolated body in a vacuum, and in Sapce you have the microwave background.
    >>>>>

    YOU sir, provided an example of two perfectly insulated bodies with emissibity and absorption at unity and drew from it a conclusion that you falsely attributed to me. Do you want to argue YOUR example? Or not? I pointed out the fallacy of the claim you made using YOUR example, and your response is to abandon your example and argue a completely different point. Pick one and stick with it.

    Mydogsgotnonose;
    You ALWAYS have to use the difference between two S-B terms because emissivity and absorptivity are operational parameters and only equal at equilibrium>>>

    Agreed. Which is why in your example each surface is losing the exact same amount of energy it is receiving and why that results in a net energy xfer of zero in accordance with SB and thermo laws. That in no way implies that a given surface at a given temperature does not radiate energy because of anything to do with the surroundings.

    Mydogsgotnonose;
    I have measured these things. Have you?
    >>>>>>>

    Yes.

  44. Stephen Wilde says:

    “There are no terms in that equation (even if you include emissivity) that pertain to how much energy is being transferred from the surface by other means, nor are there terms in that equation that pertain to how much energy is being received by any means, nor are there terms in that equation that pertain to how much radiance the body is exposed to by other bodies.”

    Quite.

    So one cannot apply S-B from a point within an atmosphere because there are no terms within that equation to deal with any of the aspects of an atmosphere that interfere with simple in/out radiative energy transfers.

    Within an atmosphere apply the Ideal Gas Law to ascertain the expected temperature of the GROUND surface.

    Outside an atmosphere apply S-B to a point at the top (or surface) of the ATMOSPHERE (difficult to determine) to ascertain the expected temperature of THAT surface.

    It must be the case that as perceived from space the Earth is at a temperature at the top of the atmosphere sufficient to radiate out exactly at the same rate as radiation comes in.

    The radiative transfers going on within an atmosphere have no effect on ground surface temperature because as per the Ideal Gas Laws that temperatutre is solely a consequence of atmospheric mass, consequent density at the ground surface and solar input.

    Whether one can measure the temperature of the sky or whether one is actually measuring the temperature of the air molecules in contact with the sensor matters not in the end because it isn’t going to affect surface temperature one jot.

    The energy content of the air only affects the volume of the atmosphere and not the temperature at the ground surface.

    The radiative fluxes up, down and laterally within an atmosphere are a consequence of the temperature of that atmosphere which is set only by pressure,the consequent density and insolation.

    Either that or N&K and the standard atmosphere are wrong.

  45. tallbloke says:

    “If I was sounding like Mosher I would have just said “wrong” and there would have been no words after that.”

    Lol. Fair comment. :)

  46. Mydogsgotnonose says:

    Disagree TB: you are correct to point out that at the exterior surface of a black body emitter, half the S-B energy goes out and half goes in but that means you have at the top of the atmosphere 80 + 40 out plus 120 in.

    And from the surface you have 160 plus the 78 SW absorbed by the atmosphere making the 238, plus 120 = 358 of which 107 etc making 248……..

  47. davidmhoffer says:

    Stephen Wilde;
    Either that or N&K and the standard atmosphere are wrong>>>

    You know very well that I am a huge advocate of N&Z. There is nothing in SB Law or the concept of “back radiation” that is in disagreement with N&Z. In fact, you’ll find that the SB constant is integral to their equations.

    What happens as a CONSEQUENCE of energy moving about in a system is entirely different from the net effect of all the feedbacks. N&Z postulate that Ts is a function of pressure and insolation. Take a closer look at their explanation though. The way they calculate “average” Ts requires that the more GHG’s and other processes there are in the atmosphere that “inhibit” energy loss, the more heat gets moved from equator to poles. In other words, GHG’s actually DO change the average temperature the way it is normally calculated. But if you calculate average temperature correctly, as N&Z showed, you get ZERO change in average temperature, but you get a more UNIFORM temperature. The net result is no change to “average” surface temperature as calculated by SB Law, no change to energy balance, but, per your own theory, the temperate zones move toward the poles, resulting in a more UNIFORM temperature.

    For a water droplet in a cloud, all that is immaterial. It has a temperature, it has an emissivity, and it radiates in accordance with those parameters.

  48. tallbloke says:

    Mydog: I don’t see how that can be possible. 240 paid in must be equalled by 240 paid out. Cant be 80 plus 40, and 120 in the back pocket for a rainy day.

    Bear with me and let me develop my argument if Hoff agrees we are on the same page.

  49. Stephen Wilde says:

    An illustration:

    At surface pressure of one bar the atmosphere is sufficiently dense to maintain a ground surface temperature averaging around 15C at current solar input.

    Changing pressure at the surface from one bar to half a bar will decrease density at the ground surface and reduce the surface temperature at the same level of solar input.

    However, increasing the volume of the atmosphere whilst leaving the surface pressure at one bar will ALSO decrease density at the ground surface and reduce the surface temperature at the same level of solar input.

    In both cases incoming energy from the sun stays the same and the outgoing radiation stays the same but the ground surface temperature is higher if there is no increase in atmospheric volume and no decrease in surface pressure.

    So if GHGs retain more energy in the air to raise the air temperature and in the process increase the volume of trhe atmosphere then that energy is DENIED to the ground surface and cannot be ADDED to the ground surface by backradiation from air to ground. The same energy cannot be in two locations at the same time.

    That is where AGW theory breaches the Laws of Thermodynamics. They want both surface AND an expanded atmosphere to be warmer simultaneously with no increase in pressure or insolation.

    Not possible.

    The surface could only get warmer if the atmosphere did NOT expand in response to more GHGs.

    The fact that the extra energy is present in the air must expand the atmosphere which reduces density at the surface thus cooling the surface to maintain top of atmosphere balance as per
    S-B. One simply substitutes less outgoing energy from the surface with more outgoing energy from the air for a zero net effect.

    S-B at top of atmosphere can only be maintained BECAUSE the Ideal Gas Law operates within the atmosphere to ensure it.

    The two Laws are simply two aspects of the same phenomenon and are mutually interdependent.

  50. davidmhoffer says:

    April 14, 2012 at 10:19 pm
    Disagree TB: you are correct to point out that at the exterior surface of a black body emitter, half the S-B energy goes out and half goes in but that means you have at the top of the atmosphere 80 + 40 out plus 120 in.
    And from the surface you have 160 plus the 78 SW absorbed by the atmosphere making the 238, plus 120 = 358 of which 107 etc making 248……..
    >>>>>>>>>>>

    At TOA there is an “average” of 240 going in, and an average of 240 going out.
    At surface there is an “average” of X going in and X going out.

    Suppose you give me $4 and I give you $2. You can argue that the net change in money between us us $2 and you would be right. But if you want to argue that as a consquence of this you have “proof” that I never gave you $2 and that you never gave me $4…well have fun with that.

  51. Stephen Wilde says:

    davidmhoffer said:

    “The net result is no change to “average” surface temperature as calculated by SB Law, no change to energy balance, but, per your own theory, the temperate zones move toward the poles, resulting in a more UNIFORM temperature.”

    I think we are in agreement. The expansion of the atmosphere is what allows the more uniform temperature by more efficiently redistributing energy within an atmosphere of greater volume.

