Radiative Transfer According to AGW: A Note from Neutrino

RADIATIVE transfer is an incredibly important topic when it comes to the earth’s climate system. It is the only way for the earth to either heat up, via absorption, or cool down, via emission.  The temperature of the earth is set by these two counter acting mechanisms. To solve this balance for the earth’s dynamic systems is incredibly difficult. But to understand the basic principles involved is fairly straight forward and central to understanding the theory of anthropogenic global warming (AGW).

Some caveats first:

Black Bodies(BB). BB’s are perfect absorbers of all incident radiation. BB’s are also the ideal or most efficient emitters of radiation. No object will emit more than a BB when at the same temperature. In the real world no object is a BB, but many objects approach BB’s over certain spectrums. For instance the sun is a good approximation of a BB.

Grey Bodies(GB). GB’s are bodies that have uniform emissivity across all wavelengths. As with BB’s no real world bodies are perfect GB’s. As such it is a useful approximation.

For example, the earth can be approximated by a GB with emissivity around 0.7 for solar wavelengths(λ < 4μm) and an emissivity near 1.0 for terrestrial wavelengths(λ > 4μm). Emissivity is a wavelength dependant quantity but approximations like the one used for the earth work very well.

Kirchhoff’s Law. This law is an equality between the absorptivity and emissivity. It holds under local thermodynamic equilibrium(LTE). The vast majority of the atmosphere is in LTE, only at high altitudes where the atmosphere is tenuous does this break down. For practical purposes we are always in LTE.

Equilibrium. Equilibrium is the state where temperatures are no longer changing. At this point a body must be emitting the same amount of energy that it is absorbing. This is a consequence of the Conservation of Energy.

For ease of calculations all bodies are assumed to be isothermal and in isolation with no conduction or convection(ie: isolated in a vacuum). Also all bodies are all solid surfaces, there is no transmission only absorption and emission. These simplifications aid in understanding the underlying pattern of radiative transfer. They are not a requirement of actually solving problems but are necessary to simplify the equations.

Radiative exchanges in a nutshell:

Kirchhoff’s equates absorptivity α and emissivity ε. Generally only ε is used when speaking of both absorptivity and emissivity.

1) α = ε

Absorption by an object is simply a function of ε. The temperature of the absorber is irrelevant to absorption. The amount absorbed(q) is a function of emissivity and the incident power(Q).

2) q = εQ

Emission from an object is a function of both ε and its temperature. The emitted power(p) is a function of emissivity, the Stefan-Boltzmann constant(σ), the temperature(T) in Kelvin raised to the fourth power and the area(A) of the emitting surface. This is the Stefan-Boltzmann Law. This law is the integration over all wavelengths of the Planck function. The Plank function describes the intensity of emitted radiation per wavelength for a given temperature.

3) p = εσT^4.A

A body will absorb radiation at a rate given by 2) and will emit radiation at a rate given by 3). For a body to be in equilibrium 2) must equal 3). If the temperature of the object is to low to allow equilibrium then the body will heat up until it reaches a temperature that satisfies this equality.

To say that a body is not emitting as much as it is absorbing is just another way as saying that the body is heating up.

4) p = q

Those are the basics as both I understand them and how they are recorded in literature. Getting these four simple concepts wrong will cause all manners of erroneous conclusions and beliefs.

This thread is opened with the intent to discuss the above compared to the understanding of other readers.

A good primer on the subject is a series of pages provided by http://hyperphysics.phy-astr.gsu.edu/hbase/bbcon.html#c1

An Example:

Example of using above basic radiative transfer equations to calculate the temperature of an object.

Body A is illuminating Body B. Body B is at a distance D from body A.

Source, body A, is represented as:

T(a) = 5778K

R(a) = 6.955*10^8m

ε = 1.0

Target, body B.

T(b) = ?

R(b) = 6.371*10^6m

D(ab) = 1.496*10^11m

ε = 0.7

Body A emits:

5) p = εσT^4.A

6) p = 1 * 5.67*10^-8W/m^2K^4 * 5778K^4 * (4 * π * 6.955*10^8m^2)

7) p = 3.84*10^26W

Power Body B intercepts:

8) Q = (πR(b)^2 / 4πD(ab)^2) * p

9) Q = (π * 6.371*10^6m^2 / 4 * π * 1.50*10^11m^2) * 3.84*10^26W

10) Q = 4.53*10^-10 * 3.84*10^26W

11) Q = 1.74*10^17W

Body B absorbs:

12) q = εQ

13) q = 0.7 * 1.74*10^17W

14) q = 1.22*10^17W

Body B emits:

15) p = q

16) p = 1.22*10^17W

Body B temperature:

17) p = εσT^4.A

18) T = (p/εσA)^.25

19) T = (1.24*10^17W / (0.7 * 5.67*10^-8W/m^2K^4 * (4 * π * 6.371*10^6m^2))^.25

20) T = 278.6K (or 5.4C)

This calculation is for the Sun/Earth system. It contains one error, the emissivity for the earth was held constant at 0.7 for both absorption of solar and emission of terrestrial radiation. The better approximation of the earth described at the top of the article would result in T = 254.8k(or -18.3C).

Of course the presence of an atmosphere would change this calculation. That is why in the real world we have a temperature of 15C, the greenhouse effect explains the difference.

By Neutrino

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This post follows  on from discussions at the thread http://jennifermarohasy.com/blog/2011/04/determining-the-total-emissivity-of-a-mixture-of-gases-containing-overlapping-absorption-bands/

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527 Responses to Radiative Transfer According to AGW: A Note from Neutrino

  1. gavin April 11, 2011 at 8:07 am #

    I like it. Congratulations!

  2. cementafriend April 11, 2011 at 11:11 am #

    Neutrino makes a few assumptions which are not correct. For a start the definition of a blackbody (neutrino BB) Perry’s Chemical Engineering Handbook (McGraw-Hill) states
    “The characteristic properties of a blackbody are that it absorbs all the radiation incident on its surface and that the quality and intensity of the radiation it emits are completely determined by its temperature”
    Note the word surface. No gas is a blackbody because it does not have a surface. Further, CO2 only absorbs and emits radiation in very narrow wavelengths. So again it is not a blackbody.
    Nasif Nahle has made some useful posts on the subject of radiation absorption/emission by CO2. I thank Jennifer for encouraging me to make the post http://jennifermarohasy.com/blog/2011/02/a-note-on-the-stefan-boltzman-equation/ which kicked off this discussion.

  3. Neutrino April 11, 2011 at 12:13 pm #

    Thank you Dr Marohasy for posting this.

    One point to clarify, the 15C referred to at the end is a reference to the current measured average global temperature.

    cementafriend,

    Dealing with a diffuse gas would add complexity but would not change the basic physics outlined above. The purpose was to simplify the physics so the basic relations could easier be seen. Dealing with a gas complicates things so the above is for solid surfaces only, sorry if that was not clear.
    It is correct that gasses are not BB’s, they are not even GB’s. But they can be approximated as such for many calculations.

    What I have written here is consistant with your quote. Set emissivity set to 1, a BB, then q = Q and p = σT^4.A. All incident is absorbed and emission is solely dependent on temperature. Which is exactly the same as your quote.

  4. Alan Siddons April 11, 2011 at 12:23 pm #

    Neutrino’s post has hardly any content. The math presented so far merely establishes that if the earth were a blackbody absorbing 4 times less than the 1368 W/m² that the sun provides, then its temperature, according to the Stefan-Boltzmann blackbody equation, would be about 278.6 K. Big deal.

    By the way, though, remember that the greenhouse effect is supposed to induce a higher SURFACE temperature than the sun can confer. According to the 1997 Kiehl-Trenberth budget, then, sunlight only amounts to 168 W/m² on the surface, which translates to 233 K or minus 40° Celsius, not minus 18. Moreover, K-T depicts a 102 W/m² loss from non-radiative heat transfer, which leaves the surface at 66 W/m². This translates to 185 K or minus 88° C. Since back-radiation brings the surface up to 390 W/m² (i.e., 66 + 324 = 390), which translates to 288 K or +15° Celsius, it’s evident that the “real” greenhouse effect raises the earth’s surface temperature by 103 degrees, not the 33 degrees that’s always cited. I await Neutrino’s explanation of how radiative transfer accomplishes such a miracle.

  5. cementafriend April 11, 2011 at 12:55 pm #

    Alan S makes a good point although he refers to K-T early paper. In the 2008 paper K-T have 40 w/m2 through the atmospheric window, 80 w/m2 latent heat and, 17 w/m2 thermals which is adds to 137 w/m2 compared with the 161 w/m2 absorbed by the surface. Now Dr Van Andel in his presentation http://climategate.nl/wp-content/uploads/2010/09/KNMI_voordracht_VanAndel.pdf (slide 26) says that the measured window is 66 w/m2 and that Trenberth wrote that he knows this (so much for his missing heat of 0.9w/m2) This then adds to 163 w/m2 more or less matching the incoming energy from the sun. Thus there is nothing left for the so-called greenhouse effect.

  6. Neutrino April 11, 2011 at 1:14 pm #

    Alan,

    The above math shows that if the earth, without an atmosphere, is placed in a 1368W/m^2 flux(which is the solar flux at the distance of 1 earth radius from the sun) then it will have a temperature of 278.6K(or 5.4C) if the earth was a BB or even an ideal GB(uniform emissivity across all wavelengths). But the earth is not an ideal GB, it has a different emissivity across the solar wavelengths as compared to the terrestrial wavelengths. With emissivity’s 0.7 and 1 respectively(again these are both just approximations, but useful ones) the earth then would have an average surface temperature of 254.8K(or -18.4C).

    For purposes of this thread bodies, the earth in this case, are assumed to be solids with no atmosphere.
    The first discrepancy that you cite of only 168W/m^2 being absorbed:
    (a note, if we are going to use numbers form the KT budget could we all please use his most recent publication. It is on page 4 here.)
    The earth(surface and atmosphere) absorbs a total of 161W/m^2+78W/m^2 = 239W/m^2 so that is the number used when doing the above calculations. It represents holding the total earth absorption the same not just the surface.
    The reason I left convection out of the above is because it complicates things. But since you pointed out that the earth loses 97W/m^2(using the 09 KT budget again) through convection and latent transfers I will make a quick comment.
    The earths temperature is not -18.4C because it is absorbing 239W/m^2 but because to shed 239W/m^2 it has to be that temperature(assuming no atmosphere again). The point being that the temperature of the body is not derived from calculating how much is arriving but how much is leaving. If you are going to use KT then the earth surface temperature is calculated solely from the 396W/m^2 surface emissions.(which leads to 15C surface temperature).

    The real greenhouse effect, no quotes needed on that, would raise the temperature of earth significantly more than 33C if it was not for convection. Convection is a negative feedback of the atmosphere, you can basically look at convection as a short circuit of the greenhouse effect. With no convection the surface would have to be radiating 396W/m^2+97W/m^2 to stay in equilibrium holding all other values constant.

    The point of simplifying the transfer to just solid bodies is to get a better understanding of the process by not being confused by clutter, after the basics are understood then complexities can be added back in.

    And the main point I want to make in response to your post is this, the temperature of an object is dictated by the emission of that object not by the absorption. Even if there are other flux’s leaving an objects surface(the latent and thermals from KT) the temperature is still just based on the current emission.

    I hope that is clear. But let’s stick to just solid surfaces in isolation for the time being please.

  7. Neutrino April 11, 2011 at 1:17 pm #

    Damn, I keep leaving a ” of of my tag try this

  8. Neutrino April 11, 2011 at 1:26 pm #

    cementafriend,

    The same point stands, temperature is not set by absorption. It is set by emission. Besides if you want to look at the full budget don’t leave out the down welling IR from the atmosphere, or the upwelling for that matter.

    Can we not deal with the incredibly complex earth system(of which the KT is just an incredibly simplified model of) but the easier to understand basics of solid bodies interacting?

  9. Alan Siddons April 11, 2011 at 1:39 pm #

    Hi Cementafriend. Please observe that the revised Trenberth chart follows the same twisted rule. A lot of people miss it. I can’t illustrate the new chart here but check for yourself.

    161 W/m² initially on the surface (231 K)
    97 W/m² lost non-radiatively
    64 W/m² surface remainder (183 K)
    333 from back-radiation
    333 + 64 = 397 W/m², the final result.

    The only difference is that 396 is radiated and 1 is now “absorbed.” Thus the surface would be 64 W/m² or 183 K, whereas the greenhouse effect makes it 396 W/m² or 289 K. The radiant “forcing” in this case, then, is 106 K.

  10. Neutrino April 11, 2011 at 1:45 pm #

    Alan,

    Same comment as before. There is no temperature associated with absorbing, temperature is only connected to emission. Strictly speaking emission is a result of temperature.

    Lets stick to the simplified example first.

  11. Alan Siddons April 11, 2011 at 2:00 pm #

    Nope, Neutrino, you cannot leave convection out because a whole new perspective emerges when it is factored in. See http://www-eaps.mit.edu/faculty/lindzen/198_greenhouse.pdf for instance.

    Surface heating by greenhouse gases must OVERSHOOT by a lot more than 33 degrees in order to compensate for non-radiative (convective and evaporative) cooling. The actual (or alleged) power of greenhouse gases is on the order of 612 W/m². That, combined with 239 from the sun, amounts to 851 W/m² in all, enough to raise the surface temperature to a blistering 350 Kelvin (believe it or not). Non-radiative cooling processes then bring that DOWN to a 288 K average. The point is that this “greenhouse effect” is far greater than the 33 degrees that you implied.

  12. Nasif Nahle April 11, 2011 at 2:15 pm #

    @Neutrino…

    For example, the earth can be approximated by a GB with emissivity around 0.7 for solar wavelengths (λ 4μm). Emissivity is a wavelength dependant quantity but approximations like the one used for the earth work very well.

    Many assumptions in one single paragraph:

    1. Earth cannot be approximated by a gray body. Earth is a gray body.

    2. Emissivity of the Earth is not solar radiation, but surface radiation. Perhaps you would mean absorptivity. In science you cannot say that absorptivity is the same as emissivity. Absorptivity is the power to absorb energy, while emissivity is the power to emit energy. The value of emissivity is equal to the value of absorptivity, but the concepts are opposed.

    3. The emissivity of the Earth is not 1, but 0.82. Here you are confused with the term. Emissivity is the potential to emit energy; it has nothing to do with the release or storage of the energy absorbed, but to the potential to emit energy. An emissivity 1 is only for ideal blackbodies and, as you have accepted, blackbodies don’t exist in real nature, consequently, the emissivity of the Earth CANNOT be the emissivity of a black body.

    4. The value you assign to the emissivity of the Earth is not real and it doesn’t work fine, as you say:

    p = 1 * 5.67*10^-8W/m^2K^4 * 5778K^4 * (4 * π * 6.955*10^8m^2)

    This equation is flawed because you’re introducing an emissivity 1 to a body A, which supposedly is the Sun. The emissivity of the Sun is not one, but 0.9875. So your calculation is wrong because it would give a power of 5.52303 x 10^17 W, while the real value, given by satellite measurements, is 3.94832 x 10^26 W.

    It seems you’re doing a “little” adjustment to make your idea fits with the numbers.

    The formula that you’re applying would give the following result, if it is to be applied as you suggest:

    p = e (5.6697 x 10^-8 W/m^2 K^4) (Tenvironment^4 – Tsurface of the Sun^4) (A)

    And the result would be -6.24032 x 10^7 W, which means that the Sun would be losing 62403200 W of energy by emitting it towards the colder space.

    If the Earth is the body B in the way of the Solar radiation, as you suggest, the Earth would be absorbing:

    p = 0.82 (5.6697 x 10^-8 W/m^2 K^4) (Tsolarsurface^4 – Tearth^4) (1 m^2) = 5.18 x 10^7 W, which is ridiculous according to observations; I mean, we would be toasted since many years ago.

    This demonstrates that you’re applying formulas that you don’t know how and where to apply.

    The scientific approach to astrophysical problems (which is the case that you have tried to offer as an example of the AGW idea) is by applying the following formula:

    GPL = QSUN / 4π (POR)^2, and the result is 1402.8 W/m^2.

    The result you provide is absolutely out of reality. You say that the Earth is absorbing/emitting
    q = 1.22*10^17W

    Which would be a power of 122222222222222222 W!!! What demonstrates that you’re not applying the correct formulas to astrophysical problems and are not careful on applying the correct science of heat transfer.

    This essay from Neutrino on AGW’s viewpoint demonstrates what I have been saying in my posts, i.e. AGW idea applies unreal numbers, unreal systems and distorts science.

    😀

  13. gavin April 11, 2011 at 2:17 pm #

    Alan; I reckon you are stepping off this basic discussion model too soon in order to win your dubious point

  14. Nasif Nahle April 11, 2011 at 2:44 pm #

    @Neutrino…

    You answered to Alan:

    Same comment as before. There is no temperature associated with absorbing, temperature is only connected to emission. Strictly speaking emission is a result of temperature.

    Absolutely wrong. Both processes are directly related to temperature. From Modest (I quote from this author because you have access to it in the Internet and you could easily find that what I am saying is true), Radiative Heat Transfer, page 66:

    “Following the same arguments as for the development of equation (1.31), augmenting the emitted flux by ε’λ and the absorbed flux by αλ, we find immediately

    αλ (T, λ, θ, ψ) = ε’λ (T, λ, θ, ψ)”

    Question to you:

    1. What does that “T” in both terms of the equation refers to?

    2. If the value for emissivity is the same value for absorptivity, and emissivity is temperature dependent, would not the absorptivity depend on the temperature of the absorber, as you say?

    😀

  15. Nasif Nahle April 11, 2011 at 2:53 pm #

    @Gavin…

    Given that this is a serious dialogue, could you be more specific on your argument to Alan?

    You said:

    Alan; I reckon you are stepping off this basic discussion model too soon in order to win your dubious point

    This thread is about science, so you are obliged to expand on your argument and to be more specific.

    If you cannot be more specific, then you are who is stepping off this discussion.

    Alan is correct. AGW idea dismisses absolutely convection and affirms, falsely, that convection is not a relevant process of heat transfer when referring to the climate of Earth. If convection was not relevant, how is it that we have rain, dripple, hail, snow, drought, etc.?

    😀

  16. Nasif Nahle April 11, 2011 at 4:23 pm #

    @Neutrino…

    Here another serious error:

    17) p = εσT^4.A

    18) T = (p/εσA)^.25

    19) T = (1.24*10^17W / (0.7 * 5.67*10^-8W/m^2K^4 * (4 * π * 6.371*10^6m^2))^.25

    20) T = 278.6K (or 5.4C)

    If you’re showing a step by step procedure, where the term (4 * π * 6.371*10^6m^2)) was taken from if it is not included in the formula of Stefan-Boltzmann that you wrote in 18)? I do know what the numbers introduced in the term are, but it is a flawed equation which does not follow the equation at 18).

    By following your “formula”, we would obtain a temperature of the Earth of

    T = (1.24*10^17 W / (0.7 * 5.67*10^-8 W/m^2 K^4 * (4 * π * 6.371*10^6 m^2))^.25

    T = (1.24*10^17 W / 3.17743 m^4)^0.25 = 14055.184 K, not 278.6 K.

    Is 14055.184 K (13782 °C) the temperature of the Earth?

    Here we are not only reading TYPOS, but an INCORRECT formula.

    Something is wrong with your formula. Why you don’t review it again?

    🙂

  17. Debbie April 11, 2011 at 4:42 pm #

    Gavin,
    I have to agree with Nasif.
    Alan has raised a valid point.
    There are variables in our atmosphere that need to be accounted for.
    There are still questions about the assumed figures and also the emissivity and radiative characteristics of gases in our atmosphere.
    Our atmosphere does not behave in the same manner as a roof on a greenhouse, neither does it behave the same as the outside of a furnace.
    There are other forces at work here and Nasif, Neutrino, Cohenite, Alan, cementafriend and others are debating what those forces might be and how they would apply to climate modeling.
    BTW Nasif, what is dripple? Maybe a typo?

  18. Nasif Nahle April 11, 2011 at 5:08 pm #

    @Debbie…

    Heh! I double typo. It should have said “drizzle”. Sorry. 🙂

    NSN

  19. gavin April 11, 2011 at 5:20 pm #

    Nasif: imo The statement “emissions is a result of temperature” is correct as it applies to the Sun. We can even use it here with a little imagination and without stretching the friendship see “Technical Emissions associated with the Mobile Model – Evaporative Emissions ” @ Google Books, but the same search leads to this page and it’s mainstream Euro engineering today-

    “Environmental problems – the greenhouse effect

    The troposphere is the lower part of the atmosphere, of about 10-15 kilometres thick. Within the troposphere there are gasses called greenhouse gasses. When sunlight reaches the earth, some of it is converted to heat. Greenhouse gasses absorb some of the heat and trap it near the earth’s surface, so that the earth is warmed up. This process, commonly known as the greenhouse effect, has been discovered many years ago and was later confirmed by means of laboratory experiments and atmospheric measurements.

    Life as we know it exists only because of this natural greenhouse effect, because this process regulates the earth’s temperature. When the greenhouse effect would not exist, the whole earth would be covered in ice.
    The amount of heat trapped in the troposphere determines the temperature on earth. The amount of heat in the troposphere depends on concentrations of atmospheric greenhouse gasses and the amount of time these gasses remain in the atmosphere. The most important greenhouse gasses are carbon dioxide, CFC’s (Chlor-Fluoro-Carbons), nitrogen oxides and methane”.

    Read more: http://www.lenntech.com/carbon-dioxide.htm#ixzz1JCCBjo8F

    http://www.lenntech.com/carbon-dioxide.htm

    The whole issue is about a heated atmosphere and how it blankets Earth

  20. gavin April 11, 2011 at 5:32 pm #

    Debbie; before rejecting Neutrino’s approach read “Perspectives on the greenhouse effect and global warming” by S M Enzler and say what bit applies to yourself hey

    Read more: http://www.lenntech.com/greenhouse-effect/global-warming-perspectives.htm#ixzz1JCI5hh8r

  21. Johnathan Wilkes April 11, 2011 at 5:32 pm #

    gavin

    When as first year student at uni. I presented something like your last comment to my lecturer about capacitors.
    Must admit I had not done any homework or study, having had a good time instead.

    He asked me what kind of muesli did I have for breakfast, meaning that all my knowledge about capacitors at that stage came from the back of the box.

    So what brand of cornflakes did you have for breakfast?

  22. gavin April 11, 2011 at 5:54 pm #

    A further defense of Neutrino on the statement “emissions is a result of temperature” see Stefan-Boltzmann’s Law and it’s 3 expressions –

    “The most important consequence of this law is that hot objects radiate energy at a rate proportional to the fourth power of their absolute Kelvin temperature”

    http://dev.physicslab.org/Document.aspx?doctype=3&filename=ThermalPhysics_Heat.xml

  23. Debbie April 11, 2011 at 6:38 pm #

    Gavin,
    I’m not rejecting Nuetrino’s approach, I’m watching this debate and learning as I go.
    I agreed with Nasif because I didn’t think your comments were appropriate or substantiated.
    Neutrino appears to be perfectly capable of looking after himself.
    I would also point out that ‘a natural greenhouse effect’ is not necessarily the same as a ‘greenhouse effect’.
    There are other variables at work here Gavin, that’s what they’re discussing/debating.

  24. cohenite April 11, 2011 at 7:14 pm #

    With all due respect to Neutrino this is a bit of an anti-climax; I was expecting some profound rebuttal to Nasif’s previous post which raised the profound point that CO2 reduces the emissivity of H2O and therefore the major positive feedback mechanism of AGW.

    It is well established that dealing with pure radiative exchange the greenhouse effect would be much higher, as Alan has noted and which Lindzen concurs:

    “As it turns out, if there were only radiative heat transfer, the greenhouse effect would warm the Earth to about seventy-seven degrees centigrade rather than to fifteen degrees centigrade. In fact, the greenhouse effect is only about 25 percent of what it would be in a pure radiative situation. The reason for this is the presence of convection (heat transport by air motions), which bypasses much of the radiative absorption.”

    The basic proof that the greenhouse effect on Earth is 33K comes from Arthur Smith’s paper which neutrino seems not to be aware of:

    http://arxiv.org/PS_cache/arxiv/pdf/0802/0802.4324v1.pdf

    Smith in turn has been critiqued by Kramm:

    http://arxiv.org/ftp/arxiv/papers/0904/0904.2767.pdf

    Who was critiqued in turn in typical laxidaisical fashion by the rabbit:

    http://rabett.blogspot.com/2009/07/krammis-eli-has-been-mining-gerhard.html

    The point about all this is that eli’s criticism of Kramm is that Kramm alegedly mistook the heating effect of 2.7u radiation as having a surface effect when it actually, according to eli, was the temperature 10-20cm beneath the surface of the moon. But according to Neutrino’s exposition the point about emissivity is that “the temperature of an object is dictated by the emission of that object not by the absorption” so the fact that 2.7u radiation penetrates to slightly below the surface is irrelevant since the emissivity of the Moon will be based on its surface and subsurface temperature.

    There, that’s off my chest, so Neutrino’s post at least gave me the opportunity to do that.

  25. Louis Hissink April 11, 2011 at 8:35 pm #

    Neutrino,

    Your explanation is incomplete – you ignore totally the massive amounts of electromagnetic energy entering into and out of the earth-system via Birkeland currents at the polar regions, and which are measured routinely in millions of amperes. CME’s and solar wind fluctuations cause surges in these currents are then are manifiest as the polar auroras. The earth also continually receives electrical currents to its surface from the ionosphere.

    Your explanation is quite correct for a rock ball in vacuo receiving only radiation. Using this model one finds thermal anomaly which is then explained by a greenhouse gas effect.

    The thermal anomaly can be better explained by the application of plasma physics to the earth system.

    The physics being used here is of Victorian era age – we have since made some progress in the field of plasma physics and the physics of the plasma universe. Add the electromagnetic domain and there is no need for a greenhouse gas effect.

    I might point out that electric currents passing through matter tend to increase the energetic state of the matter, though if you only use hydrocarbons as a source of energy, then of course that limits the number possible insights.

  26. cohenite April 11, 2011 at 10:27 pm #

    Actually Louis Neutrino is not even correct for “a rock ball in vacuo receiving only radiation”; this is what Arthur Smith attempted and was incorrect.

  27. Jennifer April 11, 2011 at 10:59 pm #

    Well I think it was good, and brave, of Neutrino to have a go at defending the dominant paradigm at this blog. Thank you.

    I am still open to posts from others in defence of AGW theory as it pertains to radiation and the greenhouse effect.

  28. Nasif Nahle April 12, 2011 at 12:24 am #

    @Gavin…

    You say:

    Nasif: imo The statement “emissions is a result of temperature” is correct as it applies to the Sun. We can even use it here with a little imagination and without stretching the friendship see “Technical Emissions associated with the Mobile Model – Evaporative Emissions ” @ Google Books, but the same search leads to this page and it’s mainstream Euro engineering today-

    The problem is not the statement “emissions is a result of temperature”, bu Neutrino’s assertion “There is no temperature associated with absorbing, temperature is only connected to emission”, which is not true.

    🙂

  29. Nasif Nahle April 12, 2011 at 1:43 am #

    I agree with Dr. Jennifer. At least, Neutrino has had the bravery of defending AGW ideas, which is nothing more than that , ideas.

    I sincerely believe that Neutrino, whoever is he, is not responsible of science distortion, but those who promote AGW pseudoscience by teaching unreal science to the public and have taken control of academies.

    NSN

  30. Schiller Thurkettle April 12, 2011 at 3:14 am #

    All of these calculations are fine and good, but matching them to measurable phenomena is problematic.

    The Earth has a molten core at approximately 7,000 F. This heat obviously conducts outwards (second law) and eliminating this factor is bound to force calculations and observations into disagreement.

  31. Nasif Nahle April 12, 2011 at 4:37 am #

    @Schiller Thurkettle…

    The Earth has a molten core at approximately 7,000 F. This heat obviously conducts outwards (second law) and eliminating this factor is bound to force calculations and observations into disagreement.

    Definitely, that load of energy conducted outwards from the Earth’s inner core must be considered to all calculations. For the planet Earth, the load of energy conducted from the inner core towards the surface is:

    q = – kA (ΔT/Δn)

    q = – (2.6296875 W / (m K)) * 1.49 × 10^14 m^2 ((5778 K – 318 K)/635700 m) = – 2.054 x 10^12 W

    Nothing negligible, huh?

    NSN

  32. Neutrino April 12, 2011 at 4:56 am #

    Alan,

    You said:
    Surface heating by greenhouse gases must OVERSHOOT by a lot more than 33 degrees in order to compensate for non-radiative (convective and evaporative) cooling.

    Which is basically what I said:
    The real greenhouse effect, no quotes needed on that, would raise the temperature of earth significantly more than 33C if it was not for convection. Convection is a negative feedback of the atmosphere, you can basically look at convection as a short circuit of the greenhouse effect. With no convection the surface would have to be radiating 396W/m^2+97W/m^2 to stay in equilibrium holding all other values constant.

    Leaving convection out gets very wrong numbers, as I said above, but it is useful to do to try and understand the basics.

  33. mkelly April 12, 2011 at 5:54 am #

    Neutrino says:

    Source, body A, is represented as:

    T(a) = 5778K

    If we are going have this discussion please include how you came up with this value. (Please donot jsut quote a book number use the formula.) I personally enjoy these discussions.

  34. Neutrino April 12, 2011 at 6:20 am #

    Nasif,

    1. An ideal GB has uniform emissivity across all wavelengths, from gamma through visible and infrared all the way to radio.
    The earth does not have uniform emissivity across every wavelength. Across the wavelengths where the earth is absorbing solar radiation(0.2μm-4μm) the earth has approximately an emissivity of 0.7. Across the wavelengths where the earth is emitting its own radiation(4μm-40μm) it has an emissivity close to 1.

    2. “In science you cannot say that absorptivity is the same as emissivity.” That is the entire point of Kirchhoff’s Law. We can say that absorptivity equals emissivity.

    3. I did not say that the earth was a blackbody. What I said was “and an emissivity near 1.0 for terrestrial wavelengths(λ > 4μm)” It is a good approximation. The number you are using(0.82) is way too low.
    Just to give a back of the envelope calculation for the earths longwave emissivity consider the oceans. The earth is about 70% oceans, the emissivity of water(measured) is 0.95-0.98. To set the lower bound assume the other 30% of the earth has an emissivity of 0.0 the total emissivity of te earth would be 0.95*0.7 +0.0*0.3 = 0.665. Since most surfaces that make up the landmass(grass, sand, trees, ice are all in the 0.8,0.9,0.95 range) all have significantly higher emissivity’s than 0.0 the overall emissivity of the earth is also higher.

    4. Again, I did not say the sun was a BB, what I did say was: “ For instance the sun is a good approximation of a BB.” . Besides your value of 0.9875 is very well approximated by 1. As you’re your value of solar power of 3.94832*10^26 is only off from mine of 3.84*10^26W (which is a measured value given by NASA) by a few percent.
    Remember this is about the general principle of radiative transfer. If you want to get into who has done the better measuremensts for all these values that’s another thread entirely.

    Yes there is a typo there(I forgot the brackets) looks like I did it on 6), 9), and 19).
    6) 4 * π * 6.955*10^8m^2 should be:
    6) 4 * π * (6.955*10^8m)^2, it is the just the surface area of a sphere 4πr^2.

    The problem with using:
    p = εσ(T(a)^4-T(b)^4).A
    to calculate the exchange of power between two objects is that both objects do not have the same emissivity(as well as the area(A)). In this case it would be better to write it all out as
    p = ε(a)σT(a)^4.A(a) – ε(b)σT(b)^4.A(b)
    As well you would need to account for the view factor between the two emitters(how much of (a)’s total power falls on (b) and vice versa)
    As you used it the formula is just the power emitted by (a) minus the power emitted by (b) assuming they have the same emissivity’s and areas. Which I think is not what you were trying to convey.

    My numbers above have an incident on the earth(Q) of 1.74*10^7W, this is not a ridiculous number. It is just 1366W/m^2. I wrote the power incident on the earth not the flux at the earth, but they are the same thing just by a factor of the surface area eclipsed by the disk of the earth.

    Yes. The earth is absorbing and emitting 1.22*10^7W, the entire earth. Or as a flux if you like that same value is 239W/m^2.

    Not one point you brought up detracts from the lead in post. My numbers are approximates, to facilitate the discussion. Is the earth shortwave emissivity really 0.7? No, but it is approximately that. So yes of course if we used a better approximation of the emissivity’s involved we would have better results. But the point of this thread was not to exactly nail down the numbers but to simplify things so the transfers could be looked at without confusing things. As such my approximations are good enough for what we are doing.

    In regards to:
    αλ (T, λ, θ, ψ) = ε’λ (T, λ, θ, ψ)
    Nasif, this is just a restatement of Kirchhoff’s Law(and a much more formal one at that). Yes the vales are slightly dependant on T. That is not the same as saying absorption is dependent on T^4. What I said still stands(bringing this up doesn’t detract from it): absorption is a function of only two things, emissivity and incident.

  35. Neutrino April 12, 2011 at 6:21 am #

    Schiller and Nasif,

    Transfer of energy from the earth’s core to the surface is negligible. That 2.1*10^12W you cite is less than 3.92mW/m^2, about one thousandths of one percent of the absorbed solar flux absorbed.

    There is a big difference when quoting flux and power, for objects the size of the earth they will be hugely different, if we were talking about an object that was only one square meter they would be the same.

  36. Neutrino April 12, 2011 at 6:22 am #

    Nasif,

    You said(partially quoting me):
    The problem is not the statement “emissions is a result of temperature”, bu Neutrino’s assertion “There is no temperature associated with absorbing, temperature is only connected to emission”, which is not true.”(your bold)

    What then is the temperature associated with an object absorbing 100W?

  37. Neutrino April 12, 2011 at 6:27 am #

    Hello mkelly,

    I should have cited a source, I apologize. Those numbers all came from NASA, Sun & Earth.

  38. mkelly April 12, 2011 at 6:48 am #

    Neutrino April 12th, 2011 at 6:27 am

    Why not enter Wien’s Law into the mix? Get you number that way.

  39. Neutrino April 12, 2011 at 6:52 am #

    These three lines all have a similar problem, they are all missing brackets around the radius’s so the power appears to just make them values in m^2 already.

    6) p = 1 * 5.67*10^-8W/m^2K^4 * 5778K^4 * (4 * π * 6.955*10^8m^2)
    9) Q = (π * 6.371*10^6m^2 / 4 * π * 1.50*10^11m^2) * 3.84*10^26W
    19) T = (1.24*10^17W / (0.7 * 5.67*10^-8W/m^2K^4 * (4 * π * 6.371*10^6m^2))^.25

    They should be:
    6) p = 1 * 5.67*10^-8W/m^2K^4 * 5778K^4 * (4 * π * (6.955*10^8m)^2)
    9) Q = (π * (6.371*10^6m)^2 / 4 * π * (1.50*10^11m)^2) * 3.84*10^26W
    19) T = (1.24*10^17W / (0.7 * 5.67*10^-8W/m^2K^4 * (4 * π * (6.371*10^6m)^2))^.25

  40. Nasif Nahle April 12, 2011 at 7:09 am #

    @Neutrino…

    I congratulate you for having the bravery of writing an essay on the undefensible AGW idea.

    Let’s start from the begining.

    You are applying the Stefan-Boltzmann equation, correct? The problems appear when you start introducing not proper quantities. For example, you assign a radius to the Sun of “R(a) = 6.955*10^8 m”. The following links will help:

    http://www.astro.uu.nl/~strous/AA/en/antwoorden/zon.html#v230

    http://hypertextbook.com/facts/2004/AmyChan.shtml

    It seems you introduced the formula to obtain the area of a sphere here:

    6) p = 1 * 5.67*10^-8W/m^2K^4 * 5778 K^4 * (4 * π * 6.955*10^8 m^2)

    You assign a radius of the solar disk when you wrote:

    R(a) = 6.955*10^8 m

    However, the number 6.955*10^8 m^2 is incorrect, if you’re considering a solar surface radius of 6.55 x 10^8 m, the second power of this number is 4.83025 x 10^17 m^2, that is R * R, not the identical value of the radius.

    The correct formula gives

    A = 4 * π * (6.955 x 10^8 m)^2 = 6.07x 10^18 m^2

    Consequently, the formula to obtain the power from the surface of the Sun would be:

    p = 1 * 5.67*10^-8 W/m^2K^4 * 5778 K^4 * (6.07 x 10^18 m^2)

    And the result of the power from the whole surface of the Sun, BY FOLLOWING YOUR FORMULA, would be 1.9885 x 10^15 W, not those “p = 3.84*10^26 W” that you obtained, even when you’re saying the Sun is a blackbody and we all know that blackbodies don’t existe in nature.

    The formula is wrongly written; consequently, I deduce that you copied and pasted results without knowing what you were doing.

    🙂

  41. Nasif Nahle April 12, 2011 at 7:33 am #

    @Neutrino…

    You say:

    1. An ideal GB has uniform emissivity across all wavelengths, from gamma through visible and infrared all the way to radio.
    The earth does not have uniform emissivity across every wavelength. Across the wavelengths where the earth is absorbing solar radiation(0.2μm-4μm) the earth has approximately an emissivity of 0.7. Across the wavelengths where the earth is emitting its own radiation(4μm-40μm) it has an emissivity close to 1.

    That’s not what you said in your post:

    “and an emissivity near 1.0 for terrestrial wavelengths(λ > 4μm)”

    You say:

    2. “In science you cannot say that absorptivity is the same as emissivity.” That is the entire point of Kirchhoff’s Law. We can say that absorptivity equals emissivity.

    You’re wrong in your interpretation of the Kirchhoff’s Law because it doesn’t refer to the concepts, but to the values. Emissivity is not the same as absorptivity. One term is the opposite to the other.

    You say:

    3. I did not say that the earth was a blackbody. What I said was “and an emissivity near 1.0 for terrestrial wavelengths(λ > 4μm)” It is a good approximation. The number you are using(0.82) is way too low.

    You say it is 0.7, the number obtained by direct observation and measurements is 0.82, which is the number I gave. Which number is higher, 0.82 or 0.7? You wrote:

    13) q = 0.7 * 1.74*10^17W

    You say:

    4. Again, I did not say the sun was a BB, what I did say was: “ For instance the sun is a good approximation of a BB.” . Besides your value of 0.9875 is very well approximated by 1. As you’re your value of solar power of 3.94832*10^26 is only off from mine of 3.84*10^26W (which is a measured value given by NASA) by a few percent.
    Remember this is about the general principle of radiative transfer. If you want to get into who has done the better measuremensts for all these values that’s another thread entirely.

    However, you assigned a total emissivity to the Sun of 1.0, which is the emissivity of the non-existent blackbodies when you wrote:

    ) p = 1 * 5.67*10^-8W/m^2K^4 * 5778 K^4 * (4 * π * 6.955*10^8 m^2)

    Double discourse?

    You say:

    The problem with using:
    p = εσ(T(a)^4-T(b)^4).A
    to calculate the exchange of power between two objects is that both objects do not have the same emissivity(as well as the area(A)). In this case it would be better to write it all out as
    p = ε(a)σT(a)^4.A(a) – ε(b)σT(b)^4.A(b)
    As well you would need to account for the view factor between the two emitters(how much of (a)’s total power falls on (b) and vice versa)
    As you used it the formula is just the power emitted by (a) minus the power emitted by (b) assuming they have the same emissivity’s and areas. Which I think is not what you were trying to convey.

    There is no problem there is you consider that the Sun is immersed into a cold environment, the space. If you dismiss the environment of the thermodynamic system on trying to know the radiative heat transfer from a thermodynamic system, you must consider the environment of such thermodynamic system, otherwise, the calculations are flawed, or biased, as you wish.

    You say:

    Nasif,

    You said(partially quoting me):
    “The problem is not the statement “emissions is a result of temperature”, bu Neutrino’s assertion “There is no temperature associated with absorbing, temperature is only connected to emission”, which is not true.”(your bold)

    What then is the temperature associated with an object absorbing 100 W?

    Again?

    αλ (T, λ, θ, ψ) = ε’λ (T, λ, θ, ψ)”

    Can you see that biiiig T?

    If you wish to know the temperature associated with an object absorbing 100 W, you MUST know the temperature of that object when it is absorbing those 100 W. It’s quite simple and your statement that absorptivity is not related to the temperature of the absorber is wrong.

    🙂

  42. Nasif Nahle April 12, 2011 at 7:54 am #

    @Neutrino…

    You should consider this formula again:

    They should be:
    6) p = 1 * 5.67*10^-8W/m^2K^4 * 5778K^4 * (4 * π * (6.955*10^8m)^2)

    That quantity is flawed also. It must be (5778 K)^4, i.e. 1.114 x 10^15 K^4

    😀

  43. Neutrino April 12, 2011 at 8:35 am #

    As pointed out by Nasif,

    The brackets are also missing on the temperature of the sun in line 6)
    6) p = 1 * 5.67*10^-8W/m^2K^4 * 5778K^4 * (4 * π * 6.955*10^8m^2)

    It should read:
    6) p = 1 * 5.67*10^-8W/m^2K^4 * (5778K)^4 * (4 * π * (6.955*10^8m)^2)

    Neither omission of brackets was reflected in the actual calculations but only in the reproductions for the post.

  44. Neutrino April 12, 2011 at 8:49 am #

    Nasif,

    A)The formulas I wrote were missing some brackets, the formula I used in excel had the correct brackets, as evidenced by the calculation result actually following from 4πr^2 not 4πr with just units squared.

    As I wrote in my response to you here, and the post here that corrects the 3 lines that have the same missing brackets. ie:
    4 * π * 6.955*10^8m^2 should be
    4 * π * (6.955*10^8m)^2

    I did not do the calculations based on 6.955*10^8m^2 but (6.955*10^8m)^2. That was a typo, you caught it and I corrected it. Evidence of this is that the result of 6), 9), and 19) all have results calculated as if the brackets were there.

    (This also applies to the 5778K just pointed out)

    B)How does ”Across the wavelengths where the earth is emitting its own radiation(4μm-40μm) it has an emissivity close to 1.” in any way contradict ”and an emissivity near 1.0 for terrestrial wavelengths(λ > 4μm)”?

    C)In regards to Kirchhoff, yes I agree it is equating the value of α and ε. Not saying that absorption is the same as emission. But the coefficient for both is the same.

    D)The earth absorbs solar wavelengths(<4μm), over that range 0.7 is a valid approximation which is easily estimated just using KT budget numbers:
    341.3W/m^2 Incident
    102.9W/m^2 Reflected
    Reflected / Incident = 1 – α
    102.9 W/m^2 /341.3 W/m^2 = 1 – α
    α = 1 – 0.3015
    α = 0.6985

    So when calculating absorption 0.7 is a close approximation, 0.82 is not.

    E)As for the sun, NASA lists the effective temperature of 5778K and emitted power of 3.846*10^24W, if you don’t like those numbers take it up with them. Again the purpose for this thread was to look at how the radiant transfers work, not what is the best estimate of those two values.

    F)There is no problem using p = εσ(T(a)^4-T(b)^4).A to calculate energy transfer as long as the emissivity’s of each body is the same, the area of each body is the same and the view factor of each body with respect to the other is the same. The only one of those three conditions that is true for the earth-sun system is the emissivity(both are near 1, again I said near its an approximation, for the respective temperatures they each are radiating at).

    G)Back to Kirchhoff again:
    αλ (T, λ, θ, ψ) = ε’λ (T, λ, θ, ψ)
    Yes, I see the T. I also see the λ, θ & ψ. Yes if we are to be absolutely correct about α and ε they both have a temperature dependence, this should be no surprise from Hottel & Lapp’s graphs. But for solids, the amount that temperature swings the values is minimal and can be ignored for approximations. Besides, that α and ε depend, slightly, on temperature does not mean the Stefan-Boltzmann equation dictates absorption. Once any temperature dependence is calculated into α then T no longer needs to be rewritten into the absorption equation, as whatever its minimal effects has already been taken into account.

    Question:
    Since you need more information to answer the question:
    What is the temperature of a body that absorbs 100W initially starting at 100K?

  45. Neutrino April 12, 2011 at 8:59 am #

    Question should have been:
    What is the equilibrium temperature of a body that absorbs 100W initially starting at 100K?

  46. cohenite April 12, 2011 at 11:12 am #

    Jennifer, my criticism of Neutrino is that he has not addressed the consequence of the point from Nasif’s earlier thread about the reduction in emissivity in the combination of CO2 and H2O and the possible ramifications of that fact for the centrepiece of AGW, positive reinforcement of CO2 heating by H2O. Until he does this blizzard of formula is a side-show.

    In respect of this sideshow Neutrino says this:

    “C)In regards to Kirchhoff, yes I agree it is equating the value of α and ε. Not saying that absorption is the same as emission. But the coefficient for both is the same.”

    This is not correct; Kirchoff only applies to a thermal equilibrium and its significance is that the sum of α and ε is 0. As Nasif has said there are 3 variables here, α and ε and the temperature of the body.

    This is where Miskolczi was subject to much criticism in respect of his central point where Aa=Ed; the argument from the pro-AGW side was that upward radiation from the surface Aa could not equal longwave down radiation from the atmosphere, Ed because Kirchoff’s law did not apply because the surface and the atmosphere were not in thermal equibrium. This is true from moment to moment but diurnally and annually they are. In this Miskolczi has been vindicated.

  47. Nasif Nahle April 12, 2011 at 11:26 am #

    @Neutrino…

    Question:
    Since you need more information to answer the question:
    What is the temperature of a body that absorbs 100W initially starting at 100K?

    I cannot guess. I’m adhered to scientific methodology. Tell me the absorptivity of that body and I will answer the question as it must be answered.

    🙂

  48. Neutrino April 12, 2011 at 11:34 am #

    What is the temperature of a body that absorbs 100W initially starting at 100K?

    Body has emissivity of 0.9

  49. Nasif Nahle April 12, 2011 at 11:35 am #

    @Neutrino…

    I won’t apply the same criterion that you applied on my typos and won’t say that your article is not worth just for having those typos; therefore, let’s adhere to the scientific knowledge. Is it correct?

    Again, you admit that the Sun is not a blackbody; however, in your calculation you introduced a solar emissivity of 1.0, which is for a blackbody. You cannot apply the Stefan-Boltzmann equation to gray bodies if you don’t know their emissivity or absorptivity.

    Talking about something else, you said The temperature of the absorber is irrelevant to absorption. The amount absorbed(q) is a function of emissivity and the incident power(Q)”. I recommend you to change this idea and adopt the scientific concept. Temperature is essential for determining both, absorptivity and emissivity coefficients.

    🙂

  50. Neutrino April 12, 2011 at 11:50 am #

    The temperature of the absorber is irrelevant to absorption. The amount absorbed(q) is a function of emissivity and the incident power(Q)
    I agree, first bolded part is a slight an overstatement.
    The point that I was trying to make is that once emissivity is calculated(or estimated) then the second half of the statement is true.

    But for real world applications the amount that emissivity varies with temperature is small enough to not come into play for solid objects over terrestrial temperatures. It certainly is a good enough approximation to say that emissivity is temperature independent as it is to say it is also wavelength independent (over certain ranges) which is what we do all the time with GB’s.

