Monday, September 26, 2011

Professor Nasif Nahle Publishes New Paper Discrediting Basis of Theory of Man-Made Global Warming

The fundamental basis of the theory of catastrophic man-made global warming is the notion that colder 'greenhouse' gases like CO2 'back-radiate' infrared capable of heating the hotter Earth surface. Professor Nasif Nahle has a new paper out explaining why this notion is false and unphysical.

Abstract: Through a series of real time measurements of thermal radiation from the atmosphere and surface materials during nighttime and daytime, I demonstrate that warming backradiation emitted from Earth’s atmosphere back toward the earth’s surface and the idea that a cooler system can warm a warmer system are unphysical concepts.


  1. Nasif's study has been useful to me because the calculations confirm my own observations, carried out with lesser quality equipment, in connection with climate-responsive design.

    What is extraordinary about all this is that it is necessary to demonstrate the non-existence of back-radiation.

    Around here, if you drive on unfenced roads in cattle country at night, you have to be careful approaching the crest of down-slopes, where for a short period your headlights are not illuminating the road. The cattle camp on the roads, because these have absorbed and stored more daytime heat, that re-radiates at night and keeps them warm. Cows know more than some climate scientists.

  2. Martin Clark said...

    "The cattle camp on the roads, because these have absorbed and stored more daytime heat, that re-radiates at night and keeps them warm. Cows know more than some climate scientists."

    So the tarmac roads and the urban heat island effect and the equatorial deserts during the day are responsible for heating the entire atmosphere, right!

    This paper by Nasif Nahle is simply an exercise in reaffirming the fallacy the the ground heats the atmosphere.

    Perception is reality. The atmosphere appears warmer near the ground, ergo the ground heats the atmosphere. Unfortunately for Nahle and all the other gatekeepers of this fallacy, the observational data proves otherwise.

    70% of the Earth is covered in oceans, which do not heat the atmosphere above. Neither does most of the land mass in the Northern and Southern hemispheres most of the year round. Neither does the any of the enormous surface area taken up by the worlds mountain ranges, agricultural, forestry, tropical rain forests, fiords, lakes, grasslands, Icecaps and polar regions.

    Just the bare earth, sand, tarmac, concrete and low lying rock at or near the Equator.

    To maintain that the ground heats the atmosphere is just nonsense. The Sun heats the atmosphere just as it heats the ground.

  3. Nahle's paper is so hilariously inept it hardly deserves a refutation. But let's just look at the first 2 pages, where Nahle demonstrates a complete lack of understanding of even the most basic principles of climate science, i.e., computing the average incoming solar radiation. The actual equation is: i = (1-a) * TSI/4 where i is mean solar radiation at the surface, a is albedo, and TSI is Total Solar Irradiance measured above the atmosphere. Nahle objects to the divide-by-4 step. He seems to be under the illusion that this is due to a factor of .49 for "thermal radiation" (huh?) plus another factor of .51 for "solar mitigation by the atmosphere" (which is ridiculous, since that's already accounted for by albedo in the real equation.)
    The real reason for the factor of 4 is simple geometry: as seen from the Sun, the earth intercepts an area of solar radiation equal to πr², because that's (obviously) the formula for the area of a circle. But the Earth's surface is spherical, and if we want to average that same amount of intercepted energy over the entire surface, we need to know the formula for the surface area of a sphere. Which is: a=4πr², exactly 4 times as much as the circular cross-section. So all of Nahle's hand-waving (and hand-wringing) is utter nonsense. He just doesn't know what he's talking about.

    1. Haha, I saw that too. Did you look just a bit further down at the equation on page 2, T = Q/e*sigma ? He then fills it in with units and numbers. Q is quite obviously a heat transfer per unit time per square area. He helpfully points out e is emittance, which is unitless both in real life and in his equation. That leaves the rather problematic sigma, which he gives the value 5.6697 x 10^-8 W/m^2 K. To me it looks like the Stefan-Boltzmann equation, P = e sigma A (T^4 - T_s^4) where P is net radiated power, A is surface area, T is the temperature of an object, and T_s is the temp of its surroundings. Q of course is simply P/A, and if we also assume T >> T_s, we get Q = e sigma T^4 or T^4 = Q/e*sigma. Sigma is Stefan's constant, 5.6703 x 10^-8 W/m^2 K^4. Obviously that was the equation that Nahle used with a few glaring typos. But more importantly, why did he use this formula? By using it, he assumes that the air is the emitter of all this energy and because the energy originally came from the sun, the air must have absorbed all the sunlight before re-emitting it. But as anyone who has witnessed a sunny day can attest, the atmosphere is mostly transparent to a broad portion of the sun's energy. You simply can not take the sun's irradiance and pretend that it came from the atmosphere. This is pretty typical of Nahle's work - even if he gets the equation right, he usually applies it to the wrong thing.