    You seem to have misunderstood the context of my commemt about N & Z.

  52. tallbloke says:

    OK, since David skipped my argument and moved on to respond to Mydog’s response to it, I’ll assume he agrees with my initial analysis.

    Is that fair David?

  53. Mydogsgotnonose says:

    Hi David: S-B only works for a vacuum because convection is a competing process for the transfer of kinetic energy from the interior of the emitter. For aluminium you need >300 °C before radiation exceeds natural convection.

    The two plates at the same temperaturre in a vacuum have zero net heat transfer because although you get bidirectional heat transfer by radiation, it completely cancels out. The way it does that is by half being absorbed and losing its identity and half bouncing back to which is added the same being emitted from the surface.

    At the surface half the excited states decay inwards and half decay out. Put the radiation shield in place which is a notional black body at the detector and you artificially create the full S-B flux.

  54. Mydogsgotnonose says:

    Hi TB, you are correct because you must consider a whole series of overlapping spherical emitters hence 198 + 40 out, 238 in.

    Entering as SW is 161 + 78 = 239 – 1 remaining = 238 which returns to space.

    The rest are accounting bit wwhich cancel out!

  55. davidmhoffer says:

    tallbloke says:
    April 14, 2012 at 9:52 pm
    David Hoffer says:
    “If it is losing energy via conductance or any other mechanism,”
    Then it isn’t a black body
    >>>>>>

    It most certainly is. Is it perfect black body? No. One has to adjust for emissivity. But it is still a black body and it still radiates based on the temperature it is at. Quick glance at your numbers, they look about right.

  56. davidmhoffer says:

    Mydogsgotnonose says:
    April 14, 2012 at 10:43 pm
    Hi David: S-B only works for a vacuum because convection is a competing process for the transfer of kinetic energy from the interior of the emitter. For aluminium you need >300 °C before radiation exceeds natural convection.>>>

    Once more you change the argument. What has the temperature at which radiance exceeds convection got to do with it? The radiance exists as a function of T and emissivity. If energy is being lost by conduction, convection, evaporation, little invisible men kidnapping joules and transporting them through miniature black holes, whatever, then one of two possibilities exist:

    1. The body is cooling at a faster rate than it would be by radiance alone, or
    2. There is an increased energy input (source is immaterial) required to KEEP it at 300K.

  57. tallbloke says:

    David: Ok, thanks, we’re agreed on numbers. I’ll let you carry on chatting with Mydog and Stephen while I attend to other matters, and then I hope you’ll drop by tomorrow to see what I have in store for you.

    So, we have an imperfect black body, with a surface which varies in emissivity and temperature all over and instead of radiating into a vacuum, it’s radiating into a roiling mess of gases, towards a corresponding shell composed of nothing which looks anything like a comparable emitter.

    I don’t think the S-B law is going to help us much down here near the surface, so we’ll circumvent that problem.

    See you tomorrow.

  58. Mydogsgotnonose says:

    Constant temperature Al in air with its full complement of GHGs, mostly water vapour. Low emissivity means that it emits much less than a BB.

    So, you can’t as in the Trenberth cartoon claim BB radiation at that temperature.

    We developed a two-colour pyrometer to measure the real temperature by fitting to the emissivity reduced Wien curve.

  59. davidmhoffer says:

    So, you can’t as in the Trenberth cartoon claim BB radiation at that temperature.>>>>

    For the last time, I said NOTHING about the Trenberth cartoon, and you’ve now postulated yet another example that has nothing to do with the issue. So aluminum has low emissivity, so what? It still radiates based on…. guess what? Temperature and emissivity.

  60. Joe Public says:

    Thanks guys for an illuminating discussions.

    It’s been like Wimbledon of the Academic World.

  61. wayne says:

    I just got a chance to read this thread so far, and Mydogsgotnonose has it pegged, in physics and reality.

    David, you should listen to him, I would be saying the same. S-B always requires a coupling of two equations or one equation using delta T and the emissivities, angles, and areas, also have to by accounted for.

    Instruments will easily form an imaginary picture of the radiative reality by removing one side of a two sided process or differentiating them, in essence casting a radiative shadow, then electronically magnifying that difference in the shadow, and will give you a reading… careful, that reading is not actually what is happening in total, don’t think they are reality, they are manufactured reality that is rarely seen in nature. Nature itself would have to provide the matter to cast the shadows which always carries an opposite and equal effect (Newton).

  62. davidmhoffer says:

    wayne;
    Huh?
    We’re discussing if “back radiation” exists or not. How to measure it is indeed a complex matter with any number of challenges to get it done accurately, I’m not suggesting otherwise. I am disputing that notion that it doesn’t exist.

  63. kuhnkat says:

    Oh, and gentlemen, you have to consider that the imperfect black body is also irradiating itself to further devalue the SB results!!

  64. Mydog, I am with you most of the way. A couple of things 1) You may not know that the radiation window is 66 w/m2 and that Trenberth knows it (see slide 26 here http://climategate.nl/wp-content/uploads/2010/09/KNMI_voordracht_VanAndel.pdf ) and has not corrected his papers because I believe he has no understanding of heat and mass transfer and does not want to let his ignorance be known. 2) you probably know if you read engineering text books on heat transfer or the section on heat & mass transfer in Perry’s Chemical Engineering Handbook that the S-B law does not apply to a gas. By definition a black body or gray body has a surface. As you say the S-B law only applies to surfaces in a vacuum. That is how Stefan originally developed his equation from empirical data some of which came from experiments by Fourier. After Stefan developed his equation scientists and engineers quickly added an emissivity factor because there is no such thing (including the sun) which is a black body which absorbs/emits radiation in all wavelengths. The latter particularly applies to gases such as CO2 or CH4 which only absorb and emit in very narrow wavelengths (CO2 about one tenth of H2O and CH4 about one fifth of CO2). 3) As a process engineer you might agree with me that the concept of albedo is something developed by climate scientists to confuse. They are just splitting the normal engineering concept of emissivity into two ill defined values -an albedo which is supposed to represent short wave radiation but what are the limits- is it 0.5 to 0.7 microns ? and a long wave emissivity which is what- 0.7 to inifinity? Some others who know even less about heat transfer regard albedo are reflected radiant energy particularly light. All of them seem to have no knowledge that emissivity of surfaces (as shown by empirical data) can vary with temperature. The absorptivity/emissivity of gases over a path length will also change with temperature. Finally, when a gas absorbs heat energy it can expand, The energy will cause an increase in kinetic energy and potential energy. The energy can be transfered to other gases and particles in the mix where chemical reactions and phase change can occur. Radiation by molecules of a gas will depend on the conditions of the surrounds. Radiation to cold space will occur close to the top of the atmosphere. Tne concept of a molecule radiating equally in all directions is wrong. At the top of the atmosphere molecules will radiate to cold space because that is the path of least resistance (or the largest driver ie the largest temperature differential)

  65. wayne says:

    Hi David. I know this is about “back-radiation”. Isn’t that covered by the simple statement that radiation from a gas emanates in all possible directions and one direction it can take is back to the matter from whence the energy came in the first place? Just remember that energy of matter cannot raise it’s own temperature. To me this is not really about “back-radiation”, bad terminology, it is about temperature differentials with a bit of emissivity tossed in for correctness.