  51. Neutrino April 12, 2011 at 12:02 pm #

    I said that the calculation you did was worthless because you actually calculated the result from the incorrectly transcribed formula. My error was not the same. Although I did transcribe the formula incorrectly on this page I had still calculated it correctly in the excel that I used to come up with the results.
    What made the results you came to originally worthless was that you had calculated results based on the transcription error, not that you had just incorrectly written it in the article.

    There is a difference between those errors, that’s not to excuse my neglect in typing the lines for this page.

  52. Nasif Nahle April 12, 2011 at 12:13 pm #

    @Neutrino…

    But for real world applications the amount that emissivity varies with temperature is small enough to not come into play for solid objects over terrestrial temperatures.

    What you say is not what it is observed in the real world, Neutrino. In the course of my professional life (many, many years) I have made calculations of emissivities of different bodies, at different temperatures; in all cases, total emissivities and total absorptivities are most sensible to changes of temperature.

    🙂

  53. Nasif Nahle April 12, 2011 at 12:22 pm #

    @Neutrino…

    I said that the calculation you did was worthless because you actually calculated the result from the incorrectly transcribed formula. My error was not the same. Although I did transcribe the formula incorrectly on this page I had still calculated it correctly in the excel that I used to come up with the results.

    However, the result was the same. A bit higher, but the same result. In your case, the wrong transcription of formulas are confusing, so they are worthless.

    In your case, you’re introducing false values, as for example the energy emitted by the Sun towards the Earth, that you assigned as if the whole load of energy emitted by the solar sphere were directed towards the Earth. This is a major mistake because the energy that could hit the Earth is not that of the whole solar sphere, but only a portion of the energy of the solar hemisphere facing towards the Earth.

    Your result is in contradiction to satellital observations and measurements because you say that the power received by the Earth is Q = 3.84*10^26 W, the same amount of the energy emitted by the whole solar sphere, i.e. p = 3.84*10^26 W. It’s a fantasy of you.

    😀

  54. Neutrino April 12, 2011 at 1:12 pm #

    If you are going to accuse me of stating something it would be wise to be accurate.

    You claim that I am saying that the power received at the earth is equal to what I say is emitted by the sun.

    To be clear this is your intent here is your quote:
    In your case, you’re introducing false values, as for example the energy emitted by the Sun towards the Earth, that you assigned as if the whole load of energy emitted by the solar sphere were directed towards the Earth.
    And continues:
    Your result is in contradiction to satellital observations and measurements because you say that the power received by the Earth is Q = 3.84*10^26 W, the same amount of the energy emitted by the whole solar sphere, i.e. p = 3.84*10^26 W. It’s a fantasy of you.

    With comments like that I wonder if you read what I post.

    So ill restate it, copied from lead post: (bold added)
    Power Body B intercepts:
    8) Q = (πR(b)^2 / 4πD(ab)^2) * p
    9) Q = (π * 6.371*10^6m^2 / 4 * π * 1.50*10^11m^2) * 3.84*10^26W
    10) Q = 4.53*10^-10 * 3.84*10^26W
    11) Q = 1.74*10^17W

    Where do I say the earth intercepts the same power the sun emits?
    Apart from the missing brackets the above is explicit. The earth intercepts Q = 1.74*10^17W, which is a fraction, an abysmally small fraction of what the sun emits.

    With you misrepresenting what I write how can we ever have an actual dialogue?

  55. Neutrino April 12, 2011 at 1:31 pm #

    It is sad, I was hoping that we could actually discuss the radiative exchange. Use some examples to see which viewpoint is valid and hopefully help some people to better understand the processes involved.

    But all you want to do is talk about the minutia. Yes nothing is a BB, it’s an approximation to make things simple. Using that approximation doesn’t invalidate the results, what it does do is make them not accurate to the nth decimal point. Yes emissivity changes with temperatures, but we can approximate it as constant just like we do with GB emissivity. Both are approximations that help in making the equations simpler but do not invalidate the results. Will my results be 100% accurate with these approximate values used? No, but so what? The point is to demonstrate the principles involved not to calculate the exact value accurately enough to send a man to the moon.

    Would you object to an example of a ball rolling down a ramp to demonstrate gravity if the friction was left out? Simplified systems are the norm in physics education, they help in conveying complex ideas. Quit getting hung up on the fine details. Once everyone agrees and understands the basics then re-introduce the complexities. The point is that absorption is εQ and emission is εσT^4.A. So what if we use ε with a value of 1. We are not calculating the actual result we are doing a simplified version of it.

  56. Nasif Nahle April 12, 2011 at 1:41 pm #

    @Neutrino…

    I didn’t want to reach to this point. I don’t like to expose people; however, you don’t leave me another chance.

    The whole calculation of you is incorrect because you are not applying the correct methodology.

    How is it that the Earth is only receiving 1368 W/m2 during daylight?

    The Stefan-Boltzmann equation is incorrectly applied. You have not used the Inverse Square Law, which is the rule that must to be applied if you wish to use the Stefan-Boltzmann equation.

    Additionaly, in order to obtain the energy received by the Earth from the Sun, the Stefan-Boltzmann does not apply because of an outer sphere volume. Got it?

    Please, Neutrino… I don’t know if it is worth to post the correct astrophysical procedure for the sake of the readers. Dr. Jennifer has the last word.

    Nasif S. Nahle

  57. Neutrino April 12, 2011 at 2:43 pm #

    Nasif,

    I am almost at a lose for words because of your last post. Are you seriously suggesting that the solar flux at earth is not 1368W/m^2?

    My rough calculation above is in no way controversial, in fact it is the same value that working scientists observe. Don’t take my word for it, go look at what NASA says. The SORCE(Solar Radiation & Climate Experiment) has a satellite up there to measure just that number. You can find their homepage here. But to make it even easier, here is a data file from them, check column 10 and tell me that the calculation that I have done here is not in sync with their numbers.

    With regards to the inverse square law and how I calculated the power the earth receives. What I did was divide the cross section of the earth, the area that it presents to the sun, by the total area a sphere represented by the earth radius from the sun. This is the percentage of the solar power that the earth intercepts. It is equivalent to doing the inverse square law, in fact it is the inverse square law just not dressed up as pretty.

    How is it that the Earth is only receiving 1368 W/m2 during daylight?
    Because that is what the flux is at the earth, not because I calculated it but because that is what is measured. It does show that I do know what I am doing, since my simple calculation, with its approximations, landed my result right in the middle of where it should be. In actuallity I believe the best current number is a little lower at around 1366W/m^2

  58. cohenite April 12, 2011 at 4:08 pm #

    Well, I thought it was important, the reduced or even non-existent feedback property of water in the over-lapping spectrum.

    Neutrino, Nasif, on the issue of emissivity, an atmospheric issue; isn’t it a matter of proportion:

    http://noconsensus.files.wordpress.com/2011/04/image2.png

  59. Nasif Nahle April 12, 2011 at 4:12 pm #

    @Neutrino…

    You say:

    I am almost at a lose for words because of your last post. Are you seriously suggesting that the solar flux at earth is not 1368 W/m^2?

    No, what I am asking you is that you must to explain why the solar flux on the outer layer of the Earth is 1368 W/m^2.

    I’m not saying it is not 1368 W/m^2

    On the remaining of your post, I do prefer to wait for the authorization of Dr. Jennifer.

    😀

  60. Nasif Nahle April 12, 2011 at 4:47 pm #

    @Cohenite…

    Very interesting. Thanks!

    Although the initial debate was on the total emissivity and absorptivity of the carbon dioxide and the effect of overlapping, Neutrino decided to create a confusion so the issue of the overlapping absorption bands was forgotten.

    The main mistake of Neutrino is that he’s calculating the total power emitted by the Sun and that it is directed in the whole towards the Earth.

    In no place he’s calculating the hemispherical emission of power, as he assures in his last post. His calculations give a false temperature of the Earth, 5.4 °C, while the real measurements by the scientists working on the issue give 293 K, 20 °C, which is the mean temperature of the Earth:

    http://hyperphysics.phy-astr.gsu.edu/hbase/solar/soldata2.html#c3

    http://books.google.com/books?id=w6Ii4r7FuskC&pg=PA7&lpg=PA7&dq=mean+temperature+of+the+earth+physics&source=bl&ots=2X3e8sHNYL&sig=mLOCbIbt2q4jZSgpQ3_osWjZDoY&hl=en&ei=RvOjTdOwMuTE0QHSiZX9CA&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBYQ6AEwAA#v=onepage&q&f=false

    http://prisma.foe.calpoly.edu/physmag.html

    http://books.google.com/books?id=uY79k7Nx-egC&pg=PA579&lpg=PA579&dq=mean+temperature+of+the+earth+physics&source=bl&ots=-7NYYoaY41&sig=ik0ltUgBo1ufzOh9aPyKWNN42Vw&hl=en&ei=J_SjTY7ANPSO0QHrjoSACQ&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBUQ6AEwADgU#v=onepage&q=mean%20temperature%20of%20the%20earth%20physics&f=false

    The errors of Neutrino show a frozen Earth, i.e. an unreal Earth, with the purpose of introduce GHE.

    Actually, when the physics procedures are well applied, we don’t need of any greenhouse effect to explain the temperature of the Earth. Oceans, as the graph in the link you provided shows, are the key for this.

    NSN

  61. Bryan April 12, 2011 at 6:17 pm #

    What Nasif Nahle is saying seems to be in line with what practitioners in heat transfer actually do.

    His present posts rely on equations and numbers gleaned from Hottel
    Without access to the same sources I cannot comment further.

    However it seems very much in line with the work of Professor Schack quoted by Gerhard Gerlich and Ralf D. Tscheuschner in Falsification Of the atmospheric CO2 greenhouse effects within the frame Of Physics.
    Perhaps Nasif has read the work of Professor Schack, any comments?

    Neutrino seems to be surprised by these matter of fact conclusions made long before the IPCC made CO2 out to be an undesirable pollutant.

    Here’s someone saying much the same as Nasif but from a different angle.

    http://www.tech-know.eu/uploads/Understanding_the_Atmosphere_Effect.pdf

  62. gavin April 12, 2011 at 8:59 pm #

    Well done Neutrino! We are now seeing the freaky arguments at the core of anti AGW pseudo science, beginning with heat from Earth’s innards and ending with fantastic ocean forces. Oh, I forgot the so impure BB’s & GB’s.

    Folks, one dosn’t start a physics degree looking for errors in estimates under say 2% or even 5%. In recent decades + or – 2% was about the best we could expect from a whole range of routine measurements.

    In fact this pedantic attitude betrays all those in the huddle and sheltering in blogsphere where almost nobody we meet has the capacity to derive physics from first principles.

    Well done Neutrino! Are you in any way involved with teaching senior math or physics?

  63. cementafriend April 12, 2011 at 9:30 pm #

    Bryan,
    The points made in the conclusion of the article by Postma are sound “We see that in every single instance of comparison, the Theory of the Greenhouse Effect appears to contradict what the Laws of Thermodynamics have to say about the exact same physical situation. This is very curious because as a scientific theory, it should be in agreement with the pre-established laws of physics. It may be possible that the Greenhouse Theory is correct, but, this would require that the Laws of Thermodynamics be not correct. However, if science does not understand the Laws of Thermodynamics, then it must be by pure coincidence that engineers have created such things as refrigerators, internal combustion engines, nuclear power plants, solar panels, and steam engines, just to name a few examples. This doesn‟t seem likely.”
    and “The conclusion of this article is very simple: there is no such thing as a radiative Theory of the Greenhouse Effect, not in real greenhouses, and certainly not in any planetary atmosphere known to man. The true role of the atmosphere, on Earth, is that it cools the ground, not warms it.”
    However, Postma gives far to much credit to radiative transfer and appears to neglect heat transfer by convection and Phase change (evaporation and condensation) at the earth surface.
    I will have to read the article again but radiation to space occurs mainly at the top of the atmosphere.

  64. gavin April 12, 2011 at 9:44 pm #

    Bryan; I fear the Postma doc rises only on blogsphere linked to sceptics

  65. spangled drongo April 12, 2011 at 10:24 pm #

    “Bryan; I fear the Postma doc rises only on blogsphere linked to sceptics”

    If you can work that out, you should be able to work out why.

    Could it be because they are the ones who:

    “have created such things as refrigerators, internal combustion engines, nuclear power plants, solar panels, and steam engines, just to name a few examples.”???

  66. Luke April 12, 2011 at 10:32 pm #

    Well I guess one could take some – what we call “measurements” – and compare that to modelled…… hmmmm

  67. Neutrino April 12, 2011 at 10:53 pm #

    Wow,

    Nasif says:
    No, what I am asking you is that you must to explain why the solar flux on the outer layer of the Earth is 1368 W/m^2.
    I’m not saying it is not 1368 W/m^2

    I did explain why that is the flux, it is the calculation of 5) through 11). My number is in total power to get flux divide by the cross section.
    1.74*10^17W / (π * (6.371*10^6m)^2) = 1.37W/m^2

    Nasif says:
    Although the initial debate was on the total emissivity and absorptivity of the carbon dioxide and the effect of overlapping, Neutrino decided to create a confusion so the issue of the overlapping absorption bands was forgotten.

    Yes the initial debate in the other thread was about overlapping emission bands, that is why this thread was opened. To discuss radiative basics. The other debate is not forgotten, as I am still waiting on a response there.

    Nasif says:
    In no place he’s calculating the hemispherical emission of power, as he assures in his last post. His calculations give a false temperature of the Earth, 5.4 °C,

    Look at the numbers, I claim with the above math that the sun emits 3.84*10^27W and the earth intercepts 1.74*10^17W. Yes the temperature is not a ‘real’ temperature of the earth, as I clearly explained in the post, did you read and comprehend what I wrote?

    Nasif says:
    The errors of Neutrino show a frozen Earth, i.e. an unreal Earth, with the purpose of introduce GHE.
    Actually, when the physics procedures are well applied, we don’t need of any greenhouse effect to explain the temperature of the Earth.

    Yes, as clearly stated in original post, the 5.4C is not a realistic temperature for the earth. It is an estimate with assumptions, one of them that the earth does not have an atmosphere. That assumption clearly changes the result. The flux at the earth of 1.37W/m^2 is a measured entity, as such that flux cannot account for a surface temperature of 15C(average). When the physics are explained a Greenhouse Effect is required.

    But again the point of this article was to simplify the math so we can discuss one specific thing, radiative transfer.

    If you believe that the calculations I have done above are truly incorrect then redo them yourself and post so we can compare. Use the starting T for the sun and preserve the ε’s I used(1.0 for sun and 0.7 for earth) so that we can compare apples to apples.(even if you do not entirely agree with those use them so we can compare the math not the initial assumptions)

  68. Neutrino April 12, 2011 at 10:57 pm #

    Correction:

    In the above post I wrote the flux at 1.37W/m^2 twice, the exponent was left off, it should have been 1.37*10^3W/m^2.

  69. Neutrino April 12, 2011 at 11:13 pm #

    Second correction:
    My number for flux at earth is 1365.9W/m^2, when I was using the 1368 number above that came from Alan here.

  70. Nasif Nahle April 12, 2011 at 11:46 pm #

    @Neutrino…

    Yes the temperature is not a ‘real’ temperature of the earth, as I clearly explained in the post, did you read and comprehend what I wrote?

    Then do it correctly. My calculations give 290 K because I apply the correct algorithms, constants and variables, without the need of torturing maths.

    Do it correctly and you’ll have correct results.

    NSN

  71. Neutrino April 13, 2011 at 12:19 am #

    Show the calculation, that was the point of this thread.

    Redo my above calculation your way so we can compare. But to be an apples to apples comparison keep the parameters the same:
    Solar power 3.8415*10^26W(Which is the same as saying the sun is at 5778K and emissivity 1)
    Earth emissivity of 0.7
    The reason in keeping the parameters the same is to see clearly the difference in the math.

    Since we are not directly talking about the ‘real’ earth(it doesn’t have an atmosphere!) don’t get hung up on the parameters. It is the math that is the discussion.

  72. Nasif Nahle April 13, 2011 at 1:02 am #

    @Neutrino…

    Here the things start incorrect. The Sun has not an emissivity of 1.0, but 0.9875, as I have been saying along this thread. On the other hand, the Earth’s total emissivity is not 0.7, but 0.82; both emissivities has been obtained from measurements.

    The surface area of emission concerning the Earth is the area of the solar sphere which exclusively is facing the Earth, consequently, it is not one half of the surface area of the sphere, but 1/4 of the surface area. The same applies to the solar surface area of the Sun.

    Given that the energy of the Sun is not concentrated by a gigantic magnifier exclusively on the Earth, it is not the whole energy emitted from 1/4 of the surface area of the Sun, but only a portion of those emissions; otherwise, the Earth would be toasted in one second.

    When we are calculating the energy incomming from the Sun, we must to consider the attenuation of the solar energy by the distance from the inner sphere to the outer sphere. Here was your first mistake because you applied it until the last calculations instead of taking it into account in the calculation of the Intensity of Solar Radiation (ISR).

    There is no one book that introduces the energy emitted by the whole surface area of the Sun. The correct procedure is calculating the solar energy emitted by the Sun taking into consideration the attenuation of the ISR. If you don’t do this, the results will be flawed.

    The correct formula to calculate the intensity of the solar radiation is:

    I = [(2hv^3/c^2) (1/e^(hv/kT) -1)] * ε sun

    🙂

  73. Nasif Nahle April 13, 2011 at 1:16 am #

    @Neutrino…

    And here is the main mistake:

    Since we are not directly talking about the ‘real’ earth(it doesn’t have an atmosphere!) don’t get hung up on the parameters. It is the math that is the discussion.

    You cannot transfer the results obtained for a hypothetical Earth to a real Earth. If you continue these calculations you will get to the point of deciding to introduce an unreal atmosphere that recycles energy, which is impossible given the Laws of Thermodynamics and other physical processes that I have been explained in my previous essays:

    http://jennifermarohasy.com/blog/2011/03/recycling-of-heat-in-the-atmosphere-is-impossible/

    In brief, you would be obliged to include unreal atmospheres, unreal processes, unreal variables, unreal constants, unreal worlds, which is what AGW proponents have been doing since the beginning of AGW idea.

    Consider the oceans, the subsurface materials, convection, conduction, clouds, etc., and you will obtain an unbiased result that doesn’t need of creating energy from the nothingness by gases with unphysical characteristics.

    It’s better, by far, to adhere ourselves to real data, experimentation, observations, etc. That is what I always do.

    NSN

  74. mkelly April 13, 2011 at 4:01 am #

    The talk of average for W/m^2 at the surface is well bogus. The artic in winter is 0 and besides part way up Everest with a W/m^2 at that latitude in summer it is still cold. W/m^2 mean very little unless associated with other factors. Heck if all you had to have is W/m^2 then the foot of the broadcast tower for a 50000 W radion station would be hot. But it is not.

    My radar range is less than a quarter of a meter square and is rated at 1000 W so it could put out near 4000 W/m^2 but it does not heat the air or much of the materials place in it only the stuff with H2O so the molecules can get excited.

    Gases dissipate heat and a there is no commerical application that uses the GHE for profit. Yet a number that use the dissipative property for profit. ie. hair dryers, fans, room heaters, etc,. If the effect of GHG is so strong it can heat a world why is there no profitable application for it?

  75. gavin April 13, 2011 at 6:55 am #

    Nasif “I apply the correct algorithms, constants and variables, without the need of torturing maths”

    Neutrino “It is the math that is the discussion”

    Nasif “Here the things start incorrect…There is no one book that introduces the energy emitted by the whole surface area of the Sun”

    Folks; I wonder if Nasif can actually do the torturing math as required by Neutrino in this thread

  76. Nasif Nahle April 13, 2011 at 6:57 am #

    @Mkelly…

    Yes, you’re right, although I wouldn’t call it “bogus”. If you have noticed, I always resort to our small meteorological station to make my calculations. Thus, my results are based on real data; local, if you wish, but real because I introduce instantaneous temperature, pressure, measurements of our local levels of CO2, etc.

    For the arctic T of 0 °C or less (suppose it is 0 °C or 273.15 K), I consider the angle of the solar quantum/waves stream striking the surface, wind velocity, humidity, etc. It’s not so easy as to apply Stefan-Boltzmann equation alone. We must to consider most of the variables and we will be always losing some variables that we cannot measure at present.

    To say that the planet Earth absorbs the solar energy as if it would be incoming from all directions with the same intensity is not correct, and it is not correct either to think that the whole load of solar energy emitted in all possible directions, as a sphere, will be intercepted by the Earth. It is scientifically erroneous. However, even NASA/NOAA fails on this issue. That is why they apply the inverse square law, which is questionable also from a strict scientific methodology. Why? Because the Earth is not a plain disk that, as you have suggested, receives the identical amount of radiant energy at every square meter of its surface.

    NSN

  77. gavin April 13, 2011 at 7:03 am #

    Jennifer; Can you consider the Postma doc for a separate thread as I reckon it too depends on a series of calcs based on initial assumptions. It could be quite interesting to see what visiting physics experts do with this one given enough time.

  78. gavin April 13, 2011 at 7:07 am #

    Nasif; please have another go at “Because the Earth is not a plain disk that, as you have suggested, receives the identical amount of radiant energy at every square meter of its surface”.

  79. Mack April 13, 2011 at 8:22 am #

    As a layman looking overall at the AGW picture, I’ve come to realise how clever the elements of scientific fraud are inbuilt into this “science”
    The biggest thing for jo public to get its head around is that the atmosphere dosn’t act as a blanket if there was no water in it despite the compelling illusion that cloud cover at night keeps heat in .
    A second huge illusion is despite the fact that in the general public’s eyes, water always cools things down ;,in combination with CO2 it dosn’t .cool the CO2 down. That right Nasif? ( I fell for that further back)
    And of course thirdly we have our illusionist Luke with his tubes of CO2 cutting out flames fooling us old guys with cameras with adjustable frequencies.
    Yes Luke science really is “magic” perpetuated by circus people like yourself.
    Old Al Gore the snakeoil salesman was onto a good thing here and will probably go down in history as the worlds greatest conman.
    Btw Luke there’s a booby prize for you at the show…The Internet Award for A Blogs Most Persistant and Irritating Troll. 🙂
    carry on everybody.

  80. cohenite April 13, 2011 at 8:26 am #

    The Postma document is fine as far as it goes but it still suffers the same omission that the previous standard of description of the greenhouse effect, Arthur Smith’s:

    http://arxiv.org/PS_cache/arxiv/pdf/0802/0802.4324v1.pdf

    Such descriptions do not describe the ocean which has the vast majority of energy storage and movement on Earth.

  81. Neutrino April 13, 2011 at 9:08 am #

    Nasif,

    Here the things start incorrect. The Sun has not an emissivity of 1.0, but 0.9875, as I have been saying along this thread. On the other hand, the Earth’s total emissivity is not 0.7, but 0.82; both emissivities has been obtained from measurements

    As I have said many times they an approximation, get over it. The measured Bond Albedo of the earth is 0.306. Both 0.7 for earth and 1.0 are excellent approximations to both your value of solar and the NASA number for earth(Heck to 1 decimal point they are exactly equal).

    The surface area of emission concerning the Earth is the area of the solar sphere which exclusively is facing the Earth, consequently, it is not one half of the surface area of the sphere, but 1/4 of the surface area. The same applies to the solar surface area of the Sun.
    Given that the energy of the Sun is not concentrated by a gigantic magnifier exclusively on the Earth, it is not the whole energy emitted from 1/4 of the surface area of the Sun, but only a portion of those emissions; otherwise, the Earth would be toasted in one second.
    When we are calculating the energy incomming from the Sun, we must to consider the attenuation of the solar energy by the distance from the inner sphere to the outer sphere. Here was your first mistake because you applied it until the last calculations instead of taking it into account in the calculation of the Intensity of Solar Radiation (ISR).
    There is no one book that introduces the energy emitted by the whole surface area of the Sun. The correct procedure is calculating the solar energy emitted by the Sun taking into consideration the attenuation of the ISR. If you don’t do this, the results will be flawed.

    Show me where I did anything close to resembling of the above. It is a misrepresentation of the above calculation.
    What I did do was calculate the total power of the sun(p = 3.84*10^26W) and then calculate how much of that total power the earth intercepts(Q = 1.74*10^17W). To do that I used the ratio of the cross section of the earths disc(π * (6.371*10^6m)^2 = 1.28*10^14m^2) and the sphere that the sun is radiating into at the distance of 1AU(4 * π * (1.50*10^11m)^2 = 2.81*10^23m^2). That ratio is 4.53*10-10 and represents the fraction of solar irradiance that the earth intercepts. It is equivalent to using the inverse square rule, as demonstrated by the fact when I calculated the flux at earth(Q / cross section) it agrees with observation to two decimal points.

    The correct formula to calculate the intensity of the solar radiation is:
    I = [(2hv^3/c^2) (1/e^(hv/kT) -1)] * ε sun

    And that is Planck’s Law which the Stefan-Boltzmann Law I used is the integration of over all wavelengths(I noted this in the opening post).
    The above formula you cite is for one single wavelength emitted by the sun, so in actuality it is not the total solar radiation but just one wavelength of it, the SB is the summation of all wavelengths.

    Do you have any actual critiques of what I did, or just critiques of misrepresentations of the above calculations?

  82. Nasif Nahle April 13, 2011 at 9:24 am #

    @Neutrino…

    Is it an actual critique to your erroneous methodology? Perhaps, if you take it that way, because it is the astrophysics equation by which we calculate the Radiative Intensity of our star and the stars of the whole universe. Real stars, not imaginary stars.

    And it is not only that, there are much algorithms that you must to learn. Unless you think that astrophysics calculations are misrepresntations.

    You calculated the emissions of the whole surface area of the solar sphere here:

    (4 * π * 6.955*10^8m^2)

    From here:

    Body A emits:

    5) p = εσT^4.A

    6) p = 1 * 5.67*10^-8W/m^2K^4 * 5778K^4 * (4 * π * 6.955*10^8m^2)

    And after that, here:

    (4 * π * 1.50*10^11m^2)

    From here:

    Power Body B intercepts:

    8) Q = (πR(b)^2 / 4πD(ab)^2) * p

    9) Q = (π * 6.371*10^6m^2/ 4 * π * 1.50*10^11m^2) * 3.84*10^26W

    Those are geometric formulas to obtain the surface area of spheres, not of disks.

    Consequently, all the remainder calculations are wrong because you’re assuming that the whole energy emitted by the solar sphere is intercepted at once by the whole surface area of the Earth.

    To do the things worst, you multiplied the result by 0.7, assuming (again) that the total absorptivity of the Earth is 0.7. It is not 0.7, but 0.82.

    You adopted 0.7 because it was convenient to fit your idea. It’s not a hair of the reality.

    😀

  83. Mack April 13, 2011 at 9:45 am #

    That second illusion….don’t worry Nasif …….the CO2 was not hot in the first place 🙂 ?

  84. Neutrino April 13, 2011 at 10:26 am #

    You calculated the emissions of the whole surface area of the solar sphere here:
    (4 * π * 6.955*10^8m^2)
    From here:
    Body A emits:
    5) p = εσT^4.A
    6) p = 1 * 5.67*10^-8W/m^2K^4 * 5778K^4 * (4 * π * 6.955*10^8m^2)

    Do you think the Stefan-Boltzmann Law is not the power radiated by a body?(Please answer this one directly as it would be central to our disagreement if your answer is No)
    Since the above is the Stefan-Boltzmann your answer would seem to be No.
    (as pointed out before there are missing brackets on the bolded part, it should read: 4 * π * 6.955*10^8m)^2)

    And after that, here:
    (4 * π * 1.50*10^11m^2)
    From here:
    Power Body B intercepts:
    8) Q = (πR(b)^2 / 4πD(ab)^2) * p
    9) Q = ((π * 6.371*10^6m^2/ 4 * π * 1.50*10^11m^2)) * 3.84*10^26W
    Those are geometric formulas to obtain the surface area of spheres, not of disks.

    Partly wrong, the one in the numerator is the surface area of a circle, and the one in the denominator is a sphere. πr^2 is area of a circle(cross section) and 4πr^2 is area of a sphere. Geometry 101.
    (as pointed out before there are missing brackets on the bolded part, it should read: (π * (6.371*10^6m)^2)/ (4 * π * (1.50*10^11m)^2))

    Consequently, all the remainder calculations are wrong because you’re assuming that the whole energy emitted by the solar sphere is intercepted at once by the whole surface area of the Earth.

    No, I am not. Look at the last post for the description of what I did, besides I have solar power as p = 3.84*10^26W and power that is incident at earth as Q = 1.74*10^17W. Does p = Q?

    Also, if the calculations are wrong then why do they agree with measurements of the flux at the earth?

    To do the things worst, you multiplied the result by 0.7, assuming (again) that the total absorptivity of the Earth is 0.7. It is not 0.7, but 0.82.
    You adopted 0.7 because it was convenient to fit your idea. It’s not a hair of the reality.

    I adopted 0.7 because it complies with the Bond Albedo of the earth reported by NASA. For those that are unaware of that term it is the total amount of solar radiation that is reflected from the earth. It is a measured quantity and is reported as 0.301 by NASA. (1-0.306 = 0.694)

  85. Neutrino April 13, 2011 at 10:28 am #

    Link to the definition of Bond Albedo above didnt work.

  86. Neutrino April 13, 2011 at 10:30 am #

    Last line of post should read:

    It is a measured quantity and is reported as 0.306 by NASA. (1-0.306 = 0.694)

  87. cohenite April 13, 2011 at 10:39 am #

    Neutrino, since you are doing a thought experiment at this time [I could be wrong] wouldn’t the geometric albedo be more appropriate than the bond albedo?

  88. Neutrino April 13, 2011 at 10:55 am #

    cohenite,

    No, I do not think so.
    The Bond Albedo is the measure of all incident that is reflected, and that is exactly what is important. If it is reflected then it cannot add to the energy budget of the earth.

    It is a thought experiment of sorts, all the inputs are real(approximates of course) but with no atmosphere.

  89. Nasif Nahle April 13, 2011 at 11:03 am #

    @Neutrino…

    Please, don’t put words in my mouth that I have never said. You ask:

    Do you think the Stefan-Boltzmann Law is not the power radiated by a body?(Please answer this one directly as it would be central to our disagreement if your answer is No)
    Since the above is the Stefan-Boltzmann your answer would seem to be No.
    (as pointed out before there are missing brackets on the bolded part, it should read: 4 * π * 6.955*10^8m)^2)

    However, the problem is that you are calculating the whole load of energy emitted by the whole surface area of the Sun and are directing it towards the Earth. That’s your mistake.

    In addition of this mistake of you, you are calculating the whole load of energy emitted by the whole solar surface area intercepted by the whole surface area of the outer sphere, not the load of energy intercepted by the Earth. Result of this: Flawed numbers.

    Quite simple.

    😀

  90. Luke April 13, 2011 at 11:04 am #

    “Also, if the calculations are wrong then why do they agree with measurements of the flux at the earth?”

    Poor Nasif.

  91. Debbie April 13, 2011 at 11:27 am #

    You know what Luke and Gavin?
    I think perhaps that Neutrino would prefer if you didn’t try to help by being rude and sarcastic.
    Otherwise he would be engaging with you.
    He’s happy enough to engage with cohenite who is at least contributing to the actual debate.
    He’s likely to have more credibility if you stay out of it.
    Sorry if that sounded rude, but most of us are actually interested in what they both have to say.
    Poor Nasif???? WTF does that even mean?
    I hope you are not being sarcastic and supercilious because English is not his first language?
    How good is your Spanish?

  92. Neutrino April 13, 2011 at 11:44 am #

    Nasif,

    I have not put words in your mouth, that is why I asked directly what you believed.

    A direct question you sidestepped once again.

    Your statement:
    The correct formula to calculate the intensity of the solar radiation is:
    I = [(2hv^3/c^2) (1/e^(hv/kT) -1)] * ε sun</i”
    (which is incorrect, as pointed out before that is the Planck function which is the intensity for each wavelength)

    Coupled with this:
    Perhaps, if you take it that way, because it is the astrophysics equation by which we calculate the Radiative Intensity of our star and the stars of the whole universe. Real stars, not imaginary stars.

    Seem to indicate you believe my use of p = εσT^4.A is not correct. So I asked you a direct question. Since you didn’t answer I will ask again:
    Do you think the Stefan-Boltzmann equation is or is not the total power radiated by a surface?

    Nasif, no matter how many times you repeat this statement does not make it true.
    However, the problem is that you are calculating the whole load of energy emitted by the whole surface area of the Sun and are directing it towards the Earth. That’s your mistake.

    My numbers(once again, they haven’t changed) are:

    Total power radiated by the sun(in all directions)
    p = 3.84*10^26W
    Total power that is incident on the earth
    Q = 1.74*10^17W

    If what you are claiming I am doing is true then those two numbers would have to be equal. Obviously they are not. I am in fact saying that the earth only intercepts only a meager 0.0000000453% of what the sun emits.

    Again, if my analysis is so flawed then how can my calculated flux at earth, 1366W/m^2, be so close to what NASA measures, 1368W/m^2?

  93. Neutrino April 13, 2011 at 11:46 am #

    If i see a direct question related to the topic I will try and answer as best I can. If I hve missed any my apoigies.

  94. Nasif Nahle April 13, 2011 at 12:16 pm #

    List of Wrong arguments of Neutrino:

    1. You admit that black bodies do not exist in nature; nevertheless, contrary to your assertion, you treat the Sun as a perfect blackbody.

    2. You admit that assuming that the Sun is a blackbody is wrong; however, you continue introducing the erroneous concept in your calculations.

    3. You calculate the energy emitted by the whole surface area of the Sun as if it was directed to the Earth. Nevertheless, what you are calculating is the energy emitted by the whole surface area of the Sun towards the whole surface area of the outer sphere. The Earth is not ubiquitous.

    4. You continue calculating the energy absorbed by the whole surface area of the sphere of Earth as if it was exposed to a sphere of solar energy; there is no night and day in your Earth, but only daylight.

    5. You admit that the final result of the temperature of the Earth, 5.4 °C is erroneous, flawed in my words; however, you resort to it to introduce a nonexistent greenhouse effect that warms up the Earth’s surface to its actual mean temperature, 290 K. In science, we call it a sophism.

    6. You suggest that the actual mean temperature of the Earth’s surface is achieved by means of the greenhouse effect, i.e. that the atmosphere is more efficient than the Sun to warm up the Earth:

    Physics calculation on the surface temperature: 290 K – 255 K = 35 K

    Neutrino’s calculation (by the Sun alone) on the surface temperature: 278.4 K – 255 K = 23.4 K

    As from Physics calculation, the Sun alone rises the temperature of the Earth up to its mean temperature, i.e. 290 K

    As from Neutrino’s calculation, the Sun alone is not capable to raise the temperature of the Earth up to its mean temperature and he has to resort to a magical atmosphere, most powerful than the Sun, to heat the surface up.

    Conversely, well done calculations gives a result that coincides with the mean temperature of the Earth’s surface, without resorting to any greenhouse effect.

    Conclusion: Neutrino’s calculations are flawed.

    Nasif S. Nahle

  95. kuhnkat April 13, 2011 at 12:51 pm #

    Luke.

    “Well I guess one could take some – what we call “measurements” – and compare that to modelled…… hmmmm”

    find that hot spot yet??

    Oh yeah, y’all don’t believe the measurements are good enough!!

  96. cohenite April 13, 2011 at 12:59 pm #

    In respect of the reduced emissivities of H2O and CO2 in the overlapping spectrums; they have been quantified here and compared with IPCC figures:

    http://wattsupwiththat.com/2011/03/02/new-paper-claims-a-value-one-seventh-of-the-ipcc-best-estimate-for-climate-sensitivity-for-a-co2-doubling/

  97. Neutrino April 13, 2011 at 1:22 pm #

    Corrections to “List of Wrong arguments”:

    1. Yes BB’s do not exist. But the sun is close to a BB, so for ease of calculations I used that approximation. As well, do you not understand how the solar temperature is measured? Solar flux is measured, total solar irradiance is calculated, and the solar temperature is calculated. The temperature that is calculated is called T(effective), and it is calculated with the approximation of a BB. If you do not like that procedure go talk to NASA and other groups that measure, calculate, and publish the numbers.

    2. I continue to use it because it is both useful and accurate. Using that approximation gives the correct solar power.(see above)

    3. See last two responses to you. In case you missed the calculated numbers, solar power = 3.84*10^26W, incident on earth = 1.74*10^17W. So, No I do not do that.

    4. Yes there is no night and day, but that does not change how much power is absorbed and only effects the distribution of the energy across the sphere. Since one of the caveats was that the bodies are isothermal this is not a part of the calculation. Again it is an approximation to ease calculation so as to demonstrate the basics. Regardless, to remain in equilibrium the absorbed power has to equal the emitted power, this equality leads to the same average temperature since rotation doesn’t affect the total absorbed.

    5. I admit that 5.4C is a bad approximation of the earth with an atmosphere. In fact it is a bad approximation of the earth with no atmosphere, which is why I mentioned it in the lead post. You are neglecting the further refinement that was introduced at the end of the post. Accounting for the difference in emissivity across solar spectrum(where absorption occurs) and terrestrial spectrum(where emission occurs) brings the temperature down to -18.3C. This is a much better estimate of the earth with no atmosphere, which is what I was calculating in the post(as stated clearly).

    6. The point of the calculation was twofold. First to demonstrate how radiative transfers work and second to show that given a purely radiative exchange the temperature of the earth would be significantly below what we experience.
    The atmosphere is not more efficient than the sun, the sun raises the earth temperature form near absolute zero all the way to 254.8K, the greenhouse effect only accounts for another last 33K.

    If you say that you can come up with a calculation similar to the above that gets the earth at an average of 15C then please post it. This thread was opened to discuss how is the proper way to calculate radiative transfers.

    The atmosphere is not magical, it is just very hard to model(whereas solid surfaces are much easier). I left it out because of that, since this was a post about basic physics. Leaving it out also it demonstrated that when not included the calculated temperature is to low(by 33C) so the obvious conclusion follows:
    If the atmosphere is included in the calculation it raises the temperature to the observed value.(ie: the greenhouse effect)

    Once again, which step is flawed and why? All the criticism you have raised so far are not actually reflective of what I calculated. As well still waiting on whether or not you believe the Stefan-Boltzmann correctly calculates power radiated from a surface.

  98. J.Hansford April 13, 2011 at 3:20 pm #

    …….Am I getting this right?

    Basically what we are dealing with here are not whole spheres… not flat disks either… but curved surfaces of partial, but dissimilar sized, parts of spheres. One curved surface radiating energy to another, smaller, curved surface… for all intents and purposes.

    Then there is the maths for the attenuation of the energy radiated out from that partial sphere.

    So without maths….
    sum of suns energy emitted from this curved surface.
    sum of the attenuation over distance of that energy until it strikes the smaller curved surface.
    sum of the absorption of that energy by this smaller curved surface.
    sum of the emitted energy from that smaller curved surface.

    Surely it’s not anymore complicated than that.

    Surely there are empirical measurements of Solar radiance… or irradiance.

    Surely there are empirical measurements of surface absorption and emissivity, or experiments that can obtain them.

    How do these empirical measurements fit the AGW Hypothesis

    …. Or haven’t we got that far yet with the settled science?

  99. Mack April 13, 2011 at 3:21 pm #

    Neutrino,
    ” If the atmosphere is included in the calculation it raises the temperature to the observed value. (ie: the greenhouse effect)”
    Luke normally accuses me of being simplistic but I find that comment of yours remarkable.
    I feel you have a long hard road ahead of you .

  100. Nasif Nahle April 13, 2011 at 4:10 pm #

    @Neutrino…

    You say:

    and it is calculated with the approximation of a BB. If you do not like that procedure go talk to NASA and other groups that measure, calculate, and publish the numbers.

    No, not to NASA because they clearly states that the equation is for calculating the attenuation of the solar sphere towards the outer sphere.

    I continue to use it because it is both useful and accurate. Using that approximation gives the correct solar power.(see above)

    You can continue using it, but modify the purpose for which the equation was developed, not for calculating the energy interferred by the Earth.

    See last two responses to you. In case you missed the calculated numbers, solar power = 3.84*10^26W, incident on earth = 1.74*10^17W. So, No I do not do that.

    You wrote it:

    Body A emits:

    5) p = εσT^4.A

    6) p = 1 * 5.67*10^-8W/m^2K^4 * 5778K^4 * (4 * π * 6.955*10^8m^2)

    7) p = 3.84*10^26W

    Bolded are mine. Now tell me, what are you calculating through that formula (in bolds)?

    Yes there is no night and day, but that does not change how much power is absorbed and only effects the distribution of the energy across the sphere. Since one of the caveats was that the bodies are isothermal this is not a part of the calculation. Again it is an approximation to ease calculation so as to demonstrate the basics.

    Neutrino, during nighttime the hemisphere in the darkness receives zero solar energy. What you are doing is to make the whole surface area of the sphere (the Earth) absorbs a portion of the energy emitted by the whole surface area of the Sun.

    Regardless, to remain in equilibrium the absorbed power has to equal the emitted power, this equality leads to the same average temperature since rotation doesn’t affect the total absorbed.

    If you admited there is night and day on the Earth, then you cannot conclude that the Earth is in thermal equilibrium.

    You are neglecting the further refinement that was introduced at the end of the post.

    I don’t think so. If you admited in your essay that the result was flawed, then it is flawed.

    The atmosphere is not more efficient than the sun, the sun raises the earth temperature form near absolute zero all the way to 254.8K, the greenhouse effect only accounts for another last 33K.

    Show your numbers; if you don’t consider the atmosphere is a blackbody, which is not, then make the atmosphere warms up the surface as if it was a more efficient primary source of energy than the Sun.

    The atmosphere is not magical

    Check!

    If the atmosphere is included in the calculation it raises the temperature to the observed value.(ie: the greenhouse effect)

    Again, it would make the atmosphere to be a more efficient radiator than the Sun.

    Once again, which step is flawed and why? All the criticism you have raised so far are not actually reflective of what I calculated. As well still waiting on whether or not you believe the Stefan-Boltzmann correctly calculates power radiated from a surface.

    All your calculations since you considered the energy emitted by the whole surface area of the Sun to be emitting towards the Earth and dismissed the outer sphere, which is not the Earth. Earth is not ubiquitous.

    I don’t think the following error is a typo:

    Power Body B intercepts:

    8) Q = (πR(b)^2 / 4πD(ab)^2) * p

    You’re introducing power to obtain power?

    Watts = (unitless) * Watts? What’s that?

    Again and for the last time. The formula is not to obtain the energy intercepted by the Earth, but to calculate the attenuation of the solar radiance towards the outer sphere.

    You must admit that your procedures are wrong, Neutrino. NASA doesn’t use that formula to calculate the solar energy intercepted by the Earth, but to know the attenuation of the solar irradiance in its way towards the outer sphere.

    🙂

  101. cohenite April 13, 2011 at 6:51 pm #

    Nasif and Neutrino; apart from the overlapping emissivity reducing effect, which I keep referring to, I think the general gist of your exchange has been well covered in this paper:

    http://www.met.hu/idojaras/IDOJARAS_vol108_No4_01.pdf

    It has been a while since I have read this 2004 Miskolczi paper which was published under the auspices of NASA and in fact was co-authored by NASA’s chief atmospheric physicist at the time and Miskolczi’s boss, Martin G. Mlynczak.

    Rereading it only confirms in my mind how defective a theory AGW is and how brilliant M is.

  102. gavin April 13, 2011 at 7:14 pm #

    Debbie; at risk of irritating someone again I will endeavor to make my points clearer. A few people here don’t know their physics in regard to this thread and where the calculus must take you. One is Nasif and I’m not sure about cohenite but I’m delighted to see it all again cause I’ve forgotten all of it over the decades.

    Now let me add that calculus as used in these arguments is all about finding shortcuts that can take us so close to the real thing and sometimes it’s enough to know what a great tool it is. I got lazy and let our instruments do the the hard yards for all parts of the spectrum in common use.

    But your faith has to depend on something that allows you to cover the same routines over and over with out asking some trumped up scientist for approval.

  103. Luke April 13, 2011 at 7:14 pm #

    Well Debs it’s pretty basic isn’t – the complex radiation models built on Nasif’s alleged incorrect physics seem to validate out on observed data. Don’t you think this is a tad surprising. If Nasif is right about a million to 1 shot. As you say WTF !

  104. Mack April 13, 2011 at 8:12 pm #

    Neutrino,
    I hope it is not too presumptuous for me to say that to do calculations with a radiative model and then chuck in an atmosphere saying that the temperature to be reached is what we have in reality,and this is totally due to a (radiative) “greenhouse effect”, screams loudly to me you have one foot in a model and the other in reality.

  105. cohenite April 13, 2011 at 9:52 pm #

    luke; Nasif is right about the reduced emissivity of H2O in the overlapping spectrum when CO2 is added; that shreds the positive feedback basis of AGW. What I see Neutrino doing is assisting in quantifying the effect. What are you doing?

  106. gavin April 13, 2011 at 10:28 pm #

    Given that Miskolczi & Mlynczak have derived the theoretical relationships that point to Greenhouse and support Hansen et al, Mack and others need to be careful with their rhetoric aimed at Neutrino and AGWers in general.

    What cohenite needs emphasize when offering the work of others is which track he supports re the thermodynamic model v the radiative model. Did we all asume this thread was about light energy and radiative transfer or not?

    Radiation at light speed from the Sun to Earth is not tripped up on it’s path in the same way outward bound IR from Earth’s surface is. Which ever way you look at it the Greenhouse Effect must be accounted for. This is where the physics realy starts.

  107. Luke April 13, 2011 at 11:21 pm #

    Cohenite – if all this is so seriously wrong at a fundamental level, why have all these atmospheric physics types, remote sensing types, and modellers been getting answers that work out?

    Cohenite – Nasif has disturbed me. My car is now invalidated on theoretical principles so I’m not driving it anymore. When he starts putting up some observational data and models it I’ll tune in ….

  108. Nasif Nahle April 13, 2011 at 11:38 pm #

    @Luke…

    That’s good! I guess you stop driving your car for the sake of the humankind and your neighbors.

    LOL!

    @Gavin…

    Radiation at light speed from the Sun to Earth is not tripped up on it’s path in the same way outward bound IR from Earth’s surface is. Which ever way you look at it the Greenhouse Effect must be accounted for. This is where the physics realy starts.

    Poor Gavin… There are more errors in this single paragraph than in the whole essay of Neutrino. Hah!!!

  109. Louis Hissink April 14, 2011 at 1:01 am #

    Luke,

    “Cohenite – if all this is so seriously wrong at a fundamental level, why have all these atmospheric physics types, remote sensing types, and modellers been getting answers that work out?”

    Easy Luke, self referencing arguments, and blind acceptance via belief.

    As Feynman pointed out, Science is about accepting the ignorance of experts.

  110. Louis Hissink April 14, 2011 at 1:06 am #

    In addition, Luke, the reason “why have all these atmospheric physics types, remote sensing types, and modellers been getting answers that work out?” is from circular reasoning.

    It goes back to the basics of the CAGW belief based on climate sensitivity, from which the modellers work from.

    Get that wrong, and the modelling card-house collapses.

  111. Alan D McIntire April 14, 2011 at 1:46 am #

    Neutrino, I appreciated your post on the theory of the greenhouse effect.