  4. I'm well aware of the geometric relationships. Nahle explains why this is a gross oversimplification and erroneous.

    So does Joe Postma

    So does Paul Clark

    So does G&T

    etc etc

  5. Based on the above comment, it's hard to believe you actually read what Nahle wrote, since Nahle doesn't actually know why the /4 operation is performed. Instead, he invents two separate /2 operations out of thin air, then proceeds to refute his own inventions.

    Joe Postma thinks the /4 operation is just fine: it's right there in his equation 7 and again in his equation 8. (Which is why he gets the right answer in equation 10).

    And Paul Clark doesn't even address the issue, at least not in the link you provided.

  6. Both Nahle and Postma show the conventional derivations used by climate scientists and then go on to show the reasons why the "average surface temp" calculation is thermodynamically incorrect.

    For instance start reading Postma on page 25

  7. @KAP...

    I am not inventing anything:

    You can see that the value given by NASA on that poster -and by Trenberth and Kevin also- is the value that I reported in my paper as an erroneous amount.

    The correct scientific procedure is as follows:

    1365 W/m^2 is the annual average of total solar radiation measured by satellites on Top Of the Atmosphere (TOA).

    682.64 W/m^2 are thermal radiation impinging on TOA.

    136.528 W/m^2 (20%) are reflected by the atmosphere, especially by clouds and dust before it touches the surface.

    136.528 W/m^2 (20%) are absorbed directly by the atmosphere, in particular by ozone, clouds and dust before it touches the surface.

    40.958 W/m^2 (6%) are scattered by molecular oxygen and nitrogen at the upper atmosphere, i.e. before it touches the surface.

    After mitigation by the atmosphere, the thermal radiation impinging on the surface is:

    Insolation = 682.64 W/m^2- 136.528 W/m^2- 136.528 W/m^2- 40.958 W/m^2= 368.626 W/m^2.

    From 368.626 W/m^2 of incident solar radiation on the surface, the surface reflects 25.8 W/m^2 (7%).

    342.8 W/m^2 are received on Earth's surface.

    The surface is formed by the cryosphere (snow and ice), the lithosphere (land), the hydrosphere (oceans) and the biosphere (living beings).

    From the solar thermal radiation received, the surface (land and oceans) absorbs 239.96 W/m^2.

    A - Total of solar thermal radiation lost directly into space before it hits on the surface:

    Ratm + Satm = 177.5 W/m^2.

    B - 136.528 W/m^2 (19.9%) are absorbed directly by the atmosphere.

    C - Total of solar thermal radiation impinging on the surface after mitigation = 368.626 W/m^2.

    Total ThR impinging on TOA = A + B + C = 177.486 W + 136.51 W + 368.626 W = 682.64 W/m^2

    Total received on the surface = 343 W/m^2.

    Total absorbed by the surface = 240 W/m^2.

    The above correct calculations explain the warming of the planet without resorting to invented greenhouse effects.

    All magnitudes were obtained from measurements.

    1365 W/m^2 of solar irradiance includes the whole electromagnetic spectrum.

    What you ignore, KAP, is that not the whole spectrum of solar radiation can be transferred as heat or work, but only a portion of it.

    You ignore also what thermal radiation is. Additionally, you are fatally confused with the concepts of static energy and dynamic energy.

  8. Run the same experiment in the parking lot at the top of Pikes Peak.

  9. Wow, epic failure. Couldn't even bother to check his calculations. Also, the conclusion defeats itself. I guess radiation cannot travel from cooler to warmer areas, right? Apparently he was detecting warm pockets of air, right? How did the radiation make it to the instrument? The measurements demonstrated it's warmer at the surface. So how did the "cooler" radiation make it there? Think about, if it were true that radiation never travels towards warmer areas, then an infrared thermometer would never be able to measure temperatures cooler than the thermometer.

    1. Wow epic failure of your comment

      Radiation is bidirectional, heat flow is one way only from hot to cold per the 2nd law. Radiation from a cold body does not heat a hot body because all of those lower energy quantum states are already filled.

  10. So once again I ask, how would an infrared thermometer measure cooler temperatures than the temperature of the thermometer? In order for it to measure the cooler temperature, it would have to move into the warmer thermometer.

    1. IR thermometers consist of a shielded thermistor, which decreases resistance if pointed at a hotter object than the thermistor, and increases resistance if pointed at a colder object than the thermistor. The device is calibrated to this in order to measure temperatures lower than the sensor.

  11. Let's try this once more. In order for an infrared thermometer to measure a temperature (i.e. the amount of infrared radiation an object emits), infrared radiation must contact the thermometer. By your reasoning, the cooler radiation can never move towards the warmer object. So if the thermometer is warmer than the object it is measuring, the infrared radiation from the cooler object would never reach it, therefore the infrared thermometer could never measure it.

    1. False

      Of course, all bodies above absolute zero emit radiation.

      Of course, radiation from a colder body reaches a hotter thermistor. The hotter thermistor increases resistance when pointed at a cooler body and therefore can measure the temperature of a cooler body.

      However, radiation from a colder body will never heat a hotter body because those "cold" lower-energy quantum states are already completely filled in the hotter body.