  66. Sparks says:

    You can’t use a Peltier to measure “Back Radiation”. Don’t be silly!

    On a cloudless night there is a greater potential difference between the ground and the sky due to the heat leaving the ground to the sky, the Peltier will give a higher reading due to the greater potential difference.

    On a cloudy night there is LESS potential difference between the ground and the sky which is dependent upon the thermal-resistivity of the clouds, this will reduce the the amount of heat leaving the ground to the sky above the clouds, the Peltier will therefor give a lower reading due to the lower potential difference.

  67. dp says:

    I’m skeptical. This can be evalutated by alternately placing a pan of hot water then a pan of ice water above the sensor. Assuming the sensor isn’t swamped by atmospheric effects and that it is protected from moving air, this simple test should reveal the radiative effects. Especially the impedance aspect.

    Somebody should probably describe the frequency response of this detector, too, to ensure it respondes to radiation from clouds. I’d like to see the extinction range of that radiation from the clouds, too and I’d be surprised it is far enough to reach the ground directly with any significant energy.

  68. sergeiMK says:

    Plenty of free data here to play with:

    http://www.nrel.gov/midc/srrl_bms/

    Someone’s already started (not well explained and has not yet removed the interdependencies) with some interesting plots:

    http://climateandstuff.blogspot.co.uk/search/label/dlwir

  69. sergeiMK says:

    The thermoelectric unit described above suffers wit a major flaw

    Conduction from top to bottom plate

    The main effect is to greatly reduce its sensitivity.

    The website at nrel (see post above) uses somewhat more sophisticated equipment!

  70. davidmhoffer says:

    sparks;
    On a cloudy night there is LESS potential difference between the ground and the sky which is dependent upon the thermal-resistivity of the clouds, this will reduce the the amount of heat leaving the ground to the sky above the clouds,>>>

    How does a cloud a few thousand meters above the ground cause the amount of heat leaving the ground to be reduced?

  71. wayne says:

    “Radiation by molecules of a gas will depend on the conditions of the surrounds. Radiation to cold space will occur close to the top of the atmosphere. The concept of a molecule radiating equally in all directions is wrong.”

    Without you asking I’ll stand corrected on that point cementafriend, for I just said that to David and it wasn’t complete. I should have been more exact, as you expounded. In net manner a gas does radiates in all directions if and only if all surrounding matter is homogeneous and at an equal temperature, a rarity. You and Mydog have done a fine job explaining the picky details.

    I also think albedo is one of the big culprits in ‘climate science speak’ I have come to realize in the last few months. I like the more engineering viewpoint that in reality there is only one term and it is emissivity/absorptivity that does vary by frequency and the characteristics of the matter under discussion. When you look at the singular emissivity by latitude bands it becomes very clear why this is so. If you view the Earth as having one “albedo” you are already way off base, ~10S-10N is always cloudy, ~10-50 has much less clouds, and 50 to the poles is once again mostly cloudy. Now recalculate the Earth’s mean temperature like N&Z did it, point by point.

  72. Ulric Lyons says:

    Mydogsgotnonose says:
    April 14, 2012 at 11:58 am

    “Ulric: the article to which you refer makes the Big Mistake of confusing impedance to IR transmission with ‘Downwelling IR’.
    There can be no heat transfer from a colder body to a hotter body.! Climate related sciences are unique in claiming there is.”

    And when the night time clouds are warmer than the surface ?

  73. Sparks says:

    davidmhoffer says:
    April 15, 2012 at 4:27 am

    “How does a cloud a few thousand meters above the ground cause the amount of heat leaving the ground to be reduced?”

    Sorry I’ll rephrase that, this will reduce the amount of heat from the ground escaping to the sky
    above the clouds
    I didn’t mean to say that this reduces the amount of heat leaving heat from the ground lol.

  74. davidmhoffer says:

    Sparks;
    Sorry I’ll rephrase that, this will reduce the amount of heat from the ground escaping to the sky
    above the clouds>>>>

    And how does it do that? How does less heat escape from the ground because there is a cloud a few thousand meters up?

  75. Hans says:

    This experiment is really interesting and it would be valuable to see some time series at different occations and and when the cloud cover is changing.
    This tread is very important and I would like to read all comments and don´t have time for a while. However, I would like to make some remarks. :

    There is a theoretical reason for “IR back radiation” to exist. The energy distribution in the troposphere is mainly governed by the second law of thermodynamics which states that the ENERGY per mass unit is trying to equalise. This means that a NET energy flux (IR radiation is one possible process) theoretically can be delivered to a colder body (but it would certainly be at a flux less than 100 W/m^2).
    See: http://tallbloke.wordpress.com/2012/01/25/hans-jelbring-an-alternative-derivation-of-the-static-dry-adiabatic-temperature-lapse-rate/

    There are massive evidence of backradiation in nature shown by meteorologal data even when there is little or no moisture in the air:
    See: https://tallbloke.wordpress.com/2012/03/16/hans-jelbring-back-radiation-and-observational-meteorologial-evidence/

    I read some comments from Mydog and agree to much but claim that IR back radiation does exist. Nature is for sure very complex most of the time and situations. It is important to really measure how big IR back radiation is in real life so we can leave the mythical +300 W/m^2 and go on with real science. Do remember that observational facts that are correctly measured are not beaten by any hypothesis or theory that contradict what has been observed.

  76. Hans says:

    wayne says: April 15, 2012 at 3:00 am

    “Hi David. I know this is about “back-radiation”. Isn’t that covered by the simple statement that radiation from a gas emanates in all possible directions and one direction it can take is back to the matter from whence the energy came in the first place? Just remember that energy of matter cannot raise it’s own temperature. To me this is not really about “back-radiation”, bad terminology, it is about temperature differentials with a bit of emissivity tossed in for correctness.”

    Hi Wayne,
    This thread deals with one form of power transmission and if it does exist or not. All the theoretical arguments and observational evidence have to be taken into account.

    The concept of emissivity is highly suspect and it will not help much to use it as an averaging concept. An homogenous substance have a well difined emissivity but used about earth or lunar surfaces there are a number of physical processes involved and emissivity becomes a variable. On moon grain size for sure is an important variable. A big obstacle in understanding “lunar surface” temperature is that electromagnetic radiation if different wave lengths sent from moon originates from differents depths. That probably also involve a “directional” influence. There is simply not any well defined surface to which the SB law can be applied correctly. The same is true about applying the SB law to gases. Still, it is a fair approximation in many applications but not at all always and each one has to be scrutinized.

  77. Hans says:

    cementafriend says: April 15, 2012 at 2:59 am

    “… you might agree with me that the concept of albedo is something developed by climate scientists to confuse. They are just splitting the normal engineering concept of emissivity into two ill defined values -an albedo which is supposed to represent short wave radiation….”

    Thank you for your valuable description which I appreciate. Albedo is well defined even if it might be hard to measure. The albedo represents the part of all solar photons that dont change energy when redirected by a planet into space again. It is a very clean concept, theoretically.