    I realize this is a simplification, like Galileo’s assumption that bodies all fall at the same speed.
    Nahle’s attacks appear to be arguing against Galileo by pointing out paper airplanes, hot air balloons, parachutes, etc, and saying that Galileo’s theory is simplistic crap. To get anywhere, we’ve got to start simple, and throw in complicating factors gradually.

    Just to clarify the various arguments in my own mind, I typed up a summary of what I understood the theory to be at “Climate Realists” some time ago. Here’s a link to my
    post.

    http://climaterealists.com/forum/viewtopic.php?f=5&t=550

    I used actual numerical examples to give myself a feel for the effects.

    With the advantage of hindsight, I didn’t need to use harmonic means, instead I could have
    again broken the atmosphere radation band into fractions, and instead of harmonic means, used the fraction escaping to space. In that case, my example would have given the
    amounts 1 *(0.3) + 1/2*(0.6) + 1/10(0.1) = 0.61 for the fraction of radiation at earth’s surface escaping to space.

  112. Nasif Nahle April 14, 2011 at 2:10 am #

    @Alan D McIntire…

    Just to be precise. Mine are not attacks to the person, starting by the fact that I don’t know who Neutrino is. Mine are claryfications on the wrongness of Neutrino’s essay and misuse of formulas that doesn’t apply to the case.

    😀

  113. Nasif Nahle April 14, 2011 at 2:15 am #

    @Alan D McIntire…

    BTW, your article at CR is all assumptions; nothing real.

    XD

  114. Neutrino April 14, 2011 at 2:45 am #

    J.Hansford,

    Basically that is it.

    Consider this as an analogy for the amount of power that the earth intercepts.

    Take a basketball. It has a total Surface Area(A), this represents the area that the sun is radiating into at a distance of 1AU.

    Now draw onto the sphere, all equidistance to each other, 1million dots. This represents the total solar power(p).

    The dots per unit area is then simply p/A. This is the flux.

    Next draw a small circle on the ball, the circle has a Cross Section(CS). This represents the area the earth eclipses on the sphere.

    The number of dots then in the circle is simply (p/A) * CS. This represents the power the earth intercepts(Q).

    So the amount of power that the earth intercepts can be written (p/A) * CS as well.

    Rewriting the equation above as Q = (CS / A) * p
    Substituting the proper geometric values for the real earth and sun we get Q = (πR(b)^2 / 4πD(ab)^2) * p
    Which is my equation 8)

    Mack,
    Yes this is overly simplistic, that was the point.
    The radiative exchange alone, between the sun and earth surface, cannot account for the temperature of the earth. If the atmosphere is included then yes the computed temperature is in agreement with measurement. Without that the calculated temperature(which is what I have done here) does not comply with actual measurements.
    How the atmosphere increases the temperature at the surface is generally referred to as the Green House Effect.

  115. Neutrino April 14, 2011 at 3:21 am #

    Nasif,

    I am truly at a loss to understand you point.

    Is the Stefan-Boltzmann used to calculate total power of a surface?
    I used p = εσT^4.A to calculate the total power the sun radiates(in all directions), do you agree or disagree with this statement?

    What you bolded (4 * π * (6.955*10^8m)^2) is simply the total surface area of the sun.
    (as pointed out before the bolded brackets I added are missing in the original)

    Are you saying that p = εσT^4.A is not the correct formula to use to calculate total power emitted?(A in the formula is Area)

    Yes during night the dark side receives zero solar power. But that fact doesn’t change how much solar power the earth receives over a 24h period. There is always on side facing the sun, and that side is absorbing q (from my calculations). For sake of simplicity one of the caveats was that the body was isothermal.

    The ‘flaw’ was that the emissivity of the earth was pegged at 0.7 across all wavelengths. This is a bad approximation but was used to do the calculations for clarity. Redoing the final calculation of the earth’s temperature with a better estimate of the emissivity over the range of terrestrial wavelengths yields a better estimate.
    Yes, using 0.7 for both emission and absorption is flawed, that is what I said. Not that the calculations were themselves flawed but one of the assumptions was. Correcting that assumption leads to a better result.

    This post is not about atmosphere, that’s why the caveats included that it is for solid bodies in isolation(ie no atmosphere)

    For what is hopefully the final time:
    I do not calculate the entire solar power and direct it all at the earth.
    I calculate the fraction of that total power which is directed at the earth.

    8) Q = (πR(b)^2 / 4πD(ab)^2) * p
    No, it is certainly not a typo.

    That formula is a Q = (CS/A) * p (CS, cross section of earth. A, area of sphere with radius of one AU)

    The flux at one AU is p/A, the attenuation as you call it, that multiplied by the cross section of the earth gets you the incident on the earth. Its that simple, its geometry. Nothing strange or untoward there.

    Besides as stated before:
    If you think the above is not how NASA or anyone else would calculate the flux at the earth why does my number 1366W/m^2 come so close to the measured value of 1368W/m^2?

  116. Nasif Nahle April 14, 2011 at 4:21 am #

    @Neutrino…

    The problem is that you’re not calculating the specific radiation emitted by a curved surface facing to the Earth, but the whole amount the Sun emits in one second, not towards the Earth, but towards the whole outer sphere>/b>.

    It’s not true that your calculations coincide with the measured value of 1368 W/m2.

    By following your calculations at:

    Body A emits:

    5) p = εσT^4.A

    6) p = 1 * 5.67*10^-8W/m^2K^4 * 5778K^4 * (4 * π * 6.955*10^8m^2)

    7) p = 3.84*10^26W

    And the formula to obtain the solar constant:

    So = E(Sun) x (R(Sun) / r)2

    we would obtain a solar constant of:

    Solar Constant (So) = 3.84*10^26 W * 0.0000215296 m^2 = 8.2673664 x 10^21 W/m^2

    Is the result from your calculation equal to 1368 W/m^2?

    NO, IT IS NOT!

    After that, you wrote:

    Power Body B intercepts:

    8) Q = (πR(b)^2 / 4πD(ab)^2) * p

    9) Q = (π * 6.371*10^6m^2 / 4 * π * 1.50*10^11m^2) * 3.84*10^26W

    10) Q = 4.53*10^-10 * 3.84*10^26W

    11) Q = 1.74*10^17W

    And the solar constant, considering the amount of energy that you say the Earth receives from the Sun, the solar constant would be:

    Solar Constant (So) = 1.74*10^17 W * 0.0000215296 m^2 = 3.7461504 x 10^12 W/m^2

    Is this result the solar constant of 1368 W/m^2?

    NO, IT’S NOT!!!

    Please, Neutrino… Correct your essay because it is absolutely flawed.

    Nasif S. Nahle

  117. Neutrino April 14, 2011 at 4:56 am #

    Nasif,

    And the formula to obtain the solar constant:
    So = E(Sun) x (R(Sun) / r)2
    we would obtain a solar constant of:
    Solar Constant (So) = 3.84*10^26 W * 0.0000215296 m^2 = 8.2673664 x 10^21 W/m^2
    Is the result from your calculation equal to 1368 W/m^2?

    A few problems:

    First, the bolded line is for flux not a power. Using my p for E(sun) is wrong as it’s a power not a flux.

    Second, the value (R(sun)/r)^2 is dimensionless. Calling it 0.0000215296m^2 is simply bad algebra. It is the dimensionless value 0.0000215296.

    Just looking at the units in your equation shows your mistake:
    3.84*10^26 W * 0.0000215296 m^2 = 8.2673664 x 10^21 W/m^2

    W*m^2 is not equal to W/m^2

    Doing the above correct with the solar flux of 6.32*10^7W/m^2 yields the value 1366W/m^2, which is the same as I have been saying all along.
    Doing it by dividing the total power the earth receives(Q) divided by the cross section(πR(b)^2) also yields 1366W/m^2.(which is how I originally calculated it)

    They are equivalent because they are doing the same calculations, just a different method.

    As a general note. If when doing any calculation the units in the formula do not add up then the formula as written is wrong. Checking the units is always a good way to do a quick check to see if you have made any mistakes.

  118. Neutrino April 14, 2011 at 5:12 am #

    Thank you Alan D McIntire,

    As a note though this post is not about the Green House Effect but rather about what the outcome would be without the Green House Effect.

    I agree simplifications can be very enlightening, slowly building in complexity is the way to go.
    But when there is confusion about the basics adding complexity just muddies the water needlessly.

  119. Nasif Nahle April 14, 2011 at 5:16 am #

    @Neutrino…

    Second, the value (R(sun)/r)^2 is dimensionless. Calling it 0.0000215296m^2 is simply bad algebra. It is the dimensionless value 0.0000215296.

    Yes, you’re right, 0.0000215296 m^2 should have been 0.0000215296.

    So the result should have been 8.2673664 x 10^21 W and 3.7461504 x 10^12 W, respectively, which makes the things worst for you.

    😀

  120. Neutrino April 14, 2011 at 5:33 am #

    That formula is for flux, if you want to talk about power then the dimensionless scaling factor should be 0.000000000453 not 0.0000215296. Found from (r/(2*d))^2. Where r is the earth’s radius and d is the distance between the sun and earth. This is basic geometry Nasif.

  121. Nasif Nahle April 14, 2011 at 5:47 am #

    And here the correct calculations that Neutrino was not able to make:

    Diameter of the Sun = 1.391 x 10^9 m

    Radius of the Sun = 6.955 x 10^8 m

    Area of the Solar disk = 1.519652 x 10^18 m^2

    Diameter of the Earth = 1.27562 x 10^7 m

    Radius of the Earth = 6.3781 x 10^6 m

    Area of the Earth’s disk = 1.278 x 10^16 m^2

    Energy Emitted by the Sun by applying correctly the Stefan-Boltzmann equation:

    Temperature of the surface of the Sun (Satellite measurement) = 5804.135 K

    The solar power emitted by the Sun in trajectory to the Earth is:

    Q/m^2= 0.9875 x (1 m^2) (5.6697 x 10^-8 W/m^2 K^4) (5804.135 K)^4 = 6.354 x 10^7 W/m^2

    What is the reason that we must introduce 1 m^2? Because to apply the Stefan-Boltzmann equation we must find the smallest surface area that resembles a flat surface. Here, Neutrino failed on the correct procedure.

    That’s the solar energy emitted by the Sun in direction to the Earth.

    And the flux of power the Earth would absorb per unit area is:

    The average flux of power inciding on the surface of the the Earth obtained by measurements is 329 W/m^2.

    Such amount of power flux causes the standard temperature of the Earth of 290 K, or 17 °C, without torturing mathematics and without resorting to any unreal “greenhouse effect”, but the power of the Sun alone.

    To calculate the attenuated amount of solar irradiance emitted towards the outer sphere, the following formula is used:

    I = E (4π x R2) sun / (4π x r2) outer sphere

    I = 6.354 x 10^7 W/m^2 (6.08 x 10^18 m^2) / (2.83 x 10^23 m^2) = 1366.34 W/m^2

    This is the solar irradiance received by the outer sphere, with planets or not.

    For calculating the incident solar irradiance (which is not constant) upon the outmost layer of the Earth, the following formula is used:

    Sk = E Sun x (R Sun / r)^2

    Sk = Solar Constant

    E = Surface Irradiance of the Sun = 6.354 x 10^7 W/m^2

    R= 6.96 x 10^8 m = Radius of the Sun

    r = 1.5 x 10^11 m = Radius of the outer sphere.

    So = Esun x (Rsun / r)^2

    So = 6. 354 x 10^7 W/m^2 * (6.96 x 10^8 m / 1.5 x 10^11 m)^2 = 1368 W/m^2

    Another methodology, simpler than the previous procedures is by applying the following formula:

    GPL = QSUN / 4π (POR)^2

    Which gives the same results.

    Nasif S. Nahle

  122. Nasif Nahle April 14, 2011 at 5:49 am #

    Here a typo:

    The solar power emitted by the Sun in trajectory to the Earth is:

    Q/m^2= 0.9875 x (1 m^2) (5.6697 x 10^-8 W/m^2 K^4) (5804.135 K)^4 = 6.354 x 10^7 W/m^2

    It should have said:

    Q/m^2= 0.9875 (5.6697 x 10^-8 W/m^2 K^4) (5804.135 K)^4 = 6.354 x 10^7 W/m^2

    Thanks!

  123. Neutrino April 14, 2011 at 7:02 am #

    Thank you Nasif for posting your calculations,

    Your calculation of solar flux at solar surface:
    Q/m^2= 0.9875 x (1 m^2) (5.6697 x 10^-8 W/m^2 K^4) (5804.135 K)^4 = 6.354 x 10^7 W/m^2

    Completely agrees with my calculation:
    7) p = 3.84*10^26W
    Dividing my p by the surface area of the sun, 4π(6.955*10^8m)^2, produces a flux of 6.32*10^7W/m^2.

    So considering the slight difference in our initial conditions these numbers agree. So I ask again why do you think my numbers are wrong if when you calculate it you get the same result?

    Your calculation for flux at earth:
    I = 6.354 x 10^7 W/m^2 (6.08 x 10^18 m^2) / (2.83 x 10^23 m^2) = 1366.34 W/m^2
    So = 6. 354 x 10^7 W/m^2 * (6.96 x 10^8 m / 1.5 x 10^11 m)^2 = 1368 W/m^2
    (I assume the difference between these two is the rounding error as they are identical calculations)

    Also completely agrees with my calculation:
    11) Q = 1.74*10^17W
    Dividing my Q by the disc of the earth, π(6.37*10^06m)^2, produces a flux of 1366W/m^2.

    Again considering the slight difference in our initial conditions these numbers agree. So I ask yet again why do you think my numbers are wrong if when you calculate it you get the same result?

    Where we part ways is this:
    What is the reason that we must introduce 1 m^2? Because to apply the Stefan-Boltzmann equation we must find the smallest surface area that resembles a flat surface. Here, Neutrino failed on the correct procedure.
    That’s the solar energy emitted by the Sun in direction to the Earth.
    And the flux of power the Earth would absorb per unit area is:
    The average flux of power inciding on the surface of the the Earth obtained by measurements is 329 W/m^2.
    Such amount of power flux causes the standard temperature of the Earth of 290 K, or 17 °C, without torturing mathematics and without resorting to any unreal “greenhouse effect”, but the power of the Sun alone.

    First, calculating for 1/m^2 is just a flux whereas my calculation was for total power. Both produce the same overall results, and neither one is better. One is a flux the other is a power, they both are correct. This is demonstrated by the fact you can convert between them by either dividing the power by total surface area or multiplying the flux by total surface area.
    There is no need for the surface to be flat, just for the area to be known.

    Second, how did you arrive at 329W/m^2, what measurement?(KT(page 4) finds that surface direct solar intensity is only 184W/m^2 of which 23W/m^2 is reflected) Since this is a thought experiment with no atmosphere how would you calculate that number?

    Third, how does 329W/m^2 result in a temperature of 290K? If we were to use the 329W/m^2 and calculate the temperature of a body emitting that much we would be at 276K via the Stefan-Boltzmann equation(using emissivity of 1).

    Could you finish the calculations off with:
    How much is absorbed by the surface.
    How much is emitted by the surface.
    What temperature of that surface.

    Since those were the interesting numbers it’s important to the discussion.

  124. Neutrino April 14, 2011 at 7:15 am #

    A note about initial numbers and using emissivity of 1 for the sun:

    Nasif and my solar statistics are slightly different(for reference all my numbers come from NASA).

    The discrepancy most likely comes from defining exactly what is the surface of the sun. Since it is a ball of gas its surface is diffuse and the exact location can be somewhat ambiguous. Same for the temperature measurements, what part of the photosphere is the correct solar temperature? We are actually ‘seeing’ down into the gas and not just the very top layer.

    As for emissivity.

    These are NASA’s numbers for the pertinent solar entities:
    Radius = 6.96*10^8m
    Total Power = 3.846*10^26W
    Temperature = 5778K

    Using those above numbers in the Stefan-Boltzmann equation we can calculate the emissivity.
    1) A = 4πr^2
    2) A = 4π(6.96*10^8m)^2
    3) A = 6.08*10^18m^2

    4) p = εσT^4.A

    5) ε = p/(σT^4.A)
    6) ε = 3.846*10^26W / (5.67*10^-8W/m^2K^4 * (5778K)^4 * 6.08*10^18m^2)
    7) ε = 1.00

    Those are NASA’s numbers, an emissivity of 1 is built into them.

  125. Neutrino April 14, 2011 at 7:22 am #

    Radius difference is most likely just a rounding issue on significant digits.

  126. Nasif Nahle April 14, 2011 at 7:31 am #

    You continue going unreal with your emissivity of 1. You cannot round up the number because you’re making the Sun a blackbody, and it is not. These links could help you to refine your calculations:

    From a site on astronomy:

    http://www.windows2universe.org/earth/climate/sun_radiation_at_earth.html:

    “If Earth were a flat, one-sided disk facing the Sun… and if it had no atmosphere… every square meter of Earth’s surface would receive 1,368 watts of energy from the Sun. Although Earth does intercept the same total amount of solar EM radiation as would a flat disk of the Earth’s radius (see figure below), that energy is spread out over a larger area. The surface of a sphere has an area four times as great as the area of a disk of the same radius. So the 1,368 W/m2 is reduced to an average of 342 W/m2 over the entire surface of our spherical planet. Another way to think of this reduction is to realize that half of Earth’s surface (the night side) is in the dark and thus receiving no solar energy at a given moment, while areas near the edges of the planet (near the poles and around dusk and dawn) are receiving reduced amounts of energy per unit area.”

    From the site http://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html:

    “Diurnal temperature range: 283 K to 293 K (10 to 20 C)”

    From the site http://hyperphysics.phy-astr.gsu.edu/hbase/solar/soldata2.html#c3:

    “Mean surface temperature 20 °C = 68 °F =293 K”

    Those are measurements, not just calculations.

    🙂

  127. Nasif Nahle April 14, 2011 at 7:38 am #

    As for your question, the flux of power absorbed by the surface, introducing real magnitudes:

    Q/m^2= 0.82 * (5.6697 x 10^-8 W/m^2 K^4) (293 K)^4 = 342.62 W/m^2

    Notice the difference? What greenhouse effect?

    🙂

  128. Neutrino April 14, 2011 at 8:06 am #

    As for your question, the flux of power absorbed by the surface, introducing real magnitudes:
    Q/m^2= 0.82 * (5.6697 x 10^-8 W/m^2 K^4) (293 K)^4 = 342.62 W/m^2
    Notice the difference? What greenhouse effect?

    Putting aside the fact that the above represents how much a surface at 293K would be emitting(not absorbing) with an emissivity of 0.82 for a second there is another potentially more serious problem.

    Namely: If the Incident flux at the earth is 1366W/m^2, that number averaged over the globe becomes 341.5W/m^2. (the area of a globe is 4x the area of the disc)

    Do you see the conflict? There are actually two in there.
    One, the surface is receiving less flux than it is emitting. 341.5W/m^2(Incident) < 342.62W/m^2(Emitted).
    You can’t have more output than input, a serious violation of Conservation of Energy.

    Two, if the incident is just 341.5W/m^2 and the emissivity is 0.82 it should only absorb 280.0W/m^2. Otherwise you are saying that absorption is not modified by absorptivity. Are you saying absorption is not modified by absorptivity?

    How do you reconcile the above two points?

  129. Hum April 14, 2011 at 8:46 am #

    Neutrino,

    I am having a real problem with your model here. This is not how AGW physicists couch the description of greenhouse warming. They state very clearly that the greenhouse effect is related to slowing the rate of cooling not increasing the temperature. Your description from you earlier post states:

    “6. The point of the calculation was twofold. First to demonstrate how radiative transfers work and second to show that given a purely radiative exchange the temperature of the earth would be significantly below what we experience.
    The atmosphere is not more efficient than the sun, the sun raises the earth temperature form near absolute zero all the way to 254.8K, the greenhouse effect only accounts for another last 33K.”

    Stated this way your model is violating the 2nd Law of Thermodynamics. The atmosphere is colder than the Earth and cannot warm it. “PERIOD”. It won’t raise it the final 33K. An atmosphere on a planet is assumed to prevent the extremes swings we see on the moon which has much higher daytime surface temps and much colder night time surface temps. Therefore our atmosphere is keeping the surface of the planet in sunlight from warming as hot as it could be during the day and keeping the surface not in sunlight from cooling as fast at night. The average temperature is what is being focused on by most scientists, but I agree with Nasif it needs to be looked on with what happens real-time while it is net absorbing daytime and net emitting nighttime. Trying to work by averages over a day really muddies what is going on when the radiative changes happen in the pico/nanosecond and the mean free path happens in a fraction of a second.

    Assuming the 2nd Law of Thermodynamics is correct(which I do) then if the greenhouse theory is correct the only thing being warmed is the atmosphere as each level gets closer to the temperature of the surface. So if we analyzed each meter of the atmosphere the first meter(closest to the surface) might be closer in temperature to the surface than it was before the greenhouse effect, the 2nd meter might be closer in temperature to the 1st meter and so on and so on. Calculations need to be done which shows how many more photons are absorbed which cause how much more collision interactions, since this is where the heat comes from in the atmosphere. Don’t forget, at the first levels since the greenhouse effect is already saturated the only additional heat that will be generated by additional CO2 will come from greater mass and air density. This along with convection and how that works with radiative effects is not factored into your model.

    The atmosphere adds complexity to any model, your model doesn’t show what the temperature would be if there were no greenhouse gases in the atmosphere. Many say it would be much hotter since CO2 and H2O emit at lower temps, and allow IR to be radiated directly to space. This then means that CO2 and H2O could have a cooling effect. All this needs to be looked at, and that is why Nasif is looking at the properties of the greenhouse gases because those properties will be part of what determines how the atmosphere as a whole radiates. But it still is a very small part of the overall movement of heat. What if increased greenhouse gases make convection and evaporation speed up to offset raditive effects?

    The problem is your toy model can’t show any of what is really going on, and saying the greenhouse gases warms the Earth is starting out admitting that your model violates the 2nd Law. You really need to to model the Earth with an atmosphere without greenhouse gases vs one with greenhouse gases vs one with a very slightly higher level of one minor .038% atmospheric gas. This model should show what changes happened radiatively, conductively and convectively. I don’t think your model will ever get there and I don’t think that comparing a blackbody without an atmosphere and making assumptions with an atmosphere is a good starting place.

    Just my 2 cents.

  130. Nasif Nahle April 14, 2011 at 8:58 am #

    Ah! Neutrino… Has you forgotten the kirchhoff’s Law? The value of the emissivity is the same than absorptivity, i.e. e = 0.82 = a = 0.82.

    Everybody knows the fraction of the surface area that is exposed to the solar emissions is 1/4. Divide the sphere in four parts and you’ll get it.

    No conflict there, Neutrino:

    Q/m^2= a * (s) (T)^4 = 0.82 (5.6697 x 10^-8 W/m^2 K^4) (293 K)^4 = 342.62 W/m^2

    The same amount.

    The problem is that you multiply and divide and the solution is not a simple rule of three.

    Two, if the incident is just 341.5W/m^2 and the emissivity is 0.82 it should only absorb 280.0W/m^2. Otherwise you are saying that absorption is not modified by absorptivity. Are you saying absorption is not modified by absorptivity?

    You just multiplied the incorrect input of 341.5 W/m^2 by 0.82. There is no violation to the first law neither to the second law. Again, the correct procedure is as follows:

    Q/m^2= e * (s) (T)^4 = 0.82 (5.6697 x 10^-8 W/m^2 K^4) (293 K)^4 = 342.62 W/m^2

    Q/m^2= a * (s) (T)^4 = 0.82 (5.6697 x 10^-8 W/m^2 K^4) (293 K)^4 = 342.62 W/m^2

    And from the link I provided:

    So the 1,368 W/m2 is reduced to an average of 342 W/m^2

    Reduced to an average… You know what it means? Okay, then 342.62 W/m^2 is not an average, although it falls inside the range from which the average was calculated.

    No conflict. Physics is correct and there is no violation to any law of thermodynamics of physics; otherwise, I would not resorted to the Stefan-Boltzmann equation.

    🙂

  131. cohenite April 14, 2011 at 10:11 am #

    Nasif and Neutrino, could you both say whether you:

    1 believe the current GAT, global average temperature, is greater then it would be if there was no atmosphere?

    2 If GAT is higher with an atmosphere do you believe it is due to the presence of H2O and CO2?

    3 If GAT is higher due to the presence of CO2 and H2O in the atmosphere is this because of their greenhouse properties; that is their capacity to absorb IR?

    Sorry to be basic but I think we are just bobbing up and down between Arthur Smith’s piece and Gerlich and Tscheuschner; I think the Miskolczi piece, his 2004 paper, is a middle ground here, but I’ll wait until you guys declare your positions.

  132. Luke April 14, 2011 at 10:14 am #

    No Louis – its’ a million to one shot. It’s not a circular argument at all unless you’re some crank who believes in electric universe theories

    You never see Nasif reference any real field obs. Telling.

    Notice how these arguments never end – we’re now back on the 2nd law ruse nonsense and saturated la la la – ta Hummer.

    When Nasif publishes his theory somewhere serious with some actual data – it’s utter rubbish. Why waste time…

  133. Luke April 14, 2011 at 10:15 am #

    Why even waste time doing solutions for the whole earth Cohers – GTA – schmat – it’s not how GCMs solve radiation balance. Not even over the target.

  134. Hum April 14, 2011 at 10:44 am #

    Luke, was I wrong? Do you really believe that IR coming from a colder body(the atmosphere is increasing the temperature of a warmer body(the Earth) by 33K? If that is true I guess the name “Warmer” for you is very appropriate.

    I thought all along that the analogy warmers use is a blanket. So is this blanket going to raise the Earths temperature, or just slow down the heat loss? Just want to make sure I am following what you are trying to model.

  135. Debbie April 14, 2011 at 11:08 am #

    Thankyou Hum,
    Your 2 cents worth is helping me to further understand what is going on here. The maths is a bit beyond me but I understand that the basic argument is what the very important variable (ie the atmosphere and the huge amount of H20) is actually doing as far as heating and cooling is concerned.
    It seems another important variable is the size and shape of the (non black) bodies Sun and Earth???
    Added to that, if the ‘settled science’ is right, then the addition of C02 is having catastrophic effects on the temperature of our planet.
    So the actual part that is being played by extra CO2 (and indeed even where the bulk of the extra comes from) definitely needs to be scrutinised.
    If Nasif is right, the models have not accounted for the properties of CO2 in our atmosphere correctly and therefore the results and projections about increases in CO2 in our atmosphere are not correct either.
    Some of it is simple.
    Quite obviously some of it is not.
    I still think that Nasif may be on the right track because he has highlighted that a lot of the ‘assumptions’ in current climate modelling are not taking in all the variables correctly.
    There is a difference between wanting something to be right or having faith in something being right (as Gavin seems to be arguing) and what may actually be right.
    Our atmosphere certainly does not behave the same as a roof over a greenhouse nor the outside of a furnace.
    Just as well. The whole place would have fried a long time ago.
    Something else (as well as those known properties about greenhouses and furnaces) is going on here.
    It seems it must be the way the atmosphere behaves?
    There’s lots of question marks because the jury is still out here 🙂

  136. Luke April 14, 2011 at 11:32 am #

    “do you really believe that IR coming from a colder body(the atmosphere is increasing the temperature of a warmer body(the Earth) ” – yep – it’s called net

    Debs – Nasif has not addressed the models actually used. What sheer nonsense. Clueless. Then we have “fried” – the old runaway ruse. For heavens sake.

  137. Nasif Nahle April 14, 2011 at 12:12 pm #

    @Cohenite…

    Thank you for your question. Here my answers:

    1 believe the current GAT, global average temperature, is greater than it would be if there was no atmosphere?

    It’s not greater with the atmosphere, but quite the contrary. It would be greater if there would not be atmosphere:

    1368 W/m^2 = 414 K = 141 °C = 285.8 °F.

    2 If GAT is higher with an atmosphere do you believe it is due to the presence of H2O and CO2?

    The presence of water vapor and carbon dioxide makes the temperature of the Earth gets lower during daylight on the hemisphere facing the. The presence of subsurface materials, oceans and water vapor maintains the atmosphere relatively warmer during nighttime than it could be without oceans.

    3 If GAT is higher due to the presence of CO2 and H2O in the atmosphere is this because of their greenhouse properties; that is their capacity to absorb IR?

    No higher GAT. During daylight a hemisphere gets warmer, while the opposite hemisphere gets colder. If oceans wouldn’t exist, the hemisphere in darkness would freeze in seconds.

    The effect of cooling of the atmosphere is quite evident:

    Without atmosphere, the temperature of the Earth would be 414 K.

    With an atmosphere, the average temperature of the Earth is 393 K.

    Sorry to be basic but I think we are just bobbing up and down between Arthur Smith’s piece and Gerlich and Tscheuschner; I think the Miskolczi piece, his 2004 paper, is a middle ground here, but I’ll wait until you guys declare your positions.

    I agree with Miskolczi findings along with the results of Gerlich and Tscheuschner. Their works have no one problem with physics.

    Nasif S. Nahle

  138. Nasif Nahle April 14, 2011 at 12:17 pm #

    Errata:

    With an atmosphere, the average temperature of the Earth is 393 K.

    I should have written:

    With an atmosphere, the average temperature of the Earth is 293 K.

    The difference between temperature with an atmosphere and without it is 293 K – 414 K = -121 K.

    NSN

  139. Neutrino April 14, 2011 at 12:52 pm #

    Nasif,

    No I have not forgotten Kirchhoff.
    Absorptivity is a simple concept, yes Kirchhoff’s relates its value to the value of emissivity but that is not the point here. For clarity I will use α when specifically talking about absorptivity and ε when specifically talking about emissivity. Yes the values of these two are the same by Kirchhoff but the action of them is based on different foundations.

    In the simplest sense Absorptivity(α) is a measure of how much Incident is absorbed. More precisely it is:
    Absorptivity(α) + Transitivity(τ) + Reflectivity(ρ) = 1
    α + τ + ρ = 1
    This in essence is a statement of the Conservation of Energy, all Incident that enters the system must be accounted for.

    Since we are dealing with a solid(no transmission) we can safely ignore and the above simplifies to:
    α + ρ = 1
    α = 1 – ρ

    Lets then apply that to your calculation: Using your value of 0.82 for absorptivity we have by definition 0.18 reflectivity.
    Looking at your numbers: 342.6W/m^2 absorbed requires by definition 75.2W/m^2 to be reflected. If there is not that amount of reflected then α + ρ ≠ 1.
    Either you have to accept that 75.2W/m^2 is reflected or violate Conservation of Energy, which one do you endorse?

    If 342.6W/m^2 is absorbed by definition the Incident has to be 417.8W/m^2?

    The point of the above is that Absorption is calculated by (α * Incident), not by the Stefan-Boltzmann equation.

    I am not sure of what rule of three you are referring to.

  140. Neutrino April 14, 2011 at 1:00 pm #

    cohenite,

    1. GAT is higher with a radiatively active atmosphere, with a radiatively inactive atmosphere it is the same as with no atmosphere. The atmosphere also acts to stabilize the temperature so there is less variation(extreme high to extreme low) with any atmosphere.
    2. Yes those are two of the radiatively active gasses in the atmosphere.
    3. Yes.

    The point of these calculations was not to establish how the Green House Effect works but to establish that without it the temperature of the earth is lower than observed.

  141. Neutrino April 14, 2011 at 1:05 pm #

    Debbie,

    How a greenhouse works and how the Green House Effect works are two different things. For good or ill we are stuck with the name Green House Effect but it doesn’t mean that it works along the same principles of a real greenhouse.

  142. Neutrino April 14, 2011 at 1:29 pm #

    Hum,

    This is not a model of the Green House Effect(GHE).

    Not to get into a discussion of the Second Law of Thermodynamics but in no way does the GHE violate it. The key component that is often missed when this comes up is net. The atmosphere radiates to the earth, but it is always receiving more radiation from the earth. So the net transfer is always earth(hotter) to atmosphere(cooler) and no violation occurs.

    Yes if I put a blanket around a corpse it would increase the temperature, only slow the cooling.

    But if you put a blanket around an alive person his or her temperature will increase. Since the person is still generating the same power but now cooling off more slowly the temperature rises. This is not a violation of any Law that I know of.

    The same thing happens with the atmosphere and the earth, the only difference is that the earth’s power source(the sun) is not inside the blanket. What allows this to work is that the atmosphere is almost transparent to solar spectrum but is somewhat opaque to terrestrial spectrums. In other words it lets the power in efficiently but does not allow the power out very efficiently. So the blanket analogy holds. And since the blanket analogy does not violate the Second Law of Thermodynamics neither does the GHE.

    Again, the calculations expressed here are quite explicitly devoid of any GHE. One point of the calculation was to demonstrate that the earth without an atmosphere is significantly colder than one with an atmosphere.

  143. Neutrino April 14, 2011 at 1:47 pm #

    Nasif,

    What is the connection between your calculated Incident(1366W/m^2 over the earth disc) and the absorbed/emitted(342.6W/m^2 over the entire sphere)?

    Phrased another way how do you go from an incident of 1366W/m^2 over the disk, or 683W/m^2 over the hemisphere or 341.5W/m^2 over the sphere to an absorbed of 342.6W/m^2?

    I see no logical or mathematical connection between your numbers. To put it in perspective:
    1366W/m^2 incident over the disc totals to 1.742*10^17W total incident.
    342.6W/m^2 absorbed over the sphere totals to 1.747*10^17W total absorbed.
    342.6W/m^2 emitted over the sphere totals to 1.747*10^17W total emitted.

    You have the earth absorbing and emitting more total power than the earth receives from the sun.

  144. Nasif Nahle April 14, 2011 at 3:32 pm #

    @Neutrino…

    1366 W/m^2 is the flux of solar power that the Earth receives at the top of the atmosphere, i.e. the outer sphere of the Earth’s atmosphere.

    342 W/m^2 is the flux of solar power absorbed by the surface and the flux of power emitted by the surface of the Earth.

    Thanks to the atmosphere, the solar flux of power decreases in its way to the surface:

    http://www.windows2universe.org/earth/climate/sun_radiation_at_earth.html

    🙂

  145. Bryan April 14, 2011 at 6:09 pm #

    I think that Hum’s observations are correct.

    Proponents of the IPCC position often think that radiation can evade the parameters of the second law.
    However radiation was tested for and is fully accounted for by Clausius.

    Alan D McIntire has an interesting post but his model would benefit with the additional real effect of every radiative interaction from a colder to a hotter object.
    In addition to radiating less they radiate at increasingly longer wavelengths.
    Never the reverse.

    When we discuss the radiation of the atmosphere; a large amount comes from the clouds.
    At the height of the clouds they radiate at around -240K.
    Considering the extensive cloud cover of the planet this accounts for most of the fictitious 33C “greenhouse effect”.
    Working back from the clouds at a height of say 4kilometers and applying the adiabatic lapse rate of 9.8K/km we arrive at the planet surface temperatures of around 15C.

    Upshot is that the radiative effects of CO2 in the troposphere though real are on such a small scale that they can be ignored for most purposes as Nasif has calculated.
    Even scienceofdoom agrees that radiative effects do not influence the lapse rate.

    The radiative effects, which unfortunately sceptics often overlook completely, are discussed in this article;

    http://www.tech-know.eu/uploads/Understanding_the_Atmosphere_Effect.pdf

  146. Neutrino April 14, 2011 at 11:36 pm #

    I realize that the atmosphere absorbs and reflects some incoming solar radiation. That is why I used the Bond Albedo to come up with the absorptivity of the earth system. The Bond Albedo is the total reflected solar radiation from both atmosphere and earth surface.

    You link clearly states this:
    The albedo of a planet (or a locale on it!) clearly affects the ability of that planet to absorb sunlight, thus converting it to heat that can warm the planet and drive its climate. Earth’s overall average albedo is about 0.31. Oceans and forests are quite dark, while deserts are lighter, and clouds, snow, and ice are very bright. Without clouds our planet’s albedo would be around 0.15, so clouds roughly double Earth’s albedo.

    Regardless of that you still didn’t answer the question. To compare apples to apples(the average surface emission/absorption to average solar incident) both must be averaged over the globe. For emission/absorption you have already done that. But the 1366W/m^2 is not the average, that would be 341.5W/m^2. Which leads back to the problems with your balance.

    This is clearly stated in your link:
    If Earth were a flat, one-sided disk facing the Sun… and if it had no atmosphere… every square meter of Earth’s surface would receive 1,368 watts of energy from the Sun. Although Earth does intercept the same total amount of solar EM radiation as would a flat disk of the Earth’s radius (see figure below), that energy is spread out over a larger area. The surface of a sphere has an area four times as great as the area of a disk of the same radius. So the 1,368 W/m^2 is reduced to an average of 342 W/m^2 over the entire surface of our spherical planet.

    Besides, you are using a source that completely agrees with the calculation I have done in the original post. From your link:
    This simplified model of Earth yields an average global temperature for our home world of 254 K (or -19Celsius or -3Fahrenheit). Such a planet would be a chilly place indeed; most or all of the water on Earth would be frozen if this were indeed the temperature of our world. In reality, Earth’s overall average temperature is roughly +15C (or +59F or 288K). As we’ll see in a bit, our atmosphere makes this planet a much more comfortable place to live!

  147. Neutrino April 14, 2011 at 11:43 pm #

    Actually the link that Nasif provided is an excellent description of the very same topic of this thread. I would recommend everyone involved to read it.

    Thank you Nasif for finding an excellent link, just bookmarked it myself.

  148. Neutrino April 14, 2011 at 11:53 pm #

    A note about Absorption.
    It is dictated by this equation.(Assuming no transmission)
    α + ρ = 1
    Absorbed = α * Incident
    Reflected = ρ * Incident

    If the above two quantities were not true then there would be an imbalance between Incident and Absorbed/Reflected. Since energy in must equal energy out(as per Conservation of Energy) the formula for Absorbed is as written above.

    Yes the value of α is slightly temperature dependant but once it has been calculated the actual equation for absorption is not temperature dependant.

    Absorption is not based on the Stefan-Boltzmann(SB), SB is a calculation of how much is emitted. I challenge you to find a single source that says differently.

  149. Hum April 15, 2011 at 12:31 am #

    Neutrino,

    You state:
    “Yes if I put a blanket around a corpse it would increase the temperature, only slow the cooling.

    But if you put a blanket around an alive person his or her temperature will increase. Since the person is still generating the same power but now cooling off more slowly the temperature rises. This is not a violation of any Law that I know of.

    On the first sentence I think you made a typo and meant to say it would not increase the temperature, only slow the cooling. Let me know if this is correct.

    In the next paragraph you are describing a reduction in cooling. The other affect you are describing is efficiency. If it operated as you describe the person radiates heat the blanket returns the heat causing the increase in temperature. The person then radiates more energy which the blanket returns causing an increase in temperature and so on and so on. Why Neutrino I think you just proved spontaneous human combustion.

    A better example is cup your hands together, your hands are warmer than they would be conducting, convecting and radiating to the air when not cupped, but they will never become warmer than your bodies core temperature.

  150. Neutrino April 15, 2011 at 2:35 am #

    Yes Hum, there should be a not in that first line.

    The reason it will not cause combustion is that it is a limited feedback not infinite. Well more precisely the feedback(which is what is raising the temperature) is an infinite series that converges on a real number.

    For example the series:
    1/2 +1/4 + 1/8 + 1/16 + 1/25 + 1/36 + 1/49 + 1/64 + 1/81 + 1/100 + 1/121 + … converges on to the value 1 even though the series itself is infinite.

    Wiki has a decent write up on Geometric Series.

  151. Nasif Nahle April 15, 2011 at 2:40 am #

    @Neutrino…

    This simplified model of Earth yields an average global temperature for our home world of 254 K (or -19Celsius or -3Fahrenheit). Such a planet would be a chilly place indeed; most or all of the water on Earth would be frozen if this were indeed the temperature of our world.

    And they are correct. Just above the paragraph you quoted, they say that:

    “1,368 W/m2 x (1 – albedo) = 4 x s x T4

    Or, using Earth’s average albedo of 0.31 and solving for T:

    T4 = [1,368 W/m2 x (0.69)] / 4s

    Which yields T = 254 K ( = -19° Celsius = -3° Fahrenheit)”

    Which is aproximately the result of your calculations.

    The problem is that you’re dismissing convection, which starts at one meter above the surface, and conduction, which is the principal and almost solitary process of heat transfer in the boundary layer surface-atmosphere.

    From those data, you will definitely conclude that the surface warms up the atmosphere, not the opposite.

    🙂

  152. Neutrino April 15, 2011 at 5:06 am #

    Nasif,

    I am truly confused reading your last post.
    So you agree my calculation done here is correct in all methods and results?

    Your comment about convection is misplaced. Yes convection matters to the atmosphere but the calculation purposely did not include any atmospheric effect. The purpose was to find what the temperature would be with no atmosphere.

    Just as the calculation was done at your link.

    Finishing the quote you posted they continue:
    In reality, Earth’s overall average temperature is roughly +15° C (or +59°F or 288 K). As we’ll see in a bit, our atmosphere makes this planet a much more comfortable place to live!

    That is exactly the point I have been trying to demonstrate with these calculations.
    Without an atmosphere the earth would be a frigid place.

    Just to be clear,
    Do you believe their calculation of -19C is accurate?
    (Their calculation used: Absorptivity of 0.69, Emissivity of 1.0, Incident flux at earth of 1368W/m^2 and absorption as the product of Absorptivity and Incident.)

  153. Bryan April 15, 2011 at 5:18 am #

    Neutrino

    …..”That is exactly the point I have been trying to demonstrate with these calculations.
    Without an atmosphere the earth would be a frigid place.”……..

    No without an atmosphere some places would reach >+100C and other places <-100C

  154. Bryan April 15, 2011 at 5:21 am #

    Neutrino

    No correction, my post above ignored the Oceans which would smooth out the extremes.

  155. Nasif Nahle April 15, 2011 at 5:42 am #

    @Neutrino…

    I am truly confused reading your last post.
    So you agree my calculation done here is correct in all methods and results?

    I cannot agree with something that is wrong. As the authors at the link I provided, they say the methodology is wrong and the result is not coincident with reality. I adhere to reality.

    Your comment about convection is misplaced. Yes convection matters to the atmosphere but the calculation purposely did not include any atmospheric effect. The purpose was to find what the temperature would be with no atmosphere.

    Are we talking scientifically or not? If yes, then my comment about convection and conduction are well placed. The calculations, if we want to be correct, must include the three processes of heat transfer.

    Just as the calculation was done at your link.

    They also say it is not correct. Besides, their temperature is not the same you obtained. You obtained a surface temperature of 5.4 K, while they obtained -18 K

    That is exactly the point I have been trying to demonstrate with these calculations.
    Without an atmosphere the earth would be a frigid place.

    This is a wrong conclusion of you and the authors because the calculations demosntrate it is the opposite:

    With an atmosphere: 293 K on the daylight surface.

    Without an atmosphere: 415 K on the daylight surface.

    Without an atmosphere: 122 K on the nighttime surface.

    Therefore, your argument at:

    Do you believe their calculation of -19C is accurate?
    (Their calculation used: Absorptivity of 0.69, Emissivity of 1.0, Incident flux at earth of 1368W/m^2 and absorption as the product of Absorptivity and Incident.)

    is nonsensical. The calculation is correct, but the conclusion is incorrect. You should learn to examine every text you read and avoid cherrypicking. The authors say just before the calculations:

    This next section delves into the math of that energy balance calculation. These calculations are supplied for those who are interested in them, but this is not required knowledge for this course. Feel free to skim over this section, and to take our word for it with regards to the results if you are not interested in the details of the mathematics.

    After their calculations, they say:

    This simplified model of Earth yields an average global temperature for our home world of 254 K (or -19° Celsius or -3° Fahrenheit). Such a planet would be a chilly place indeed; most or all of the water on Earth would be frozen if this were indeed the temperature of our world. In reality, Earth’s overall…”

    Their calculations are okay, their conclusion about the disparity regarding reality is okay, the problem is the conclusion which contradicts what they had say before in the same article:

    Although Earth does intercept the same total amount of solar EM radiation as would a flat disk of the Earth’s radius (see figure below), that energy is spread out over a larger area. The surface of a sphere has an area four times as great as the area of a disk of the same radius. So the 1,368 W/m2 is reduced to an average of 342 W/m2 over the entire surface of our spherical planet. Another way to think of this reduction is to realize that half of Earth’s surface (the night side) is in the dark and thus receiving no solar energy at a given moment, while areas near the edges of the planet (near the poles and around dusk and dawn) are receiving reduced amounts of energy per unit area.

    As I read this contradiction, I proceeded to calculate by means of other methodologies and found the same conclusions than other scientists have obtained through other methodologies, i.e. the atmosphere cools the Earth.

    Finally, convection is a heat transfer process that works as a conveyor of energy.

    Expect more on this from me in future essays.

    🙂

  156. Neutrino April 15, 2011 at 5:45 am #

    Yes Bryan, without an atmosphere there would be wild extremes in temperature. I would guess the ocean would be mostly frozen.

    The purpose of the calculation is what would be the average temperature be.

  157. mkelly April 15, 2011 at 6:48 am #

    The earth with no atmosphere is very similar to the moon. So the temperatures would be similar. Moon 107 C during day and -153 C at night.

  158. Neutrino April 15, 2011 at 7:01 am #

    Nasif,

    You are splitting hairs that don’t exist. Both myself and the authors of that page clearly stated that the calculated numbers were for an earth without an atmosphere. For that reason the results are wrong for an earth that does in reality have an atmosphere. But the methodology is correct for the stated purposes.

    So the question was not if they, and myself, correctly calculated the real earth temperature, we didn’t, but if we correctly calculated the temperature of an earth without an atmosphere.

    Are our calculations correct for an earth without an atmosphere?
    (You seem to be saying yes they are)

    Yes we are talking scientifically. If there is no atmosphere there is no convection. So for the calculations done by both myself and your links authors convection is irrelevant.

    Reread both my piece and theirs. We have slightly different flux’s(1368 for them 1366 for me) and albedo’s(0.3 for me and 0.31 for them) so it is not a surprise that our numbers are not exactly the same. But my calculation came to -18.3C and theirs was -19C. That to me is fairly close agreement.

    You quote me and then add:
    “That is exactly the point I have been trying to demonstrate with these calculations.
    Without an atmosphere the earth would be a frigid place.
    This is a wrong conclusion of you and the authors because the calculations demosntrate it is the opposite:
    With an atmosphere: 293 K on the daylight surface.
    Without an atmosphere: 415 K on the daylight surface.
    Without an atmosphere: 122 K on the nighttime surface.

    Aside from the question of where the day/night temperatures come from and ignoring the spatial distribution of temperatures across the hemisphere a straight average yields -4.7C.
    What is your point again? Since the average measured temperature is 15C you seem to be agreeing a no atmosphere planet would be colder.
    How is it a wrong conclusion that -18.3(or -19C) or even a -4.7C world is a frigid place?

    You go on to quote me again and respond:
    “Do you believe their calculation of -19C is accurate?
    (Their calculation used: Absorptivity of 0.69, Emissivity of 1.0, Incident flux at earth of 1368W/m^2 and absorption as the product of Absorptivity and Incident.)
    is nonsensical. The calculation is correct, but the conclusion is incorrect.

    It is not nonsensical; it is a calculation of a different earth. One without an atmosphere. Yes our earth does have an atmosphere so obviously we should not expect the results to agree with observations placing average temperature at around 15C. It is a conclusion for the earth without an atmosphere and is quite appropriate for that earth.
    The point is that the atmosphere accounts for the difference between our calculations and the observed global temperature.

    I am glad you think the calculations are correct.
    This implies that you now believe that Absorption is the product of Incident and Absorptivity.

    No, knowing how to do these kinds of calculations is not required but if you want to claim that the average global temperature is explained with no GHE then you are required to do them.

    Yes, these are simplified models. They tell us interesting things. One of which is that to know the correct average temperature for the earth we must include the atmosphere. It also gives us some insight into how much the atmosphere effects the final temperature.(ie: 15C- -18.3C = 33.3C accounted to atmospheric effects)

    You quote: (your bold)
    “<Another way to think of this reduction is to realize that half of Earth’s surface (the night side) is in the dark and thus receiving no solar energy at a given moment, while areas near the edges of the planet (near the poles and around dusk and dawn) are receiving reduced amounts of energy per unit area.
    As I read this contradiction, I proceeded to calculate by means of other methodologies and found the same conclusions than other scientists have obtained through other methodologies, i.e. the atmosphere cools the Earth.

    Where exactly is the contradiction in the above quote? Its simple geometry, the point directly beneath the sun will experience the full flux of 1368W/m^2, as you progressively move away from that point the flux will be lower and lower because the angel of incidence is changing. At the terminus of the hemisphere it goes down to zero, clearly on the reverse side the flux is zero.
    So on average the earth receives 342W/m^2 but doesn’t receive that at all points at all times. Seems to be a rather obvious conclusion. What exactly do you think they are contradicting?

    How on earth can you conclude that the atmosphere cools the earth? The above calculation clearly shows that without an atmosphere the earth would be -18.3C compared to the currently observed of roughly 15C.
    Besides to conclude that the atmosphere cools the earth you have to first establish that the earth is warmer with no atmosphere, which is to say do the above calculations. Unfortunately for your point of view they disagree with you.

  159. gavin April 15, 2011 at 7:04 am #

    Guys c’mon please! An Earth without an atmosphere? What about an ocean hey

    One way or another we have a blanket effect and you can’t escape that fact.

  160. Graeme m April 15, 2011 at 7:07 am #

    I don’t understand the math but this is a fascinating exchange. I don’t wish to sidetrack the thread, but I don’t follow this:

    Although Earth does intercept the same total amount of solar EM radiation as would a flat disk of the Earth’s radius (see figure below), that energy is spread out over a larger area. The surface of a sphere has an area four times as great as the area of a disk of the same radius. So the 1,368 W/m2 is reduced to an average of 342 W/m2 over the entire surface of our spherical planet. Another way to think of this reduction is to realize that half of Earth’s surface (the night side) is in the dark and thus receiving no solar energy at a given moment, while areas near the edges of the planet (near the poles and around dusk and dawn) are receiving reduced amounts of energy per unit area.

    Given that only a hemisphere faces the sun’s light, why is a sphere used to calculate the actual energy? And at the edges of that hemisphere I’d have imagined that much of the light would be reflected rather than absorbed.

  161. Graeme M April 15, 2011 at 7:10 am #

    Whoops never mind, you’ve just explained it…

  162. gavin April 15, 2011 at 7:10 am #

    BTW those extreme max min temps are only skin deep even with out an atmosphere.

  163. Nasif Nahle April 15, 2011 at 7:33 am #

    @Neutrino…

    You are splitting hairs that don’t exist. Both myself and the authors of that page clearly stated that the calculated numbers were for an earth without an atmosphere. For that reason the results are wrong for an earth that does in reality have an atmosphere. But the methodology is correct for the stated purposes.

    Wrong! The authors say:

    If Earth were a flat, one-sided disk facing the Sun… and if it had no atmosphere… every square meter of Earth’s surface would receive 1,368 watts of energy from the Sun.

    Why to lie, my dear Neutrino? Because you are wrong in your calculations.

    😀

  164. Nasif Nahle April 15, 2011 at 7:44 am #

    And they continue saying:

    As you can imagine, as sunlight passes through our atmosphere, some of it is scattered and absorbed, reducing the amount that actually reaches the ground</b. We'll take up that issue in a bit, but for now we'll continue to simplify our discussion by assuming an airless Earth.

    You see? You are misinterpreting to the authors, who admit that the calculation, similar to your calculation but well done, not like yours, is incorrect:

    This simplified model of Earth yields an average global temperature for our home world of 254 K (or -19° Celsius or -3° Fahrenheit). Such a planet would be a chilly place indeed; most or all of the water on Earth would be frozen if this were indeed the temperature of our world. In reality, Earth’s overall average temperature is roughly +15° C (or +59°F or 288 K).

    Please, don’t come that the authors agree with your calculations. You wrote:

    This calculation is for the Sun/Earth system. It contains one error, the emissivity for the earth was held constant at 0.7 for both absorption of solar and emission of terrestrial radiation. The better approximation of the earth described at the top of the article would result in T = 254.8k(or -18.3C).

    No, that is not the error, the error is your unrealistic calculations. They are flawed, that’s why you obtained a frozen Earth. The Earth is not frozen and we don’t need of any greenhouse effect to keep it warm, but only the power of the Sun and the action of our oceans.

    😀

  165. Neutrino April 15, 2011 at 7:47 am #

    Nasif,

    Why do you persist in saying only half the story.

    Your quote goes on to say:
    Although Earth does intercept the same total amount of solar EM radiation as would a flat disk of the Earth’s radius (see figure below), that energy is spread out over a larger area. The surface of a sphere has an area four times as great as the area of a disk of the same radius. So the 1,368 W/m2 is reduced to an average of 342 W/m2 over the entire surface of our spherical planet.

    There is no incongruity here, the sun radiates onto the earth. The total incident is expressed by the flux at earth multiplied by the area of the disc of the earth. Averaging that flux over the globe does not change the total. This is simple geometry Nasif.

    What lie are you insinuating I have made?

    I’ll ask this again since you seem to in one breath say the calculations are correct then the next that they are wrong.

    Are the calculations that I have done valid?
    (and by extension those of your linked article since they are the same as mine)

  166. Neutrino April 15, 2011 at 7:58 am #

    Nasif,

    My calculations are identical to theirs. The only differences are two slight changes in parameters. I used a Bond Albedo of 0.3 while they used one of 0.31 and my flux is 1366W/m^2 compared to their 1368W/m^2.

    We both calculated that without an atmosphere the earth would be frozen. I calculated -18.3C they calculated -19C.

    Both of our calculations are unrealistic only in the sense that the earth does have an atmosphere. But we both were calculating what the earth’s temperature would be if there was no atmosphere. As such the calculations are quite realistic for the purpose they were done.

  167. Nasif Nahle April 15, 2011 at 8:05 am #

    @Neutrino…

    You are using the same tactics, you are not saying true. You say:

    (Their calculation used: Absorptivity of 0.69, Emissivity of 1.0, Incident flux at earth of 1368W/m^2 and absorption as the product of Absorptivity and Incident.)

    However, their calculations are:

    1,368 W/m2 x (1 – albedo) = 4 x s x T4

    Or, using Earth’s average albedo of 0.31 and solving for T:

    T4 = [1,368 W/m2 x (0.69)] / 4s

    Which yields T = 254 K ( = -19° Celsius = -3° Fahrenheit)

    That is not absorptivity, but 1 (nonexisten blackbody emissivity) minus albedo. Absorptivity is obtained by experimentation, not by deduction; precisely in the same way that Hottel, Leckner, Lapp, Sarofim, Ludwig, etcetera, etcetera, obtained it.

    That’s the reason by which the authors obtained a wrong result, despite their arithmetics is okay, from their calculations. The Earth is not a blackbody and the absorptivity measured with appropiate technology gave a result of 0.82. They made the same mistake than you did but theirs was not so evident. Period!

    Regarding the contradiction, one would be blind if one don’t see it:

    The authors say:

    As you can imagine, as sunlight passes through our atmosphere, some of it is scattered and absorbed, reducing the amount that actually reaches the ground. We’ll take up that issue in a bit, but for now we’ll continue to simplify our discussion by assuming an airless Earth.

    And after their calculations with a wrong conclusion, they say:

    This simplified model of Earth yields an average global temperature for our home world of 254 K (or -19° Celsius or -3° Fahrenheit). Such a planet would be a chilly place indeed; most or all of the water on Earth would be frozen if this were indeed the temperature of our world. In reality, Earth’s overall…”

    Why you falsely say that their calculations are identical to yours when your result is quite different from theirs? It only happens in your dreams:

    The authors result is Which yields T = 254 K ( = -19° Celsius = -3° Fahrenheit)

    Your result is 20) T = 278.6K (or 5.4C)

    Hah!

    😀

  168. Mack April 15, 2011 at 8:45 am #

    Nasif, 14th 12.12pm to cohenite.
    ” If the oceans wouldn’t exist, the hemisphere in darkness would freeze in seconds”
    Thats a very chilling thought.
    Is this with a normal earth atmosphere? Because if it is, so much for that nice warm blanket of 380ppm CO2 🙂 But oh we have to wait for a “net” effect. Maybe a couple of million years.
    Aaahahahahahahahahahahaha

  169. Nasif Nahle April 15, 2011 at 9:03 am #

    @Mack…

    Thats a very chilling thought.
    Is this with a normal earth atmosphere? Because if it is, so much for that nice warm blanket of 380ppm CO2 🙂 But oh we have to wait for a “net” effect. Maybe a couple of million years.
    Aaahahahahahahahahahahaha

    Of course with a “normal” atmosphere. No oceans, no water vapor, no clouds, no thermostat, no… etcetera, and tha carbon dioxide “blanket” that additionaly creates energy from the nothingness and besides recycles it. But yes! You’re right! We have to wait for the “net” effect in many million years, so the blanket theory is confirmed. 🙂

    NSN

  170. Mack April 15, 2011 at 9:07 am #

    OK ,
    My mistake ,we’ve got a model with no atmosphere.
    I wonder how long it would take to freeze otherwise.

  171. Mack April 15, 2011 at 9:23 am #

    But such an animal dosn’t exist because this is reality.
    And this is what these AGWers need to come in to.

  172. Neutrino April 15, 2011 at 9:41 am #

    Nasif,

    The term in their formula (1 – albedo) is the absorptivity. The 1 does not represent a BB. Albedo is the reflectivity of the planet.
    Absorptivity + Reflectivity = 1
    1- Reflectivity = Absorptivity
    Which is exactly what they wrote above. And yes Albedo is obtained by experiment, NASA measures it and finds it to be 0.306. Where do you get your value of 0.82 from?

    Reposting your quote:(my bold)
    As you can imagine, as sunlight passes through our atmosphere, some of it is scattered and absorbed, reducing the amount that actually reaches the ground. We’ll take up that issue in a bit, but for now we’ll continue to simplify our discussion by assuming an airless Earth.

    Where is the contradiction there? They clearly say they are ignoring any atmospheric effect as well as admitting that the atmosphere, if they included it, would have an effect on the calculation.

    Once again Nasif,
    My calculated results:
    20) T = 278.6K (or 5.4C)
    This calculation is for the Sun/Earth system. It contains one error, the emissivity for the earth was held constant at 0.7 for both absorption of solar and emission of terrestrial radiation. The better approximation of the earth described at the top of the article would result in T = 254.8k(or -18.3C).

    With this being what was described at the top of the article:
    For example, the earth can be approximated by a GB with emissivity around 0.7 for solar wavelengths(λ 4μm).
    Their calculated results:
    This simplified model of Earth yields an average global temperature for our home world of 254 K (or -19° Celsius or -3° Fahrenheit).

    How are these different in any meaningful way? (given the slightly different parameter values)

  173. Bryan April 15, 2011 at 9:43 am #

    Neutrino

    As mkelly says;

    The earth with no atmosphere is very similar to the moon. So the temperatures would be similar. Moon 107 C during day and -153 C at night.

    Rotation of Earth 1day!
    Moon rotation day/night 14 days = 28 days.

    Add in Oceans!
    Add in an atmosphere! mainly non IRradiating N2
    Addi in clouds!

    Where is the CO2 effect?

    minus CO2, a frozen snowball as you say – get real!

  174. Neutrino April 15, 2011 at 9:44 am #

    Second quote from My Calculated results got cut in half, here it is again hopefully complete:

    For example, the earth can be approximated by a GB with emissivity around 0.7 for solar wavelengths(λ 4μm).

  175. Neutrino April 15, 2011 at 9:46 am #

    Odd looks like the ‘greater than’ and ‘less than’ signs are messing it up, here it is agin without them:

    For example, the earth can be approximated by a GB with emissivity around 0.7 for solar wavelengths(λ less than 4μm) and an emissivity near 1.0 for terrestrial wavelengths(λ greater than 4μm).

  176. Neutrino April 15, 2011 at 9:49 am #

    Bryan,

    Where in this entire thread have I said that?
    The point is a calculation with no atmosphere which would logically set the baseline for calculating how having an atmosphere changes the temperature.

    Nowhere did I say that if just CO2 was removed then the planet would freeze.

  177. Neutrino April 15, 2011 at 10:05 am #

    To get this back onto the actual discussion of the calculations:

    Where we left off was with Nasif’s numbers for flux and absorption. They were:
    So = 6. 354 x 10^7 W/m^2 * (6.96 x 10^8 m / 1.5 x 10^11 m)^2 = 1368 W/m^2
    Q/m^2= a * (s) (T)^4 = 0.82 (5.6697 x 10^-8 W/m^2 K^4) (293 K)^4 = 342.62 W/m^2

    From Nasif’s radius we can get the surface area of both the disc and the globe.
    Radius of the Earth = 6.3781 x 10^6 m
    area of disc = π * (6.3781*10^6m)^2 = 1.278*10^14m^2
    area of globe = 4 *π * (6.3781*10^6m)^2 = 5.112*10^14m^2

    If we sum up the total power of these three quantities we have:
    Total Incident:
    1368W/m^2 * area of disc = 1.748*10^15W
    Total Absorbed:
    342.62W/m^2 * area of globe = 1.751*10^15W

    How do you have the earth absorbing more power than it is receiving from the sun? (not even taking into account it isn’t absorbing all the incident anyways)

  178. Debbie April 15, 2011 at 10:22 am #

    but, but, but,
    Our AGWers say that their models and projections are ‘reality’.
    Do we have two realities perhaps?
    Maybe we have 2 theories and they need to be tested against ‘reality’?
    And Luke, before you jump up and down, reality does not mean ‘published’.

    I get that we have to figure out what happens without the atmosphere there because then we can figure out the difference when we add the atmosphere.

    I don’t think anyone disagrees that the atmosphere protects us from the extremes of heating and cooling.

    But, if the assumptions about how the added factor of the atmosphere behaves are incorrect, then the projections aren’t correct either.

    The ‘net effect’ of added CO2 won’t be correctly projected if the basic assumptions are not right.

    So the theories will need to be tested against reality and if it takes 1000’s of years to know for sure well……hmmm????

    I now have to ask:
    1)why are politicians etc trying to scare the life out of us over this stuff? and
    2)Why can’t people like Nasif and Neutrino continue to keep figuring this out until it does match up with ‘reality’?

    And Luke, before you jump all over me again, the reason many of us ‘clueless non- scientific types’ (as you call us) are interested and want to know more, is because of question number one.

  179. Mack April 15, 2011 at 10:33 am #

    Neutrino,
    What Nasif has clearly pointed out to me (and I hope you) by saying that the planet would freeze in seconds on the unlit side is the total, enormous,overpowering,all encompassing effect that water has in controlling the earths temperature, dwarfing anything you might say about about our trace natural gas. So in effect by bringing this model to our attention you have only dug yourself into a deeper pit.

  180. Nasif Nahle April 15, 2011 at 10:37 am #

    @Neutrino…

    You say:

    The term in their formula (1 – albedo) is the absorptivity. The 1 does not represent a BB. Albedo is the reflectivity of the planet.

    No, not so simple. Absorptivity is defined as:

    α = [(∫ 0 ∞) (∫ 2π1) α’λ (r, λ) (T, λ, si) Iλ (r, λ, si) cos θi dΩi dλ] / [(∫ 0 ∞) (∫ 2π1) Iλ (r, λ, si) cos θi dΩi dλ]

    Nothing similar to your definition of absorptivity.

    You and the authors deduced (wrongly) the proportion of power absorbed by discounting reflectivity. It’s not the correct procedure to calculate absorptivity, unless you don’t know what absorptivity is, because it gives biased results.

    Absorptivity of any material MUST to be measured, not deduced. And this is the problem not only of you, but of the whole GH idea.

    😀

  181. Nasif Nahle April 15, 2011 at 10:43 am #

    @Neutrino…

    If we sum up the total power of these three quantities we have:
    Total Incident:
    1368W/m^2 * area of disc = 1.748*10^15W
    Total Absorbed:
    342.62W/m^2 * area of globe = 1.751*10^15W

    You have not understood a single word of this science. Sorry… Don’t you see that /m^2? Do you know what it means? I guess you don’t.

    🙂

  182. Neutrino April 15, 2011 at 10:47 am #

    Debbie,
    I get that we have to figure out what happens without the atmosphere there because then we can figure out the difference when we add the atmosphere.

    That is exactly the point, I couldnt agree more.

  183. Nasif Nahle April 15, 2011 at 10:53 am #

    @Debbie…

    2)Why can’t people like Nasif and Neutrino continue to keep figuring this out until it does match up with ‘reality’?

    The problem with Neutrino and the people of his kind is that they invent something and then try to make it matches with reality. They cannot do that because they have to invent formulas, variables, blackbodies, constants (like the famous 5.35), etc.

    On the other hand, scientists observe a real phenomenon, produce a provisional explanation, continue their observations and conduce experiments to falsify or verify their hypothesis. If reality does not support their explanation, scientists look for more real data until they reach a theory, which is a reliable explanation of the observed phenomenon, according to the theory of truth.

    Here, you can see the difference between pseudoscience (represented by Neutrino) and science (represented by this humble scientist).

    Pseudoscience use some scientific terms to argue over a constructed idea.

    Science use observed phenomena, real events, evidence, experiments, etc. to obtain explanations about real phenomena, not to tax citizens. 🙂

    In brief, pseudoscience got an idea and try to find support from reality. Science got data from reality and produces an explanation on that reality.

    NSN

  184. Neutrino April 15, 2011 at 11:09 am #

    Nasif,

    Yes, it is that simple, it is a measured quantity. Shine light on something and measure how much light is reflected, that in essence is what the Bond Albedo measurement is. Now if you wanted to calculate what the absorptivity is from basic physical properties your formula may be correct. But if you are actually measuring it then that is how you do it.
    The equality again:
    Absorptivity + Reflectivity + Transmisivity = 1
    Since we are dealing with solids this equality simplifies to:
    Absorptivity + Reflectivity = 1
    It is a basic relationship. It is a fundamental relationship. And apart from you it is an undisputed relationship.

    By definition if the Incident is not Reflected it has to be Absorbed, where do you think it goes if it neither absorbed nor reflected? Basic Conservation Laws dictate that relationship, are you saying that absorptivity + reflectivity does not equal 1?

    You have not understood a single word of this science. Sorry… Don’t you see that /m^2? Do you know what it means? I guess you don’t.

    Yes, that W/m^2 means it is the effect for just one square meter, what I did is add all the square meters up involved in the earth system. What I calculated there, based on your numbers, is what the total earth is intercepting and absorbing. Or are you saying that the number you calculated were not for the entire globe? Since that was the question I assumed that was what your answer represented.

    What then is the total incident power the earth receives and the total power the earth absorbs?

  185. Nasif Nahle April 15, 2011 at 11:15 am #

    @Neutrino…

    This is plain wrong:

    Yes, it is that simple, it is a measured quantity. Shine light on something and measure how much light is reflected, that in essence is what the Bond Albedo measurement is. Now if you wanted to calculate what the absorptivity is from basic physical properties your formula may be correct. But if you are actually measuring it then that is how you do it.
    The equality again:
    Absorptivity + Reflectivity + Transmisivity = 1
    Since we are dealing with solids this equality simplifies to:
    Absorptivity + Reflectivity = 1

    Again, absorptivity is defined by:

    α = [(∫ 0 ∞) (∫ 2π1) α’λ (r, λ) (T, λ, si) Iλ (r, λ, si) cos θi dΩi dλ] / [(∫ 0 ∞) (∫ 2π1) Iλ (r, λ, si) cos θi dΩi dλ]

    Got it???

    😀

  186. Neutrino April 15, 2011 at 11:20 am #

    Calling what I calculated pseudoscience is a lark.
    Nasif objects to the use of a approximation to a BB, get over it, science makes that approximation all the time. Besides the final result being colder than observed temperatures does not hinge on approximating the earth’s emission by a black body. As my line 20) clearly show, even when approximating the earth by a ideal GB the temperature is only 5.4C, still 10C lower than actual observed measurements.

    Not one of the formulas that were used was invented in the sense Nasif claims. I did not pick a temperature that I ‘wanted’ the earth to be and then build a calculation around it, what I did is use freely available data about the real world and standard physics formulas that describe how things interact. The end result is cold, very cold. That result makes me neither happy nor sad. It simply is a fact of life we live with.

    Nasif,

    Are you seriously saying that if I shine a light on an object, measure how much is reflected I cannot calculate the absorbed component?

    If you are, answer this for me:
    100W laser directed at a surface.
    Measured 30W reflected.
    Where did the other 70W go?

  187. cohenite April 15, 2011 at 11:25 am #

    I previously asked both N’s this question:

    “Nasif and Neutrino, could you both say whether you:

    1 believe the current GAT, global average temperature, is greater then it would be if there was no atmosphere?”

    Nasif kindly replied:

    “It’s not greater with the atmosphere, but quite the contrary. It would be greater if there would not be atmosphere:

    1368 W/m^2 = 414 K = 141 °C = 285.8 °F.”

    Neutrino kindly replied:

    “1. GAT is higher with a radiatively active atmosphere, with a radiatively inactive atmosphere it is the same as with no atmosphere. The atmosphere also acts to stabilize the temperature so there is less variation(extreme high to extreme low) with any atmosphere.”

    Neutrino’s position is consistent with Arthur Smith’s:

    http://arxiv.org/PS_cache/arxiv/pdf/0802/0802.4324v1.pdf

    Arthur says this: “Observed parameters for Earth prove that without infrared absorption by the atmosphere, the average temperature of Earth’s surface would be at least 33 K lower than what is observed.”

    Arthur describes the GAT without an atmosphere as Tave(t) and the radiative temperature due to the greenhouse atmosphere as Teff (t) and says;

    “this average temperature Tave(t) is always less than or equal to the effective thermal radiation temperature Teff (t), so T 4ave is less than or equal to T 4eff;” T 4ave and T4eff are the Stefan-Boltzman derived emissive temperatures. Arthur shows various planet’s Tave and Teff with only Mars having an equivalent Tave and Teff [Table 1].

    We have already seen in this marathon with the 2 N’s that without convection the greenhouse effect would be much larger on Earth; that being the case, given that Mars has an atmosphere almost entirely composed of the greenhouse gas CO2 and no convection, why does it have equivalent Tave and Teff if the greenhouse effect works as Arthur Smith describes?

  188. Neutrino April 15, 2011 at 12:04 pm #

    Mack,

    Nasif has made no substantiated claim that the earth would freeze in seconds if it were not for the water. At the rate the earth is radiating, it would take days to release the equivalent of the stored energy in the atmosphere. This calculation is not exactly analogous to how long it would take to ‘freeze’ but is given to demonstrate how much energy is in the system.

    The lower 10km of the atmosphere is roughly 3*10^21g.(about 75% of the total mass of the atmosphere)
    Dry air has a heat capacity of 1.0035J/(gK).
    The temperature profile goes from 15C at surface to -50C at 10km.(take -20C as a rough average)
    Using the above rough estimates the first 10km of the atmosphere contains(very roughly):
    3*10^21g * 253K * 1.0035J/(gK) = 7.62*10^23J

    To radiate all that energy away at the current rate the earth emits would be:
    Current emitted power = 1.22*10^17W
    7.62*10^23J / 1.22*10^17W = 6.25million seconds = 72days

    Now this is an incredibly rough estimate. It suffers from many problems but gives a rough guide as to how much energy is in the system and how long it would take to radiate it all away, and this is just the atmosphere which contains the least of the land/ocean/atmosphere system. If memory serves the ocean is over 90% of the hear capacity of the planet.

    Some problems the above calculation suffers from:
    The heat capacity of air is for dry air, moist air would be higher.
    The average temperature is just a rough guess in the middle of the range, lower altitude temperatures should be weighted more because the air is denser there.
    The surface emission is dependent on its temperature so as it emitted energy its temperature will go down slowing the rate of loss of energy.

  189. Neutrino April 15, 2011 at 12:16 pm #

    cohenite,

    I appreciate that you want to talk about how the GHE works but until a temperature can be established for the no atmosphere condition then we would just be going in circles. Since that calculation sets the baseline.
    We already are going in circles here just trying to do the much simpler no atmosphere calculation.

  190. Alan D McIntire April 15, 2011 at 12:21 pm #

    I think that 1368 watts /m^2 = 414 K is an overestimate.

    For a blackbody, 1000K radiates at 56,790 watts/m^2,

    (1368/56,790)^0.25 = 0.39396 implying 394 K, not 414 K.

    If earth’s emissivity is 0.95, average for the oceans, we divide that by 0.95 to the 1/4 power or

    394/0.987= 399 K

  191. Mack April 15, 2011 at 1:00 pm #

    Debbie,
    There was a comment by Neutrino to you on 14th of April 1.o5pm.
    ” How a greenhouse works and how the Greenhouse Effect works are TWO DIFFERENT things. For good or ill we are stuck with the name Greenhouse effect but it does’t mean that it works along the same principle of a greenhouse.”
    This is pure DOUBLE dutch designed to pull the wool over your eyes.
    Not only is Neutrino trying to unshackle himself from laws of thermodynamics related to the surface of the earth he is now also trying to unshackle himself from convection.

  192. Luke April 15, 2011 at 1:11 pm #

    Assures Mack pers. Comm.

  193. Mack April 15, 2011 at 1:51 pm #

    Neutrino,
    I’m beginning to get where you’re coming from now.
    As a physicist ( I presume ) you have a “Greenhouse Effect” in the radiation world which is seperate from the real world.

  194. Neutrino April 15, 2011 at 2:05 pm #

    Mack,

    That a real greenhouse works by preventing convection is common knowledge, or at least it should be. That the Green House Effect works by radiative exchanges in an atmospheric column of decreasing temperatures should also be common knowledge. Convection in the GHE actually works to limit the effect. They do not work on the same principles but do have a common name because of a misconception as to how actual greenhouses work from many many years ago. We are stuck with the name now.

    The GHE is not separate from the real world, it is a consequence of the real world.

    But again, this thread was started with the intention of looking at basic radiative exchanges. The GHE is complex problem. I have always thought it best to start with simple processes and then build up the complexity rather than try and explain complex interactions without understanding the basics. How anyone can argue for or against the GHE without a solid understanding of basic radiative physics is beyond me. The example calculation done here is just that basic radiative physics.

    One other point about my beliefs, I do not believe in the GHE for any other reason than it is a consequence of the equations. I neither like it nor dislike it, it is something that simply is.

  195. cementafriend April 15, 2011 at 2:30 pm #

    Cohenite, 11.25AM not sure what you are getting at and I may have missed something- however, photographs from the Mars Rover etc show dust storms which means there are winds and convection. When there is an atmosphere there will be natural convective heat transfer (need to know the Nusselt number and Prandtl number) from the surface to the gas in the atmosphere. Winds increase the convective heat transfer (this then includes a Reynolds number). That is called forced convection. Any atmosphere on a planet will result in distribution of heat from the “day” sun warmed side to the “night” cool side. This makes the surface temperature temperature around the planet more even than a similar planet in the same location without an atmosphere. When there is an atmosphere gravity also has an effect on temperatures in the atmosphere which is determined as the lapse rate.
    As I put in an earlier post by definition a blackbody has a surface. The Stefan-Boltzman equation for blackbodies (or even gray bodies) can not be applied to a gas.

  196. cohenite April 15, 2011 at 2:41 pm #

    cementafriend; my mistake; I was thinking of water but did not include it, so the conclusion should be no convection on Mars due to water vapour.

  197. Mack April 15, 2011 at 3:12 pm #

    Neutrino,
    Thankyou for your reply and clarifying for me the'”greenhouse effect”.
    However my logic tells me that some back radiation (if there is any, or if you have found it)of CO2 on this planet will not have the slightest impact on the earth’s temperature.
    I’m sorry for butting in with the complex stuff I guess it’s just the nature of the internet.
    Proceed with your simple model.

  198. Nasif Nahle April 15, 2011 at 3:25 pm #

    @Neutrino…

    Nasif has made no substantiated claim that the earth would freeze in seconds if it were not for the water.

    It is substantiated on any book you read on climate.

    LOL!

    😀

  199. Neutrino April 15, 2011 at 3:25 pm #

    Just go give some perspective to people reading:

    What I have been trying to explain is standard physics. Don’t take my word for it, since I am an unknown here my word most likely does not carry a lot of weight anyways. I had hoped the logic and consistency of the math and physics involved would be sufficient.

    So lets look for outside opinions, Universities seem to be a good place to start. Searching the internet for online course material I found the following in just a few minutes:
    (it actually took me longer to write this and make all the hyperlinks than it did to track the links down):

    Columbia University
    Harvard University
    Massachusetts Institute Technology
    Iowa State University
    University of Washington

    This information is readily available for anyone interested to find it. All these institutions come to the same basic conclusion in their courses.
    That the temperature of earth would be frigid, ≈-18C, if not for an atmosphere.
    And they all use the same basic physics, as I outlined them at the start of this thread, to arrive at that conclusion.

  200. Neutrino April 15, 2011 at 3:27 pm #

    And it looks like I should have taken just a bit longer making the links….

    Columbia University
    Harvard University
    Massachusetts Institute Technology
    Iowa State University
    University of Washington

  201. Mack April 15, 2011 at 3:34 pm #

    Well simple model with some complex problems!

  202. Nasif Nahle April 15, 2011 at 3:35 pm #

    @Neutrino…

    You’re again lying… You say:

    Nasif,

    Are you seriously saying that if I shine a light on an object, measure how much is reflected I cannot calculate the absorbed component?

    In the first place I didn’t say that, but since you’re asking, if you shine a light on an object and measure how much is reflected, you are not calculating the absorbed component, but just the reflected light.

    Through this erroneous methodology, you are dismissing many physical effects which happens with quantum/waves.

    What you’re doing is creating a fantasy world. Again, the definition of absorptivity is as follows:

    α = [(∫ 0 ∞) (∫ 2π1) α’λ (r, λ) (T, λ, si) Iλ (r, λ, si) cos θi dΩi dλ] / [(∫ 0 ∞) (∫ 2π1) Iλ (r, λ, si) cos θi dΩi dλ]

    If you’re a physicist, as you say, derive a formula from the above equation and then calculate the real absorptivity of the Earth.

    It is not rare that you resort to erroneous formulas because you accept that you don’t know what the total emissivity is. If you ignore, by your own words, what the total emissivity is, you ignore also what the total absorptivity is.

    😀

  203. Luke April 15, 2011 at 3:42 pm #

    In the cotton field EBEX experiment – bottom of this URL http://scienceofdoom.com/2010/07/31/the-amazing-case-of-back-radiation-part-three/

    In the figure Upward and downward radiation measurements, EBEX 2000, Kohsiek (2007)

    It would be fun for our greenhouse doubters to explain the long wave readings? Presumably they don’t add up ? why?

    Hope Mack and Debs don’t get disturbed.

  204. Nasif Nahle April 15, 2011 at 3:44 pm #

    @Neutrino…

    But again, this thread was started with the intention of looking at basic radiative exchanges. The GHE is complex problem. I have always thought it best to start with simple processes and then build up the complexity rather than try and explain complex interactions without understanding the basics. How anyone can argue for or against the GHE without a solid understanding of basic radiative physics is beyond me. The example calculation done here is just that basic radiative physics.

    Demonstrate that the following equation supports the GHE:

    q/m^2 = 0.82 (5.6697 x 10^-8 W/m^2 K^4) (293 K)^4

    It is the formula to calculate the amount of energy absorbed by the surface introducing real measurements.

    You cannot demonstrate through this correct formula that the GHE is here applied to get the result because the temperature of 293 K is the temperature derived from the amount of energy absorbed by the Earth that has been measured and averaged.

    The remainder of the “Earths without atmosphere would be”, are just speculations without scientific basis.

    😀

  205. Nasif Nahle April 15, 2011 at 3:55 pm #

    @Neutrino…

    I am truly confused reading your last post.
    So you agree my calculation done here is correct in all methods and results?

    Tell me where I said that. You’re lying again. Your calculations are absolutely wrong and I have demonstrated it.

    You ignore what total emissivity is and you admitted it.

    You’re calculating the whole solar power emitted by the total surface area of the Sun towards the whole outer sphere.

    For the readers see what Neutrino calculated, the outer sphere is the whole sphere formed by the orbit of the Earth around the Sun. Up, Down, Left, Right, towards everywhere.

    You are misusing the Stefan-Boltzmann equation.

    You are introducing a false absorptivity of the Earth which is contrary to measurements.

    You are considering the Sun is a blackbody, when you have admited that blackbodies don’t exist.

    Your calculations give a faked temperature of the Earth that has nothing to do with the real average temperature of our planet:

    20) T = 278.6K (or 5.4C)

    278.6 K is not the same temperature that the Earth has, i.e. 293 K in average.

    My calculations give the real temperature because I am basing my numbers in real measurements.

    😀

  206. Neutrino April 15, 2011 at 3:59 pm #

    Yes, if I measure 30W reflected I have not directly measured the absorbed. I have calculated it based on a basic identity in physics.

    Through this erroneous methodology, you are dismissing many physical effects which happens with quantum/waves.

    Either the photon is absorbed or reflected there are no other mysterious interactions(yes QED is mysterious but the interaction is clear in this case)

    Nasif, your maintaining that absorptivity is not dictated by:
    Absorptivity + Reflectivity = 1

    Should be the death nail to any credibility you have here. That is a fundamental equation of Conservation of Energy. This is such a basic tenet of modern science I am actually at a loss of how to respond to your denial of it.

    q/m^2 = 0.82 (5.6697 x 10^-8 W/m^2 K^4) (293 K)^4
    It is the formula to calculate the amount of energy absorbed by the surface introducing real measurements.

    The Stefan-Boltzmann Law, which is what you wrote above, is the summation of the Planck Function. The Planck function describes emission from a hot body. It is not a formula that can calculate absorption.
    That you also keep maintaining this should shred any remaining credibility you have on these topics.

    I challenge you to find one scholarly reference that says the SB can be used to calculate the amount of energy absorbed.

    Since we just added a new page I don’t want the links to get lost, here they are again. All five of them clearly in no uncertain terms contradict what Nasif is trying to assert.
    Columbia University
    Harvard University
    Massachusetts Institute Technology
    Iowa State University
    University of Washington
    You will need to scroll down on a few of the links to get to the relevant sections.

    As a clarification to all: I have never claimed I was anything, including a physicist.

  207. Nasif Nahle April 15, 2011 at 4:00 pm #

    @Neutrino…

    We both calculated that without an atmosphere the earth would be frozen. I calculated -18.3C they calculated -19C.

    Sooorry, but you’re lying again. You calculated that the temperature of the Earth, after all of your erroneous calculations, is

    Body B temperature:

    17) p = εσT^4.A

    18) T = (p/εσA)^.25

    19) T = (1.24*10^17W / (0.7 * 5.67*10^-8W/m^2K^4 * (4 * π * 6.371*10^6m^2))^.25

    20) T = 278.6K (or 5.4C)

    278.6 K is not a freeze temperature!!! Hah!

    Neutrino, why to lie when what you wrote in your post is still there? Do you think that the readers won’t go to read your mistakes again?

    😀

  208. Graeme M April 15, 2011 at 4:02 pm #

    I must admit that for me this argument while instructive does seem to be rather circular.

    If I understand Neutrino’s simple demonstration correctly, can I pose a question for each to answer? No references to ‘real’ situations, no arguments about which equations to use etc, just a simple answer to a purely theoretical question. I ask this so that I can see what the fundamental difference in outcome is.

    Imagined system for demonstration purposes.

    A sphere of identical size to the earth, devoid of atmosphere, which exactly follows earth’s orbit but is placed exactly opposite earth. In all respects, it shares earth’s rotational and orbital periods etc.

    This sphere can, for the purposes of this demonstration, be approximated by a Grey Body with emissivity of 0.7 for solar wavelengths(λ 4μm) – REGARDLESS of whether these numbers are correct.

    Using the known measurements for the sun and its radiation, what would be the average surface temperature of the sphere?

    Please show the known measurements that you introduce.

  209. Graeme M April 15, 2011 at 4:04 pm #

    Sorry, should have read:

    Imagined system for demonstration purposes.

    A sphere of identical size to the earth, devoid of atmosphere, which exactly follows earth’s orbit but is placed exactly opposite earth. In all respects, it shares earth’s rotational and orbital periods etc.

    This sphere can, for the purposes of this demonstration, be approximated by a Grey Body with emissivity of 0.7 for solar wavelengths(λ 4μm) and an emissivity of 1.0 for terrestrial wavelengths(λ > 4μm) – REGARDLESS of whether these numbers are correct.

    Using the known measurements for the sun and its radiation, what would be the average surface temperature of the sphere?

    Please show the known measurements that you introduce.

  210. Nasif Nahle April 15, 2011 at 4:04 pm #

    @Neutrino…

    The Stefan-Boltzmann Law, which is what you wrote above, is the summation of the Planck Function. The Planck function describes emission from a hot body. It is not a formula that can calculate absorption.
    That you also keep maintaining this should shred any remaining credibility you have on these topics.

    Whatever… You cannot demonstrate the formula is not valid and it seems you’re avoiding to answer my question.

    You’re a scientist, a physicist?

    Well, derive the formula from the definition of absorptivity:

    α = [(∫ 0 ∞) (∫ 2π1) α’λ (r, λ) (T, λ, si) Iλ (r, λ, si) cos θi dΩi dλ] / [(∫ 0 ∞) (∫ 2π1) Iλ (r, λ, si) cos θi dΩi dλ]

    And answer my question:

    Demonstrate that the following equation supports the GHE:

    q/m^2 = 0.82 (5.6697 x 10^-8 W/m^2 K^4) (293 K)^4

    And for your knowledge, science is not based on scholarly, but on reality… Heh!

    If my calculations and the measurements from the real world agree, then your scholars and you are wrong.

    😀

  211. Nasif Nahle April 15, 2011 at 4:18 pm #

    @Graeme M…

    Using the known measurements for the sun and its radiation, what would be the average surface temperature of the sphere?

    It is exactly the number I and scientists are saying, i.e. 293 K, without the imaginary GHE:

    Known data (real measurements and averages):

    Solar flux power striking on the top of the Earth’s atmosphere = 1368 W/m^2

    q or Solar flux power striking on the Earth’s surface = 342 W/m^2

    σ or Stefan-Boltzmann constant = 5.6697 x 10^-8 W/m^2 K^4

    α or Absorptivity and emissivity of the Earth (measured and averaged) = 0.82

    T = ??

    T = (q / σ * α)^(1/4) = [(342 W/m^2) /(5.6697 x 10^-8 W/m^2 K^4 * 0.82)]^(1/4) = 292.86 K, or 293 K by rounding up the number.

    As you can see, the atmosphere attenuates the intensity of the flux of power from the Sun, and the average temperature of 293 K is produced by the solar power striking on the surface alone. It is not the product of any GHE, it is imaginary.

    NSN

  212. Nasif Nahle April 15, 2011 at 4:24 pm #

    @Graeme M…

    If you introduce 0.7, which is a unreal absorptivity, you would obtain:

    T = (q / σ * α)^(1/4) = [(342 W/m^2) /(5.6697 x 10^-8 W/m^2 K^4 * 0.7)]^(1/4) = 304.67 K, or 305 K by rounding up the number. And this happens without any “greenhouse effect”, but by the solitary effect of the solar radiation.

    NSN

  213. Neutrino April 15, 2011 at 4:29 pm #

    Nasif you have rattled off too many questions to answer easily, besides you still are avoiding my questions.

    So ill just start with the last question, if you want me to answer more ask them in succession in a give and take.
    And answer my question:
    Demonstrate that the following equation supports the GHE:
    q/m^2 = 0.82 (5.6697 x 10^-8 W/m^2 K^4) (293 K)^4

    That is the SB equation, as such it is calculating how much flux is emitted by a surface at 293K with emissivity 0.82. By itself it does not confirm or deny the GHE. It simply is a calculation.

    My question:
    If you think it is calculating absorption produce a reference stating that the SB can calculate how much is absorbed at a surface.

    I also have a request request:

    You wrote:(quoting me and then responding)
    ”We both calculated that without an atmosphere the earth would be frozen. I calculated -18.3C they calculated -19C.
    Sooorry, but you’re lying again. You calculated that the temperature of the Earth, after all of your erroneous calculations, is

    20) T = 278.6K (or 5.4C)
    278.6 K is not a freeze temperature!!! Hah!
    Neutrino, why to lie when what you wrote in your post is still there? Do you think that the readers won’t go to read your mistakes again?

    Please reread the post, that is not where I finished, immediately following line 20) is:
    This calculation is for the Sun/Earth system. It contains one error, the emissivity for the earth was held constant at 0.7 for both absorption of solar and emission of terrestrial radiation. The better approximation of the earth described at the top of the article would result in T = 254.8k(or -18.3C).

    Claiming that I am lying is a serious charge and should be backed up by something concrete, I clearly finish the post with a statement that with no atmosphere the earth’s temperature would be -18.3C.

  214. Neutrino April 15, 2011 at 4:38 pm #

    Graeme,

    That is the original calculation I started this thread off with 😉

    All my variables came from NASA forearth and sun

    You can see the same calculation at all 5 of these sites:
    Columbia University
    Harvard University
    Massachusetts Institute Technology
    Iowa State University
    University of Washington

    as well as the one Nasif linked earlier:
    href=”http://www.windows2universe.org/earth/climate/sun_radiation_at_earth.html”>here

  215. Neutrino April 15, 2011 at 4:39 pm #

    damn hyperlinks.

    here

    Need some sleep, until tomorrow. Goodnight.

  216. Neutrino April 15, 2011 at 4:41 pm #

    lol definately time for sleep

    Link to article Nasif linked

  217. Nasif Nahle April 15, 2011 at 4:55 pm #

    @Neutrino…

    20) T = 278.6K (or 5.4C)

    No where this variable in the links you provided.

    http://www.windows2universe.org/earth/climate/sun_radiation_at_earth.html

    This is not either:

    7) p = 3.84*10^26W

    Neither this:

    14) q = 1.22*10^17W

    Neither this:

    ε = 0.7

    Besides, all of your links are about blackbodies radiation, and you know that the Sun and the Earth are not blackbodies. Well… you know blackbodies don’t exist. Agree?

    And reality says you’re wrong.

    😀

  218. Nasif Nahle April 15, 2011 at 4:56 pm #

    @Graeme M…

    Please, read my answer to your question above the smoke curtain that Neutrino placed.

    Thanks!

    NSN

  219. cohenite April 15, 2011 at 7:26 pm #

    luke, you link to SoD’s thread which was dominated by the irritating Mark who even I can out-argue.

  220. Graeme M April 15, 2011 at 9:58 pm #

    Hmmm… Interesting. So in the case of our demonstration system, the average temperature of the sphere’s surface is as follows:

    Neutrino – 279k (6C)
    Nasif – 305k (32C)

    That’s quite a difference for the simple system. Now I’m going to have to re-read this and actually try to follow the maths. Or can someone else summarise in a few words why the difference?

    By the way, I don’t follow the idea of ‘average’ here. Is it relevant? After all, there won’t be any place on the surface of the sphere that will actually be that temperature (except very briefly twice per day). On average in fact any one place will either be very hot for most of the day, or very cold for most of the night. What would be more informative is what our respective N’s calculated max and min temps for the surface, wouldn’t it?

    Regardless, Neutrino’s objective was to set the scene so to speak. His post is intended to show the basic principles involved in radiative transfer so we can better evaluate the theory of anthropogenic global warming (AGW). Now, clearly he and Nasif differ on these basic principles.

    What is that difference? In as few words as possible?

  221. Nasif Nahle April 15, 2011 at 11:45 pm #

    @Graeme M…

    Look at the Moon, Graeme, and you’ll find who is doing the things correctly.

    Moon has not atmosphere and it is at the same distance than the Earth from the Sun. The Moon is receiving the same amount of power from the Sun than the Earth; however:

    Moon’s temperature on its side facing the Sun: 107 °C or 380.15 K

    Moon’s temperature on its night hemisphere: -153 °C or 120.15 K

    Moon has not atmosphere, so nobody can argue it has a GHE.

    Now compare it with my results for an Earth without atmosphere:

    Earth’s Temperature during daytime: 141.02 °C or 414.17K

    Earth’s Temperature during nighttime: -135 °C or 138.15 K

    With its atmosphere, the temperature of the Earth drops down to 293 K. Therefore, the atmosphere ATTENUATES the solar irradiance striking on the surface, not increases the energy absorbed by the surface.

    Things are very clear. Why the confusion? The GHE is killed by physics and thermodynamics.

    😀

  222. Neutrino April 16, 2011 at 12:05 am #

    Graeme,

    Minor correction, my final result for the above calculation was -18.3C.

    The difference lies in only two points. Well really only one, how to calculate the absorbed component.

    First, Nasif is rejecting the use of (1- reflectivity) to calculate absorptivity. It is well documented that the earth’s reflectivity(albedo) is approximately 0.3 which leads to absorptivity of 0.7. Every source agrees on this, except Nasif.

    Second, Nasif has divorced absorption from the Incident(I). He claims that the SB calculates the absorption(it does not). Absorptivity(α) is the fraction of Incident that is absorbed(that is the definition of the term). Absorbed is then simply I * α. It is no more complicated than that.
    In Nasif’s calculation the Incident is irrelevant since the only variable is T in his calculation.

    I am still waiting on Nasif to find a single source to supports his belief that the SB calculates absorption. I do not expect he will ever answer the request for a source since there does not exist any. He simply is wrong.

    In regards to your concern over the idea of an average. Yes it is problematic and actually an interesting discussion could follow on just that. But until we settle on the average calculation I fear the more sophisticated approach would be futile.

  223. Neutrino April 16, 2011 at 12:07 am #

    Nasif,

    Literally for the last time, read my post. My final calculated number is -18.3C, that you continuously keep misrepresenting this speaks volumes.

    No, the links do not contain the numbers of total power emitted by the sun and total power incident on the earth. I choose to calculate these values because I believed the geometry would be easier for people to visualize. It is equivalent to the approach of calculating the flux directly. This is evidenced by all arriving at the same value of about 255K.
    To do this they all generally do not calculate all the intermediate steps either.

    For example the MIT link does not introduce any intermediate values, just derives the formula for T.
    Equating emitted to absorbed we find that:
    Te = ((So(1 − αp)) / 4σ)^.25 (1)
    Note ‘a’, the radius of the Earth, does not appear.