  78. Konrad says:

    An interesting experiment, however there may be some improvements possible.

    I am currently building a new down welling IR experiment using peltier cooling. However I am taking a different approach. I have attached a peltier chip to an aluminium plate with a matt black surface. (320 x 230mm x 1.5mm) The hot side of the chip is cooled by a water block, ice water and a fish tank pump. The plate is insulated on all sides except the top face with EPS foam. The top surface is covered with IR transparent cling film 10mm over the surface of the plate. Silver based heat sink paste is used to connect probe thermometers to the underside of the plate.

    The procedure is to shield the experiment from DWLWIR and cool the plate to below the average temperature of the lower troposphere. The peltier chip and water pump will then be switched off and the plate exposed to the night sky. The resultant temperature response will then be compared to similar initial conditions and subsequent illumination by a halogen light source equal to 300 w/m2.

    Does anyone want to place bets on what will happen? Willis? Joel? Mosher? My next post may be titled “silence of the scams” ;)

  79. davidmhoffer says:

    konrad;
    The resultant temperature response will then be compared to similar initial conditions and subsequent illumination by a halogen light source equal to 300 w/m2.>>>>

    You do understand that the downward IR varies dramatically by latitude, season, humidity and many other factors? That the chance of you conducting your experiment in a place and time representative of the average for the globe is about zero?

  80. tallbloke says:

    When I asked ‘scientist of doom’ for links to empirical measurements of the radiation looking down at the ocean surface and up into the air he prevaricated and told me the instruments were very expensive. When I pushed further and asked why some of the many billions of dollars sunk in climate research hadn’t produced a comprehensive dataset he didn’t answer.

    Where’s the data?

    The excellent contributions to this thread point to an answer. A comprehensive dataset would reveal that the attempt to treat Earth as a black body radiative calculation problem is a failure.The way in which radiation and convection is supposedly coupled in the models is a botched up nonsense. The neatly accounted figures of the Keihl-Trenberth energy budget diagram are an illusion.

    ‘CementaFriend is right that radiation to space is asymmetric with radiation back towards the surface, and not only because of the ‘atmospheric window’. As he said:

    “Radiation by molecules of a gas will depend on the conditions of the surrounds. Radiation to cold space will occur close to the top of the atmosphere. The concept of a molecule radiating equally in all directions is wrong. At the top of the atmosphere molecules will [predominantly] radiate to cold space because that is the path of least resistance (or the largest driver ie the largest temperature differential)”

    Therefore the radiative greenhouse effect cannot account for the surface temperature of Earth. Couple this with the observation that longwave radiation promotes evaporation of the ocean surface but cannot penetrate to heat its bulk and it becomes clear that the principle cause of the elevation of the temperature of the ocean bulk which maintains surface temperature cannot be ‘back radiation’ which although it exists as part of the radiative flux, is more a symptom of temperature than its cause, and is much smaller in magnitude than the figures on the K-T diagram indicate, as Mydog has shown, and which no longer features on NASA’s energy budget diagram anyway.

    The much larger effect is that of the pressure gradient and the higher near surface air density, and the limiting effect that surface pressure has on the ocean’s ability to lose heat.This, coupled with the resulting asymmetric throughput of solar energy, is what causes the temperature near the surface to be higher, and the temperature at high altitude to be lower than the ‘effective radiating temperature’ equivalent to the incoming energy. It is a redistribution of that energy, not a creation of ‘extra energy’, as has been mistakenly/mischievously claimed by those seeking to discredit the hypothesis we are working with here at the Talkshop.

    Scientific truth always prevails, and the better we can understand and characterise it, the faster it will prevail.

  81. tallbloke says:

    Konrad: Excellent! Please let me publish the details and results here for you.
    Will you be continuing with the pressure experiments too?

    Hoff: “You do understand…the chance of you conducting your experiment in a place and time representative of the average for the globe is about zero”

    You do understand that the performance of a standalone experiment result has merit in establishing the intitial evidence supporting a hypothesis don’t you?

  82. Bryan says:

    The radiation per molecule is released in a random direction.
    The radiation per volume however must go from denser to less dense.

    Gravity, temperature and second law are all in agreement.
    For equal adjacent volumes of gas at at the same temperature the radiative effects are self cancelling.

    The radiative effects are already naturally included in the bulk thermodynamic quantities such as heat capacity of whatever gas mixture in being considered.
    To do radiative transfer calculations without taking this into account will lead to gross over calculation of energy transfer.

  83. davidmhoffer says:

    You do understand that the performance of a standalone experiment result has merit in establishing the intitial evidence supporting a hypothesis don’t you?>>>

    His hypothesis is that the result will not be a specific number. He could measure thousands of points on the earth surface and not get that number and the number could still be perfectly correct. A valid hypothesis requires predicting what the value will be for a given set of conditions.

    I predict that on the day he does his experiment, the temperature at his local will not be 288K.

    I’ll more than likely be correct, and that fact is entirely useless to prove what the average temperature of the earth is.

  84. davidmhoffer says:

    http://coen.boisestate.edu/vsridhar/files/2011/10/S_1AFM_longwave_2002.pdf

    http://journals.ametsoc.org/doi/abs/10.1175/1520-0450(1999)038%3C0474%3AAIPFEE%3E2.0.CO%3B2

    http://coldregionsresearch.tpub.com/SR04-1/SR04-10039.htm

    http://adsabs.harvard.edu/abs/1996JCli….9..646G

    http://www.atmos-chem-phys-discuss.net/3/5099/2003/acpd-3-5099-2003-print.pdf

    http://gewex.org/2009Conf_gewex_posters/Trentmann_G3-6.pdf

    Let me know when you’re done wading through those.
    I might add, that there are a number of world reknown physicists (ex, richard lindzen) who have written considerable volumes on the flaws in the IPCC reports and CAGW theory. To my knowledge, not a single one of them has challenged the notion of “back radiation”. They challenge only the estimated feedbacks in terms of magnitude and sign. Odd that giants of physics like Lindzen and Choy and Spencer and so many others with clear expertise in the field who regularly challenge the foundations of CAGW don’t bother with this argument about the existance of back radiation that seems to obvious to commenters here.

  85. Sparks says:

    davidmhoffer says:
    April 15, 2012 at 7:01 am

    And how does it do that? How does less heat escape from the ground because there is a cloud a few thousand meters up?

    I didn’t mean less heat escapes from the ground because there are clouds (lazy choice of words at 5am).
    What I should have said was, because of the low pressure-system associated with cloud formation and as heat flows naturally from hot to cold and considering that temperature is a potential function of the heat flow this pressure system restricts the heat flow and reduces the thermal potential difference around the peltier, which will give a lower reading of induced thermoelectric voltage.

    Heat Flow = Thermal Potential Difference [over] Thermal Resistance.

    The reading or induced thermoelectric voltage from the pelter is in response to a temperature difference and not a response to heat being emitted from clouds as “back radiation”.

  86. tallbloke says:

    David Hoffer says:
    “His hypothesis is that the result will not be a specific number. He could measure thousands of points on the earth surface and not get that number”

    Yes Hoff, but if it’s a long, long way from the expected number, then it is significant. I suggest to Konrad he also performs the experiment under full cloud cover to see what he gets then. We can then compare ratios with the less well controlled system described in this post too.