    And then solves it, getting:
    Te depends only on αp and So. Putting in numbers for the Earth we find
    that:
    Te = 255K

    (note: αp they are using this to denote the albedo. I had used α to denote absorptivity)

    Harvard begins with a power and then converts it to a flux:
    The total radiation ES emitted by the Sun (temperature TS = 5800 K) per unit time is given by the radiation flux sTS4 multiplied by the area of the Sun:
    Es = 4 πRs^2σTs^4 (7.8)
    where RS = 7×105 km is the Sun’s radius. The Earth is at a distance d = 1.5×108 km from the Sun. The solar radiation flux FS at that distance is distributed uniformly over the sphere centered at the Sun and of radius d ( Figure 7-9 ):

    Fs = Es/(4πd^2) = σTs^4Rs^2 / d^2 (7.9)

    7.8 is my 6), 7.9 is my 14)/Disc of the earth. These calculations are the same as mine, not in every step but in content.

    The point being is not that my procedure was mirrored step by step, but that the final result is the same. This is because my calculation is equivalent to theirs. If you cannot see the equivalence then I’m not sure how to explain it to you, in essence it is just geometry.

    Yes, all the links used an absorptivity of 0.7(approximately, Harvard uses 0.72 for example). That is the value of (1 – Albedo). And all the links approximate the earth’s emissivity as 1.0(the number is actually not in the calculation they are using the SB with no emissivity, which is the same as saying it is 1)

    Using a BB as an approximation is not evil, wrong, deceitful or anything else. It is an approximation that is all. Approximations are done every day in every field in science, why do you object so strenuously here?

  224. Nasif Nahle April 16, 2011 at 12:14 am #

    @Graeme M…

    Look at the Moon, Graeme, and you’ll find who is doing the things correctly.

    Moon has not atmosphere and it is at the same distance than the Earth from the Sun. The Moon is receiving the same amount of power from the Sun than the Earth; however:

    Moon’s temperature on its side facing the Sun: 107 °C or 380.15 K

    Moon’s temperature on its night hemisphere: -153 °C or 120.15 K

    Moon has not atmosphere, so nobody can argue it has a GHE.

    Now compare it with my results for an Earth without atmosphere:

    Earth’s Temperature during daytime: 141.02 °C or 414.17K

    Earth’s Temperature during nighttime: -135 °C or 138.15 K

    With its atmosphere, the temperature of the Earth drops down to 293 K. Therefore, the atmosphere ATTENUATES the solar irradiance striking on the surface, not increases the energy absorbed by the surface.

    Things are very clear. Why the confusion? The GHE is killed by physics and thermodynamics.

    @Neutrino…

    I won’t answer any of your posts until you answer my questions:

    Well, derive the formula from the definition of absorptivity:

    α = [(∫ 0 ∞) (∫ 2π1) α’λ (r, λ) (T, λ, si) Iλ (r, λ, si) cos θi dΩi dλ] / [(∫ 0 ∞) (∫ 2π1) Iλ (r, λ, si) cos θi dΩi dλ]

    Demonstrate that the following equation supports the GHE:

    q/m^2 = 0.82 (5.6697 x 10^-8 W/m^2 K^4) (293 K)^4

    Why are you evading to answer them?

    😀

  225. Neutrino April 16, 2011 at 12:41 am #

    Nasif,

    The definition of absorptivity is simple from a measurement perspective
    Absorptivity = Reflectivity – 1

    What you have written in no way contradicts that. Your definition is the precise definition(judging by the look of it, making no claim as to its veracity) if your want to calculate the value from basic physics. I do not have the ability or the need to do that here. And neither should you. That absorptivity is commonly calculated from the above identity is standard practice, as evidenced by each of the five links to academic courses material.

    As I said before that equation neither proves nor disproves the GHE. It is simply a calculation of the emitted flux from an object at 293K with emissivity 0.82. Since 293K does not represent any acknowledged current measurement of the average surface temperature of the earth it really doesn’t have any bearing on the discussion. NASA’s reference sheet has the average temperature at 15C(288K). As well any current measurement of the earth’s emissivity is going to be in the range of 0.9-1.0 so again using 0.82 has really no bearing in the discussion. As an example to this, water and ice which constitute the bulk of the earth’s surface have an emissivity in the range 0.95-0.985.

    Could you produce a reference that supports the use of SB to calculate absorption? Without that you really do not have any foundation to your position.

  226. Nasif Nahle April 16, 2011 at 2:22 am #

    @Neutrino…

    The definition of absorptivity is simple from a measurement perspective
    Absorptivity = Reflectivity – 1

    That is why you got flawed results.

    Answer my questions and you’ll get the truth about absorptivity. Here my questions are:

    Derive the formula from the definition of absorptivity:

    α = [(∫ 0 ∞) (∫ 2π1) α’λ (r, λ) (T, λ, si) Iλ (r, λ, si) cos θi dΩi dλ] / [(∫ 0 ∞) (∫ 2π1) Iλ (r, λ, si) cos θi dΩi dλ]

    Demonstrate that the following equation supports the GHE:

    q/m^2 = 0.82 (5.6697 x 10^-8 W/m^2 K^4) (293 K)^4

    You’re running in circles. Answer those questions; otherwise you’ll insist on your erroneous calculations.

    🙂

  227. mkelly April 16, 2011 at 2:37 am #

    Area of sphere 4(pi)r^2
    area of disc (pi)r^2

    This is some of the GHE radiative transfer problem I object to. You divide by 4 because of the sphere formula but say the sun shines on a disc. You should not use the the sphere formula when there is no sunshine on half.

    SOD has an example on his blog of mesurements taken in Saskachewan(sp) Canada showing a sun radiance of 700 W/m^2 with an absorbtionof .84.

    This would lead to a temperature of 75 C at noon on a clear day. But the actual temperature measured was 40 C. The graphs also show the up and down IR. But it shows that the atmosphere has kept the earth cooler than it otherwise would have been.

    My heat transfer book has a chart of the W/m^2 at various angles for the earth at the surface it starts at 1061 for 90 deg and goes down to 41 for 5 deg. That amount travels with the sun as it goes from 23.5 N to S.

    Lastly no surface can warm itself. The atmosphere get nearly all its energy from the earth so the atmosphere cannot warm the earth.

    The mass of the atmosphere adds heat we know cause we have a lapse rate. And we have something in science called STP, standard temperature and pressure. At 1 atm we have 0 C. So even if the the GHE was involved it would be 15 deg max not 33. If you take positions on the same latitude but different elevations the higher elevation gets more sun light but is cooler in general.

  228. Neutrino April 16, 2011 at 2:48 am #

    Nasif,

    This argument is dead.

    Absorptivity = 1- Reflectivity is standard physics.
    The Stefan-Boltzmann Law is standard physics.

    That I, or anyone else, can use the above to find absorptivity or power/flux emitted is standard physics.
    That the SB is not used for calculating absorption is standard physics. Every link I have provided in this thread documents this(Even the link that you provided did as well).

    I don’t mean this to be an appeal to authority but you are contradicting basic physics without one shred of evidence.

    Unless you can present some modicum of evidence that your view is supported by science I see no point moving forward. I have provided references, now it is your turn.

  229. Nasif Nahle April 16, 2011 at 3:26 am #

    @mkelly…

    Your book is correct on adding the effect of the angle of incidence of solar power on a hemispherical surface. This is the reason by which I introduce only instantaneous variables.

    For example, we had yesterday an average incident solar power flux of 230 W/m^2. The peak was afternoon, at 19 h UT, with a flux of almost 435 W/m^2 (measurements by net radiometers and pyrgeometers), which gives a temperature of 37 °C (311 K). The calculation of temperature coincides with the temperature we had at 19 h UT:

    http://www.timeanddate.com/weather/mexico/monterrey/historic

    Our coordinates are 25º 48´ North latitude and 100º 19′ West longitude, and we are at an altitude of 513 meters above sea level.

    AGW and its GHE are pseudoscience.

    🙂

  230. Graeme M April 16, 2011 at 7:38 am #

    I am not sure Neutrino that I understand why in your answer to me you referred back to your calculation using more accurate figures for the earth. The example I gave was not earth. I used your example values for a GB with emissivity around 0.7 for solar wavelengths(λ 4μm). You and Nasif got different answers, I am trying to find out exactly why without going back through the whole post.

    I also still struggle with the usefulness of averages. Using the moon example, there is no average temperature anywhere at all so it’s a fiction to suggest it, even if it is a useful mathematical method.

    The fact is, the moon’s surface in the sun is very hot, 107C, and in the dark very cold, -153C (if those numbers are correct). So the reality is that without an atmosphere, at the time the sun is shining, the surface is simply very hot. Which is what Nasif suggests. The idea of an average removes the fact.

    The atmosphere and the oceans, regardless of any GHE, must in sum lead to a narrowing of the range of possible temperatures. Perhaps as Nasif suggests, it attenuates the effects of the presence or lack of solar radiation. And if the effect is such that in the presence of solar radiation, the surface is cooler than it would otherwise be, what does that suggest an atmosphere and an ocean does?

  231. Luke April 16, 2011 at 8:18 am #

    “AGW and its GHE are pseudoscience.” Nobody is listening Nasif – you’re some unpublished kook here. Who cares. It’s just noise on the vine.

    I still await your alternative energy balance explanation of a single day of radiometer readings with your “new” theory”. How about a road test. LOL !

  232. Neutrino April 16, 2011 at 8:20 am #

    Graeme,

    I referred you back to my original calculation because the end result(-18.3C) was for an earth absorbing with absorptivity of 0.7 across solar wavelengths and an emissivity of 1.0 across terrestrial wavelengths. The 5.4C result(line 20) is for an earth with uniform emissivity of 0.7 across both spectrums.

    Nasif and I arrive at different answers because he believes that Absorption is governed by the SB equation and I believe it is governed by the simple multiplication of Incident*Absorptivity. Also I think there is a difficulty on his part understanding the geometry of the model.

    Every reference supplied so far in this thread support my conclusion, not Nasif’s.

    Nasif also has no reference that supports his claim that SB calculates absorption. He keeps saying things like scientists agree but never seems to mention which ones or point out where they stated their agreement.

    Yes the atmosphere modifies the temperature, it reduces the swings from high to low. The use of average is a simplification that is useful in calculating totals nothing more. Until an average can be accurately calculated theactual cannot. Why? Because the average is just a simplification of the actual.

  233. Graeme M April 16, 2011 at 8:55 am #

    Thanks Neutrino. So now we have:

    Neutrino – 254.7k (-18.3C)
    Nasif – 305k (32C)

    And this is because Nasif derives absorption from SB while you believe that it is derived from multiplying incident by absorptivity.

    Much clearer, thanks.

  234. Neutrino April 16, 2011 at 9:10 am #

    Yes.

    In fact, Nasif’s calculation of temperature is invariant with respect to solar flux. His temperature is set before he calculates absorption. His calculation for absorbed is only dependant on absorptivity and temperature. This alone should demonstrate he is doing something wrong.

    How would his calculation change if the solar flux doubled? Or halved? The only way it could change is by choosing a new arbitrary T. But since the point of the calculation is to actually calculate T he is in a circular loop. For Nasif to calculate the temperature he must already know the temperature.

  235. cohenite April 16, 2011 at 10:08 am #

    “Absorptivity = Reflectivity – 1″; shouldn’t that be Absorptivity = 1 – Reflectivity? That typo aside the main area of contention [at this stage!] is, as Neutrino states, ” He [Nasif] claims that the SB calculates the absorption(it does not).”

    SB is written as E = σT^4. SoD says it is more accurate to write it as:

    E = εσT^4

    Where ε is the emissivity; this makes sense because the total energy, E, an object has is measured by its emission and temperature; the emission is the energy leaving and the temperature is the energy staying; unless the object has its own heating source, which the sun has, and to a far lessor extent the Earth has, the total energy the object has will be a measure of what it has absorbed or gained from another source[s].

    It would seem therefore that the 2 N’s are talking about different sides of the same coin. Having said that I am still trying to wade through Nasif’s formula for absorptivity:

    α = [(∫ 0 ∞) (∫ 2π1) α’λ (r, λ) (T, λ, si) Iλ (r, λ, si) cos θi dΩi dλ] / [(∫ 0 ∞) (∫ 2π1) Iλ (r, λ, si) cos θi dΩi dλ]

    A bit of help in deciphering that would be nice Nasif!

  236. Neutrino April 16, 2011 at 10:28 am #

    Yes, cohenite.
    Thanks for catching that, I transposed the line. It should be Absorptivity = 1 – Reflectivity.

    It really is not two sides of the same coin. Nasif is claiming that the SB can calculate absorption. Nowhere in the literature does it say that.
    And besides, it is illogical. Absorption is a measure of how much of the Incident is absorbed, as such the Incident must be part of the equation. What Nasif is doing by claiming that SB can calculate absorption is divorcing the Absorbed value from the Incident.

    Technically I believe that the SB is actually just E = σT^4, adding ε to it extends the equation to GB’s as well as the ideal BB theoretical starting point.

  237. el gordo April 16, 2011 at 10:38 am #

    I too await Nasif’s road test and is it be to much to ask that it be presented in simpler language. Otherwise it’s a wink in the dark.

  238. Neutrino April 16, 2011 at 10:49 am #

    I am posting this again to just emphasize how odd Nasifs claims are:

    Total surface are of the earth: 5.11*10^14m^2
    Total area earth blocks of solar radiation: 1.28*10^14m^2
    Flux at earth distance: 1368W/m^2
    Flux Absorbed at earth: 342.6W/m^2

    The total power the earth intercepts is then the area the earth blocks * the flux at the earth distance.
    And the total power absorbed is the total surface area of earth * absorbed flux.

    Power Incident = 1.28*10^14m^2 * 1368W/m^2 = 1.748*10^14W
    Power Absorbed = 5.11*10^14m^2 * 342.6W/m^2 = 1.751*10^14W

    So unless Nasif has somehow changed the geometry of the earth he has it Absorbing more than it has as Incident from the sun.

  239. Nasif Nahle April 16, 2011 at 11:01 am #

    @Neutrino…

    You continue saying lies on me:

    “He [Nasif] claims that the SB calculates the absorption(it does not).”

    Show me where I said that.

    Don’t start trolling, please.

    I ask you again:

    Here the definition of absorptivity:

    α = [(∫ 0 ∞) (∫ 2π1) α’λ (r, λ) (T, λ, si) Iλ (r, λ, si) cos θi dΩi dλ] / [(∫ 0 ∞) (∫ 2π1) Iλ (r, λ, si) cos θi dΩi dλ]

    Derive the formula. And…

    Demonstrate that the following equation supports the GHE:

    q/m^2 = 0.82 (5.6697 x 10^-8 W/m^2 K^4) (293 K)^4

    😀

  240. Nasif Nahle April 16, 2011 at 11:03 am #

    @Neutrino…

    Now you are trying to place your ignorance on me. You say that I calculated:

    Power Incident = 1.28*10^14m^2 * 1368W/m^2 = 1.748*10^14W
    Power Absorbed = 5.11*10^14m^2 * 342.6W/m^2 = 1.751*10^14W

    It is you who are doing those erroneous calculations.

    😀

  241. Nasif Nahle April 16, 2011 at 11:06 am #

    @Cohenite…

    A bit of help in deciphering that would be nice Nasif!

    It is the definition of absorptivity. I’m still waiting for Neutrino derives the formula.

    NSN

  242. Nasif Nahle April 16, 2011 at 11:09 am #

    @Neutrino…

    This reveals your true face, Neutrino. You say:

    Nasif also has no reference that supports his claim that SB calculates absorption. He keeps saying things like scientists agree but never seems to mention which ones or point out where they stated their agreement.

    Show the readers where I said that.

    Again,

    Here the definition of absorptivity:

    α = [(∫ 0 ∞) (∫ 2π1) α’λ (r, λ) (T, λ, si) Iλ (r, λ, si) cos θi dΩi dλ] / [(∫ 0 ∞) (∫ 2π1) Iλ (r, λ, si) cos θi dΩi dλ]

    Derive the formula. And…

    Demonstrate that the following equation supports the GHE:

    q/m^2 = 0.82 (5.6697 x 10^-8 W/m^2 K^4) (293 K)^4

    Sorry, Neutrino, but if you cannot answer simple questions, don’t lie about what I’m saying.

    😀

  243. Nasif Nahle April 16, 2011 at 11:21 am #

    @elgordo…

    I too await Nasif’s road test and is it be to much to ask that it be presented in simpler language. Otherwise it’s a wink in the dark.

    In simple words, the amount of solar power that the Earth receives on top of its atmopshere is 1368 W/m^2.

    However, the amount of solar power striking on the surface of the Earth is not 1368 W/m^2, but 342 W/m^2.

    What does it mean? That the atmosphere, far from heating up the Earth, it cools it by attenuating the amount of energy that strikes on the Earth’s surface. From this site, http://www.windows2universe.org/earth/climate/sun_radiation_at_earth.html we can read:

    The surface of a sphere has an area four times as great as the area of a disk of the same radius. So the 1,368 W/m2 is reduced to an average of 342 W/m2 over the entire surface of our spherical planet.

    And you can read the average Earth’s temperature, which coincides exactly with my calculations, from:

    http://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html

    http://hyperphysics.phy-astr.gsu.edu/hbase/solar/soldata2.html#c3

    And many other sites.

    I hope this answers your question.

    😀

  244. Nasif Nahle April 16, 2011 at 11:25 am #

    @elgordo…

    The mathematical definition of absorptivity is a test to Neutrino, so I cannot tell you what the derived formula is, until he derives it.

    Sorry.

    NSN

  245. Neutrino April 16, 2011 at 12:00 pm #

    Nasif,

    Yes to be fair you have never actually explicitly said that the SB calculates absorption. But what you have repeatedly done is use the SB to calculate absorption.

    Evidence of this is:

    Nasif Nahle April 14th, 2011 at 7:38 am
    As for your question, the flux of power absorbed by the surface, introducing real magnitudes:
    Q/m^2= 0.82 * (5.6697 x 10^-8 W/m^2 K^4) (293 K)^4 = 342.62 W/m^2
    Notice the difference? What greenhouse effect?

    Clearly states you are calculating absorbed, the above is the SB equation.

    Nasif Nahle April 14th, 2011 at 8:58 am
    You just multiplied the incorrect input of 341.5 W/m^2 by 0.82. There is no violation to the first law neither to the second law. Again, the correct procedure is as follows:
    Q/m^2= e * (s) (T)^4 = 0.82 (5.6697 x 10^-8 W/m^2 K^4) (293 K)^4 = 342.62 W/m^2
    Q/m^2= a * (s) (T)^4 = 0.82 (5.6697 x 10^-8 W/m^2 K^4) (293 K)^4 = 342.62 W/m^2

    Appears from above you are calculating first emitted then absorbed using the same equation, the above is the SB equation.

    Nasif Nahle April 15th, 2011 at 3:44 pm
    Demonstrate that the following equation supports the GHE:
    q/m^2 = 0.82 (5.6697 x 10^-8 W/m^2 K^4) (293 K)^4
    It is the formula to calculate the amount of energy absorbed by the surface introducing real measurements.

    Clearly says the above formula is calculating absorbed, the above is the SB equation.

    Nasif Nahle April 15th, 2011 at 4:18 pm
    T = (q / σ * α)^(1/4) = [(342 W/m^2) /(5.6697 x 10^-8 W/m^2 K^4 * 0.82)]^(1/4) = 292.86 K, or 293 K by rounding up the number.
    As you can see, the atmosphere attenuates the intensity of the flux of power from the Sun, and the average temperature of 293 K is produced by the solar power striking on the surface alone. It is not the product of any GHE, it is imaginary.

    Implies that the T is calculated from the Incident flux, the above is the SB equation.

    Please so I do not misrepresent you: Do you believe the SB calculates absorption? Yes/No
    If No, what are you doing in the above quotes?

  246. el gordo April 16, 2011 at 12:11 pm #

    Thanks Nasif, it gives me something to take away and ponder.

  247. Neutrino April 16, 2011 at 12:16 pm #

    It is a pointless test. Absorption can be easily calculated from Incident and Reflected components. And that is how every link supplied here has done it. If Nasif objects to me using a measured Absorptivity then he should also object to Nasif using a measured Absorptivity.

    Nasif Nahle April 12th, 2011 at 7:33 am
    You say it is 0.7, the number obtained by direct observation and measurements is 0.82, which is the number I gave. Which number is higher, 0.82 or 0.7? You wrote:

    Nasif Nahle April 13th, 2011 at 1:02 am
    Here the things start incorrect. The Sun has not an emissivity of 1.0, but 0.9875, as I have been saying along this thread. On the other hand, the Earth’s total emissivity is not 0.7, but 0.82; both emissivities has been obtained from measurements.

    Nasif Nahle April 15th, 2011 at 8:05 am
    That’s the reason by which the authors obtained a wrong result, despite their arithmetics is okay, from their calculations. The Earth is not a blackbody and the absorptivity measured with appropiate technology gave a result of 0.82. They made the same mistake than you did but theirs was not so evident. Period!

    Nasif Nahle April 15th, 2011 at 4:18 pm
    α or Absorptivity and emissivity of the Earth (measured and averaged) = 0.82

    So I have to ask, what is the source for the 0.82?

  248. Nasif Nahle April 16, 2011 at 12:25 pm #

    @Neutrino…

    You say:

    Evidence of this is:

    Nasif Nahle April 14th, 2011 at 7:38 am
    “As for your question, the flux of power absorbed by the surface, introducing real magnitudes:
    Q/m^2= 0.82 * (5.6697 x 10^-8 W/m^2 K^4) (293 K)^4 = 342.62 W/m^2
    Notice the difference? What greenhouse effect?”

    Clearly states you are calculating absorbed, the above is the SB equation.

    Nope! I’m not calculating absorptivity. I am using the formula as it must be. What I am calculating is absorbed Q/m, which is absorbed flow of power, not absorptivity, unless you cannot understand how to read formulas.

    You say:

    Nasif Nahle April 14th, 2011 at 8:58 am
    “You just multiplied the incorrect input of 341.5 W/m^2 by 0.82. There is no violation to the first law neither to the second law. Again, the correct procedure is as follows:
    Q/m^2= e * (s) (T)^4 = 0.82 (5.6697 x 10^-8 W/m^2 K^4) (293 K)^4 = 342.62 W/m^2
    Q/m^2= a * (s) (T)^4 = 0.82 (5.6697 x 10^-8 W/m^2 K^4) (293 K)^4 = 342.62 W/m^2”

    Appears from above you are calculating first emitted then absorbed using the same equation, the above is the SB equation.

    What I wrote is quite differernt from what you said that I had written; you said:

    Power Incident = 1.28*10^14m^2 * 1368W/m^2 = 1.748*10^14W
    Power Absorbed = 5.11*10^14m^2 * 342.6W/m^2 = 1.751*10^14W

    And I never made those calculations, but you.

    You say:

    Appears from above you are calculating first emitted then absorbed using the same equation, the above is the SB equation.

    You’re again dismissing Kirchhoff’s Law… Heh!

    You say:

    Nasif Nahle April 15th, 2011 at 3:44 pm
    “Demonstrate that the following equation supports the GHE:
    q/m^2 = 0.82 (5.6697 x 10^-8 W/m^2 K^4) (293 K)^4
    It is the formula to calculate the amount of energy absorbed by the surface introducing real measurements.”

    Clearly says the above formula is calculating absorbed, the above is the SB equation.

    Which introduces emissivity because the Earth is not a blackbody (which don’t exist) and has a limited absorptivity. From Pitts and Sissom’s book, page 304:

    “Since we are considering gray surfaces, E may be expressed as eσT^4…” 😀

    Consequently, your argument is false. Besides, you ignore the side the terms are and what we are calculating in a formula. The left side of the formula, Neutrino, is what we are calculating. The right side of the formula is the known variables and constants. Heh!

    In the formula that I used, q/m^2 is the power flux, not the absorptivity. To calculate absorptivity, we must calculate it considering some variables and constants. Derive the formula from the definition of absorptivity:

    α = [(∫ 0 ∞) (∫ 2π1) α’λ (r, λ) (T, λ, si) Iλ (r, λ, si) cos θi dΩi dλ] / [(∫ 0 ∞) (∫ 2π1) Iλ (r, λ, si) cos θi dΩi dλ]

    You say:

    Nasif Nahle April 15th, 2011 at 4:18 pm
    “T = (q / σ * α)^(1/4) = [(342 W/m^2) /(5.6697 x 10^-8 W/m^2 K^4 * 0.82)]^(1/4) = 292.86 K, or 293 K by rounding up the number.
    As you can see, the atmosphere attenuates the intensity of the flux of power from the Sun, and the average temperature of 293 K is produced by the solar power striking on the surface alone. It is not the product of any GHE, it is imaginary.”

    Implies that the T is calculated from the Incident flux, the above is the SB equation.

    Are you implying something else in your mind? The formula is for calculating the temperature caused by a given load of energy. So you see that it is incorrect? 😀

    You say:

    Please so I do not misrepresent you: Do you believe the SB calculates absorption? Yes/No
    If No, what are you doing in the above quotes?

    Not only misrepresenting me, but lying on me.

    Again, the Stefan-Boltzmann, as it is written, calculates the power flux, whatever it is what you “believe” or not.

    I have answered your nonsensical questions. Now, answer my two simple questions:

    Here the definition of absorptivity:

    α = [(∫ 0 ∞) (∫ 2π1) α’λ (r, λ) (T, λ, si) Iλ (r, λ, si) cos θi dΩi dλ] / [(∫ 0 ∞) (∫ 2π1) Iλ (r, λ, si) cos θi dΩi dλ]

    Derive the formula. And…

    Demonstrate that the following equation supports the GHE:

    q/m^2 = 0.82 (5.6697 x 10^-8 W/m^2 K^4) (293 K)^4

    😀

  249. Neutrino April 16, 2011 at 12:30 pm #

    In this case simple is incorrect:

    In simple words, the amount of solar power that the Earth receives on top of its atmopshere is 1368 W/m^2.
    However, the amount of solar power striking on the surface of the Earth is not 1368 W/m^2, but 342 W/m^2.
    What does it mean? That the atmosphere, far from heating up the Earth, it cools it by attenuating the amount of energy that strikes on the Earth’s surface.

    This missing the point of the solar flux and the reduction of it by one quarter. Perhaps it is just a problem of not understanding the geometry.

    This link from MIT has a graphic at the top of page 5 that may help.

    The 1368W/m^2 is not attenuated by the atmosphere(in this model we are discussing there is no atmosphere). The 1368W/m^2 of solar flux from the disc is averaged over the 4xlarger surface area of the sphere and becomes 342W/m^2.

  250. Debbie April 16, 2011 at 12:30 pm #

    Luke,
    Could that possibly be because the measuring instruments and resultant computer modelling are ‘calibrated’ or ‘assuming’ that the emissivity of CO2 is 1?
    As you repeatedly point out, I’m not a scientist, but that is one possible answer to your question when asked in relation to the discussion that is going on here.
    Just an observation.

  251. Nasif Nahle April 16, 2011 at 12:32 pm #

    @Neutrino…

    You say:

    It is a pointless test. Absorption can be easily calculated from Incident and Reflected components. And that is how every link supplied here has done it. If Nasif objects to me using a measured Absorptivity then he should also object to Nasif using a measured Absorptivity.

    It would be pointless if you would know how to do it. The reason you don’t answer it is because you know that your arguments are against science and by doing it you, by yourself, will be demonstraring that AGW and its GHE are pseudoscience.

    Answer the questions…

    😀

  252. Nasif Nahle April 16, 2011 at 12:33 pm #

    @Neutrino…

    This missing the point of the solar flux and the reduction of it by one quarter. Perhaps it is just a problem of not understanding the geometry.

    LOL!!!

  253. Neutrino April 16, 2011 at 12:34 pm #

    “Please so I do not misrepresent you: Do you believe the SB calculates absorption? Yes/No
    If No, what are you doing in the above quotes?
    Not only misrepresenting me, but lying on me.
    Again, the Stefan-Boltzmann, as it is written, calculates the power flux, whatever it is what you “believe” or not.

    The question was about Absorbed flux:
    Yes or No does the SB calculate the Absorbed flux?

    From the quotes provided it appears the answere is Yes.

  254. cohenite April 16, 2011 at 1:11 pm #

    How is absorbed flux different from power flux?

  255. Nasif Nahle April 16, 2011 at 1:31 pm #

    @Cohenite…

    It’s Neutrino’s confusion. First he said that I was calculating absorptivity; now he says I’m calculating “absorbed flux”… I’m calculating the absorbed power, which can be translated to energy, not the flux of power or the absorptivity.

    Absorptivity is calculated through a well developed formula that Neutrino doesn’t know, so he’s distracting the dialogue to other issues.

    First he must answer my questions. He’s defending his essay, not me.

    NSN

  256. Nasif Nahle April 16, 2011 at 1:35 pm #

    @Neutrino…

    The question was about Absorbed flux:
    Yes or No does the SB calculate the Absorbed flux?

    No! I’m not calculating absorbed flux, either SB equation calculates absorbed flux. SB equation is used to calculate the absorbed power.

    🙂

  257. Neutrino April 16, 2011 at 2:18 pm #

    Technical note:
    Power is a measure in Watts(W)
    Flux is a measure in Watts/meter^2(W/m^2)

    All of Nasif’s calculations have been in W/m^2 so they are flux’s.
    Whereas all my original calculations were in W’s so they were power’s.

    To switch between them just multiply or divide by the surface area involved.

    The SB as written εσT^4.A is calculating a power and εσT^4 is calculating a flux. In both cases it is an emitted number.

    The question was always about absorption, power or flux it doesn’t matter. It was never about absorptivity.

    You just wrote: (my bold)
    SB equation is used to calculate the absorbed power.

    Thank you Nasif for clearly saying that you are using the SB to calculate Absorbed power.(technically flux, but close enough)

    You quoted me and added:(my bold)
    “Please so I do not misrepresent you: Do you believe the SB calculates absorption? Yes/No
    If No, what are you doing in the above quotes?
    Not only misrepresenting me, but lying on me.

    So how is it that I was lying?

  258. Neutrino April 16, 2011 at 2:21 pm #

    Note about the geometry:
    The flux that is incident on earth is for a plane yet when we calculate the average incident/absorption/emission for the earth surface we are doing so from a sphere. To account for this the Solar flux is divided by 4, this is the ratio of the area of a disc to a sphere.

    This can be seen from the diagram at MIT(page 5) and the following caption.
    Figure 4: The spinning Earth is imagined to intercept solar flux over a disk
    and radiate terrestrial energy away isotropically from the sphere. Modified from
    Hartmann, 1994.

    As well as the from a diagram from Harvard(fig 7-9) and the following text.
    This solar radiation flux FS is intercepted by the Earth over a disk of cross-sectional area pRE2 representing the shadow area of the Earth ( Figure 7-9 ). … The mean solar radiation flux absorbed per unit area of the Earth’s surface is FSpRE2(1-A)/4pRE2 = FS(1-A)/4.

  259. cohenite April 16, 2011 at 3:01 pm #

    Neutrino, your distinction between a flux and power is correct but the issue [for me] is that the flux is an averaged amount; the problem with that in determining the radiative properties of the Earth have been dealt with in this paper:

    http://pielkeclimatesci.files.wordpress.com/2009/10/r-321.pdf

    The problem may be summed up thus: (A + B)^4 > A^4 + B^4; that is applying SB to the Earth’s emissivity will be different depending on whether all points are added and averaged and then cubed or whether each point is cubed first and then added. The differences can be huge and represent the primary problem with Arthur Smith’s attempt to prove a greenhouse temperature.

    If I understand Nasif right he is concerned with A^4 + B^4 while you are dealing with (A + B)^4.

  260. Nasif Nahle April 16, 2011 at 3:02 pm #

    @Neutrino…

    Hey! You’re forgetting to answer my questions. You cannot continue if you don’t answer my questions. It is supposed you’re defending your essay, not me.

    Answer my questions, Neutrino… Whatever you say about MIT, Harvard, etc. is not what I’m asking you. Answer my questions!

    😀

  261. Nasif Nahle April 16, 2011 at 3:43 pm #

    The problem with AGW fantasy is that they are reasoning in a very, very wrong way.

    The measured flux of solar power striking on top of the Earth’s atmosphere is 1368 W/m^2. That’s a measurement made by satellite instruments. It’s reality:

    http://www.eoearth.org/article/Solar_radiation

    From Materials Technology Limited:

    http://www.drb-mattech.co.uk/uv%20map.html

    “Solar Constant – The solar constant is the amount of energy received at the top of the Earth’s atmosphere on a surface oriented perpendicular to the Sun’s rays (at the mean distance of the Earth from the Sun). The generally accepted solar constant of 1368 W/m2 is a satellite measured yearly average.”. Bolds are mine.

    From NASA site:

    http://earthobservatory.nasa.gov/Features/SORCE/sorce_03.php

    If that were all there was to the Earth’s radiative balance, scientists studying the Sun would have probably long since moved on to another climate-related problem. Analyzing the Sun and its affects on climate, however, is further complicated by the fact that the amount of radiation arriving from the Sun is not constant. It varies from the average value of the TSI—1,368 W/m2—on a daily basis. Bolds are mine.

    However, AGW proponents have changed this measurement and say that the energy received on top of the Earth’s atmosphere is 342 W/m^2. And the fantasy starts here because they say that the solar irradiance striking the surface is 161 W/m2:

    http://chriscolose.files.wordpress.com/2010/03/kiehl4.jpg

    However, the average of measured flux of solar power on the surface is 342 W/m^2!!! Not 161 W/m^2.

    How AGWers could explain that here, in Monterrey, we are receiving around 400 W/m^2 of insolation at 19 h UT???

    This is the origin of AGW fantasy. They change measurements, variables, places, constants, etc., and say that the temperature of the Earth caused by the solar radiation is 20) T = 278.6K (or 5.4C)

    😀

  262. wayne April 16, 2011 at 8:40 pm #

    Nasif, hate to correct here, I too disagree as you do with most of AGW silly science, but, you said:
    “How AGWers could explain that here, in Monterrey, we are receiving around 400 W/m^2 of insolation at 19 h UT???”

    That is the same reading, 400 W/m^2, if is was noon and your latitude was between 36ºN and 37ºN at noon. But you must be further south for it 19 h UT so the angle to the sun would be the same (ignoring eccentricy and the exact day). What I am saying is, the 161 is a cosine weighted mean. The real reading measured at a point and time, of course, is always much greater except when the sun is low near the horizon. At noon directly under the sun you have to take that 161 and multiply by pi. Any thing else also cosine adjust from the zenith. Just didn’t want you to get of track by forgetting a factor.

    ( and I know, that 161 W/m^2, though maybe accurate (probably not but close), is a very mangled number indeed that has no tie to anything you or I here on this real Earth would ever experience of even measure without undoing all of the averaging and smearing around to and over the night time too, for T&K’s diagram’s Earth is lit from all sides all of the time!! That would be a weird sun wouldn’t it, shaped like a hoola-hoop, ) ☺

  263. Louis Hissink April 16, 2011 at 9:01 pm #

    It might be useful to consider other sources of energy into the earth-system than the obvious radiation one, the electromagnetic route(EM).

    Limiting discussion to radiation results in rhetorical excesses, but if the EM route is added, new insights become possible. Using EM principles the observed thermal imbalance is explicable. Ignoring it causes the proposition of a greenhouse gas effect.

  264. Luke April 16, 2011 at 9:26 pm #

    Gee Wayne – isn’t it incredible that climate modellers haven’t thought of that ! Wow. They’re just so silly.

  265. wayne April 16, 2011 at 9:39 pm #

    Errata:

    Sorry Nasif, not pi, how about four… but you should get my basic point. I’ve tried before to work that whole process backwards using K&T’s figures to what an actual reading should be at some given point on the globe so comparison could be made, cosine adjusted, and can’t seem to get it, but, that is a good question and point!

  266. Neutrino April 17, 2011 at 12:33 am #

    Cohenite,

    Yes I agree the using the average temperature can be problematic. It is the fourth power of temperature that is really averaged to get total power emitted.

    As an example to this effect:
    T______T^4
    10_____10000
    11_____14641
    9_______6561
    12_____20736
    8_______4096
    Averages of above:
    10_____11207

    What the above shows is that the fourth power is taken of the average T then the result(first line) will be to low from the actual averaged powers(last line).

    Using the average temperature will always underestimated the total power. As such the calculated temperature of -18.3C(by myself and every reference provided in this thread) is the maximum that a sphere could have and be radiating the required power(1.22*10^17W) to stay in equilibrium. If there are any variations in the temperature around the mean(which there will be) then the average temperature must be lower than that to still maintain the same total emitted power.

  267. Nasif Nahle April 17, 2011 at 12:42 am #

    @Wayne…

    Thanks for your comment.

    Yes, you are correct. The number I reported is an instantaneous measurement on ground at 19 h UT.

    I don’t know if K&T made a mistake or if they did it intentionally, but they took the value of the energy striking the surface as if it were the energy on top of the atmosphere. Then they went on discounting the atmospheric effects until making that the energy stiking on the surface was 161 W, which is absolutely false.

    NSN

  268. Neutrino April 17, 2011 at 12:55 am #

    In regards to the geometry:

    To expand on what wayne said.

    The 1368W/m^(which is both a measured and calculated value) is the flux at the earth. As such only one point on the globe will be receiving that at any given time, everywhere else will have a reduced flux.

    The point directly underneath the sun(ignoring tilt for this description) will be noon on the equator. Every other point on the day side of the planet will be getting a portion of that 1368W/m^2. To actually calculate the correct flux the cosine of the angle relative to the center point is used.

    So*cos(ө) = local flux.

    At the center cos(0) = 1 and that point gets the full flux.
    At the terminus or edge cos(90) = 0 and that point gets zero flux.
    All points inbetween receive between 1368W/m^2 and zero.

    The average of all points on the daylight hemisphere is going to be 1368W/m^2 / 2 or 684W/m^2.

    But of course the planet has two sides, both the daylight and nighttime, so the overall average flux would have to be further divided by 2 to get the previously mentioned 342W/m^2.

    This is the average flux at TOA for every place on earth. It is not the flux you could go outside and measure at any given time of day or night. If you were to make continual measurement day night, summer winter at each point on earth and then average them together this is the number you would come up with.

    The KT budget has this flux correct, it is then attenuated by both the being absorbed and reflected in the atmosphere. The surface again reflects a portion with only a final 161W/m^2 absorbed at the surface.

  269. Nasif Nahle April 17, 2011 at 1:14 am #

    The biggest lie said is that the average of solar energy on top of the atmosphere is 342 W/m2. 342 W/m2 is the average of the solar power striking on the surface, not on top of the atmosphere.

    And the lies continue by double discounting the effect of the atmosphere by attenuating the energy incoming from the Sun.

    That is the reason AGW always has unreal frozen planets.

    The sources are clear:

    The measured flux of solar power striking on top of the Earth’s atmosphere is 1368 W/m^2. That’s a measurement made by satellite instruments. It’s reality:

    http://www.eoearth.org/article/Solar_radiation

    From Materials Technology Limited:

    http://www.drb-mattech.co.uk/uv%20map.html

    “Solar Constant – The solar constant is the amount of energy received at the top of the Earth’s atmosphere on a surface oriented perpendicular to the Sun’s rays (at the mean distance of the Earth from the Sun). The generally accepted solar constant of 1368 W/m2 is a satellite measured yearly average.”. Bolds are mine.

    The solar power striking on top of the atmosphere is 1368 W/m2 (measured) and it is attenuated by the atmosphere making the surface receives an average of 342 W/m2 (also measured).

    Those calculations to make the solar energy coincides with AGW idea is pure, pure pseudoscience.

    NSN

  270. Mack April 17, 2011 at 2:27 am #

    Nasif has pointed out that the model Neutrino is talking about is right there in front of us,ie the moon.
    You guys seem to know what the temps are each side of the moon so why don’t you just multiply them up proportionately to the size of the earth and then do whatever the hell you want to do to call that an average.
    Or is this being totally oversimpliistic.

  271. Nasif Nahle April 17, 2011 at 2:32 am #

    @Mack

    No, it’s not simplistic; on the contrary, it is a reliable procedure.

    I have writen a brief on the errors of AGW idea, so you can see whay I affirm, with solid basis, that AGW is pseudoscience. Here it goes:

    Pseudoscientific assumptions and procedures on AGW idea:

    1. Real world says that the annual average energy received on a grid of 1000 squares of top of the atmosphere is 1368 W/m^2.

    AGW uses some intricate forced geometric calculations that CANNOT BE USED ON AVERAGES and gives a total amount of energy on top of the atmosphere of 342 W/m^2.

    2. Real world says that the average of the energy received by the surface is 342 W/m^2.

    AGW takes the average energy striking on the surface -not far enough very high above our heads on top of the atmosphere- as if it was the energy received on top of the atmosphere. With this, AGW says that the energy emitted by the Sun is attenuated through the space between the Earth and the Sun from 9.66 x 10^25 W to 342 W/m^2, on the very top of the Earth’s atmosphere.

    3. Real world says that the energy received on top of the atmosphere incomes through the atmosphere and is attenuated by the atmosphere from the average of 1368 W/m^2 down to the average of 342 W/m^2.

    AGW says that the average of solar energy received on top of the atmosphere is attenuated by the angle of incidence from 1368 W/m^2 to 342 W/m^2 BEFORE IT ENTERS THE ATMOSPHERE!

    4. Real world says that the energy striking on the surface varies depending on the angle of incidence of the solar radiation and that by the rotation of the Earth, the flux of solar power is continuous. Therefore, the average flux of solar power striking on the surface, after attenuation by the atmosphere, is 342 W/m^2, which causes an average global temperature of ~20 K. No need to invent a nonexistent greenhouse effect.

    AGW says that the average of energy received by the surface of the Earth, after attenuated by the atmosphere, is 161 W/m^2, which causes an average global temperature of – 42.3 °C. They have to invent an inexistent recycling of heat in the atmosphere, a fabulous storage of heat by the atmosphere, a multiplication of the energy in the atmosphere and, what is more weird, they have to invent a blanket covering the whole Earth… These constitute the pseudoscience of AGW and lukeAGW.

    Nasif S. Nahle

  272. Alan D McIntire April 17, 2011 at 2:48 am #

    As I stated previously on another post, I’m a lukewarmer, but Nasif Nahle has a point regarding temperatures at various points on earth being above freezing regardless of a greenhouse effect.

    Wiith no greenhouse, a flux of 1368 watts/m^2, and an ocean covered surface with an emissivity of 0.95:

    At sunrise, the sun’s angle is 0 degrees, at noon 90 degrees, at sunset 0 degrees again, so the amount of radiation the equator is getting is 1368 watts/meter^2 cos x. Integrate that cos x from sunrise, pi/2, to sunset, -pi/2, and you get 2/pi for the average radiation. Halve that for the nighttime period where the equator gets no radiation from the sun, and you get an average surface flux of
    1368/3.14159 watts/m^2, divide that by 0.95 for average ocean emissivity, and you get an effective average temperature of (458.366/390.7)^0.25*288K = 299.73 K with NO greenhouse effect.

    Half of the surface area of the earth is within 30 degrees of the equator. Take a point
    30 degrees from the equator, and you’d get a flux of Cos 30 degrees* flux at equator, for
    a temperature of (0.866*458.366/390.7)^0.25* 288K = 289.14 K, still above 288K, the presumed average for the earth.

    Take a point 45 degrees from the equator, at the latitude of southern France, and you get
    (0.7071*458.366/390.7)^0.25 *288 =274.86 K, barely above freezing.

    Even with NO greenhouse effect, over 70% of an ocean covered earth would be above 0 C.

  273. Nasif Nahle April 17, 2011 at 4:02 am #

    @Alan D McIntire…

    I’m not a lukewarmer; I am a scientist.

    With no greenhouse effect, the reality is that the average temperature of the Earth’s surface is 293 K, or 20 °C.

    Without atmosphere, well… see at the Moon.

    NSN

  274. Neutrino April 17, 2011 at 5:29 am #

    Alan,

    I agree even with the no atmosphere model some parts of the sphere would be above freezing.

    I also agree the integration of the flux around the equator is So/π but there are a few errors how you calculate your temperature. Just looking at water.

    The average Incident flux in the band around the equator,
    Q =1368W/m^2 / π
    Q =435W/m^2

    Absorbed flux in this band, (using KT09 average surface absorptivity, 0.875)
    q = 435W/m^2 * Absorptivity
    q = 435W/m^2 * 0.875
    q = 381W/m2

    Emitted flux is equal to Absorbed flux.

    Temperature of the band emitting above flux,
    T = (p / εσ)^.25
    T = (381W/m^2 / (0.95*5.67*10^-8W/m^2K^4))^.25
    T = 290K = 16.9C

    For a band at latitude 30degrees you are correct another scaling factor of 0.866, cos(30), needs to be applied to the incident.(about Sydney Australia)

    The average Incident flux in the band around the 30degree latitude,
    Q = 1368W/m^2 * 0.866 / π
    Q = 377W/m^2

    Absorbed flux in this band,
    q = 377W/m^2 * Absorptivity
    q = 377W/m^2 * 0.875
    q = 330W/m2

    Emitted flux is equal to Absorbed flux.

    Temperature of the band emitting above flux,
    T = (p / εσ)^.25
    T = (330W/m^2 / (0.95*5.67*10^-8W/m^2K^4))^.25
    T = 280K = 6.6C

    For a band at latitude 45degrees another scaling factor of 0.7071, cos(45), needs to be applied to the incident.(Southern France)

    The average Incident flux in the band around the 45degree latitude,
    Q = 1368W/m^2 * 0.7071 / π
    Q = 308W/m^2

    Absorbed flux in this band,
    q = 308W/m^2 * Absorptivity
    q = 308W/m^2 * 0.875
    q = 269W/m2

    Emitted flux is equal to Absorbed flux.

    Temperature of the band emitting above flux,
    T = (p / εσ)^.25
    T = (269W/m^2 / (0.95*5.67*10^-8W/m^2K^4))^.25
    T = 266K = -7.2C

    Yes the equator will be above freezing but southern France would be an ice cube. Sydney about in the middle of those two ranges.

    The freeze line for just water would occur at about 38.1degrees which is right around Washington DC.

    For a band at latitude 38degrees another scaling factor of 0.787, cos(38), needs to be applied to the incident.

    The average Incident flux in the band around the 45degree latitude,
    Q = 1368W/m^2 * 0.787 / π
    Q = 343W/m^2

    Absorbed flux in this band,
    q = 343W/m^2 * Absorptivity
    q = 343W/m^2 * 0.875
    q = 300W/m2

    Emitted flux is equal to Absorbed flux.

    Temperature of the band emitting above flux,
    T = (p / εσ)^.25
    T = (300W/m^2 / (0.95*5.67*10^-8W/m^2K^4))^.25
    T = 273K = 0.0C

    Using the overall earth Absorptivity of 0.7 would lower all the above numbers by about 14C each.

  275. Nasif Nahle April 17, 2011 at 5:34 am #

    No matter how AGW distorts numbers and torture mathematics, the reality (measurements) is that the surface average temperature of the Earth is 293 K (20 °C), and the average solar power on top of the atmosphere is 1368 W/m2.

    😀

  276. Neutrino April 17, 2011 at 6:22 am #

    I really do not know how to address Nasif’s misconceptions.

    That at the distance the earth is from the sun, 1.50*10^11m, there is a flux of 1.37*10^3W/m^2 is both measured and calculated.

    That the profile of the earth(what it blocks of the suns output) is a cross section of πR^2, 1.28*10^14m^2, is also a completely uncontroversial statement. It’s just geometry.

    Combining these two numbers we can calculate the total power that the earth intercepts,
    1.28*10^14m^2 * 1.37*10^3W/m^2 = 1.72*10^17W

    That the earth has a total surface area of 4πR^2, 5.10*10^14m^2, is also simple geometry.

    Dividing the total power intercepted by the total surface area logically gives us the average flux for the earth.
    1.72*10^17W / 5.10*10^14m^2 = 342W/m^2

    There is no magic, sleight of hand, attenuation from atmosphere or angle of incidence in this number. It is simply the average over the whole earth surface of the total power the earth receives.

    The 1.37*10^3W/m^2 value is what the earth receives on its daylight side, averaging that over the entire globe is both useful(allows us to calculate average temperature) and appropriate(does not change the total energy budget).

    As well he continually maintains that the earths average temperature is 20C when all organizations measureing that number put it at 15C.

    Reference NASA for all the above numbers.

  277. Graeme M April 17, 2011 at 7:51 am #

    Neutrino, I have no maths background to really follow much of this, but I do have a question. I understand why you are dividing by 4, as you say simple geometry. But why is that valid?

    What is the solar power at any point on a flat disk located one kilometer away from the TOA? Is that the 1368 number? Regardless, let’s assume it is. That is the power at any point one wishes to measure it. Thus if I measure it at 1mm above the TOA it will be the same at any location, regardless of the shape of the TOA, surely.

    Once it enters the atmosphere it is affected by such things as angles of incidence, reflectivity, absorption, whatever.

    But what AT the TOA do we need to make a calculation? What happened in that 1mm?

  278. Bryan April 17, 2011 at 8:20 am #

    Neutrino

    From your own highly reduced asumptions;

    Emitted flux is equal to Absorbed flux.

    Temperature of the band emitting above flux,
    T = (p / εσ)^.25
    T = (381W/m^2 / (0.95*5.67*10^-8W/m^2K^4))^.25
    T = 290K = 16.9C

    Now enters water vapour => clouds, raindrops => real radiating surfaces that emit across all wavelengths around their characteristic temperature.
    All factors adding to the real insulating properties of our atmosphere and oceans.

    No need to build up a narrative that relies on the IR line spectral properties of CO2 which compose 0.03% of the atmosphere in the so called greenhouse effect.

  279. Mack April 17, 2011 at 8:47 am #

    Thankyou Nasif,
    Well that about sums it up.
    btw. I wonder what Neutrino was doing yesterday.
    Taking recordings and measurements you think? 🙂

  280. Neutrino April 17, 2011 at 8:47 am #

    Graeme,

    What you say is almost exactly right,

    Pretend as you say the flat disk 1km in front of the earth, just large enough to completely block any sunlight from the earth.

    Yes the flux at any point on that disc is 1.37*10^3W/m^2, the solar constant(So).

    This flux is not strictly speaking the ToA flux.
    Since the area below this disc is actually 2x larger a straight translation of it to every point on the hemisphere would be a massive increase to the total power hitting the earth.
    To avoid this error the So is divided by 2 to get the average ToA for each point under the disc while perserving the total power.

    Averaging the ToA for the entire sphere the flux becomes So/4.

    If you want to know exactly what is the flux at any given location then you will need to do some math with cosines and such to get the angles of incidence to see how much that flux spreads out.

    Strictly speaking the above numbers for hemisphere and globe are outside any effects of the atmosphere, they are all ToA values. Any effect of the atmosphere to attenuate the ToA flux gets applied after the ToA calculation.

    Its just the total solar power intercepted by earth averaged over the entire globe.

    I hope that helps.

  281. Neutrino April 17, 2011 at 8:48 am #

    Bryan,

    Actually No. We do not need to know any of those complications for the purpose of calculating the temperature of the earth without an atmosphere.
    By definition there is no atmosphere.
    Now if we were calculating the actual temperature of the planet then those considerations would come into play. One of the considerations that needs to be added is CO2, it’s a small component but it does effect the balance.

  282. Luke April 17, 2011 at 9:05 am #

    Rubbish Nasif – it’s not how GCMs solve radiation – and you well know it. It’s done explicitly for each grid box. Why are we hung up on this whole indicative average notion.

    What a total misrepresentation and many here have swallowed it.

    And as I have said – let’s assume you are – correct – now do the energy budget against obs for one location ! The cotton field above for example.

  283. el gordo April 17, 2011 at 9:39 am #

    ‘….and many here have swallowed it.’

    Not me bro, I’m hanging out for the open thread so we can talk about the weather. More fun than splitting irrelevant hairs on this thread.

  284. Alan D McIntire April 17, 2011 at 9:40 am #

    I’d like to point out that my prior link where I addressed zero and negative greenhouse effects applies to Trenberth’s model. Note that 67 watts is supposedly absorbed directly into the atmosphere. Presumbly that’s converted into radiation by clouds, dust particles, etc which are radiating in roughly the same wavelength’s as they absorb radiation from earth’s surface.

    Assuming that is the case, then plugging in the negative greenhouse effect, we get

    {(67/(x+1)) absorbed in atmosphere + 168 hitting earth’s surface}*(X+1) = 492

    67 + 168 + 168X = 492, so X = 1.53 effective atmosphere greenhouse effect. Note
    that 67/235, or 28.5% of incoming sunlight has a net zero effect acting on it.
    Only 71.5% of the solar radiation gets through to earth’s surface.

    Applying Beer’s law I = I0 * e^(-.242). Presumably a 1% increase in water vapor in the atmosphere would increase this to -.244 , indicating an additional negative feedback not addressed previously.

  285. Nasif Nahle April 17, 2011 at 10:34 am #

    @Graeme…

    Neutrino says:

    This flux is not strictly speaking the ToA flux.

    When he refers to the flux of solar power striking the TOA.

    However, the reality is that it is, strictly speaking, the TOA average flux obtained from measurements:

    http://www.eoearth.org/article/Solar_radiation

    From Materials Technology Limited:

    http://www.drb-mattech.co.uk/uv%20map.html

    “Solar Constant – The solar constant is the amount of energy received at the top of the Earth’s atmosphere on a surface oriented perpendicular to the Sun’s rays (at the mean distance of the Earth from the Sun). The generally accepted solar constant of 1368 W/m2 is a satellite measured yearly average.”

    From NASA site:

    http://earthobservatory.nasa.gov/Features/SORCE/sorce_03.php

    If that were all there was to the Earth’s radiative balance, scientists studying the Sun would have probably long since moved on to another climate-related problem. Analyzing the Sun and its affects on climate, however, is further complicated by the fact that the amount of radiation arriving from the Sun is not constant. It varies from the average value of the TSI—1,368 W/m2—on a daily basis.

    I apologize for repeating those links, but they are talking about the reality of measurements, not just calculations.

    Mathematics is an esential tool for science to explain our theories about how nature behaves. Mathematics doesn’t create reality.

    All the best,

    NSN

  286. Debbie April 17, 2011 at 11:32 am #

    Luke,
    It appears you are slightly missing the point here.
    Nasif is actually questioning the methodology and the assumptions.
    One possible (repeat possible) answer to your question about the cotton field example is that the calibration of the instruments have been set to ‘assume’ that CO2 has an emissivity of 1.
    The resultant modelling could be ‘assuming’ the same.
    That is just an observation.
    BTW, because of your repeated comments about Nasif’s ‘credibility’ via publication and also his reputation, my curiosity got the better of me and I googled him.
    His creds look perfectly OK to me. He also has a great deal of credibility outside of ‘blogosphere’
    I was not able to give Neutrino the same recognition because he must be using a pseudonymn.
    BTW Neutrino, that’s not a problem, it wasn’t me questioning people’s credibility.

  287. cementafriend April 17, 2011 at 11:52 am #

    Nasif, you have been very patient with Neutrino. It is typical of AGW believers to extract some simple rule and then pretend it applies to a complex system such as heat transfer. For example assuming that CO2 is the only important gas in the atmosphere and that water vapor and clouds composed of water droplets and ice particle do not matter.
    A small point with regard to radiant energy for the sun the following recent paper http://www.leif.org/EOS/2010GL045777.pdf from Kopp and Lean give a value of 1361 W/m2. They claim this to be accurate to 2 decimal places which of course is nonsense. No one accurately knows emissivity of the sun and its actual surface area (the present area is based on a visual projection).
    Further the distance from the sun varies over time. The main point is that the new measurement is considerably lower than the accepted value you give.
    Another point in the presentation of Van Andel to KNMI in Sept 2010 he says that Trenberth knows that the radiation window is 66 W/m2 and not 40 W/m2 see here http://climategate.nl/wp-content/uploads/2010/09/KNMI_voordracht_VanAndel.pdf . The K-F-T paper of 2008 is nonsense and should never have been published (there is nothing new from the earlier paper and the figures are based on wrong assumptions).
    An average gobal temperature is meaningless. It gives no imformation on climate, climate changes and weather in important regional areas as said many times in papers in which Roger Piekle Snr is co-author. Energy budgets are also nonsense because of the overwhelming heat capacity of oceans.

  288. Luke April 17, 2011 at 2:43 pm #

    “One possible (repeat possible) answer to your question about the cotton field example is that the calibration of the instruments have been set to ‘assume’ that CO2 has an emissivity of 1”

    Nuh and how silly !

    Debs – a long thread and not even how GCMs solve the problem. Yawn.

    If Nasif – seriously believes his revolutionary thesis – get published or be overlooked by history. Just think the world remains unaware of his remarkable finding…. could be a gamer changer- but we’re reliant on the IPCC reading and believing Jen’s blog.

  289. Bryan April 17, 2011 at 6:23 pm #

    Neutrino

    ….”earth without an atmosphere.
    By definition there is no atmosphere.”…….

    Why do IPCC proponents argue that to prove a “greenhouse effect” you must remove the entire atmosphere?.

    Surely the test should be that CO2 and H2O in the gaseous phase do not radiate in the IR should be sufficient, all other features will remain the same.

    What effect would this have on our supposed 15C average surface temperature?

    I would suggest very little, perhaps a little higher if anything.

  290. Mack April 17, 2011 at 9:27 pm #

    Nasif,
    It appears to me from what you have said is that the AGWers are just doing a sleight of hand by confusing us between the surface of the earth and the “surface” of the atmosphere.

  291. kuhnkat April 18, 2011 at 7:20 am #

    Luke,

    “Rubbish Nasif – it’s not how GCMs solve radiation – and you well know it. It’s done explicitly for each grid box.”

    Except the solution for each grid box depends on the solution for surrounding grid boxes!! If it does not it is meaningless.

    Which came first, the chicken or the egg?? HAHAHAHAHAHAHAHAHAHAHAHAHA

  292. Luke April 18, 2011 at 8:54 am #

    Still stupid KuhnKat – the whole argument here about averaging not being appropriate is well known. Which is why GCMs don’t do that. Fundamentally the solution is at grid box level.

  293. Graeme M April 18, 2011 at 12:54 pm #

    Somewhat topical, particularly the radiative calculations. And while not topical in the context of this thread, and while this is an oldish paper, his conclusions re the effects of CO2 are interesting, tho perhaps not as interesting as his conclusions with respect to water vapour forcings and the GHE of CFCs and HCFCs…

    http://journalofcosmology.com/QingBinLu.pdf

  294. Graeme M April 18, 2011 at 12:56 pm #

    That should read “water vapour feedbacks”… Don’t want to be accused of not having a clue what I am writing about.

    Of course – I don’t.

  295. cohenite April 18, 2011 at 1:59 pm #

    luke, doesn’t your head hurt from lying? All GCMs use a variety of GAT, the average global temperature; GAT has no bearing on the Earths radiative Balance. And can you point to one GCM which factors in the point noted by Nasif that CO2 reduces the radiative properties of H2O in the overlapping spectrum?

  296. Nasif Nahle April 18, 2011 at 5:31 pm #

    @Mack…

    Sorry for the delay on answering messages; a new family member has just arrived yesterday. My ninth grandchild. 🙂

    You say:

    Nasif,
    It appears to me from what you have said is that the AGWers are just doing a sleight of hand by confusing us between the surface of the earth and the “surface” of the atmosphere.

    That is exactly what they are doing. The average of all grids on top of the atmosphere is 1368 W/m2, which Lean et al have changed it to 1360 W/m2. I hope they don’t continue dropping the number until it coincides with K-T diagrams… LOL!

    Thus, 1368 W/m2 is the solar power received on top of the atmosphere. They say it is 342 W/m2, which is a lie.

    Those 342 W/m2 represent the power received on the surface of the Earth, i.e. 100 km below the top of the atmosphere (ToA).

    That’s the reason by which they always show an unreal frozen Earth.

    NSN

  297. cohnite April 18, 2011 at 6:12 pm #

    Congratulations Nasif.

  298. Luke April 18, 2011 at 9:13 pm #

    “All GCMs use a variety of GAT”

    Cohenite – are actually that dense ? So after all this you know nothing. How amazing.

    Can you show where one GCM radiation model is wrong? Nasif hasn’t.

  299. cohenite April 18, 2011 at 9:16 pm #

    “Can you show where one GCM radiation model is wrong?”

    http://rossmckitrick.weebly.com/uploads/4/8/0/8/4808045/mmh_asl2010.pdf

  300. gavin April 18, 2011 at 10:44 pm #

    Perhaps Cohenite could open in another thread with his full explanation of how this Mckitrick linked article can help us deal with the warming question after considering the process in detail.

  301. gavin April 18, 2011 at 10:58 pm #

    Folks; a good practitioner in perspective always grounds his thinking with an eye to the rear whist searching for that suitable yardstick that enables a near perfect calibration on the horizontal surface.

    http://glacierchange.wordpress.com/2011/04/07/ochsentaler-austria-rapid-glacier-retreat/

  302. Luke April 18, 2011 at 11:17 pm #

    Not an answer to my specific question Cohers.

  303. cohenite April 18, 2011 at 11:55 pm #

    It’s a dumb question luke; all the models are wrong to some extent and some of the modellers are even honest enough to admit it:

    http://journals.ametsoc.org/doi/full/10.1175/2007JTECHO546.1

    I like Pinker, she seems reasonable. The McKitrick paper shows that the role of water in the radiative process which AGW relies on can’t be right; McKitrick showed this against all the models and all the data; given the general gist of this debate between Neutrino and Nasif I thought it was appropriate. If you haven’t the wit or honesty to concede that, that is your problem not mine.

  304. Neutrino April 19, 2011 at 12:39 am #

    That basic physics lead to an earth with no atmosphere being quite frigid is well basic.

    To reject this conclusion some of the basic principles need to be wrong. Nowhere in this thread has anyone given any substantiated rebuttal of the original four points that led this thread off.

    1) α = ε
    2) q = εQ
    3) p = εσT^4.A
    4) p = q

    Nasif has claimed that the absorbed(2) is calculated by using the SB(3) but has offered no proof of this.

    There are two good reasons to not accept Nasif’s assertion:

    One, it is not a logical statement.
    The SB is the integration of the Planck function, the Planck function is the emission profile of emitted radiation. It is clearly a function that deals with emission not absorption.
    Besides, if absorbed were actually calculated as Nasif maintains then No ideal GB could ever be heated by incident radiation. Why? Because by Nasif’s method absorbed always equals emitted. He is using the same formula to calculate both.
    Additionally, using Nasif’s method the Incident is not in the equation. In effect he is calculating an absorbed value regardless of the Incident.

    Two, it is not supported by the literature.
    He has not one single reference that supports his claim. Counter that with five universities, as well as his own link, all calculating Absorbed in the same way as above.

    As to the issue that developed in the thread regarding Absorptivity.
    That Absorptivity can be calculated from the Albedo is both common practice(again all five universities plus Nasif’s own link did this) but also logical. If 30% of the Incident power is reflected(this is NASA’s measurement) then 70% of the power is absorbed by definition(assuming no Transmittance). Incidentally, even if Nasif maintains that doing this is not appropriate then he still has the issue that it does define an upper limit to the possible value of Absorptivity. If 30% is reflected then at most 70% can be absorbed, this is simple arithmetic. If Nasif maintains his assertion that (Absorptivity = 1 – Reflectivity) is wrong or that at the very least it doesn’t represent the upper limit on the value of Absorptivity then he has to address the Conservation of Energy, since he has more energy leaving the system than is entering.

    In regards to the geometry. Dividing the total solar power across the entire surface of the globe is not sleight of hand, it is obvious. Or at least it should be. That Nasif maintains the ToA value is equal to So is bizarre. Since the purpose of this calculation was to obtain the average temperature the simplest way to demonstrate that Nasif is wrong is to just simply look at the diurnal cycle.

    If the earth is facing away from the sun 12h of every 24h period then it is obvious that the average Incident cannot be So. The earth can only receive So for at most half of the time, for the other 12h the earth is receiving 0W! A further refinement is that even on the daylight side not every point is subject to So. In actuality every point is subject to So*cos(ө). Averaging this over the daylight side the Incident becomes So/2. Extending this average to the nighttime side as well it becomes So/4. It’s not sleight of hand, it’s called math.

    Not liking the result of the calculation is not a good reason to reject it, in fact it is a terrible reason to reject it. If anyone can find an error in the above assertions then that would be a valid reason to reject the conclusion, as yet no one has proffered one.

  305. Nasif Nahle April 19, 2011 at 4:33 am #

    @Neutrino…

    You say:

    That basic physics lead to an earth with no atmosphere being quite frigid is well basic.

    Absolutely wrong…

    Nasif has claimed that the absorbed(2) is calculated by using the SB(3) but has offered no proof of this.

    Show me where I said that… Do not resort to lies when you’re losing a debate.

    The definition of absorptivity is as follows:

    α = [(∫ 0 ∞) (∫ 2π1) α’λ (r, λ) (T, λ, si) Iλ (r, λ, si) cos θi dΩi dλ] / [(∫ 0 ∞) (∫ 2π1) Iλ (r, λ, si) cos θi dΩi dλ]

    Derive the formula and you have what I really said.

    😀

  306. Nasif Nahle April 19, 2011 at 4:35 am #

    That Absorptivity can be calculated from the Albedo is both common practice(again all five universities plus Nasif’s own link did this) but also logical.

    It could be said by the pope or whoever you wish. Science says it’s wrong and that’s all.

    😀

  307. Luke April 19, 2011 at 5:02 am #

    Still ducked it Cohers. Unable to answer specifics Cohenite widens the debate.

  308. mkelly April 19, 2011 at 6:54 am #

    Comment from: Neutrino April 17th, 2011 at 12:55 am

    The 1368W/m^(which is both a measured and calculated value) is the flux at the earth.

    So at TOA the temperature is 394 K. And it cools on the way down to the surface. I agree the atmosphere cools us from the heat of the sun.

  309. gavin April 19, 2011 at 6:56 am #

    Folks; Cohenite needs to explain how the applied combinations treatment in that last link can be used to debunk the more conventional physics and calculus as used here by Neutrino and most AGWers who know anything about the science process

  310. gavin April 19, 2011 at 6:58 am #

    mkelly; have another go, please

  311. cohenite April 19, 2011 at 12:15 pm #

    Buzz off gavin.

    luke, I forget your purpose here is to make mischief; you want me to say that Neutrino is right and Nasif is wrong; Neutrino has laid out how he thinks SB should be applied, in terms of emission not absorption, Nasif wants his formula for absorptivity explained by Neutrino; I don’t understand the formula and have said so; do you understand it?

    In terms of this by Neutrino: “Besides, if absorbed were actually calculated as Nasif maintains then No ideal GB could ever be heated by incident radiation. Why? Because by Nasif’s method absorbed always equals emitted.” But that is what Kirchoff says is it not?

    I have said that this discussion is similar to the Arthur Smith discussion where the flaw was in the averaging which I also pointed out to Neutrino; this is germane because where averaging may establish some basic ground-rules in an atmosphereless scenario it is completely inappropriate in an atmosphere where there is radiative effect; that is, establishing the radiative principles in an atmosphereless situation is contradicted by the operation of the radiative principles in an atmosphere. I have given the reason as simply as I can in this: (A + B)^4 > A^4 + B^4, which I’m sure you don’t understand the significance of luke.

    But anyway, if you don’t believe me take your mate SoD’s word:

    http://scienceofdoom.com/2010/06/03/lunar-madness-and-physics-basics/

    Your mate SoD concludes: “The right way to calculate a planet’s average radiation is to calculate it for each and every location and average the results. The wrong way is to calculate the average temperature and then convert that to a radiation. In the case of the earth’s surface, it’s not such a noticeable problem.”

    I disagree that “it’s not such a noticeable problem” on Earth and Lubos can explain why:

    http://motls.blogspot.com/2008/05/average-temperature-vs-average.html

  312. Neutrino April 19, 2011 at 1:18 pm #

    cohenite,

    No, that is not what Kirchhoff says.
    Kirchhoff says that the value of emissivity equals the value of absorptivity. What governs absorption is different than what governs emission. They are both scaled by the same factor but the equations for both are not the same.

    As for the averaging of temperatures opposed to the emissions. Yes, averaging the emissions is the better way to do it. No surprise there, but what doing a simple average for temperature gives us is the maximum possible temperature to emit a given flux. So in the context of calculating what the temperature the planet would be with no atmosphere it is useful. It tells us that the planet would be at least 33C colder.

  313. Neutrino April 19, 2011 at 1:21 pm #

    (tried to post this several times earlier, maybe the links are causing problems)

    It is difficult to have a discussion with someone when they persist in claiming that they did not make certain statements, and then accuse me of lying when pointing it out.

    Nasif said:
    Show me where I said that… Do not resort to lies when you’re losing a debate.
    And earlier Nasif said:(quoting me then adding, my bold)
    “The question was about Absorbed flux:
    Yes or No does the SB calculate the Absorbed flux?
    No! I’m not calculating absorbed flux, either SB equation calculates absorbed flux. SB equation is used to calculate the absorbed power.

    Nasif is quite explicit that he is using the SB to calculate Absorbed. This is also backed up from a previous post where I quoted him actually using SB to calculate Absorbed many times). His confusion over power vs. flux aside(Nasif has always been calculating flux not power) he is quite clearly stating that SB can be used to calculate an Absorbed component.

    With regard to Absorptivity.
    Yes if the pope calculated it via (Absorptivity = 1 – Reflectivity) he would be doing valid science, as supported by every link provided. It is a basic Identity of optics. If Nasif wants to go and derive the absorptivity of a surface, in this case the earth, he is free to do that. I would rather just use a basic Identity of Conservation of Energy to quickly calculate it from another measured quantity. Neither my argument not Nasif’s relied upon a derived value for absorptivity, as can be seen here for Nasif.
    That it is standard practice to use the above Identity is evidenced by the use of it by all the references cited.

  314. Mack April 19, 2011 at 2:27 pm #

    Nasif Nahle says ( April 16 th 3.43pm)
    “The measured flux of solar power striking THE TOP OF THE EARTH’S ATMOSPHERE is 1368w/sq m.
    That’s a measurement made by satellite instruments. It’s reality:”
    So Neutrino this is an undisputed, measured, FACT. You can’t get away from it.
    Any model you may have, with or without an atmosphere, must commence with this FACT that it is 1368w/sq m at the top of the atmosphere.
    Neutrino, you can sit on this site talking maths and physics until you’re blue in the face but it won’t change this FACT.
    Going from this figure down to your figure of 161w/sq m (I think it was) because of an atmosphere is to put it mildly “an inversion of the truth”
    It is just as well you have a mask on pulling this little ploy of atmosphere – no atmosphere.
    You remind me of a person with the pea and thimbles.
    You have the pea (surface(s) ) covered over with showers of equations and moved around the table.

  315. cohenite April 19, 2011 at 3:02 pm #

    Thank you Neutrino: “No, that is not what Kirchhoff says. Kirchhoff says that the value of emissivity equals the value of absorptivity. What governs absorption is different than what governs emission. They are both scaled by the same factor but the equations for both are not the same”

    I must confess absorptivity, absorptance and absorbance exceed my subtlety threshold but let’s stick absorptivity [and emissivity] and look at Kirchoff which defines the connection between absorptivity and emissivity as:

    “The absorptivity αλ is the ratio of the energy absorbed by the wall to the energy incident on the wall, for a particular wavelength. This will be proportional to αλEbλ(λ,T) where Ebλ(λ,T) is the intensity of black body radiation at wavelength λ and temperature T. The emissivity of the wall is defined as the ratio of emitted energy to the amount that would be radiated if the wall were a perfect black body. That will be ελEbλ(λ,T) where ελ is the emissivity at wavelength λ”

    So the measure of absorptivity is αλEbλ(λ,T)

    The measure of emissivity is ελEbλ(λ,T).

    What am I missing?

  316. Mack April 19, 2011 at 3:59 pm #

    Btw.
    Congrats Nasif, More excitment . 🙂

  317. Graeme M April 19, 2011 at 4:44 pm #

    OK, I know I’m labouring what seems a fairly basic point, but this post was about establishing basics.

    I am still troubled by the dividing of the solar constant of 1368w/m^2 by 4. I have read the various references Neutrino has offered and think I broadly understand. All appear to agree on that division as required to convert a value of total power incident on a disk of diameter the same as Earth to total power received on a sphere that rotates.

    But this is where I run into a sort of cognitive issue.

    Assuming no sphere hanging in space at point x. If I measure the solar power at that point, it is on average 1368w/m^2. If I place a flat board there facing the sun, the sunward side receives 1368w/m^2, the dark side none. The total power received by that board is only the amount received by the sunward side. If I wish to average it for the total area of that board, I divide by 2. That sounds sensible, yes?

    I interpret this to mean that any point in line of sight to the sun is receiving on average 1368w/m^2.

    I can then consider matters of reflectivity and absorption/emmissivity and using the equations calculate the temperature of the surface of that board. And I can calculate an average for the board’s total surface area.

    However, I start with the number of 1368w/m^2 BEFORE I start doing any calculations. Is that not correct? Because any point in line of sight must receive the full power of the sun, regardless of the actual shape of the object on which that point exists. All physical effects of angles of incidence, surface properties and so on must be applied to that full power.

    So, why is it different for our sphere? A point at .0001mm ABOVE the surface of an atmosphereless sphere is the same as any point in a vacuum at that distance from the sun it seems to me.

    Therefore, the total power received by our atmosphereless earth must be 1368w/m2 over the area of our hemisphere in any given period. Once we derive that, we can apply whatever formulas and parameters necessary to calculate temperature.

    So, what am I missing?

  318. Mack April 19, 2011 at 5:11 pm #

    Nasif,
    As a matter of interest,what year roughly would those satellite readings of solar flux have come into existance?

  319. wayne April 19, 2011 at 8:00 pm #

    Mack, you will find the answer about solar flux in relation to satellite readings here:
    http://eospso.gsfc.nasa.gov/eos_homepage/for_scientists/atbd/docs/ACRIM/atbd-acrim.pdf

    Within it show the ERB (nimbus7), ACRIM I, ACRIM II, ERBS previous plots. I find this statement especially curious…

    “The direct correlation of luminosity and solar activity is a major discovery from
    the solar cycle results. It agrees in sense with that predicted from the
    coincidence of the “Little Ice Age” climate anomaly and the “Maunder Minimum”
    of solar activity during the 16th and 17th centuries. (Eddy 1976) The average
    temperature decreased by between 0.5 and 1.5 Kelvin (k) in response to what
    models indicate was a total irradiance decrease of between 0.25 and 1.0 %.”.

    … being that since current TSI is said to be but a bit over 1361 we very well may have misassumed solar secular invariability due to the wide “absolute accuracy” ranges of these instruments. Read this manual, much withn should surprise you! At least you will then know about how these measurements are actually made.

  320. wayne April 19, 2011 at 8:13 pm #

    Nasif, congrats gramp…. I so enjoy mine but I can’t even imagine a ninth!

  321. Mack April 19, 2011 at 9:15 pm #

    Wayne,
    Thanks for the link and there’s a lot of stuff there 🙂
    What gets me that this figure of 13 hundred something has been around since 1902! measured by water flow! They got 1346 w/sq m. then.
    Nasif says the latest figure is 1360w/sq m.
    It’s simply unbelievable that science has had that figure established for so long and yet nobody ( or scientist) has ever thought that if the AGW people claim it’s cut down to about 160w/sq m. that’s a hell of a lot of attenuation. Get the dimmer onto a lightbulb and turn it down by that proportion and it would be damn near out.
    However it’s more like they’re treating us like mushrooms. 😉 🙂

  322. Neutrino April 19, 2011 at 11:42 pm #

    Graeme,

    Almost there.
    Your flat board is a good starting point, so to measure the average power we would divide by 2 as you say.

    The difference between a flat board and a hemisphere is the curvature. Look at the link Nasif provided, it has a good diagram at the bottom that helps visualize the difference at 0, 30 and 60 latitude. The sun’s strength is only 1368W/m^2 at the equator at noon, everywhere else it is less because of the angle. So if I was standing on 60latitude I would only be receiving 50% of the So because the sun is no longer directly overhead. Averaging this reduced flux over the hemisphere we end up with So/2.
    What I always thought was a simpler way of describing it(but saying the exact same thing) is that the total surface area of your board is only πR^2 whereas the surface area of the hemisphere is 2πR^2. The total power the board receives is spread out over the larger area of the hemisphere yielding a flux of 684W/m^2, So/2.

    So to account for the curvature of the front of the board, or the larger total area, take So/2 as the average flux on the frontside. As you said earlier to account for the backside divide by 2 again and that gets us to the So/4 value for average global ToA.

    Saying it this way may help, a point 0.0001mm above the surface, ToA, is not the same everywhere on the frontside of the sphere(but it is the same everywhere on a flat disc). At noon on the equator that point is directly between the sun and earth and therefore receives the full So flux. Looking directly up at a point 80latitude is not directly between the sun and earth so only receives a portion of the So flux(So*cos(ө)).

  323. Neutrino April 20, 2011 at 12:06 am #

    cohenite,

    What you are missing there is that ‘T’ in both equations only refers to the source.

    Absorption is dealing with an Incident that was emitted somewhere else, the Ebλ(λ,T) in αλEbλ(λ,T) does not refer to itself but to the source of the Incident power.
    Emission is the opposite, in this case the Ebλ(λ,T) in ελEbλ(λ,T) is a function calculating emitted power from its own temperature. This then is the power used as the Incident in the calcultaion of absorption.

    Id like to point out the first part of your quote:
    The absorptivity αλ is the ratio of the energy absorbed by the wall to the energy incident on the wall, for a particular wavelength.
    This is stating what I originally started with at the beginning of this post. This is my line 2). And it completely contradicts Nasif using the SB to calculate absorption.

  324. Neutrino April 20, 2011 at 12:24 am #

    Mack,

    I have no problems with the fact of that measurement, I have acknowledged it many times and even calculated it myself in this thread.

    What I do have a problem with is asserting that the total So is received at the ToA everywhere. This is a logically inconsistent statement. Simply because it is night half the time! Therefore just as a matter of principle the average ToA cannot be So.

  325. mkelly April 20, 2011 at 1:19 am #

    Comment from: gavin April 19th, 2011 at 6:58 am

    mkelly; have another go, please

    I have no idea what you are talking about.

  326. Graeme M April 20, 2011 at 6:45 am #

    Thanks Neutrino. I have no difficulty with the geometric aspect of translating coverage to a sphere, that though is not my point. I will try just once more to explain my question, but it’s possible I can’t frame it well enough to explain.

    IF we have a sphere in a vacuum, there is no atmosphere to attenuate the sun’s power. Thus there can be no effect of attenuation at any point on the sunlit side of the sphere. I assume the matter of the sun weakening because it is not overhead is an atmospheric effect, it can’t surely be a physical effect of distance can it?

    Put another way, to measure the solar flux at the equator with the sun overhead, or at the almost rim of the hemisphere, in effect is nothing more than a matter of location. If I were to absent the sphere instantaneously and measure the flux at those points, it must surely have the same power allowing for the slight difference in physical distance. To argue otherwise is to suggest the sun’s power varies enormously with every point in the larger sphere.

    Or put still another way, using my physical locations of equator and rim as fixed locations relative to the sun and the sphere’s orbit, allow the sphere to move on in its orbit by 1000km. What will the solar flux be at those locations? There is now no sphere there, so any measuring device should show largely similar values allowing for any attenuation by the small relative distance.

    My point is that surely any point in sunlight must get the full power of the sun relative to that points location, regardless of its physical arrangement (sphere, board, or cube). Once it strikes that surface, we can apply discounting effects for angles, absorptivity etc in calculating the temperature. But we can’t discount the power first before we factor in the constraints.

    On a sphere with an atmosphere, we would of course have reflective, refractive and absorptive effects from the atmosphere (and so angles etc come into play). That would have to be applied to the full power of the sun as it is received at the TOA (assuming theoretically that the TOA is viewed as a solid boundary).

  327. Neutrino April 20, 2011 at 7:52 am #

    Graeme,

    I think I see what you are getting at. Look again at the graphic at the bottom of the page where you can select the latitude from this site.

    If the sun is directly overhead, at equator, the flux of 1368W/m^2 is spread out over 1 square meter on the ground so the flux received at that point is 1368W/m^2. But at 60latitude that same flux of 1368W/m^2 is now spread out over 2 square meters on the ground so the flux received at this point is a flux of just 684W/m^2. This is the attenuation of the cos(ө) induced by geometry and has nothing to do with the atmospheric attenuation.

  328. gavin April 20, 2011 at 8:26 am #

    Guys; imo Neutrino has been too polite all the way through with his encouragement of us in dealing with the basics. A good lecturer should have thrown his wooden blackboard duster at the back of the class many times. Oh what patience!

    When fiddling with stuff we don’t understand I suggest it’s worth using imagination and trying a few ideas then see what fits any evolving concepts but it’s a strictly an individual process as we each have preconceived structures well established in our grey matter. Rules we must follow hey.

    After many years working for industry & science I have used many short cuts while forgetting most of the math and physics that went with a particular technology. Last week in retirement I quickly trued an apprentice made set square with my hammer and dolly while thinking the next customer doesn’t need to know how I adjusted the riveted joint so brutally. From experience, it helps to know how the other person thinks in all enterprises but somebody probably passed that metal work project in the interest of child development.

    IMO In this thread we have suffered somewhat as people considered Neutrino’s outline so a few thoughts about that if I can recall.

    The blue planet has been abused with Black Body and Grey Body thinking but its valid on the basis of schoolkid science that some unfortunately missed. Shine a well focused torch on a desk globe or a disk in a dark lab and consider the shadows. Does it matter the colour of either ? What matters is the power of the lamp and the distance. Optics was one of my better subjects.

    This globe could be rotated a few degrees and photographed over and over till everyone agreed it was enough to see its average colour but what about its heat in a vacuum?

    Colour can be an illusion to the naked eye. As a budding teenage artist I read lots by Winsor & Newton on the history of pigments and manufacture of mediums. As a young adult I became a research assistant at ICI labs in the science of Dulux for Aus. Conditions. Hence this wild notion on the problem of albedo, how do we define the average colour of your classic redhead in prime condition but with lots of freckles?

    Now let’s make this profound statement. By general agreement, our average redhead has a skin colour called “white” because there are no blackheads with freckles and we can’t tell his or her core temperature from their flaming red hair.

    We can differ on my “white” skin from reflected light since it’s actually a very pale “pink” after closer inspection and so a “blue” world wins.

    Nasif, lets welcome the next gen with open arms. Btw I’ve lost count after twenty odd

  329. gavin April 20, 2011 at 9:07 am #

    Further, on the blue globe in black space concept I have to say a lot of practical instruments depend on the black body idea from physics.

    With furnace engineering to photography we are shooting bits of light onto or from particles while using a detector that mimics an eye. So convenient for using the more developed parts of our grey mater structures. After a few weeks studding Kodak processing plants making everything for film including xray and IR emulsions I started amateur photography with some zeal.

    I bet some of us have spent decades playing with camera bodies, lenses, hoods , filters and film only to be dismayed by their obsolescence as digital techniques overwhelmed us. Now consider how one particle must equal the yes or no case as we go down the path with a box of switches. All is not what it seems when dealing with grey matter!

  330. Graeme M April 20, 2011 at 9:09 am #

    Thanks again neutrino, I see what you are saying. The lightbulb came on afetr I posted – thinking it over on the drive to work the w/m^2 bit suddenly jumped out at me – I was just thinking of the power as a point value rather than an average over a unit area.

  331. Nasif Nahle April 20, 2011 at 9:46 am #

    @Graeme…

    This is the answer given by Neutrino to your question:

    If the sun is directly overhead, at equator, the flux of 1368W/m^2 is spread out over 1 square meter on the ground so the flux received at that point is 1368W/m^2. But at 60latitude that same flux of 1368W/m^2 is now spread out over 2 square meters on the ground so the flux received at this point is a flux of just 684W/m^2. This is the attenuation of the cos(ө) induced by geometry and has nothing to do with the atmospheric attenuation.

    But is not true. 1368 W/m2 is the power flux the Earth receives, NOT ON THE SURFACE, as Neutrino Gray says, but on the top of the atmosphere. It is a measured quantity.

    The ozone layer, the water vapor, the clouds, dust and aerosols attenuates it in the following way:

    1368 W/m^2 / 1.35 reflected by the atmosphere and Earth’s surface = 1013.3 W/m^2

    1013.3 W/m^2 / 1.20 absorbed by the atmosphere = 844.4 W/m^2

    From this power, the surface only absorbs a power of 692.41 W, from which 41.54 W/m^2 are emitted from the surface towards the outer space and 207.7 W are dissipated by conduction, oceanic currents and friction.

    Therefore, the effective load of power emitted towards the atmosphere is 443.17 W/m^2, which causes a surface temperature of 297.34 K, or 24.2 °C.

    The remainder of his answer falls by its own weight.

    NSN

  332. Nasif Nahle April 20, 2011 at 9:49 am #

    @Gavin…

    Nasif, lets welcome the next gen with open arms. Btw I’ve lost count after twenty odd

    A future full of lies and control over people? No, thanks!!! I do prefer truth, science, freedom.

    😀

  333. gavin April 20, 2011 at 11:00 am #

    Nasif; it’s their world I’m concerned about not ours. I could ask what is your religion but it;s enough to say I’m afraid your rhetoric has roots back to a time when we had “heretics” and victims of that state could be executed, even drawn and quartered in public.

    “Truth, science and freedom” are just what you make it. In my language there are no absolutes except for the concepts we may have for disciplines such as Physics and “Black Bodies”.

    Have you ever been a serious photographer?

    Any picture can save a lot of words as we illustrate and communicate. But we do have color blindness with some and thats wort thinking about re individualized messages however I doubt it changes the message tone to any degree. Now think of the illusion as we watch out TV. It’s a learned process at the receiver end and clever tricks at the other.

    When was science not an art hey

  334. spangled drongo April 20, 2011 at 11:35 am #

    gav,

    I think you should stick to adjusting old precision instruments with the back of the axe and leave the science to others. [And I don’t necessarily criticize the BoA method just maybe not on precision instruments]

  335. cohenite April 20, 2011 at 11:36 am #

    gavin says this: “Truth, science and freedom” are just what you make it. In my language there are no absolutes except for the concepts we may have for disciplines such as Physics and “Black Bodies”.

    Whether gavin understands what he is saying or not, and on past experience certainly not, this is just a description of the tabula rasa approach to human existence favoured by such twits as Foucault. The ramifications for science is that there are no certainties [did you think that one through gavin since the AGW idiots say the science is settled] and no final proofs; life is an illusion and reality determined by the subjective experience of the dominant, hegemonic view; how new age!

    Getting back to reality;

    Neutrino has kindly responded to my query about absorptivity and emissivity: αλEbλ(λ,T) and ελEbλ(λ,T) are not equivalent because the temperature, T, for absorptivity is that of the incident source, the sun, while T in the emissivity equation is that of the emitting source. Neutrino argues that Nasif is wrong for determining absorptivity T from emissivity T.

    I would like Nasif to comment on this but for me there is an immediate confounding of Neutrino’s position; this is to do with my previous comment that conclusions based on an atmosphere free concept cannot be extrapolated to an atmosphere. In an atmosphere there is backradiation contributing to the incident T on the absorptivity of the surface; however this incident T from backradiation is in fact from the surface; in this circumstance calculating emissivity from the surface is a measure of the backradiation returning as absorptivity incident T.

  336. Mack April 20, 2011 at 11:45 am #

    Neutrino,
    Even with an atmosphere the real world must get close enough to that 1360w/sq m at midday at midlatitudes.
    To extrapolate a model such as this to determine what the temp of the real world should be in my mind is too simplistic and ridiculous. .

  337. Neutrino April 20, 2011 at 11:49 am #

    In response to Nasif,

    No one is arguing that the solar constant(So) is 1368W/m^2. But that is not the same as the average value of ToA for the planet as a whole.

    The simplest way to refute this is the day/night cycle. Since when the earth is turned away from the sun there is effectively 0W/m^2 incident flux the average ToA can be at most So/2. Arguing against this is incredibly strange.

    This thread was about basic transfers, the atmosphere was left out on purpose. But a quick look at Nasif’s numbers is illuminating.

    Total flux entering:
    1368W/m^2

    354.7W/m^2 reflected by the earth surface and atmosphere
    168.9W/m^2 absorbed by the atmosphere
    692.4W absorbed by the earth surface(assuming this is supposed to be W/m^2 and is simply a typo)

    Total absorbed or reflected by the earth surface and atmosphere:
    354.7W/m^2 + 168.9W/m^2 + 692.4W/m^2 equals
    1216W/m^2

    What happened to the other 156W/m^2?
    If it is not reflected or absorbed by the earth or atmosphere then where did it go? Nasif has more energy entering the system than he has accounted for with interactions within the system.

    As well now Nasif seems to be implying that the temperature of the earth surface is 24C. This is only accomplished by calculating temperature using an emissivity of 1 for the 443.17W/m^2 he says the surface is emitting towards the atmosphere. Strangely this calculation ignores the 41.54W/m^2 he says is emitted from the surface to space directly.

    Once again Nasif’s numbers just don’t make any logical sense.

  338. Neutrino April 20, 2011 at 12:03 pm #

    Mack,

    Yes at midday on the equator the ToA will be So, at mid latitudes(≈30degrees) it will be 87% of So. But you also have to remember for the entire night at all latitudes ToA will be effectively 0W/m^2.

    The calculation is simplistic, that was the point, but that does not render it useless.

  339. Neutrino April 20, 2011 at 12:11 pm #

    cohenite,

    The point I was trying to make was that the flux or power used when calculating absorption is from the source. The temperature of the absorber does not come into play, except for any effect it has on the absorptivity term.

    The ‘T’ in the equation as written αλEbλ(λ,T) has nothing to do with the temperature of the absorber. It is the temperature used to calculate the emission Eb from whatever source is emitting it.

  340. Mack April 20, 2011 at 12:58 pm #

    I’m talking about the SURFACE of a real world Neutrino.

  341. cohenite April 20, 2011 at 1:03 pm #

    Neutrino: “The temperature of the absorber does not come into play, except for any effect it has on the absorptivity term.”

    That was my point; what is being absorbed from backradiation is what was emitted and what was emitted is based on the temperature of the surface.

  342. Mack April 20, 2011 at 1:23 pm #

    Ignore that last comment Neutrino I find myself argueing in circles between the reality we have on this earth and that simple ball of rock in space you have for a model.

  343. Mack April 20, 2011 at 2:39 pm #

    The fundamental flaw with your whole argument here Neutrino is that as far as the earth’s temp is concerned the oceans must be included as part of the atmosphere when you include an atmosphere into your model.

  344. Nasif Nahle April 20, 2011 at 4:18 pm #

    @Neutrino…

    If it is not reflected or absorbed by the earth or atmosphere then where did it go? Nasif has more energy entering the system than he has accounted for with interactions within the system.

    As well now Nasif seems to be implying that the temperature of the earth surface is 24C. This is only accomplished by calculating temperature using an emissivity of 1 for the 443.17W/m^2 he says the surface is emitting towards the atmosphere. Strangely this calculation ignores the 41.54W/m^2 he says is emitted from the surface to space directly.

    Once again Nasif’s numbers just don’t make any logical sense.

    Now I see your arithmetics is highly deficient.

    You are adding when what I did was dividing… LOL!

    😀

  345. Nasif Nahle April 20, 2011 at 4:22 pm #

    @Mack…

    The fundamental flaw with your whole argument here Neutrino is that as far as the earth’s temp is concerned the oceans must be included as part of the atmosphere when you include an atmosphere into your model.

    In my calculations the oceans are included, that is the reason that the absorptivity that I introduce in the formulas is the average of hydrosphere, biosphere, cryosphere and lithosphere. The average, after measurements is 0.82, not the unreal 1-t.

    As you can see, Mack, AGW is based on pure speculations, not in reality.

    NSN

  346. Nasif Nahle April 20, 2011 at 4:28 pm #

    @Neutrino,

    The ‘T’ in the equation as written αλEbλ(λ,T) has nothing to do with the temperature of the absorber. It is the temperature used to calculate the emission Eb from whatever source is emitting it.

    Is it your derivation of the definition of absorptivity? Or is it a condensation of the definition? Answer the question and in the same post, derive the formula to obtain absorptivity, that you did never answer. Here again the definition, now derive the formula:

    α = [(∫ 0 ∞) (∫ 2π1) α’λ (r, λ) (T, λ, si) Iλ (r, λ, si) cos θi dΩi dλ] / [(∫ 0 ∞) (∫ 2π1) Iλ (r, λ, si) cos θi dΩi dλ]

    I remember you that you wrote the “article”, so you must answer the questions of the readers.

    😀

  347. Nasif Nahle April 20, 2011 at 4:33 pm #

    @Neutrino…

    As well now Nasif seems to be implying that the temperature of the earth surface is 24C. This is only accomplished by calculating temperature using an emissivity of 1 for the 443.17W/m^2 he says the surface is emitting towards the atmosphere. Strangely this calculation ignores the 41.54W/m^2 he says is emitted from the surface to space directly.

    Again with more lies… Neutrino, show where I said that… Please, Neutrino, to lie is not a systematic methodology.

    And here:

    Nasif is quite explicit that he is using the SB to calculate Absorbed. This is also backed up from a previous post where I quoted him actually using SB to calculate Absorbed many times). His confusion over power vs. flux aside(Nasif has always been calculating flux not power) he is quite clearly stating that SB can be used to calculate an Absorbed component.

    More lies about what I said. It seems you’re desperate; take it easy, Neutrino…

    Derive the formula for absorptivity, which is the formula I use to calculate absorptivity, from the definition of absorptivity and stop saying things about what I said or not, which evidently have been created by your mind:

    α = [(∫ 0 ∞) (∫ 2π1) α’λ (r, λ) (T, λ, si) Iλ (r, λ, si) cos θi dΩi dλ] / [(∫ 0 ∞) (∫ 2π1) Iλ (r, λ, si) cos θi dΩi dλ]

    😀

  348. Nasif Nahle April 20, 2011 at 5:11 pm #

    @Cohenite…

    Neutrino has kindly responded to my query about absorptivity and emissivity: αλEbλ(λ,T) and ελEbλ(λ,T) are not equivalent because the temperature, T, for absorptivity is that of the incident source, the sun, while T in the emissivity equation is that of the emitting source. Neutrino argues that Nasif is wrong for determining absorptivity T from emissivity T.