    Penn State’s radiometer is giving them results like the ones in Ulric’s linked course page.

    Hoff said:
    “giants of physics like Lindzen and Choy and Spencer”

    One two three all together now…

    APPEAL TO AUTHORITY!

    Lol. :)

    Anyway, who trusts theoretical physicists to know how instruments are constructed and operate better than process engineers like Mydog who actually construct them? Not me.

    Seriously though, answer these simple questions:
    If there is 333W/m^2 downwelling, and 398W/m^2 upwellling, which way is the heat flowing and how much of it is there?

    And if there’s 65W/m^2 downwelling and 130W/m^2 upwelling and 138 buzzing around sideways?

  87. Stephen Wilde says:

    I’ll just pull this back to a point I made above, the significance of which seems to have been lost in the wider discussion.

    The S-B Law must be applied from outside the atmosphere and from that perspective every planet with an atmosphere must always stabilise at a surface temperature which results in energy in equalling energy out. If it failed to do so there would be no gaseous atmosphere.

    If one then applies the Ideal Gas Law within the atmosphere I aver that it is that Law involving a variable atmospheric volume which ensures that the correct amount of energy is delivered at the correct rate to achieve equilibrium of energy in and energy out at the top of the atmosphere.

    In effect the operation of the Ideal Gas Law ensures that the atmosphere expands to the degree necessary to eliminate any thermal effect at the surface from more GHGs.

    Relating that to this thread the fact is that the equilibrium temperature of the atmosphere is controlled by expansion and contraction and it is the temperature of the molecules of air right in front of the sensor that is being measured and not a downward flow of radiation from higher up.

    That temperature being set by pressure and insolation at top of atmosphere and not by radiation coming down from above.

  88. davidmhoffer says:

    Spencer is a theoretical physicist? You may want to check up on that.

  89. Malaga View says:

    tallbloke says: April 14, 2012 at 8:45 pm
    On Earth we have x joules of energy leaving the surface as latent heat of evaporation, y joules leaving as long wave radiation, and z joules leaving by thermal convection plus any others I forgot.

    PLUS any heated particles lost from the exosphere…

    Earth’s exosphere
    The main gases within the Earth’s exosphere are the lightest gases, mainly hydrogen, with some helium, carbon dioxide, and atomic oxygen near the exobase. The exosphere is the last layer before outer space. Since there is no clear boundary between outer space and the exosphere, the exosphere is sometimes considered a part of outer space.

    http://en.wikipedia.org/wiki/Exosphere

    Even when there are no ongoing meteor showers, the Moon is surrounded by a gaseous halo called the lunar “exosphere,” says Wilson. Consisting of a just a few hundred atoms per cubic centimeter, the exosphere is barely more than a vacuum. The solar wind blows it into a long tail much like a comet’s. It points away from the Sun and extends for hundreds of thousands of kilometers. The giant tail is so rarefied that it’s completely invisible to the unaided eye even when the Earth passes through it once a month around the time of the New Moon.

    http://science.nasa.gov/science-news/science-at-nasa/2000/ast26oct_1/

  90. Stephen Wilde says:

    “At dawn, before the sun rose but the sky above was turning blue, the temperature of the steel sheet rose BEFORE the local air temperature.
    My claim is that was backradiation from WATER VAPOUR in the clear sky above.”

    More likely, scatter through the atmosphere from the sun even before it came above the horizon ?

    Should we be distinguishing between

    i) Scattering of solar shortwave input as it progresses down through the atmosphere and

    ii) Longwave radiation emitted by warmed molecules in the atmosphere itself ?

    Even beneath cloud cover if there is any light at all there is lots of scattered solar energy that has penetrated through the clouds. That continues to add energy to any surfaces whilst the clouds suppress cooling by reducing convection and increasing humidity to decrease evaporation.

    Wouldn’t that affect the temperature recorded by the sensor in daytime ?

    At night time the ground keeps emitting energy whilst clouds suppress convection and decrease evaporation so is the sensor recording simply the accumulation in the air of energy previously held in the ground ?

    If a cloud passes over a cold ground surface with resultant warming of the ground surface is not the necessary energy coming up out of deeper ground which has not yet been cooled by surface radiation before the cloud came along ?

  91. tallbloke says:

    MV: Leif frequently regales us with the analogy that the amount of particle mass involved in solar proton events and such is roughly equivalent to the weight of a well fed chicken. Th Exosphere is very tenuous and the amount of heat capacity is negligible, although the temperatures on a per molecule or atomic basis can be very high.

    The goings on in the upper atmospheric layers are of high interest electrically and chemically, but in terms of contribution to Earth’s energy budget directly, not so much.

  92. tallbloke says:

    davidmhoffer says:
    April 15, 2012 at 11:02 am

    Spencer is a theoretical physicist? You may want to check up on that.

    He’s both a theoretical and instrumental physicist. Having read a couple of his posts on his experiments with pointing low cost sensors at the sky from insulated boxes, I’m not so sure he has thought around all the angles though. He seems to set off from assumptions which may not be a solid as he believes them to be.

    No harm in us adopting a sceptical stance and trying to work things out for ourselves anyway. I think you are attacking something of a straw man here. Radiation buzzing around in the atmosphere certainly exists and no-one here denies it. The questions are more to do with the way ‘back radiation’ is characterised, measured and modeled. I think the idea of disregarding the portion of the flux which cancels out Mydog’s ‘Prevost energy’) is reasonable and sensible. We’ll see what is left over at night in one instance when Konrad performs his experiment.

  93. SteveB says:

    davidmhoffer says:

    appeal to authority……appeal to authority………appeal to authority……..appeal to authority.

    david, in addition to perhaps getting a tighter grip on the science, you might take a minute or two and firm up your grip on the logical fallacies; what they are and what constitutes them.

    From the nizkor project: (http://nizkor.org/features/fallacies/appeal-to-authority.html)

    Description of Appeal to Authority
    An Appeal to Authority is a fallacy with the following form:

    Person A is (claimed to be) an authority on subject S.
    Person A makes claim C about subject S.
    Therefore, C is true.
    This fallacy is committed when the person in question is not a legitimate authority on the subject. More formally, if person A is not qualified to make reliable claims in subject S, then the argument will be fallacious.

    This sort of reasoning is fallacious when the person in question is not an expert. In such cases the reasoning is flawed because the fact that an unqualified person makes a claim does not provide any justification for the claim. The claim could be true, but the fact that an unqualified person made the claim does not provide any rational reason to accept the claim as true.

    When a person falls prey to this fallacy, they are accepting a claim as true without there being adequate evidence to do so. More specifically, the person is accepting the claim because they erroneously believe that the person making the claim is a legitimate expert and hence that the claim is reasonable to accept. Since people have a tendency to believe authorities (and there are, in fact, good reasons to accept some claims made by authorities) this fallacy is a fairly common one.

    Since this sort of reasoning is fallacious only when the person is not a legitimate authority in a particular context, it is necessary to provide some acceptable standards of assessment. The following standards are widely accepted:

    The person has sufficient expertise in the subject matter in question.
    Claims made by a person who lacks the needed degree of expertise to make a reliable claim will, obviously, not be well supported. In contrast, claims made by a person with the needed degree of expertise will be supported by the person’s reliability in the area.