    I would like Nasif to comment on this but for me there is an immediate confounding of Neutrino’s position; this is to do with my previous comment that conclusions based on an atmosphere free concept cannot be extrapolated to an atmosphere. In an atmosphere there is backradiation contributing to the incident T on the absorptivity of the surface; however this incident T from backradiation is in fact from the surface; in this circumstance calculating emissivity from the surface is a measure of the backradiation returning as absorptivity incident T.

    Neutrino doesn’t know what he’s talking about. He said that he didn’t know what the total emissivity is, so he doesn’t know what the total absorptivity is.

    The definition of absorptivity clearly states that the absorptivity depends on several variables, one of which is temperature:

    α = [(∫ 0 ∞) (∫ 2π1) α’λ (r, λ) (T, λ, si) Iλ (r, λ, si) cos θi dΩi dλ] / [(∫ 0 ∞) (∫ 2π1) Iλ (r, λ, si) cos θi dΩi dλ]

    Obvioulsy, the absorptivity depends on the temperature of the absorber system, not of the source of power, and here Neutrino is blatantly wrong.

    I have seen in the Internet the wrong procedure to calculate absorptivity by substracting the transmissivity from the absorptivity of a black body. Even, the formula to calculate the absorbed radiant energy by substracting transmissivity is wrong. The real formula, which is not to calculate the absorptivity of a system, but of opacity is as follows:

    1- e^((4π*K*s)/λ0)

    Given that k for the carbon dioxide acquires a very low magnitude, it is not so opaque to thermal radiation. The latter is coincident with its total emissivity of almos zero, as experiments have demonstrated (Hottel, Leckner, etc., etc.).

    Also, Neutrino’s definition of total absorptance is wrong. He defines it as αλEbλ(λ,T) and says that T is not for temperature. I would like that he say what that “T” is for… 😀

    The definition of total absorptance is as follows:

    α’λ (T, λ, θ, ψ)

    If it is a hemispherical surface, the definition changes:

    α’λ (Iin, T)

    But the temperature T is always present in any definition of absroptance.

    By the way, absorptance, which is the definition of “absorptivity” that Neutrino wrongly copied, is not a surface property.

    NSN

    😀

  349. Neutrino April 21, 2011 at 1:08 am #

    cohenite,
    Quick correction:
    That was my point; what is being absorbed from backradiation is what was emitted and what was emitted is based on the temperature of the surface.
    The radiation coming from the atmosphere is emitted from its own temperature. Yes it originally came from the earth surface but it was absorbed and thermalized in the atmosphere. It is then emitted from there based on the temperature of the atmosphere not the surface.
    Yes the temperature of the atmosphere at very low altitudes will be very close to that of the surface but it will not necessarily be the same.

  350. Neutrino April 21, 2011 at 2:02 am #

    Mack,
    I realize this is an overly simplified example. But until a calculation can be done for the simple scenario doing one for the more complex one is doomed to failure form the beginning. Yes oceans effect the calculation.

    As for your question about surface flux,
    Yes near the equator at high noon the incident will be close to So. But again the point is not what the instantaneous value is but the average.
    NOAA has a program to measure, amongst other things, the down welling solar flux. The furthest south point they measure is in Mississippi, check the instantaneous measurements for one day in midsummer last year and then also the averaged values for the entire year. On the single day the value reaches above 1000W/m^2 but averaged over a year it is approximately 200W/m^2.

    (having problems with the direct links to the observation pages but you can get there from the main link to the SURFRAD page)

  351. Nasif Nahle April 21, 2011 at 2:04 am #

    @Neutrino…

    Yes the temperature of the atmosphere at very low altitudes will be very close to that of the surface but it will not necessarily be the same.

    If the measured instantaneous temperature of the ground in my location, at this precise moment, is 48.1 °C, what the temperature of the air “very close to the surface”, let’s say 1 cm above the surface, is?

    😀

  352. Neutrino April 21, 2011 at 2:17 am #

    Addressing Nasif’s:

    I was giving him the benefit of the doubt about the math.

    Now I see your arithmetics is highly deficient.
    You are adding when what I did was dividing… LOL!

    So if he actually did mean that the earth and atmosphere together reflect 1013.3W/m^2 then the math is even stranger.

    Total Entering
    1368W/m^2(same as before)
    1013.3W/m^2 reflected by the earth surface and atmosphere
    844.4W/m^2 absorbed by the atmosphere
    692.4W absorbed by the earth surface(assuming this is supposed to be W/m^2 and is simply a typo)

    Total absorbed or reflected by the earth surface and atmosphere:
    1013.3W/m^2 + 844.4W/m^2 + 692.4W/m^2 equals
    2550.1W/m^2

    So now the total flux Nasif has in the system is nearly twice what he says is entering. Heck even just the sum of what is absorbed in the atmosphere and surface exceeds the total input without even considering the reflected component.

    Nasif again calls me a liar(how many times is this now?)
    Quoting me then adding:
    “As well now Nasif seems to be implying that the temperature of the earth surface is 24C. This is only accomplished by calculating temperature using an emissivity of 1 for the 443.17W/m^2 he says the surface is emitting towards the atmosphere. Strangely this calculation ignores the 41.54W/m^2 he says is emitted from the surface to space directly.
    Again with more lies… Neutrino, show where I said that… Please, Neutrino, to lie is not a systematic methodology.

    Since the temperature of an emitting object is determined by p = εσT^4 it is only a matter of algebra to figure out what emissivity Nasif used to calculate the above temperature.
    1) p = εσT^4
    2) ε = p / σT^4
    3) ε = 443.17W/m^2 / (σ * (297.34K)^4)
    4) ε = 1

    So ill ask again: How is this misrepresenting, or lying about, Nasif’s statements?
    The only way anyone can get a temperature of 24C(297.34K) from a flux of 443.17W/m^2 is to use an emissivity of 1. Adding the other 41.54W/m^2 emitted into the calculation and it would drive the emissivity to an unrealistic 1.09 which is why I assumed he did not include that in the calculation, again giving him the benefit of the doubt.

    And again Nasif accuses me of lying(actually this is the same one he keeps recycling).
    Quoting me and then adding:
    “Nasif is quite explicit that he is using the SB to calculate Absorbed. This is also backed up from a previous post where I quoted him actually using SB to calculate Absorbed many times). His confusion over power vs. flux aside(Nasif has always been calculating flux not power) he is quite clearly stating that SB can be used to calculate an Absorbed component.
    More lies about what I said. It seems you’re desperate; take it easy, Neutrino…

    Here is where he actually said it:(my bold)
    No! I’m not calculating absorbed flux, either SB equation calculates absorbed flux. SB equation is used to calculate the absorbed power.

    Aside from his confusion over flux/power he is quite explicit in saying that absorbed is calculated using the SB.

    And then Nasif brings up this red herring again:
    Neutrino doesn’t know what he’s talking about. He said that he didn’t know what the total emissivity is, so he doesn’t know what the total absorptivity is.

    In another thread I stated that I did not know what the value of a quantity was, I did not state that I was unaware of what the concept of the entity was. Besides it is generally good practice to admit when you do not know an answer. But to link the two statements is quite an overreach. I do know what the total absorptivity of the earth system is. It is calculated to be approximately 0.7 by using the measured value of the Bond Albedo.

    Derive the formula for absorptivity, which is the formula I use to calculate absorptivity, from the definition of absorptivity and stop saying things about what I said or not, which evidently have been created by your mind:
    α = [(∫ 0 ∞) (∫ 2π1) α’λ (r, λ) (T, λ, si) Iλ (r, λ, si) cos θi dΩi dλ] / [(∫ 0 ∞) (∫ 2π1) Iλ (r, λ, si) cos θi dΩi dλ]

    Asking me to derive the value is unnecessary for the purposes of this thread. I have a calculated value for it that comes from a measured quantity. Yes I wrote this article(no scare quotes needed Nasif) but nowhere does the arguments laid out hinge on a derived value of absorptivity.

    Incidentally, I have also agreed(several times) that the value of absorptivity is temperature dependant. But the ‘T’ in the equation αλEbλ(λ,T) (which cohenite supplied not me) is a variable in the emission Eb not a variable for the absorptivity or absorbed. Absorbed depends on α which, as pointed out before, has a dependence on its temperature and the Incident flux or power. In the case of the above formula the Incident is Ebλ(λ,T) so the ‘T’ clearly applies to the temperature of the emission source.

  353. cohenite April 21, 2011 at 10:02 am #

    Nasif, I understand that; my point was that not all of the incident at the surface was from the sun and that was the difference between Neutrino’s scenario and what actually happens on the planet.

  354. Nasif Nahle April 21, 2011 at 10:37 am #

    @Neutrino…

    You’re worsening your arithmetics. You say that I said:

    1368W/m^2(same as before)
    1013.3W/m^2 reflected by the earth surface and atmosphere
    844.4W/m^2 absorbed by the atmosphere
    692.4W absorbed by the earth surface(assuming this is supposed to be W/m^2 and is simply a typo)

    Total absorbed or reflected by the earth surface and atmosphere:
    1013.3W/m^2 + 844.4W/m^2 + 692.4W/m^2 equals
    2550.1W/m^2

    Have not you seen still your mistakes? You’re falling in a deep pit, Neutrino.

    You say:

    Nasif again calls me a liar(how many times is this now?)

    As many as you lie about me. Show me where I said that I obtained the total absorptivity from the Stefan-Boltzmann equation. Could you? Isn’t it a lie you said against me?

    And you continue with your practice when you say:

    Here is where he actually said it:(my bold)
    “No! I’m not calculating absorbed flux, either SB equation calculates absorbed flux. SB equation is used to calculate the absorbed power.”

    Absorbed power is the same as total absorptivity, from your standpoint? You had mentioned a solitary expression “flux”; flux of what?

    What that T means, after all? What that W in the expression W/m2 is meant for?

    You say:

    Asking me to derive the value is unnecessary for the purposes of this thread. I have a calculated value for it that comes from a measured quantity. Yes I wrote this article(no scare quotes needed Nasif) but nowhere does the arguments laid out hinge on a derived value of absorptivity.

    But it is quite significant that you derive the formula from the definition of total absorptivity because you are saying that I calculated absorptivity from the Stefan-Boltzmann equation, when actually I calculate total absorptivity from the equation derived from the definition of total absorptivity.

    See how you’re falling in the pit created by yourself?

    😀

  355. Nasif Nahle April 21, 2011 at 10:40 am #

    @Cohenite…

    Nasif, I understand that; my point was that not all of the incident at the surface was from the sun and that was the difference between Neutrino’s scenario and what actually happens on the planet.

    Sorry… Now I understood what you were saying.

    NSN

  356. Neutrino April 21, 2011 at 11:03 am #

    Nasif,

    Reread my direct questions to you. The question has always been about Absorbed not Absorptivity. Absorbed power or Absorbed flux doesn’t matter, the point was how do you calculated the absorbed component.

    I have never said you calculated the absorptivity anywhere, as you have always maintained you are using a measured value.

    Just to be clear, find one quote from me where I assert you calculate Absorptivity from the SB. You will not be able to because I never asserted it.

    If my interpretation of your above breakdown of So moving through the atmosphere is incorrect then explicitly state your position. As you have written it it is ambiguous. Twice now I have tried to understand what you wrote, but no matter how I look at those numbers they do not add up to So.

    What is your explicit breakdown then?
    If the first line:
    1368 W/m^2 / 1.35 reflected by the atmosphere and Earth’s surface = 1013.3 W/m^2
    does not mean that 355W/m^2 is reflected(my first take on it) or that 1013W/m^2 is reflected(my second take on it) then what exactly does it mean?

  357. gavin April 21, 2011 at 8:33 pm #

    Been thinking for some time Nasif has borrowed from biology

    Task; find THE Absorptivity and Emissivity definitions by common use in engineering, chemistry, biology, acoustic engineering etc via links that relate to IEEE publications

    See first “Engineering Thermodynamics” R K Rajput 2010 – Google Books

    TYPICAL (from remote sensing technology)

    “A conceptual model for effective directional emissivity from nonisothermal surfaces” IEEE explore

    http://ieeexplore.ieee.org/xpl/freeabs_all.jsp?arnumber=789646

    “The new definition of effective emissivity of non-isothermal rough surface and its approximate expression for continuous canopy vegetation” Chen IEEE explore

  358. gavin April 21, 2011 at 10:04 pm #

    Try adding “climate” and “CO2” to “emissivity”- IEEE also “absorptivity” – IEEE

    “Overlap Emissivity of CO2 and H2O in the 15-µm Spectral Region” IEEE explore

    sample abstract

    “Radiative transfer in model atmospheres including CO2, water vapor, ozone, and clouds can be explored by computer simulation for the purpose of predicting the mean temperature at the surface of the earth. Where two or more gases have overlapping absorption lines, the overlap emissivity determined from the band-averaged expressions is a possible source of uncertainty. This method has been tested in a 250-cm−1 region around 15 µm where the CO2 absorption overlaps the tail of the water vapor rotation band. Published high resolution spectra of these gases enable the overlap effects to be calculated for small spectral intervals and summed over the band. This calculation was made and compared with the value given by the approximate treatment for the same gas concentrations, pressure, and temperature. Agreement was within 10 percent, indicating that the more detailed spectral calculation is unnecessary”.

    For cohenite; a very interesting description with lots of graphs see “Carbon Heat Trapping: Merely a Bit Player in Global Warming” by Richard J. Petschauer, Senior Member IEEE.

    http://climateclash.com/2011/01/09/g5-carbon-heat-trapping-a-critique/

    for those who wondered, IEEE publications were mandatory reading for a wide range of radio spectrum engineering and often the last stop for regulators.

  359. Nasif Nahle April 22, 2011 at 1:30 pm #

    @Neutrino…

    What is your explicit breakdown then?
    If the first line:
    1368 W/m^2 / 1.35 reflected by the atmosphere and Earth’s surface = 1013.3 W/m^2
    does not mean that 355W/m^2 is reflected(my first take on it) or that 1013W/m^2 is reflected(my second take on it) then what exactly does it mean?

    It’s simple; it is a division, not a subtraction. It seems you cannot differentiate between those simple arithmetic procedures.

    🙂

  360. Nasif Nahle April 22, 2011 at 1:31 pm #

    @Gavin…

    It seems you also ignore what total emissivity and total absorptivity are… LOL!

    😀

  361. Nasif Nahle April 22, 2011 at 1:34 pm #

    @All…

    I have demonstrated that my calculations of total emissivity and total absorptivity of carbon dioxide are based on the experimental/observational work of scientists. Therefore, whatever AGWers say on trying to discredit my calculations are doing it against REAL science. In science we know these practices as antiscience.

    AGW is a myth.

    NSN

  362. spangled drongo April 22, 2011 at 1:43 pm #

    gav,

    Good link and one I can subscribe to:

    “Recently he had trouble understanding how a rise of carbon dioxide, or any substance, in the atmosphere from such low values as 0.035 percent could have such an overly large influence on the Earth’s future temperature as some predict. So he decided to read and learn about this from the perspective of an individual with a technical background, but not in the field of climate. After a few infrared measurements on summer nights showed that the amount of heat being radiated from the atmosphere was much less than some climate models predicted, he began an intensive study resulting in his own computer simulations based on available atmospheric data and well-known laws of infrared physics. While weather predictions and long-term climate are very complex and beyond the author’s expertise, he feels the single issue of heat absorption and radiation due to carbon dioxide is much simpler, well understood, and better modeled and measured as proposed here. For reasons explained in the report, he went from being unknowledgeable to skeptical to now very doubtful about a harmful future temperature rise due to increased carbon dioxide levels.

    “None of the work done by the author related to this report was funded by any organization, company or person, besides the author who paid personally for the infrared measuring device and the fees for the carbon dioxide and water vapor spectral transmittance computer calculations.”

    Richard J. Petschauer Edina, Minnesota January, 2008

  363. cohenite April 22, 2011 at 7:44 pm #

    I too endorse SD’s endorsement of gav’s endorsement of the Petschauer paper, and the comments following.

  364. gavin April 22, 2011 at 7:46 pm #

    SD; I suggest you don’t get carried away by by that climate Doc since it’s not published nor peer reviewed beyond blogsphere. Also we know the author is not from a climate science background. We can guess though his stuff was sent to Senator James Inhofe and others as part of a larger anti AGW campaign.

    So, what have we? A well written essay by an old timer who could have once been a sparky or a plumber then got smart with early computers hence that IEEE tag.

    As you are still around SD, while many others are out on the coast somewhere over Easter I can share this. While watching a recently recorded edition of Spooks I was struck by this somewhat familiar statement from the lead character. “I’m here to check your security systems”. Now imagine what’s that all about?

    For these purposes we can substitute “control systems” from the industrial environment then I can say “I’m here to check your instruments” which is why you should read more on “absorptivity and emissivity” via IEEE papers on methods, instruments etc that link to modern climate science and radiative energy transfer. It’s everyone’s chance to catch up on surface thermodynamics too.

  365. gavin April 22, 2011 at 7:56 pm #

    Cohenite; Sorry I missed you there in between so let’s add another recommendation that you read those following comments on Climate Clash carefully. Unfortunately my concentration is now too weak to do them justice but I reckon they are significant to this thread.

  366. Neutrino April 23, 2011 at 4:23 am #

    Nasif’s response makes no logical sense.
    It’s simple; it is a division, not a subtraction. It seems you cannot differentiate between those simple arithmetic procedures.
    Either the total reflected is 355W/m^2 or it is not. It’s not a question of division or subtraction. The above doesn’t address the question asked.

    The ozone layer, the water vapor, the clouds, dust and aerosols attenuates it in the following way:
    1368 W/m^2 / 1.35 reflected by the atmosphere and Earth’s surface = 1013.3 W/m^2
    1013.3 W/m^2 / 1.20 absorbed by the atmosphere = 844.4 W/m^2
    From this power, the surface only absorbs a power of 692.41 W,

    No matter how the above is read it is not coherent, the numbers simply do not add up. For it to be coherent the total input would have to be fully accounted for, it isn’t.

    His lack in actually responding makes it virtually impossible to discuss anything with him.

    Then he adds this grandiose statement:
    @All…
    I have demonstrated that my calculations of total emissivity and total absorptivity of carbon dioxide are based on the experimental/observational work of scientists. Therefore, whatever AGWers say on trying to discredit my calculations are doing it against REAL science. In science we know these practices as antiscience.
    AGW is a myth.

    Besides the fact this thread has nothing to do with carbon dioxide he is declaring something that he has not done.

    The entire point of engaging Nasif on my part has been that he has not been using experimental or observational work of scientist correctly. He has found a few equations and tables and then grotesquely misused them.

    Case in point from this thread, his continued use of the Stefan-Boltzmann Law to calculate Absorbed flux or power(I said Absorbed Nasif, not Absorptivity).
    Quoting him directly: SB equation is used to calculate the absorbed power.

    He has yet to produce a single reference that would support the above assertion. The simple reason why he cannot do this is because it does not exist.
    Beyond that he has not even articulated why from a logical point he could use the SB the way he does. He has not done this because it is an illogical assertion. The SB is an integration of the Planck emission spectra.

    Anti-science is not supporting your argument, which is something Nasif has done in spades.

  367. Nasif Nahle April 23, 2011 at 4:00 pm #

    @Neutrino…

    Nasif’s response makes no logical sense.
    “It’s simple; it is a division, not a subtraction. It seems you cannot differentiate between those simple arithmetic procedures.”
    Either the total reflected is 355W/m^2 or it is not. It’s not a question of division or subtraction. The above doesn’t address the question asked.

    You are who doesn’t know how to distinguish between a subtract and a division. I wrote:

    1368 W/m^2 / 1.35 reflected by the atmosphere and Earth’s surface = 1013.3 W/m^2

    1013.3 W/m^2 / 1.20 absorbed by the atmosphere = 844.4 W/m^2

    From this power, the surface only absorbs a power of 692.41 W.

    The only thing you have to do is to multiply the quotient by the divisor… LOL!!!

    1.35 X 844.4 W = ?????

    1.20 X 692.41 W = ?????

    LOL!

    😀

  368. Nasif Nahle April 23, 2011 at 4:07 pm #

    Let me show you, Neutrino, how to solve the problem…

    The quotient of the first division is 1013.3 W/m2…

    The divisor is 1.35…

    Multiply 1013.3 W/m2 x 1.35 = 1368 W/m2!!!!!! Wow!!!

    Now… The quotient of the second division is 844.4 W/m2…

    The divisor is 1.2…

    Now, multiply the divisor by the quotient…

    844.4 W/m2 x 1.2 = 1013.3 W/m2!!!! Wow Again!!!

    You see? That is what you should have do. LOL!

    😀

  369. Nasif Nahle April 23, 2011 at 4:11 pm #

    @Neutrino…

    You say:

    So I have to ask, what is the source for the 0.82?

    Unless you’re living in a fantasy world, you should understand what the concept “measurement” means.

    😀

  370. Nasif Nahle April 23, 2011 at 4:15 pm #

    @Neutrino…

    Ah, Neutrino… Lying again… You say:

    Here is where he actually said it:(my bold)
    “No! I’m not calculating absorbed flux, either SB equation calculates absorbed flux. SB equation is used to calculate the absorbed power.”

    But you said:

    Comment from: NeutrinoApril 15th, 2011 at 3:59 pm

    The Stefan-Boltzmann Law, which is what you wrote above, is the summation of the Planck Function. The Planck function describes emission from a hot body. It is not a formula that can calculate absorption.

    What did you say? LOL!

    😀

  371. Nasif Nahle April 23, 2011 at 4:20 pm #

    @Neutrino…

    You say:

    He has yet to produce a single reference that would support the above assertion. The simple reason why he cannot do this is because it does not exist.

    It is evident that you don’t read books on heat transfer.

    ANSWER THESE QUESTIONS, don’t evade! You have not answered a single one of my questions:

    What the following formula is for?

    Q = e (A) (σ) (T^4)

    Go on with your answer.

    Now tell us, what are the units obtained by using the above formula?

    Go on with your answer.

    Finally, tell us what those units are describing.

    Go on with your answer.

    😀

  372. gavin April 23, 2011 at 8:58 pm #

    I say Neutrino wins this debate hands down and this Nasif needs to refresh his gas physics and clean up his rhetoric before tackling climate science again

  373. gavin April 24, 2011 at 1:37 am #

    Ooops gas? I meant “radiative” physics sorry

  374. Nasif Nahle April 24, 2011 at 2:02 am #

    @Gavin…

    I say Neutrino wins this debate hands down and this Nasif needs to refresh his gas physics and clean up his rhetoric before tackling climate science again

    If Neutrino has not answered a solitary question, it only means that Neutrino doesn’t know the physics of heat transfer.

    Besides, he is incompetent to analyse simple arithmetics.

    So you opinion is just a fantasy, like AGW idea.

    😀

  375. Nasif Nahle April 24, 2011 at 2:21 am #

    Just only see his erratic arithmetics:

    He wrote his “analysis”:

    So if he actually did mean that the earth and atmosphere together reflect 1013.3W/m^2 then the math is even stranger.

    Total Entering
    1368W/m^2(same as before)
    1013.3W/m^2 reflected by the earth surface and atmosphere
    844.4W/m^2 absorbed by the atmosphere
    692.4W absorbed by the earth surface(assuming this is supposed to be W/m^2 and is simply a typo)

    Total absorbed or reflected by the earth surface and atmosphere:
    1013.3W/m^2 + 844.4W/m^2 + 692.4W/m^2 equals
    2550.1W/m^2

    However, what I wrote was as follows:

    1368 W/m^2 / 1.35 reflected by the atmosphere and Earth’s surface = 1013.3 W/m^2

    It is a division to know the energy left after reflection by the atmosphere.

    Neutrino had have to multiply, not to add. LOL!

    The correct analysis would be 1.35 * 1013.3 W/m^2 = 1368 W/m^2

    I wrote:

    1013.3 W/m^2 / 1.20 absorbed by the atmosphere = 844.4 W/m^2

    Again, a division to know how much energy is left after the amount left from reflection is applied.

    The correct analysis is 844.4 W/m^2 * 1.20 = 1013.3 W/m^2

    And again, 1013.3 W/m^2 * 1.35 = 1368 W/m^2

    Wow!!! Neutrino doesn’t know simple basic arithmetics!!!

    Then I wrote:

    From this power, the surface only absorbs a power multiplied by 0.82, i.e. 692.41 W/m^2,

    That means that, from the energy left after absorption and reflection by the atmosphere, the surface only absorbs 692.41 W/m^2 from the incident power of 844.4 W/m^2.

    See Neutrino’s ignorance on simple basic arithmetics?

    Then I wrote:

    from which 41.54 W/m^2 are emitted from the surface towards the outer space and 207.7 W/m^2 is dissipated by conduction, oceanic currents and friction.

    692.41 W/m^2 – 41.54 W/m^2 – 207.7 W/m^2 = 443.17 W/m^2

    Now, the additions:

    443.17 W/m^2 + 41.54 W/m^2 + 207.7 W/m^2 = 692.41 W/m^2 … Hehehe!

    Again, Neutrino doesn’t know simple basic arithmetics and he is incompetent to do correct analysis of physics procedures.

    😀

  376. Nasif Nahle April 24, 2011 at 2:25 am #

    Now I will stand here waiting for him answers my questions… LOL!

    😀

  377. Neutrino April 24, 2011 at 3:09 am #

    Nasif,

    The question I asked was not about how to do simple arithmetic but what your simple arithmetic represented.
    Are you asserting that 355W/m^2 or 1013W/m^2 are reflected?

    You just added this:
    Then I wrote:
    From this power, the surface only absorbs a power multiplied by 0.82, i.e. 692.41 W/m^2,
    That means that, from the energy left after absorption and reflection by the atmosphere, the surface only absorbs 692.41 W/m^2 from the incident power of 844.4 W/m^2.

    So it appears you are saying that the earth surface there is an incident of 844W/m^2, of which 692W/m^2 are absorbed.
    Correct?
    If this is what you are saying then you are agreeing with the method that I presented in the original article for calculating absorbed. Line 2) q = εQ.

    The problem with the above though is that if 844W/m^2 is the Incident flux at the earth surface and 692W/m^2 of it is absorbed then what happens to the other 152W/m^2?
    Since you have already accounted for what was reflected from the surface then where does the 152W/m^2 go? Which comes back to my original point about your breakdown, it does not add up.

    You continue with the accusations of lying:
    Ah, Neutrino… Lying again… You say:
    Here is where he actually said it:(my bold)
    “No! I’m not calculating absorbed flux, either SB equation calculates absorbed flux. SB equation is used to calculate the absorbed power.”
    But you said:
    Comment from: NeutrinoApril 15th, 2011 at 3:59 pm
    The Stefan-Boltzmann Law, which is what you wrote above, is the summation of the Planck Function. The Planck function describes emission from a hot body. It is not a formula that can calculate absorption.
    What did you say? LOL!

    What I did say, and have always been saying, is that the SB Law does not calculate Absorbed. What you quoted above seems to confirm that’s what I am saying.

    As to your questions:
    What the following formula is for?
    Q = e (A) (σ) (T^4)

    What you have written is the SB, it is the same formula as my number 3). As I have always maintained it is a formula to calculate the amount of emitted power from a surface.

    Now tell us, what are the units obtained by using the above formula?
    As written it is in units of Watts(W).

    Finally, tell us what those units are describing.
    Those units describe how much energy per second leaves the surface.

    Just go to any university site and you can easily verify this. As an example here is the Georgia State relevant site.

  378. Neutrino April 24, 2011 at 3:23 am #

    Just to be clear,

    Any budget has to sum to the total input. If it doesn’t then the budget is not valid.

    As an example:
    Total Incident = I
    Atmosphere+Surface Reflects = X
    Atmosphere Absorbs = Y
    Surface Absorbs = Z

    X+Y+Z has to equal I, if it does not then a mistake has been made.

  379. Nasif Nahle April 24, 2011 at 6:47 am #

    @Neutrino…

    The question I asked was not about how to do simple arithmetic but what your simple arithmetic represented.
    Are you asserting that 355W/m^2 or 1013W/m^2 are reflected?

    Now you’re on the correct way to arithmetics…

    Peixoto and Oort, Physics of Climate, pp. 94 and 366; so don’t invent that those are my assertions. 🙂

    So it appears you are saying that the earth surface there is an incident of 844W/m^2, of which 692W/m^2 are absorbed.
    Correct?
    If this is what you are saying then you are agreeing with the method that I presented in the original article for calculating absorbed. Line 2) q = εQ.

    No, it doesn’t appear that I am saying that the Earth’s surface incident is 844 W/m^2… It is what authors say that is the incident solar direct normal irradiance on the surface of the Earth:

    http://solardat.uoregon.edu/SolarRadiationBasics.html

    Therefore about 1000 w/m2 of the incident solar radiation reaches the earth’s surface without being significantly scattered.”

    http://edmall.gsfc.nasa.gov/inv99Project.Site/Pages/science-briefs/ed-stickler/ed-irradiance.html

    S~ 1000 W/m2 (Clear day solar insolation on a surface perpendicular to incoming solar radiation. This value actually varies greatly due to atmospheric variables.)

    http://ocw.tudelft.nl/fileadmin/ocw/courses/SolarCells/res00026/CH2_Solar_radiation.pdf

    The value of 1000 W/m2 was incorporated to become a standard. This value of the irradiance is close to the maximum received at the earth’s surface.

    You say:

    The problem with the above though is that if 844W/m^2 is the Incident flux at the earth surface and 692W/m^2 of it is absorbed then what happens to the other 152W/m^2?

    Now your arithmetics fails again… Why you didn’t go to kindergarten?

    Read what I wrote in my post:

    …from which 41.54 W/m^2 are emitted from the surface towards the outer space and 207.7 W/m^2 is dissipated by conduction, oceanic currents and friction.

    692.41 W/m^2 – 41.54 W/m^2 – 207.7 W/m^2 = 443.17 W/m^2

    Now, the additions:

    443.17 W/m^2 + 41.54 W/m^2 + 207.7 W/m^2 = 692.41 W/m^2 …

    You’re quite lost with your arithmetics, Neutrino…

    LOL!

    You wrote:

    What you have written is the SB, it is the same formula as my number 3). As I have always maintained it is a formula to calculate the amount of emitted power from a surface.

    “Now tell us, what are the units obtained by using the above formula?”
    As written it is in units of Watts(W).

    “Finally, tell us what those units are describing.”
    Those units describe how much energy per second leaves the surface.

    Please, go back to your university instead of recommending me to go to mine. You don’t know what those units describe… You’re lost, Neutrino… You say:

    “Finally, tell us what those units are describing.”
    Those units describe how much energy per second leaves the surface.

    YOU ARE WRONG! Watts is not units of flux of energy per second… You are wrong! You don’t know physics!

    Watts (W) is units for power… ONLY POWER. It doesn’t denotes time… It is TIMELESS, neither flux… BECAUSE WE ARE NOT EXPRESSING AREA OR LINEAR FLUX. You are wrong, Neutrino. It’s evident you know nothing of physics. This small mistake puts you on a very compromised place of credibility. It is basic physics!!!

    😀

  380. Nasif Nahle April 24, 2011 at 7:11 am #

    Following Gavin’s and Luke’s “arguments”… POOR NEUTRINO’S UNIVERSITY OF GEORGIA STATE… 😀

  381. Neutrino April 24, 2011 at 8:53 am #

    Nasif said:
    YOU ARE WRONG! Watts is not units of flux of energy per second… You are wrong! You don’t know physics!
    Watts (W) is units for power… ONLY POWER. It doesn’t denotes time… It is TIMELESS, neither flux… BECAUSE WE ARE NOT EXPRESSING AREA OR LINEAR FLUX. You are wrong, Neutrino. It’s evident you know nothing of physics. This small mistake puts you on a very compromised place of credibility. It is basic physics!!!

    First. I did not say that a Watt is ”units of flux of energy per second”. What I did say was:
    Those units(Watts) describe how much energy per second leaves the surface.”.
    A flux has units of W/m^2 so the use of the word “flux” as you used it is inappropriate.

    Saying that Power(W) doesn’t denote time is bizarre as the definition of 1Watt is 1Joule/second. A Watt is an amount of energy(J) per unit time(s).

    Just as a demonstration of this look at the units of the equation:
    Q = e (A) (σ) (T^4)
    e: dimensionless quantity
    A: m^2
    σ: J / ((K^4) (m^2) s)
    T: K
    Put that all together and we get:
    Q = (m^2) * (J / (K^4 * m^2 * s)) * (K^4)
    Q = J * (K^4) * (m^2) / ((K^4) * (m^2) * s)
    Q = J (K^4) (m^2) / ((K^4) (m^2) s)
    And that leaves us with:
    Q = J / s
    Which is simply a Watt(W).

    I agree this is all extremely basic physics that anyone can open a text book or go online to confirm.

  382. Nasif Nahle April 24, 2011 at 12:44 pm #

    @Neutrino…

    You said:

    how much energy per second

    All has been said.

    You said:

    A flux has units of W/m^2 so the use of the word “flux” as you used it is inappropriate.

    Exactly! But you’re saying it after I said it. Sorry…

    You wrote:

    σ: J / ((K^4) (m^2) s) Bolds are mine…

    Where you got that “s” if you wrote:

    Q = e (A) (σ) (T^4)?

    A new S-B constant?

    You say:

    And that leaves us with:
    Q = J / s
    Which is simply a Watt(W).

    But… Watt is not energy, but power… And you said:

    Those units describe how much energy per second leaves the surface. Bolds are mine.

    Which is absolutely incorrect…

    Much of the online info is wrong… so it is better to read good books on science.

    😀

  383. Nasif Nahle April 24, 2011 at 1:07 pm #

    @Neutrino…

    The final blow to your arithmetics:

    You wrote your new Stefan-Boltzmann constant in the following way:

    Put that all together and we get:
    Q = (m^2) * (J / (K^4 * m^2 * s)) * (K^4)
    Bolds are mine…

    That is… Your wrote J/(K^4 * m^2 * s) as the new Stefan-Boltzmann constant. BUT YOU ARE WRONG AGAIN. It should have been written in the following way:

    W = J/s

    In the expression W/m^2 K^4, substituting W by J/s, it MUST be written as follows:

    (J/s) / (m^2 K^4)

    That expression will give you a false result with false units:

    (J m^2 K^4) / s

    So you cannot eliminate K^4 and your result will be K^8

    Additionally, you will not be able to eliminate m^2, so your result will have m^4.

    And your units, in the final result, would be expressed in (J m^4 K^8) / s

    That’s why I told you have invented a new S-B constant… Sorry, Neutrino… these are not just typos…

    Anyway, and sincerely, I congratulate with you because you have had the courage to write an essay on AGW idea. 🙂

    NSN

  384. Neutrino April 24, 2011 at 1:11 pm #

    Once again I am at a loss as to how to respond to Nasif.

    But… Watt is not energy, but power… And you said:
    Those units describe how much energy per second leaves the surface. Bolds are mine.
    Which is absolutely incorrect…

    This is a stunning point we have arrived at. Nasif is actually trying to argue that a Watt is not a Joule per second.

    Do not take my word for it, here is the National Institute of Standards and Technology definition of the SI unit Watt. (Table 3 Line 7)

    power, radiant flux watt W J/s m2•kg•s-3

  385. Neutrino April 24, 2011 at 1:22 pm #

    Try again Nasif,

    Stefan-Boltzmann Constant

    Q = (m^2) * (J / (K^4 * m^2 * s)) * (K^4)

    As written has units J/s. I realize it is not always easy to read inline equations but all the m’s and K’s cancel in the above as it is written.

    And also this article was not about the AGW idea as you call it but radiative transfer. Which of course is involved in the GHE which AGW is an enhancement of.

  386. Nasif Nahle April 24, 2011 at 1:37 pm #

    @Neutrino…

    More lies on what I said…

    Watts is not units for energy per second, but for power. Do not get more confused…

    Do not take my word for it, here is the National Institute of Standards and Technology definition of the SI unit Watt. (Table 3 Line 7)

    “power, radiant flux watt W J/s m2•kg•s-3”

    You see why I said many things are wrong in the internet? Here the correct units for power:

    http://en.wikipedia.org/wiki/Power_(physics)

    http://www.aps.org/policy/reports/popa-reports/energy/units.cfm

    http://www.smpspowersupply.com/powerunits.html

    http://www.eformulae.com/engineering/units_of_power.php

    http://www.convert-me.com/en/convert/power

    UNITS FOR FLUX OF ENERGY…

    http://spacemath.gsfc.nasa.gov/weekly/6Page84.pdf

    http://www.gordonengland.co.uk/conversion/heatflux.htm

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

    Etcetera…

  387. Nasif Nahle April 24, 2011 at 1:43 pm #

    @Neutrino…

    Watts = J/s = cal/s = POWER

    Joule = cal = W*s = ENERGY

    Watt/m^2 = (J/s) / m^2 = Flux of power

    J/m^2 = (W*s) / m^2 = radiant flux or energy flux.

    You cannot say that W is energy because it is POWER; and you said:

    Those units(Watts) describe how much energy per second leaves the surface.”

    See your confusion?

  388. Nasif Nahle April 24, 2011 at 2:00 pm #

    From NASA site at http://spacemath.gsfc.nasa.gov/weekly/6Page84.pdf:

    Total Energy – Joules and ergs – The total amount of energy in various forms (kinetic, potential, magnetic, thermal, gravitational)

    Power – Watts, Joules/second or ergs/second – the rate at which energy is produced or consumed in time. Power = Energy/Time

    Flux – Watts/meter2, Joules/sec/meter2 or ergs/sec/meter2 – the rate with which energy flows through a given area in given amount of time: Flux=Power/Area

    I hope it is clear now. Power is not energy, either energy flux per second.

    😀

  389. Neutrino April 24, 2011 at 2:13 pm #

    This is pointless.

    I say a Watt is Energy per second. W = J/s.

    Nasif cries foul saying I am wrong.
    Watts is not units for energy per second, but for power. Do not get more confused…

    Nasif then produces references that say a Watt is Joule per second. W = J/s.

    Which is exactly where we started with what I said.

    To add to the insanity Nasif adds:
    You cannot say that W is energy because it is POWER; and you said:
    “Those units(Watts) describe how much energy per second leaves the surface.”
    See your confusion?

    No, all I see is your inability to read, what you are quoting of me does not say power is energy, it quite clearly says power is energy per unit time.

    I have no idea what you are objecting to.

  390. Nasif Nahle April 24, 2011 at 2:46 pm #

    @Neutrino…

    You call my arguments insane, I call your arguments ignorance.

    Your essay is about the power absorbed by the Earth and you conclude with an erroneous number:

    T = 278.6K (or 5.4C)

    And it is the product of ignorance about reality. The world doesn’t need a GHE to be maintained warm (I’m rounding up the numbers):

    1368 W/m^2 is the flux of POWER received on top of the atmosphere.

    From this amount of POWER per square meter, 35% is reflected by the atmosphere:

    Power per square meter (flux) entering towards the surface after reflected by the atmosphere = 1368 W/m^2 / 1.35 reflected by the atmosphere and Earth’s surface = 1013 W/m^2

    Power per square meter (flux) entering towards the surface after absorption by the atmosphere = 1013 W/m^2 / 1.20 = 844 W/m^2

    844 W/m^2 strikes the surface, but only 692 W/m^2 are absorbed by the surface.

    From those 692 W/m^2 of power absorbed by the surface, 41 W/m^2 are lost towards the space (atmospheric window). Therefore, only 651 W/m^2 are left in the surface.

    From those 651 W/m^2, 208 W/m^2 are dissipated by conduction, oceanic currents and friction; therefore, only 443 W/m^2 are left in the surface as effective power absorbed.

    This amount of power causes a temperature of 297 K, which is in accordance with the standard average of temperature of the Earth.

    Who needs, again, who needs a GHE, a warming of a warm surface by a colder system, etc.?

    These are correct calculations, not the calculations made by Neutrino, which gave an imaginary temperature.

    😀

  391. Nasif Nahle April 24, 2011 at 3:06 pm #

    @Readers…

    I have given references that confirms that the solar irradiance striking on the surface of the Earth is around 1000 W/m^2

    http://solardat.uoregon.edu/SolarRadiationBasics.html

    “Therefore about 1000 w/m^2 of the incident solar radiation reaches the earth’s surface without being significantly scattered.”

    http://edmall.gsfc.nasa.gov/inv99Project.Site/Pages/science-briefs/ed-stickler/ed-irradiance.html

    “S~ 1000 W/m^2 (Clear day solar insolation on a surface perpendicular to incoming solar radiation. This value actually varies greatly due to atmospheric variables.)”

    http://ocw.tudelft.nl/fileadmin/ocw/courses/SolarCells/res00026/CH2_Solar_radiation.pdf

    “The value of 1000 W/m^2 was incorporated to become a standard. This value of the irradiance is close to the maximum received at the earth’s surface.”

    And thus my three books on astrophysics, my four books on Ecology, my two books on Climatology say.

    The same is by direct measurements of the solar irradiance on groung by radiometers, pyrometers and pyrgeometers adjusted to measure the solar irradiance EXCLUSIVELY.

    The question is, why Trenberth and Kiehl, the IPCC, Neutrino and all AGWers say the solar irradiance on the surface of the Earth is 161 W/m^2?

    Why this big lie?

    The oceans alone are the main systems that absorb 98% of the absorbed solar irradiance of the whole hemisphere and, additionaly, they are the main systems that store and distribute power. Oceans continue emitting power during nighttime, and it is the mechanism by which the atmosphere on the night side of the Earth is kept warm.

    If we adhere ourselves to the reality of this world, there is no need of greenhouse effects, neither backradiations, etc.

    The solar irradiance received during daytime is given by the incident angle of solar quantum/waves streams and the attenuating factors of the atmosphere.

    Then… Why to lie?

    NSN

  392. gavin April 24, 2011 at 6:17 pm #

    Guys; this is so easily settled by a bit of a history lesson given ‘Power = Work divided by Time’ for all engine comparison when “Horsepower” had to be translated into engineering terms for early steam engines, and later automotive piston engines, turbines and electric motors.

    I can recall the old illustration used to define Horsepower where a typical English draft horse lifted a given weight up a well shaft in a certain time. Engines were likewise rated by “piston speed” and with electric motors we had a measure “Torque” relating foot pounds with RPM however each country had their own horse to start with.

    Its worth reading how standards for power evolved, none the least is the American SAE system that still hangs about after metrification elsewhere.

    Sorry Nasif; you must be a very young bloke in this Power game

  393. Mack April 24, 2011 at 6:54 pm #

    Gavin,
    I think for a moment Nasif maybe was thinking that power from the sun was timeless, (reality) .

  394. Mack April 24, 2011 at 7:09 pm #

    …is timeless

  395. Nasif Nahle April 24, 2011 at 7:23 pm #

    @Gavin…

    Sorry Nasif; you must be a very young bloke in this Power game

    Sorry Gavin… You don’t know elemental physics. Watts has no units of time, whatever you say.

    Show me where the units of time are included in the expression 1 W.

    😀

  396. Nasif Nahle April 24, 2011 at 7:24 pm #

    have given references that confirms that the solar irradiance striking on the surface of the Earth is around 1000 W/m^2

    http://solardat.uoregon.edu/SolarRadiationBasics.html

    “Therefore about 1000 w/m^2 of the incident solar radiation reaches the earth’s surface without being significantly scattered.”

    http://edmall.gsfc.nasa.gov/inv99Project.Site/Pages/science-briefs/ed-stickler/ed-irradiance.html

    “S~ 1000 W/m^2 (Clear day solar insolation on a surface perpendicular to incoming solar radiation. This value actually varies greatly due to atmospheric variables.)”

    http://ocw.tudelft.nl/fileadmin/ocw/courses/SolarCells/res00026/CH2_Solar_radiation.pdf

    “The value of 1000 W/m^2 was incorporated to become a standard. This value of the irradiance is close to the maximum received at the earth’s surface.”

    And thus my three books on astrophysics, my four books on Ecology, my two books on Climatology say.

    The same is by direct measurements of the solar irradiance on groung by radiometers, pyrometers and pyrgeometers adjusted to measure the solar irradiance EXCLUSIVELY.

    The question is, why Trenberth and Kiehl, the IPCC, Neutrino and all AGWers say the solar irradiance on the surface of the Earth is 161 W/m^2?

    Why this big lie?

    The oceans alone are the main systems that absorb 98% of the absorbed solar irradiance of the whole hemisphere and, additionaly, they are the main systems that store and distribute power. Oceans continue emitting power during nighttime, and it is the mechanism by which the atmosphere on the night side of the Earth is kept warm.

    If we adhere ourselves to the reality of this world, there is no need of greenhouse effects, neither backradiations, etc.

    The solar irradiance received during daytime is given by the incident angle of solar quantum/waves streams and the attenuating factors of the atmosphere.

    Then… Why to lie?

    😀

  397. gavin April 24, 2011 at 9:02 pm #

    see mJ = (mW/cm2) x (Area in cm2) x (Time in sec) etc

    http://www.grc.nasa.gov/WWW/k-12/Numbers/Math/Mathematical_Thinking/sun12.htm

  398. gavin April 24, 2011 at 9:55 pm #

    Nasif would have us believe people such as these good folk

    http://ieeexplore.ieee.org/xpl/freeabs_all.jsp?arnumber=5652453

    http://www.arm.gov/measurements/irradswbbdiffdn

    and anyone else working on the physical basis for climate change within the IPCC framework are wasting their time

  399. Nasif Nahle April 25, 2011 at 1:49 am #

    @Gavin…

    Those articles are pure speculation. REALITY is that about 1000 W/m2 strikes on the surface of the Earth, believe it or not… IT IS A MEASUREMENT, that is REALITY, not speculations.

    😀

  400. Nasif Nahle April 25, 2011 at 1:51 am #

    @Gavin…

    Besides, those links supports what I say about induced emission. LOL!

    😀

  401. Nasif Nahle April 25, 2011 at 1:54 am #

    @Gavin…

    And 137 mW/cm^2 is 1370 watt / (m^2)

    Hah!

    😀

  402. kuhnkat April 25, 2011 at 7:36 am #

    Gavin,

    “…and anyone else working on the physical basis for climate change within the IPCC framework are wasting their time”

    Since they are just now discovering that the sun is not always a good blackbody I think they ARE wasting their time until they do a better job of measuring the actual output of the sun instead of trying to play Climate Model games with some illusionary STANDARD output that may have never actually happened!!!

  403. gavin April 25, 2011 at 9:43 am #

    kuhnkat: “until they do a better job of measuring the actual output of the sun” THEY?

    Nasif: “what I say about induced emission” WHAT do you say?

    Guys; don’t get me wrong about laser physics cause its not what I’m on about other than seems it’s a big area of instrument and measurement systems research these days and you should too should wonder why.

    Nasif; IMO you haven’t answered Neutrino on the fundamental question of radiative energy doing work in a given time. The only thing you are consistent in is knocking AGW as a theme and that’s not playing physics.

    Please drop those terms like “reality or truth as opposite to lie” until you can show how you derived them from a recognized math process. Thanks

  404. Nasif Nahle April 25, 2011 at 3:18 pm #

    @Gavin…

    Nasif: “what I say about induced emission” WHAT do you say?