    Determining whether or not a person has the needed degree of expertise can often be very difficult. In academic fields (such as philosophy, engineering, history, etc.), the person’s formal education, academic performance, publications, membership in professional societies, papers presented, awards won and so forth can all be reliable indicators of expertise. Outside of academic fields, other standards will apply. For example, having sufficient expertise to make a reliable claim about how to tie a shoe lace only requires the ability to tie the shoe lace and impart that information to others. It should be noted that being an expert does not always require having a university degree. Many people have high degrees of expertise in sophisticated subjects without having ever attended a university. Further, it should not be simply assumed that a person with a degree is an expert.

    Of course, what is required to be an expert is often a matter of great debate. For example, some people have (and do) claim expertise in certain (even all) areas because of a divine inspiration or a special gift. The followers of such people accept such credentials as establishing the person’s expertise while others often see these self-proclaimed experts as deluded or even as charlatans. In other situations, people debate over what sort of education and experience is needed to be an expert. Thus, what one person may take to be a fallacious appeal another person might take to be a well supported line of reasoning. Fortunately, many cases do not involve such debate.

    The claim being made by the person is within her area(s) of expertise.
    If a person makes a claim about some subject outside of his area(s) of expertise, then the person is not an expert in that context. Hence, the claim in question is not backed by the required degree of expertise and is not reliable.

    It is very important to remember that because of the vast scope of human knowledge and skill it is simply not possible for one person to be an expert on everything. Hence, experts will only be true experts in respect to certain subject areas. In most other areas they will have little or no expertise. Thus, it is important to determine what subject area a claim falls under.

    It is also very important to note that expertise in one area does not automatically confer expertise in another. For example, being an expert physicist does not automatically make a person an expert on morality or politics. Unfortunately, this is often overlooked or intentionally ignored. In fact, a great deal of advertising rests on a violation of this condition. As anyone who watches television knows, it is extremely common to get famous actors and sports heroes to endorse products that they are not qualified to assess. For example, a person may be a great actor, but that does not automatically make him an expert on cars or shaving or underwear or diets or politics.

    There is an adequate degree of agreement among the other experts in the subject in question.
    If there is a significant amount of legitimate dispute among the experts within a subject, then it will fallacious to make an Appeal to Authority using the disputing experts. This is because for almost any claim being made and “supported” by one expert there will be a counterclaim that is made and “supported” by another expert. In such cases an Appeal to Authority would tend to be futile. In such cases, the dispute has to be settled by consideration of the actual issues under dispute. Since either side in such a dispute can invoke experts, the dispute cannot be rationally settled by Appeals to Authority.

    There are many fields in which there is a significant amount of legitimate dispute. Economics is a good example of such a disputed field. Anyone who is familiar with economics knows that there are many plausible theories that are incompatible with one another. Because of this, one expert economist could sincerely claim that the deficit is the key factor while another equally qualified individual could assert the exact opposite. Another area where dispute is very common (and well known) is in the area of psychology and psychiatry. As has been demonstrated in various trials, it is possible to find one expert that will assert that an individual is insane and not competent to stand trial and to find another equally qualified expert who will testify, under oath, that the same individual is both sane and competent to stand trial. Obviously, one cannot rely on an Appeal to Authority in such a situation without making a fallacious argument. Such an argument would be fallacious since the evidence would not warrant accepting the conclusion.

    It is important to keep in mind that no field has complete agreement, so some degree of dispute is acceptable. How much is acceptable is, of course, a matter of serious debate. It is also important to keep in mind that even a field with a great deal of internal dispute might contain areas of significant agreement. In such cases, an Appeal to Authority could be legitimate.

    That being said, rather than simply tick off immagined logical fallacies, answer the statements given directly.

  94. sergeiMK says:

    As I posted above-
    Why not use the data here – If you think it is faked then contact them and tell them why!

    http://www.nrel.gov/midc/srrl_bms/

    If you do not want to do the calcs then try asking this person fo
    do the analysis you want.

    http://climateandstuff.blogspot.co.uk/search/label/dlwir

    I find it amazing that few seem to believe the results from real instruments

  95. Richard111 says:

    Yes, a valid point by Stephen Wilde at 11:31 am above.

    You can see recorded levels of daylight under heavy cloud here: http://www.milfordweather.org.uk/solar.php

    When I next try my experiment I will note the precise time and compare with the official rise time.
    As for radiation from the upper air reaching the ground with enough energy to warm up 11.5 kg of sheet steel by a couple of degrees in less than 30 minutes… well, I dunno. :)
    My garden is about 400 metres due north from the sun sensor.

  96. tallbloke says:

    SergeiMK: “I find it amazing that few seem to believe the results from real instruments”

    It’s not so much that Sergei.. What we’re looking at is the question of causality and power. If the purported effect of DLWIR is being cancelled by an equal amount of ULWIR, then the fact that the net flux is upwards means the causality issue raises itself.

    Is there more LW radiation buzzing about at Earth’s higher than Moon average surface temperature because:
    a) Gases raised to higher temperatures by a warm ocean radiate more strongly? or because
    b) Gases radiating more strongly raise air temperatures and warm the ocean from the top down?

    Everything I know about thermodynamics tells me a) is more likely the case. If you have a go at heating a swimming pool with a hairdryer, you’ll see what I mean.

  97. Donald Mitchell says:

    I did not see any mention of surface conditions of the aluminum plates. If the outer surfaces are bare aluminum, they can act as very good mirrors for infrared at wavelengths greater than three to five microns. If this is the case, painting the with a titanium dioxide paint designed for high emissivity at these wavelengths could increase the voltages significantly.

    The emissivity or reflectance from materials at non visible wavelengths is often greatly different from their visible characteristics. Several decades ago, I worked with a device which had major problems that were caused by black anodized aluminum parts which were acting as mirrors for fairly near infrared.

  98. Malaga View says:

    TB: roughly equivalent to the weight of a well fed chicken

    Good to know that Leif works in the internationally agreed WFC units rather than American KFC units :-)

    the temperatures on a per molecule or atomic basis can be very high

    However, the RWFC unit [Roasted Well Fed Chicken] is a new one to me :-) Is that RWFC per second? :-)
    ….
    Thanks for the feedback… not sure I am totally convinced…
    A distinct lack of data makes it impossible to argue otherwise…
    But I will keep my eyes open for any confirmation [or otherwise]…
    ….
    I did stumble upon this regarding Venus [which is NOT Earth and is off topic]:

    The “induced magnetotail” that points away from Venus in the direction of the earth is a teardrop-shaped plasma structure filled with “a lot of little stringy things” that was first detected by NASA’s Pioneer Venus Orbiter in the late 1970s. In 1997, Europe’s Solar and Heliospheric Observatory (SOHO) Satellite showed that the tail stretched some 45,000,000 kilometres into space, more than 600 times as far as anyone had realized and almost far enough to “tickle” the earth when the two planets are in line with the sun.

    http://www.thunderbolts.info/tpod/2008/arch08/080220venustail.htm

    Their embedded graphic does illustrate why the Venus transit in June 2012 could be very interesting…

    Their referenced article in the New Scientist is more on topic [my highlights]

    NASA’s Pioneer Venus Orbiter first found the tail in the late 1970s. Around 70 000 kilometres from the planet, the spacecraft detected bursts of hot, energetic ions, or plasma. The tail exists because ions in Venus’s upper atmosphere are bombarded by the solar wind, a stream of plasma that blows out from the Sun.

    http://www.newscientist.com/article/mg15420842.900-science–planets-tail-of-the-unexpected.html

    Therefore:
    Can we expect some incoming Roasted Well Fed Chicken from Venus later this year?
    Can we expect some earthquakes when these Roasted Well Fed Chicken hit the Earth?

    Serious questions:

    When does the “induced magnetotail” of Venus “tickle” planet Earth?

    How does the “induced magnetotail” of Venus “tickle” planet Earth?

  99. tallbloke says:

    Donald Mitchell: Thanks for that, a point for Konrad to consider in his experimental setup.

    M.V.: Well off topic you rascal. :)
    Anyway, not too long to wait until we find out.

  100. tallbloke says:

    S. Fred Singer says:

    “Now let me turn to the deniers. One of their favorite arguments is that the greenhouse effect does not exist at all because it violates the Second Law of Thermodynamics — i.e., one cannot transfer energy from a cold atmosphere to a warmer surface. It is surprising that this simplistic argument is used by physicists, and even by professors who teach thermodynamics. One can show them data of downwelling infrared radiation from CO2, water vapor, and clouds, which clearly impinge on the surface. But their minds are closed to any such evidence.”

    Read more: http://www.americanthinker.com/2012/02/climate_deniers_are_giving_us_skeptics_a_bad_name.html#ixzz1s778uIRz

  101. wayne says:

    “One can show them data of downwelling infrared radiation from CO2, water vapor, and clouds, which clearly impinge on the surface. But their minds are closed to any such evidence.”

    Sure TB, but as I explained above that the “reading” of the “data” is a manufactured reality. You can, by numerous methods, as the peltier case above, create devices to block the inverse and always larger effect of temperature and radiation to get a reading on “back-radiation”, I realize that, but in the pure logic form as I see it, this is not measuring the real reality. NET radiation is the real reality.

    When you stand back and look at each of these devices and the way they work it usually plays on temperature differences at some level.

    Dr. Singer didn’t go quite far enough in explaining denialists various positions which in a sense shows his set position. One thing that irks me to no end is someone speaking of one side of a two side physical process with nary a mention of what is occurring on the flip side. Physics equations have equal signs and to separate effects on one side of an equation and inferring that the other side does not change is simply wrong.

    Here’s a short example. Ground 288K and somewhere above the temperature is 255K. There is a given ‘x’ amount of radiative transfer always occurring in this static state. You double the concentration of a GHG that absorbs more radiation and you try to infer that the rate of radiation from the surface will decrease “without a change in the temperature differential or possibly emissivity”. That by core physics is wrong and that is were people trying to do that with “back-radiation” is also wrong. Now, if the temperature above does change, like the hot spot in the mid troposphere that never appeared, then I also would agree that the rate from the surface would decrease, the temperatures dictate that.

    This peltier case is a great example, at the core it is measuring temperature differences and cloud cover changes but I also realize that above the clouds, between the clouds and space, an opposite effect is also occurring to some degree, it has to by very simple physic principles in equations with equal signs (without getting into further lengthy examples).

    That is how I see most of these discussions and I guess that puts one of my feet in the denialist camp. GHGs cool the atmosphere, not the opposite. All of these processes are always occurring simultaneous and each at the rate dictated by the local matter (surface, clouds, temperatures, pressures) that dictates each processes possible rate, and to split one out one without acknowledging the flip side is where most of confusion seems to creep in.

  102. tallbloke says:

    Wayne, I completely agree. I do wonder about some of prof Singer’s pronouncements. Last year he made one about climategate where he said the data was truncated “to hide the decline in temperature”. Not “to hide the decline in the proxy temperature”. Now maybe it was just sloppy wording, or a ‘senior moment’ or maybe he was trying to slide a message to his audience, I just don’t know.

  103. wayne says:

    TB, got you there. Let’s just assume at first those were both slips. I hate to paint anyone in a certiain light just to, usually, find it is me that is misunderstanding them or injecting my own assumptions! :-) (but that was a good read)
    I too often write ideas too brief or a wrong word and that ends up leaving gapping holes for misunderstanding. Happens all of the time.

  104. tjfolkerts says:

    Mydogsgotnonose, why the obsession with “Prevost Exchange Energy”?

    This rule was formulated before the 2nd Law of thermodynamics, and before photons were understood. Historically, it is an important development, but this principle is ignorant of the advances made in physics over the last 200+ years.

    Could you define (in an equation) the principle of “Prevost exchange energy”?

  105. tjfolkerts says:

    This is a cool backyard experiment, but if you want to do it right, why not simply use instruments that have been designed and refine specifically for this purpose?

    http://en.wikipedia.org/wiki/Pyrgeometer

  106. tchannon says:

    tkf,

    That is not the same thing and is a classic mental leap. Given you jumped sideways a few days ago when Roger ask an awkward question related to this, I’m wondering what happens next.

    The peltier device is measuring energy flow, a very different thing from a proxy device playing subset based on assumption and only half. (also notoriously defective devices)

    More or less the peltier device is showing the radiation balance.

    The end result is unlikely to be any surprise but actually doing things hands on is good.

    Hans showed data on much the same. In essence under clear sky there is very little incoming radiation even though the dominant water vapour is still present.

  107. tjfolkerts says:

    To intensify the effect, you might try

    1) painting the surface to increase the IR emissivity. Aluminum has an emissivity of ~ 0.1, while many paints (even if not black) are around 0.9. This will make the Peltier much more effective at picking up changes in IR

    2) adding a reflector like this to aim at particular parts of the sky.

    http://www.solarcooking.org/plans/funnel.htm

  108. sergeiMK says:

    tchannon says: April 16, 2012 at 8:37 pm

    tkf,

    That is not the same thing and is a classic mental leap. Given you jumped sideways a few days ago when Roger ask an awkward question related to this, I’m wondering what happens next.
    The peltier device is measuring energy flow, a very different thing from a proxy device playing subset based on assumption and only half. (also notoriously defective devices)
    More or less the peltier device is showing the radiation balance.
    =============
    The PIR devices on the site I referenced above is spec’d as:
    Temperature Dependence: ±1% over ambient temperature range -20 to +40°C.
    Linearity: ±1% from 0 to 700 Wm-2.
    Response time: 2 seconds (1/e signal).
    Cosine: better than 5%.
    Mechanical Vibration: tested up to 20 g’s without damage.
    Calibration: blackbody reference

    It uses a thermopile in its measurement!!

    One can measure the balance between for example weights using a double pan balance (the method of this blog) or by measuring individually and accurately the two weights.the methods of
    the Solar radiation Research Laboratory, Both methods may work but the more accurate is that done with calibrated instrumentation,
    As I pointed out the peltier device suffers from thermal conduction through the peltier junctions which will reduce the thermal difference and hence reduce the output voltage.

  109. tchannon says:

    I agree sergei, far from ideal.
    As shown the peltier is being used for voltage output whereas it ought to be loaded. The point is more this is a very direct power measurement of energy flow, intuitive. There’s little problem in coupling the bottom to the ground, not so easy to take a forward radiation measurement.

  110. ge0050 says:

    davidmhoffer says:
    April 15, 2012 at 7:01 am
    And how does it do that? How does less heat escape from the ground because there is a cloud a few thousand meters up?

    question: If the energy content, joules per cubic meter of the sky a few thousand meters up is greater on a cloudy night than on a clear night, will this affect the surface temperature?

    It would seem to me that it would, through radiation, conduction and convection.

    My understanding is that energy is transferred from colder objects to warmer objects all the time – both by radiation and conduction. This is very easy to demonstrate from the Maxwell–Boltzmann distribution that this is true for conduction..

    However, at the same time there is a greater transfer of energy from warmer objects to colder objects. Thus the NET energy transfer is always from warmer objects to colder objects. This net energy transfer is “heat”.

    The confusion I read in some of the posts is the use of the term “heat” to refer to ANY energy transfer, without distinguishing that heat only refers to NET energy transfer.

    The M-B distribution could not exist if cold molecules did not routinely increase the kinetic energy (temperature) of warm molecules, while at the same time reducing their own kinetic energy (temperature). Otherwise the distribution would collapse around the mean.

    However, it is more common for warm molecules to increase the kinetic energy of cold molecules, thus net energy transfer is from warm to cold. (statistical thermodynamics). In the case of conduction this can be calculated from the probability of various collisions based on their kinetic energy between molecules and results in the M-B distribution.

    Confusion results because radiation is often represented as “different” than conduction, in that radiation allows energy to flow from cold to warm, while conduction does not. As a result climate science holds that only radiation can warm the surface via GHG. This is a nonsense.

    The M-B distribution shows that energy flows from cold to warm via conduction as well as via radiation. In other words, there is nothing unique about GHG and radiation. If GHG and back radiation warms the planet, then so does non GHG and conduction.

    http://en.wikipedia.org/wiki/Maxwell%E2%80%93Boltzmann_distribution

  111. Bryan says:

    Very well expressed ge0050 !
    the only thing that I would add is that heat is also capable of doing thermodynamic work whereas backradiation is not.

  112. mkelly says:

    ge0050 did you read the link you provided.

    “The Maxwell–Boltzmann distribution applies to ideal gases close to thermodynamic equilibrium with negligible quantum effects and at non-relativistic speeds.”

    How exactly does this apply to solid earth or non gas photons, especially when not anywhere near thermodynamic equilibrium.

    By the way how does the last low kinetic energy molecule that transferred energy to a higher energy molecule tell all the other low energy molecules they cannot do any more transferring. How do the molecules know exactly how many of them can do something to fit into the M-B curves?

  113. RKS says:

    FAO Tallbloke.

    Not sure where else to post this.

    Just saw this on DM science comments page.

    Is this true or defamatory?

    >>>>>>Tallbloke is a web content administrator for a northern university. His other day job is to promote the university’s environmental policy. Seriously.
    – George, Durham, 17/4/2012 17:31

    >>>>>>Tallbloke knows no more about climate science than you do. By his own admission he is ‘an engineer and history of science graduate’ (whatever one of those is). What is it about his blog that impresses you? The out of context cherry picked data? The pretty graphs with no annotation or scale information? Or just that it tells you what you want to hear?
    – George, Durham, 17/4/2012 15:57

    Your excellent blog is getting a very public slagging off.

  114. tallbloke says:

    RKS: George is just jealous, because I get to publish on a blog with a better audience than the newspaper page he rants on. ;)

  115. Wayne Job says:

    Richard111, Place a mirror at 10,000 Ft focused on your plate and it will light up like day time, well before sunrise. Perhaps what you are seeing is reflected parts of the electromagnetic spectrum that cause heat or indeed a lens effect created by our atmosphere that would do a similar thing. Many things have simple answers that are easily explained by existing science.

    Heating stuff from radiation from colder things is very difficult to mathematically explain. The back radiation thing from cold to hot and its magnification times three in water vapour seems to be an invention to prove global warming and is an inbuilt feature of the models.

    Look further afield and your answer may be an important key to the understanding of our planet.

  116. tallbloke says:

    Sorokhtin, 2001a: Greenhouse effect: Myth and reality. Vestnik Russian Academy of Natural Sciences 1:8-21..

    “According to our estimates, convection accounts for 67%, water vapor condensation in troposphere accounts for 25%, and radiation accounts for about 8% of the total heat transfer from the Earth’s surface to troposphere.”

    “Thus, convection is the dominant process of heat transfer in troposphere, and all the theories of Earth’s atmospheric heating (or cooling) first of all must consider this process of heat (energy)– mass redistribution in atmosphere (Sorokhtin, 2001a, 2001b; Khilyuk and Chilingar, 2003, 2004).”

  117. paulinuk says:

    It’s been stated here several times by several people that the only atmospheric gases that radiate thermal energy back to the surface are the greenhouse gases. It’s true that GHG’s are the only ones able to absorb infrared to the greatest extent but I would like to remind people of what Thermal radiation is according to Wikipedia:

    “Thermal radiation is electromagnetic radiation generated by the thermal motion of charged particles in matter. All matter with a temperature greater than absolute zero emits thermal radiation. The mechanism is that bodies with a temperature above absolute zero have atoms or molecules with kinetic energies which are changing, and these changes result in charge-acceleration and/or dipole oscillation of the charges that compose the atoms. This motion of charges produces electromagnetic radiation in the usual way. However, the side spectrum of this radiation reflects the wide spectrum of energies and accelerations of the charges in any piece of matter at even a single temperature.”

    So according to the definition, when molecules of gas bump into each other and they suddenly change direction or start spinning violently, this will cause the electrons and protons within the molecule to emit thermal radiation because they have been accelerated, irrespective of what type of molecule it is.
    This is exactly what goes on within the sun: Due to nuclear fusion taking place, the Suns’ plasma has accelerating electrons and protons of mainly hydrogen and helium emiting thermal radiation.
    The confusion arises, it seems, around some molecules ie the GHG’s H2O,Carbon dioxide, Methane etc which have dipole oscillations which allow them to absorb/ emit radiation from the Earth’s surface in the infra red region. These dipole oscillations will give the molecule internal motion, and not directly translational or rotational motion.
    Also, all air molecules get some of their energy from collisions at the surface or from other air molecules.
    Therfore, unless someone can point to experiments or measurements that show otherwise, Nitrogen, Oxygen, Hydrogen in the atmosphere must be assumed to emit loads of unaccounted for Thermal radiation back to the surface.

  118. […] plucked this comment off the Hockey Schtick, as it ties in well with a recent post which generated a lot of comment. That post was looking at back radiation over dry areas. This one […]

  119. […] An example of a cloud detector appeared in this article http://tallbloke.wordpress.com/2012/04/14/yes-virginia-back-radiation-delivers-measurable-heat-just-… […]