    Read my essays on this blog. I say many things about total emissivity that it seems you don’t undertand.

    Nasif; IMO you haven’t answered Neutrino on the fundamental question of radiative energy doing work in a given time. The only thing you are consistent in is knocking AGW as a theme and that’s not playing physics.

    It’s your opinion, just that. I have talked about REAL science, not fantasies created by numerology. I have not left a single Neutrino’s question without an answer; instead, Neutrino has not answered questions attained to his essay. I remember you that I’m not defending an essay, but Neutrino; therefore, Neutrino is who MUST answer our questions.

    Regarding the energy doing work, are you against the concept or what?

    😀

  405. gavin April 25, 2011 at 6:18 pm #

    Nasif; you just keep on jumping around.

    Induced Emission = “Stimulated Emission” as in lasers where the induced photon has a phase and direction relationship with amplification

    http://www.thefreedictionary.com/Stimulated+emission+of+radiation

    Total Emissivity = The RATIO of the total amount of radiation emitted. Max = Black Body =1

    Process/industrial instruments and controls handbook By Gregory K. McMillan, Douglas M. Considine – Google Books com.au

    Sorry, try again hey

    “Numerology” is for the birds or something but not me so don’t come back with the high and mighty again.

    “Regarding the energy doing work” ? Mate; when just in the workforce by a week they gave me a loan of their factory 15″ shifter on a daily basis so I could earn my keep helping with the steam heated drying process and a whole lot of other ancient paper making machines. Since I was a tall skinny bloke then it was necessary to quickly learn how to undo huge nuts and bolts that had ceased after being previously tightened by our part time weight lifters. Our boss there was a former railway loco tire fitter. We all learned Newtonian physics the hard way.

    Applied physics became my specialty in a wide range of science and technology from then on. Trouble shooting advanced measurement systems during and after their implementation was something I did on short notice most of the time and often for visiting consultants. I doubt very much you could join such a team after struggling with your reasoning and rhetoric aimed at other professionals in the field.

    Apologies to Neutrino but I had to say it eventually

  406. mkelly April 26, 2011 at 1:51 am #

    Gavin and Neutrino name one profitable product on the market that uses the GHE caused by CO2. A blanket, coat, hat, house wrap, etc.

    If I manufacture a product claiming that I can get back more energy than put in as measured by W/m^2 would you by it?

  407. Nasif Nahle April 26, 2011 at 3:25 am #

    @Gavin…

    You’re off topic… LOL!

    The problem here is not induced emission, which applies to everything in the know universe, so the idea of constructing lasers. LOL!

    The point on this thread is that the numbers of Neutrino and AGWers are plainly wrong because they don’t coincide with REAL measurements.

    Starting from the incident solar radiation on top of the atmosphere, which IS NOT 342 W/m2, BUT 1368 W/m2. The incident solar radiation on the surface is ~1000 W/m2, NOT 161 W/m2.

    AGW numbers are unreal, unphysical, pseudoscientific. Period!

    😀

  408. Neutrino April 26, 2011 at 4:26 am #

    Nasif, do you understand the concept of an average?

    The solar constant is 1368W/m^2, but that is not the average ToA value. For 12hours every day the earth is turned away from the sun and receives 0W/m^2. Just by this fact alone average ToA cannot be 1368W/m^2.

    As for what gets to the surface of the earth. I recommend looking at the SURFRAD site from NOAA. Pick a site and look at the measured Downwelling Solar. For Example:

    Goodwin Creek December 1 2010:
    Downwelling peaks at just 600W/m^2.
    That was the peak for the day but what was the average?

    Goodwin Creek July 1 2010:
    Downwelling peaks at over 1100W/m^2.
    That was the peak for the day but what was the average?

    Goodwin Creek 2010:
    December average looks to be around 100W/m^2, July average looks to be around 275W/m^2. The average for the year seems to be approximately 200W/m^2.
    How do you explain these actual measurements?

    The ToA is an average value not the instantaneous peak value at highnoon in the summer.

  409. Nasif Nahle April 26, 2011 at 4:27 am #

    Just to be clear on real physics:

    Considering a broad collision process whereby a large number of particles or agents randomly and repeatedly interact in pairs, with prearranged conservation law(s).

    The stationary single-particle distribution function capitalizes on entropy-like functional.

    This condition amounts to a factorization property related to the binary collision law, from which the correct calculation of phase space directly goes after.

    Thanks for reading!

    😀

  410. Nasif Nahle April 26, 2011 at 4:29 am #

    Therefore, Neutrino is wrong, absolutely wrong.

    NSN

  411. Nasif Nahle April 26, 2011 at 4:32 am #

    @Neutrino…

    It’s instantanous measurements made by satellites on top of the atmosphere for the solar power striking the outer layer of the atmosphere and by instruments on the surface for the solar power striking the surface of the planet… You’re lost Neutrino.

    😀

  412. Neutrino April 26, 2011 at 4:38 am #

    Ill ask again Nasif,

    What is the average value of ToA?

    Since it is night time for half of the time what is the satellite measurement during that time period?

  413. Neutrino April 26, 2011 at 4:40 am #

    mkelly,

    There are no commercial products, profitable or otherwise, that use nuclear fusion. Does that mean that nuclear fusion is not real?

    I can’t go to the hardware store and buy a device that takes advantage of the relativistic effects of length contraction. Does that mean relativity is wrong?

    The earth does not “get back more” energy than the input of 1.22*10^17W from the sun.

  414. Nasif Nahle April 26, 2011 at 5:02 am #

    @Neutrino…

    Since it is night time for half of the time what is the satellite measurement during that time period?

    Since the human body is only alive by ~70 years with a temperature of around 37 °C and it is dead for the eternity with a temperature of 0 °C, what is the average temperature of the human body?

    You cannot average night with day because you don’t have the variable present at night on the illuminated hemisphere; you’re doing pseudophysics. You can average only when the solar beams are striking on a hemisphere. Additionally, the opposite hemisphere is receiving the same amount of solar irradiance. You are wrong, Neutrino.

    1368 W/m2 is the MEASURED flux of solar power received by the top of the atmosphere ON THE SIDE FACING THE SUN. You are mixing pears with apples:
    The net solar power received by the outer sphere is:
    Pnet = 1404.14 W/m^2 x 2.812 x 10^23 m^2 = 3.95 x 10^26 W
    Which coincides perfectly with the satellite measurements. No matter how you twist the numbers and invent pseudophysics, the amount of solar power received on each square meter of the outer sphere, where the Earth is placed is 1368 W, whatever you think.

    On the other hand, we have here pyrgeometers, radiometers and pyrometers measuring the amount of solar power received on ground. Specifically adjusted to detect solar irradiance. The instantaneous measurement at this location and hour of the day is 604.28 W/m^2.

    Now tell me, if it is an instantaneous measurement and you say that the energy received on top of the atmosphere is 342 W/m^2, where the remainder excess of power came from? You see how yours is pseudophysics?

    😀

  415. Neutrino April 26, 2011 at 5:25 am #

    This truly is an insane conversation.

    The concept of an average is not pseudo-science. Over a 24h period any place on earth is facing the sun only a portion of that, as such at most that particular point can only receive 1368W/m^2 during the day and 0 during the night.
    So the average cannot be 1368W/m^2.

    I truly cannot believe I am having an argument about what an average is with someone who claims they are a scientist.

    For a final time,
    The average of 342W/m^2 is for a 24h period over the entire year. Not an instantaneous measure in mid afternoon during spring.

  416. gavin April 26, 2011 at 7:33 am #

    “You cannot average night with day” oh yeah; we’ve been measuring various conditions to average for quite a while now and some fall under the term “ambient” What about the sea?

    “you don’t have the variable present at night on the illuminated hemisphere” sure; but can we have this as a mathematical expression? On the other hand why do we have to deal with this lot

    “The stationary single-particle distribution function capitalizes on entropy-like functional” etc that could be grounded by some reading here

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

    but is better placed in these circles

    http://math.nyu.edu/seminars/probability_seminar.html

    And whois “D”

    Will the real Nasif please stand up

  417. Neutrino April 26, 2011 at 7:39 am #

    Last ditch effort to demonstrate the average value of ToA:

    Picture a day on the equator, at noon the sun is directly overhead(ө = 0) every 20min the earth rotates 5degrees and because of this the ToA flux is attenuated by cos(ө). At night the sun is behind the earth so no flux is received.

    Below is the ToA flux every 20min for a spot on the equator starting at high noon:
    Time________Angle__Cos(ө)__Flux W/m^2
    12:00-12:20 000 1.00 1368.00
    12:20-12:40 005 1.00 1362.79
    12:40-13:00 010 0.98 1347.22
    13:00-13:20 015 0.97 1321.39
    13:20-13:40 020 0.94 1285.50
    13:40-14:00 025 0.91 1239.83
    14:00-14:20 030 0.87 1184.72
    14:20-14:40 035 0.82 1120.60
    14:40-15:00 040 0.77 1047.95
    15:00-15:20 045 0.71 967.32
    15:20-15:40 050 0.64 879.33
    15:40-16:00 055 0.57 784.65
    16:00-16:20 060 0.50 684.00
    16:20-16:40 065 0.42 578.14
    16:40-17:00 070 0.34 467.88
    17:00-17:20 075 0.26 354.06
    17:20-17:40 080 0.17 237.55
    17:40-18:00 085 0.09 119.23
    18:00-18:20 090 0.00 0.00
    Night time
    06:00-06:20 270 0.00 0.00
    06:20-06:40 275 0.09 119.23
    06:40-07:00 280 0.17 237.55
    07:00-07:20 285 0.26 354.06
    07:20-07:40 290 0.34 467.88
    07:40-08:00 295 0.42 578.14
    08:00-08:20 300 0.50 684.00
    08:20-08:40 305 0.57 784.65
    08:40-09:00 310 0.64 879.33
    09:00-09:20 315 0.71 967.32
    09:20-09:40 320 0.77 1047.95
    09:40-10:00 325 0.82 1120.60
    10:00-10:20 330 0.87 1184.72
    10:20-10:40 335 0.91 1239.83
    10:40-11:00 340 0.94 1285.50
    11:00-11:20 345 0.97 1321.39
    11:20-11:40 350 0.98 1347.22
    11:40-12:00 355 1.00 1362.79

    The average flux, over a day, received by a point on the equator is the sum of the flux each 20min divided by 72(number of 20min intervals in a 24hour period). The result is 435W/m^2.

    So at any given time the average is 435W/m^2, but the instantaneous can range from 1368W/m^2 all the way down to zero.

    Doing this for every latitude, ie averaged over the entire globe, yields a value of 342W/m^2.

    This isn’t rocket science, it’s just simple geometry. It’s neither controversial nor pseudo-science, it simply is an average of the total flux received.

  418. Nasif Nahle April 26, 2011 at 10:19 am #

    @Neutrino…

    The concept of an average is not pseudo-science. Over a 24h period any place on earth is facing the sun only a portion of that, as such at most that particular point can only receive 1368W/m^2 during the day and 0 during the night.
    So the average cannot be 1368W/m^2.

    The Sun shines over your head during nightime also… LOOOL!!!

    That’s the root of your falacy.

    Satellite measurements are not averaged over the whole surface of the Earth on a daily basis, but on a yearly basis.

    No matter how much you twist and torture mathematics; satellites measured the solar irradiance striking on top of the atmosphere and it is 1368 W/m^2 during daylight, not 342 W/m^2.

    😀

  419. Nasif Nahle April 26, 2011 at 10:22 am #

    Get real:

    The solar irradiance striking on the surface has been measured also. Do not think it is the amount of power absorbed by the surface, but the amount of power received by the surface after attenuation by the atmosphere in a clear day and with solar irradiance beams striking perpendicularly on the surface:
    http://solardat.uoregon.edu/SolarRadiationBasics.html

    “Therefore about 1000 w/m^2 of the incident solar radiation reaches the earth’s surface without being significantly scattered.”

    http://edmall.gsfc.nasa.gov/inv99Project.Site/Pages/science-briefs/ed-stickler/ed-irradiance.html

    “S~ 1000 W/m^2 (Clear day solar insolation on a surface perpendicular to incoming solar radiation. This value actually varies greatly due to atmospheric variables.)”

    http://ocw.tudelft.nl/fileadmin/ocw/courses/SolarCells/res00026/CH2_Solar_radiation.pdf

    “The value of 1000 W/m^2 was incorporated to become a standard. This value of the irradiance is close to the maximum received at the earth’s surface.”

    LOL!

  420. Nasif Nahle April 26, 2011 at 10:27 am #

    And more real:

    The measured flux of solar power striking on top of the Earth’s atmosphere is 1368 W/m^2. That’s a measurement made by satellite instruments. It’s reality:

    http://www.eoearth.org/article/Solar_radiation

    From Materials Technology Limited:

    http://www.drb-mattech.co.uk/uv%20map.html

    “Solar Constant – The solar constant is the amount of energy received at the top of the Earth’s atmosphere on a surface oriented perpendicular to the Sun’s rays (at the mean distance of the Earth from the Sun). The generally accepted solar constant of 1368 W/m2 is a satellite measured yearly average.”. Bolds are mine.

    From NASA site:

    http://earthobservatory.nasa.gov/Features/SORCE/sorce_03.php

    If that were all there was to the Earth’s radiative balance, scientists studying the Sun would have probably long since moved on to another climate-related problem. Analyzing the Sun and its affects on climate, however, is further complicated by the fact that the amount of radiation arriving from the Sun is not constant. It varies from the average value of the TSI—1,368 W/m2—on a daily basis. Bolds are mine.

    However, AGW proponents have changed this measurement and say that the energy received on top of the Earth’s atmosphere is 342 W/m^2. And the fantasy starts here because they say that the solar irradiance striking the surface is 161 W/m2:

    http://chriscolose.files.wordpress.com/2010/03/kiehl4.jpg

    LOL!

  421. Nasif Nahle April 26, 2011 at 10:30 am #

    And the formula of insolation at any place of the Earth, on the surface (Land and oceans) is:

    I = Solar irradiance upon the surface * (Cos Z)

    Now make your calculations.

    😀

  422. Nasif Nahle April 26, 2011 at 10:36 am #

    The fallacy of Neutrino and AGWers is that they divide a load of power that must not be divided because it is a quantity received on the surface area of the outer sphere.

    Neutrino thinks that the outer sphere is the surface of the Earth, then his pseudoscience. The formula to obtain the amount of solar power striking on the surface of the Earth (LAND AND OCEANS) is:

    I = Solar irradiance upon the surface * (Cos Z)

    Not on the top of the atmosphere where the solar constant doesn’t change. From there the origin of the fallacious argument of Neutrino and AGWers.

    I know that Neutrino doesn’t know what the formula is going on because he puts the sky on the floor and the floor on the sky.

    NSN

  423. Nasif Nahle April 26, 2011 at 10:50 am #

    I call your attention to the following article just for you, the kind readers, see the fallacy of Neutrino:

    http://www.eoearth.org/article/Solar_radiation

    Please, read the graph on “Total Solar Irradiance: Original Data (top) and Composite (bottom)”.

    Take note of the numbers on the vertical axis, which correspond to the solar irradiance on top of the atmosphere year by year.

    Do you see any 342 W/m2 that Neutrino, AGWers and Trenberth-Kiehl reports as the ANNUAL average solar radiation on top of the atmosphere?

    NO!

    Do you see, by a chance the number “calculated” by Neutrino of 239 W/m2?

    NO!

    So, here you can see very clearly Neutrino’s and AGWers’ paeudoscience.

    The average is on a yearly basis, not on an hourly basis and the angle of incidence has nothing to do with it because it is not insolation, but total solar irradiance on top of the atmosphere. We could say also that it is the solar constant for the whole outer sphere.

    NSN

  424. wayne April 26, 2011 at 11:32 am #

    Nasif:

    “The Sun shines over your head during nightime also… LOOOL!!!”

    That’s the first thing that caught my eye too reading neutrino’s last ditch effort!
    Whoa… that’s too much!

    It seems person’s today calling themselves scientists have been trained far too much on computers and equations with little commonsense and logic to know what is actually happening in the actual real world that the equations represent.

    Nasif, take these with a grain of salt… they are a couple of energy budgets using KT97 and TFK09’s data for the overall averages of reflection (albedo), irradiance at the surface, and measured TOA LW upwelling radiation. These are very rough, just a first stab at a cosine weighted across 24 hour view but they give you a closer idea of what actually occurs with Trenberth’s numbers.

    http://i56.tinypic.com/avc5g.jpg : KT97
    http://i53.tinypic.com/ir6lie.jpg : TFK09

    They need some work:
    (o) Need a better estimate of the actual evaporation at nighttime.
    (o) The exact time that the minimum and maximum temperatures are reached might need to be moved 30 minutes or one hour sooner.
    (o) Window radiation is strictly gauged by the surface temperature and I don’t know if this is correct, but, the average s hould be close anyway.
    (o) …

    What do you think? Better?

  425. wayne April 26, 2011 at 11:36 am #

    Nasif, I figure in this case a picture is worth a MILLION words, at least.
    LOL! ☺

  426. L.J. Ryan April 26, 2011 at 3:41 pm #

    Neutrino;

    You are trying to minimize the solar input component relevance to earth’s temperature while manipulating the actual data to maximize a nonexistent atmospheric forcing component. To illustrate your error and build on Nasif Nahle’s point, consider you time plot data. Employing solely GHG physics and affiliated models, 1368 W/m^2 confers what surface temperature…specific ToA time:

    Time________Angle__Cos(ө)__Flux W/m^2
    12:00-12:20 000 1.00 1368.00

    To further the understanding of all interested readers, please provide a like plot data for back radiation and associated surface temperatures at all corresponding ToA intervals.

  427. Nasif Nahle April 26, 2011 at 3:57 pm #

    @Wayne…

    By far, much better! Please, continue working on this. As you say, it’s just a matter of refinement.

    The atmospheric window fluctuates according to the materials of the surface and the energy retained by the surface, the subsurface materials and the oceans; however, the coefficient that you used on your calculations is correct because it fluctuates according to the absorbed solar irradiance and the hour of day. It’s not correct to argue a constant value for the atmospheric window during the 24 hours.

    Regarding some constants taken from Trenberth’s-Kiehl’s essays, they used the generally accepted values, although they adjusted them a bit to make their cipher coincided with their idea and to recover those 50 W/m2 lost that they still have not found anywhere… Heh!

    Congratulations, Wayne! Your’s is a comprehensible assessment. 🙂

    NSN

  428. wayne April 26, 2011 at 5:47 pm #

    Nasif, I can tell through your general words you are a good scientist, you take the right approach and thanks for your kind words.

    “It’s not correct to argue a constant value for the atmospheric window during the 24 hours.”

    Exactly. Hope I can find enough real data in the jungle of the net to actually do that. Miskolczi is next. That is going to be a bit harder since he has a series of simultaneous equations that must also be met along with the parameters. His approach seems more at a pure planetary level, and as he said, there are really only three known measurements available for use as input to an energy budget, all other numbers are derived or only estimated. In fact, those spreads are more on a the same viewpoint that Miskolczi takes as an atmospheric physicist and it makes hugely more sense, and is in fact, it is the correct viewpoint of what actually occurs. Thousands of radiosondes show it.

    “Regarding some constants taken from Trenberth’s-Kiehl’s essays, they used the generally accepted values, although they adjusted them a bit to make their cipher coincided with their idea and to recover those 50 W/m2 lost that they still have not found anywhere… Heh!”

    Got ya. If you notice those spreads have nowhere “backradiation”, it is not even necessary and only totally confuses the subject. Boy, we both know that and can see the confusion crammed into so many fertile minds wanting to know the science behind this CO2 hubbub.

    If you, or anybody, ever come across any pertinent data and also knowing the source, please pass it. I am seeking it.

    Here’s some examples:
    (o) Closer estimate of the global mean diurnal temperature range, maybe even by season.
    (o) Closer estimate of how much evaporation occurs during the nighttime (that was one of my wild guesses I had to use, I guessed 1/3, may be much smaller)
    (o) Closer estimate of what time during daytime thermals peak. Same with evaporation. Is it surface temperature driven, irradiance driven, or a known ratio combination.

    It’s amazing but reading climate related papers, articles, blogs for over a year I have never come across such data. Science sure has changed or at least seems it has if such simple data is not available.

  429. gavin April 26, 2011 at 6:54 pm #

    Nasif says “The atmospheric window fluctuates according to the materials of the surface and the energy retained by the surface” ? Who else hey. Now see this

    http://earthobservatory.nasa.gov/Features/RemoteSensing/remote_04.php

    Anything other than air is outside the atmospheric window

    “power” that must not be divided because it is a quantity received on the surface area of the outer sphere

    Matey; your POWER is not the energy from the Sun at Earth’s surface if it’s not a work done in some time

    Insolation: Solar Radiation Striking the Surface

    http://edmall.gsfc.nasa.gov/inv99Project.Site/Pages/science-briefs/ed-stickler/ed-irradiance.html
    I = S cos Z

  430. wayne April 26, 2011 at 9:03 pm #

    gavin,

    Nasif said: “The atmospheric window fluctuates according to the materials of the surface and the energy retained by the surface …”

    Yeah, I see the missing word and just mentally insert it to make Nasif correct, would do the same for you. The missing word is ‘radiance’ and it goes between the word ‘window’ and ‘fluctuates’. No big deal. The atmospheric radiation window on a line-by-line basis only changes if the atmosphere radiative response (optical thickness) somehow changes, but, the radiance upwelling from the surface in the window does what Nasif said, it varies with the materials of the surface and the energy retained by the surface (temperature) on a point by point basis.

  431. mkelly April 27, 2011 at 12:23 am #

    Comment from: Neutrino April 26th, 2011 at 4:40 am

    mkelly,

    There are no commercial products, profitable or otherwise, that use nuclear fusion. Does that mean that nuclear fusion is not real?

    I can’t go to the hardware store and buy a device that takes advantage of the relativistic effects of length contraction. Does that mean relativity is wrong?

    The earth does not “get back more” energy than the input of 1.22*10^17W from the sun.

    I am sorry you did not read the question correctly nor respond with a product. If you cannot grasp the underlying implication of the question again I am sorry.

  432. Neutrino April 27, 2011 at 2:08 am #

    mkelly,

    I did not respond with a product because the lack of a viable product does not mean that the GHE is not real. It could also mean that is not practical to build a device that exploits it.
    Hence why the lack of devices exploiting nuclear fusion does not mean that nuclear fusion is not real but rather that it is impractical with today’s technology.

    If you presented me with a device you claimed produced more energy than it consumed I would be highly skeptical of it. But this is in no way analogous to the GHE. The GHE does not produce any energy. The earth emits(produces) virtually the same quantity of energy as it absorbs(consumes).

    But if you must have a product, try a survival blanket. It is not the same as the GHE, it is using reflected instead of radiated energy, but works on the same general principle. Neither the GHE nor a survival blanket create any energy but the person or planet that is surrounded by one will be at a higher temperature than they otherwise would be.

  433. L.J. Ryan April 27, 2011 at 3:42 am #

    Neutrino;

    You said:

    “try a survival blanket. It is not the same as the GHE, it is using reflected instead of radiated energy, but works on the same general principle. Neither the GHE nor a survival blanket create any energy but the person or planet that is surrounded by one will be at a higher temperature than they otherwise would be.”

    Yes both the earth and the person heat their respective surrounding layers . The earth radiates at 255K the poor sap under the survival blanket radiates at 310K. The earth’s layer, according to GHG physics, is 113% efficient, raising it’s temperature to 288K. The highly reflective blanket can only muster efficiencies in the 90. Maybe, our freezing victim should exhale under his blanket..forcing out a bit more energy.

  434. gavin April 27, 2011 at 9:24 am #

    Wayne; I don’t agree with your view that a word is missing above because it’s just a piece of unproofed rhetoric from some group of writers who are most likely paid to disrupt climate science as part of someone else’s game.

    IMO these types of omissions, and other line by line blunders are far too frequent and the question remains who are the other amateurs here?

    Now what you can do is try to find those key words with “radiance” added elsewhere on the net in association with “Nasif” and come back with all the places where it is written up properly.

    Clue; Power = Work divided by Time

  435. Neutrino April 27, 2011 at 9:39 am #

    L.J.,

    If you are going to compare ‘efficiencies’ then compare apples to apples.

    The thermal blanket reflects about 90% of the incident radiation back, the atmosphere radiates about 84% of surface radiation back towards the surface.

    Or if you want to compare temperature ratios actually compare temperature ratios.

  436. wayne April 27, 2011 at 12:28 pm #

    Neutrino, you are so showing how much of a scientist you are not, careful. I’ve never conversed on Jennifer Marohasy’s site before but it is almost like there are some serious pseudo-science flying here.

    You said: “The thermal blanket reflects about 90% of the incident radiation back”

    First, there is no thermal blanket, no reflecting 90% backwards. That is why Einstein said “all physics is local”. Locally, by the lapse rate, (you do know what that is don’t you) anywhere within the atmosphere up unto the tropopause it is always colder. And since it is always colder the higher you go, by the Stefan-Boltzmann equation of gray bodies (it is also equally valid on an individual sub-band basis), as you yourself showed above by q = ε σ ( T(a)^4 – T(b)^4 ) all energy flows from lower warmer to higher colder areas, layer by layer and you can make those layers as small as you want. And since when within the troposphere the mean free path of our greenhouse gases is on the order of meters. There is no raining down watts upon to the surface, there are ALWAYS more going upward, and, that by physics means there is only a mild ~60-70 W/m^2 flowing from the surface toward space. See your Kiehl-Trenberth diagrams. It is the 390-324=66 W/m^2 in the case of the 1997 paper. It’s estimated 396-333=63 W/m^2 in their 2009 paper. Just because they draw huge red arrows on their over simplified diagram DOES NOT mean that what it is making you imagine is actual, it is not.

    Come on Neutrino, if you are really a scientist, be a scientist. If your over your head in a discussion like this, just be honest and say so, I forgive easily.

    The “the atmosphere radiates about 84% of surface radiation back towards the surface.” you said is also not actual but I have already explained that above.

    Now, show your true colors, are you a scientist or not, do you know physics or not?

    BTW: No matter in this universe can be warmer, at all, none, by it’s own radiation. Physical fact.

  437. L.J. Ryan April 27, 2011 at 12:36 pm #

    Neutrino:

    Ok, so walk me through the math:

    Terrestrial emissions: 240 W/m^2 x .84 = 202 W/m^2 re-radiated toward the surface added to solar input 240 W/m^2 sums to 442 W/m^2…is this correct Neutrino?

    Human emissions: 524 W/m^2 x .90=472 W/m^2 reflected back toward the body, does this get added to the emissions summing to 996 W/m^2? Do we need consider absorptivity of human skin? Does our freezing victim be need be concerned with fever or spontaneous combustion should coverage continue too long?

  438. Nasif Nahle April 27, 2011 at 1:03 pm #

    @Gavin…

    Your chit-chat is of kindergarted…

    Now what you can do is try to find those key words with “radiance” added elsewhere on the net in association with “Nasif” and come back with all the places where it is written up properly.

    Clue; Power = Work divided by Time

    It cannot be that you, being so old, are trying to argue on this issues.

    Power = W = J/s

    I have repeated this once and once again… Don’t you read my posts?

    😀

  439. Nasif Nahle April 27, 2011 at 1:06 pm #

    @Gavin…

    Nasif says “The atmospheric window fluctuates according to the materials of the surface and the energy retained by the surface” ? Who else hey. Now see this

    And it is true. You cannot have the same energy flowing freely to the space (atmospheric window) if you have a different amount of power absorbed by the surface. It is not a fixed quantity. LOL!

    😀

  440. Nasif Nahle April 27, 2011 at 1:12 pm #

    @Neutrino…

    If you presented me with a device you claimed produced more energy than it consumed I would be highly skeptical of it. But this is in no way analogous to the GHE. The GHE does not produce any energy. The earth emits(produces) virtually the same quantity of energy as it absorbs(consumes).

    I’ll show you one. See the diagram… Energy out from nothingness:

    http://chriscolose.wordpress.com/2008/12/10/an-update-to-kiehl-and-trenberth-1997/

  441. Nasif Nahle April 27, 2011 at 1:14 pm #

    @Neutrino…

    The thermal blanket reflects about 90% of the incident radiation back, the atmosphere radiates about 84% of surface radiation back towards the surface.

    Thermal blanket… Another fabrication of AGW machine. Where is it and what is made of?

    😀

  442. Nasif Nahle April 27, 2011 at 1:23 pm #

    @Wayne…

    Your calculations are correct. Yesterday and today, I made some measurements of the insolation in my location because the maximum temperatures (weather, not climate) were so high, 38.6 °C yesterday and 42 °C today.

    The instruments on ground, which are adjusted to detect only the solar irradiance on the surface, gave measurements at 20 hrs UT of 512 W/m^2, yesterday, and 728 W/m^2, today.

    We cannot divide those measurements by anything neither make averages because those are instantaneous measurements on the ground, i.e. direct measurements.

    I see your work very valuable; therefore, I ask you to continue working on it.

    Thanks!

    NSN

  443. Nasif Nahle April 27, 2011 at 1:26 pm #

    @L.J. Ryan…

    You’re correct on your questioning to Neutrino. I will repeat your question so it doesn’t be ignored by him.

    L. J. Ryan wrote:

    “Neutrino:

    Ok, so walk me through the math:

    Terrestrial emissions: 240 W/m^2 x .84 = 202 W/m^2 re-radiated toward the surface added to solar input 240 W/m^2 sums to 442 W/m^2…is this correct Neutrino?

    Human emissions: 524 W/m^2 x .90=472 W/m^2 reflected back toward the body, does this get added to the emissions summing to 996 W/m^2? Do we need consider absorptivity of human skin? Does our freezing victim be need be concerned with fever or spontaneous combustion should coverage continue too long?”

    NSN

  444. wayne April 27, 2011 at 2:43 pm #

    Nasif,

    “We cannot divide those measurements by anything neither make averages because those are instantaneous measurements on the ground, i.e. direct measurements.”

    Exactly, and I should have mentioned for others that very caveat. Those figures are also adjusted and just averages (longitude wise). Always adjusted assuming the 31.3% albedo (Trenberth’s 97 estimate) and in the later case, 29.9%. Also, those are at the equator, any latitude cosine adjustments would have to be made if comparing actual measurements north or south. So, the 729.9 W/m^2 shown during the hours just before and after noon on a dry clear-sky day should be reading somewhat below 729.9 / (1 – 0.313) or less than 1062 W/m^2 by measurement, near noon, on the equator, for the atmosphere itself, in those conditions, would not be absorbing as much direct solar radiation as the average shows in column C either (no clouds).

    That spreadsheet was so incredibly simple to make I’m surprise someone else hadn’t made it, wondering what the nighttime might be like in those numbers so often tossed around. Feel free to use those, pass, or just make your own spreadsheet so you can adjust the parameters to test other data yourself.

    For instance: I was looking at what the AMSU instruments ( http://discover.itsc.uah.edu/amsutemps ) at sea-level are showing and their equivalent temperature has always hovered about 294.75K ±0.25K for global average, not 288 as Trenberth assumes as the mean global average temperature of the surface, so, just change it and see the effect. Since we are now near the bottom of the range of the last nine years, I think I’ll check that out too. What would the OLR then need to be to compensate. Those can be a great tool but remember what they are, and, what they are not.

    ( Noticed the mention of New Braunfels earlier, ahhh, the Guadalupe!! Fun!! )

    .

  445. gavin April 27, 2011 at 7:13 pm #

    “The atmospheric window fluctuates according to the materials of the surface -” Which dumb ass writer made that one up for you Nasif?

    The atmospheric window fluctuates because of crap in the air not crap on the ground you twit.

    Read the chapter “Greenhouse Gases and Climate Change” by Jeremy Colls 2002 on Google Books

  446. gavin April 27, 2011 at 7:44 pm #

    Wayne: When Nasif imposes on my age it can be an advantage cause I recall good experiments in Physics that were fully discussed in some class or other. One of those was the ranging exercise where a laser was bounced off the moon. Another was finding weak radio signals from a crippled space vehicle.

    Both cases I suggest fly in the face of your recent statements along these lines “all energy flows from lower warmer to higher colder areas, layer by layer and you can make those layers as small as you want. And since when within the troposphere the mean free path of our greenhouse gases is on the order of meters. There is no raining down watts upon to the surface”

    Some of my associates formaly working at tracking stations could have a bit to say about radiation cold to hot etc.

  447. L.J. Ryan April 28, 2011 at 2:07 am #

    gavin:

    The question is NOT whether radiation is isotropic but rather does lower energy radiation increase the energy of higher energy. So bouncing a laser off the moon or detecting weak orbital signal, is not controversial. To say, however, the lower energy colder atmosphere increases the energy of the higher energy warmer surface is very controversial. Yes, I know radiative transfer equations blah.. blah…blah. But, as Neutrino has failed to establish, it is a process which exist alone within GHG physics. Or maybe gavin, you can offer a spontaneous process which cold increases temperature of warm, or low energy increases higher energy, or by reflection /re-radiation a body increase it’s own energy. This should be a simple task for such an experienced learned, gainfully associated proponent.

    So gavin, when you come to the realization such extemporaneous mechanisms do not exists, do you question your dogma?

  448. Neutrino April 28, 2011 at 3:57 am #

    The insistence that emissions from a lower temperature object cannot affect a higher temperature object are truly bizarre. A surface absorbs(based on its absorptivity) any incident radiation. All it knows is that a photon hit it, the surface does not know what temperature is associated with that photon. It simply either absorbs or reflects it.

    Since you like real numbers rather than theory lets use the SURFRAD network, which everyone can look at.
    So we are all on the same page lets all use the same data as well, July 2 2010 Goodwin Creek MS.(nothing special about July 2 except it appears to be a cloudless day there)
    http://www.srrb.noaa.gov/cgi-bin/surf_check?site=good&mos=July&day=2&year=2010&p5=dpir&p6=upir&p8=rns&ptype=gif

    No averages, just total sums.
    In that 24hour period recorded by SURFRAD a 1m^2 section of ground has:
    Total Solar energy absorbed: 24.0MJ
    Total energy Emitted: 38.9MJ

    The surface is emitting 62% more energy than it is absorbing from solar radiation alone. How can you account for this imbalance? Either the surface is rapidly cooling, which is not supported by the temperature plots, or it is absorbing energy from another source.

    The only other available source is the downwelling longwave radiation from the atmosphere.
    Total IR energy incident: 31.8MJ

    Thermal blankets are simply an IR reflective material, how is that a fabrication?

  449. wayne April 28, 2011 at 5:58 am #

    gavin, you’re right. I missed putting in the word “net” two places. Do you not have enough scientific knowledge to insert them for me?
    I think this might be related to Nasif’s mention of apparent age. Scientists to scientists talk loose sometimes knowing the idea without passing each sentence through a committee for tit-for-tat correctness. My last plea, please.

    I’m so impressed that Nasif can translate from Spanish to English so clearly… I’ll overlook the inherent gender and language structure slips (and even a missing parentheses). Didn’t even know who Nasif was two weeks ago but I can recognize a good scientific mind when I read their words.

    gavin, so you will know a little of me, I was a dedicated AGWer for years through ignorance. Didn’t have the time at that moment to get trained in planetary atmosphere science. So, I read the news and am always concerned that we follow the proper and best path we can, but always correct in science. That is where I took the turn in the last year or two. It’s a scientifically deep subject.

    Can radiation go from cold to warm. Of course, of course. Everywhere, all of the time. I assume you know that. But L.J. Ryan is correct, that is not the question. Does the net energy ever flow to something colder, never. That is a pointless question to argue, it is built into the Stefan-Boltzmann equations itself.

    I consistently see two concepts colliding. One is that of above, colder to warmer, forget it, can’t and does not happen. The second is can the atmosphere act as an insulator. Now that IS a real and deep question that needs to be correctly answered. I must admit, that tends to scramble my scientific mind at times. That brings us exactly to this point, radiative transfer. I’ve read Kiehl and Trenberth’s papers. I’ve downloaded and have read about 500 others so far. Oh, Miskolczi’s papers too. That realm is where the real scientific discussion is happening.

    If Miskolczi’s paper is correct, we may not have so big of a problem and maybe none at all. When you step back and look at the big picture, they are all correct, Kiehl, Trenberth, Miskolczi, it all narrows down to whether the mean IR radiative window’s frequencies optical thickness is being affected by us humans. If it is, we do have problems. If it actually doesn’t, then not so much. CO2 is increasing and on one hand it will be God’s gift to mankind and on the other hand might cause various problems. The science goes on.

    It doesn’t seem by your words that you have taken the time to read all of these papers, Trenberth’s and Miskolczi’s, a few times each, to understand this deep question. If you do try to read Miskolczi’s I have a couple of pointers.

    First, I too was taken aback by the complexity of these papers. I had to read them multiple times, the number of variables is challenging, some of the data graphs hard to understand at first, and I don’t like the way Miskolczi structured his paper. Must be because he is Hungarian. But in real science that doesn’t matter one iota. Years ago I bought the 1300 page book “GRAVITY” on Einstein’s general relativity and I still am not very versed even though I have written Horizon type solar system simulators complete with relativity corrections. Read, gavin, read and step up to next layer up when discussing, drop the word picking. Talk substance please.

  450. wayne April 28, 2011 at 7:28 am #

    Neutrino, are you still stuck on individual photons. You are talking points that without detailed data is meaningless it seems. Read my comment to gavin, much of the same applies. You are going in circles (I’ll tire quickly if you insist the discussion sticks here).

    I do see your point though, I think, on an overall basis. You seem to keep pressing home that if an upper cold layer absorbs more radiation than it is emitting it’s temperature will rise. No problems there, commonsense. Will that then warm the warmer layers below? Well yes, and very locally not by itself.

    If that now warmer upper layer is still colder than the lower one, it by itself, the colder layer, will not be able to warm the lower warmer layer (SB), BUT, since the differential of the two layers is now smaller and by SB the upward energy transfer from the lower will be less so that warmer layer will also, over some short time, equalize and be warmer. The big point is that it cannot get rid of as much of it’s own energy to the cooler one above as much as before (SB again).

    And can that, by the same logical process, layer by layer going downward, propagate down to the surface? Of course and you don’t need to go layer-by-layer to see that. An overall SB from the surface to that now-warmer upper layer will get exactly the same answer as a layer-by-layer integration. But, by the time it propagates to the surface, the surface’s temperature would not change at all, zip, zero, the temperatures of the layers between would change upward. I can integrate that numerically for you as an example if you don’t know how to do it yourself.

    Is that your point of so many thousands of words above? Yes or no, please keep this civil and short whenever possible. Don’t want overstay our welcome here. (BTW, thanks, Dr. Marohasy, have enjoyed this very important subject here)

    That is a whole lot of words that I normally would not delve into for thermal equalization, equipartition between the degrees of freedom, and Fourier heat transfer handles all of that in the tropopause case with a whole lot of less words. That is since the entire troposphere is in local thermal equilibrium (LTE) at any and all points. Radiation in the frequencies outside of the radiative window act exactly as simple conduction but with a faster (speed of light) and longer (few meters) reach. That is why speaking of individual photos is pointless and a huge waste of words.

    SIDE NOTE, MODS ?: Does anyone know why my sig and email refuse to be remembered as it does on any other wordpress site, I have to type them in each and every comment. Cookies are on for jennifermarohasy.com.

    .

  451. wayne April 28, 2011 at 7:52 am #

    Neutrino… missed this earlier.

    You said:
    “The surface is emitting 62% more energy than it is absorbing from solar radiation alone.”

    Your just don’t get it do you, the real science. Must still be looking at Trenberth’s horrible diagram. Solar radiation is our ONLY energy source (but ~0.03 W/m2 from mainly interior unstable isotopes).

    See how comical such a statement that is you made? No such thing is REALLY, ACTUALLY happpening EXCEPT on Trenberth’s graphic !!! Stop Neutrino, stop, check your science logic !

    .

  452. Neutrino April 28, 2011 at 8:33 am #

    wayne,

    You seem to have missed the link, here it is again:
    http://www.srrb.noaa.gov/cgi-bin/surf_check?site=good&mos=July&day=2&year=2010&p5=dpir&p6=upir&p8=rns&ptype=gif

    It is from the NOAA SURFRAD network. They are measurements taken on the 2nd of July 2010 in Goodwin Creek Mississippi.

    It clearly shows that the total emitted radiation from the surface is 38.9MJ while the total solar absorbed over the same period is 24.0MJ. Heck even the total solar Incident is only 30.3MJ which is still not enough to account for the surface emissions.
    These numbers are not from the KT budget, they are direct measurements.

    Since the emitted exceeds the absorbed by nearly 15MJ the only two logical conclusions are:
    The surface is rapidly cooling.
    Or there is another input of energy to the surface.

    Since the temperature plot for that day does not support the first possible conclusion we are only left with the second. Clearly the internal energy leftover from the planet formation and nuclear decay are not enough.

    Unless you can explain it in some other way the only viable solution is that the downwelling emissions from the atmosphere make up the deficit.

  453. gavin April 28, 2011 at 9:19 am #

    Wayne; since I started helping those officially monitoring our emissions back in the early 1960’s it’s unlikely I’m about to do another refresher course with Johnny come lately’s in my retirement from all such R&D, routines etc. but thanks for the invite. As an illustration, in those days there were few competent with gear used to mind industry as we went into large scale automation of many well established processes. My job could include a calibration of everything from a water level gauge to a large mag flow meter or perhaps some nuclear device, even xrays, all analogue systems of course.

    However it’s the field of electronics that burnt me out most and these days I struggle to duplicate a mobile phone number without error and that’s usually a missing digit. I also need to leave IT to my next gen thank you.

    But beware, I spent my last decades in employment in active management and compliance monitoring of both new and old RF systems as we gained the newer technologies. Guess what, it meant a lot of research on a day to day basis to gain the confidence of my peers in other organisations. Typical projects were interferences, impacts on health and communications security. Modelling is my second nature, particularly when dealing with remote customers

  454. wayne April 28, 2011 at 11:11 am #

    Neutrino,

    http://www.srrb.noaa.gov/cgi-bin/surf_check?ptype=gif&site=bond&date=22-jun-2010&p5=dpir&p6=upir&p8=rns

    That’s on the summer solstice. (I’ll use the link, thanks) So what. That shows, being a day’s and position on the earths data for a day, exactly what my (really Kiehl and Trenberth’s) two spreadsheet shows only there they are the global averages over multiple years and using the mean albedo.

    The difference of upwelling minus downwelling IS the average ~62 W/m^2 and net flux is always upward in normal conditions (rarely clouds can ACTUALLY be warmer than the surface and you can get a momentary reversal of the normal flux). Miskolczi shows this rare effect does actually occur over long periods at the poles. But I still don’t see your point.

    Are you trying to convert me? ☺

    I do have one HUGE, HUGE advantage over both you and gavin, I’ve been where you are for years, I’ve done that, read that, said that. I know everything you now say and maybe will say from your words above. Sorry, it’s tiresome old hat and doesn’t vaguely resemble science.

    The big problem is you seem not scientifically minded enough to clamp your view temporarily and see MY POINT, NASIF’S POINT, the opposite side of this conversation. You seem to refuse. A conversation is a two way street, both sides must at least be able to understand in depth and acknowledge the other side’s viewpoint to proceed any further.

    I’ve given you enough to chew on, so, I’m back out of this for a while. Got better things to do.

    @ gavin — seems you’ve had a productive and enjoyable life, so have I, ran an entire plant for decades, who knows, I just maybe could be your senior, but to me none of that means anything. If a bright twenty year old came here talking proper science he might put both of us to shame and that would be just fine with me, I would listen. (I hate authority, and please don’t believe anything I have said for what authority I might have projected through my words. If you ever learn something from me, do it because of the content and that you find it is correct. Also, I’m human and do make mistakes on occasion. ☺ )

    .

  455. L.J. Ryan April 28, 2011 at 11:30 am #

    Neutrino;

    You said:
    The only other available source is the downwelling longwave radiation from the atmosphere.

    This assumes nothing else effects temperature. Before considering alternate methods for increasing temperature, lets get back to the questions I asked, Nasif Nahle reposted and which you’ve avoided.

    Terrestrial emissions: 240 W/m^2 x .84 = 202 W/m^2 re-radiated toward the surface added to solar input 240 W/m^2 sums to 442 W/m^2…is this correct Neutrino?

    Human emissions: 524 W/m^2 x .90=472 W/m^2 reflected back toward the body, does this get added to the emissions summing to 996 W/m^2? Do we need consider absorptivity of human skin? Does our freezing victim be need be concerned with fever or spontaneous combustion should coverage continue too long?

    Any interested readers should note your evasion, and that you do not feel confident enough to discuss the case on its merits.

  456. gavin April 28, 2011 at 11:37 am #

    Wayne; it occurred to me that in all my years modeling radio systems I never once bothered to check if a transmitting or receiving antenna was hotter or colder than the other. Its enough to say neither microwave nor longwave care about an equilibrium of temperatures either way.

    In fact while trailing HF nets on emergency frequencies we could sometimes get signals right round the Globe a couple of times while using a a normal whip despite all the air temp variations on the way. Those icy nights are remembered well enough too. Now consider how those TV signals from a tower on a mountain at say some 40 deg south where ice regularly forms on metal gets to your nice warm home down in the valley. Controlled radiation at any frequency must be a Godsend

  457. gavin April 28, 2011 at 12:06 pm #

    “Does our freezing victim be need be concerned with fever or spontaneous combustion should coverage continue too long?”What a nong Q.

    1 Having achieved near perfect spontaneous combustion in fluids and gases with and without teams of engineers and physicists I can say LJ’s body won’t burn in any condition ie no grey matter.

    2 Living systems including Gaia all have feedback mechanisms.

    3 We need to know whois dead or alive in this situation

  458. cohenite April 28, 2011 at 12:28 pm #

    That the ground can emit more than it is receiving from the solar source is a point in time effect as Wayne notes and has been shown in Geiger’s text:

    http://books.google.com.au/books?id=KaJHBv9FbYIC&printsec=frontcover&dq=Geiger%E2%80%99s+%E2%80%9CThe+Climate+Near+The+Ground%E2%80%9D&source=bl&ots=2vWksoLUjX&sig=uzVQx1QlpuuIgcLfqjFI7x7-Eas&hl=en&ei=t_1oTJy5B5DCcYLg3awP&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBQQ6AEwAA#v=onepage&q=Geiger%E2%80%99s%20%E2%80%9CThe%20Climate%20Near%20The%20Ground%E2%80%9D&f=false

    Figures 11.1. 11.2 and 11.3 show that over a day or year the amount emitted and received equate.

  459. L.J. Ryan April 28, 2011 at 1:45 pm #

    gavin;

    Dodging questions and throwing insults, sure signs of a loosing argument. So gavin, can you or can you not offer a spontaneous process which cold increases temperature of warm, or low energy increases higher energy, or by reflection /re-radiation a body increase it’s own energy? Man up gavin, answer the question.

    Any interested readers should note your evasion and recognize the shallowness of your argument.

  460. cementafriend April 28, 2011 at 4:45 pm #

    Cohenite, the book you refer to is dated and takes no account of convection.
    Have you read Dr Van Andel’s paper “Note on the Miskolzci Theory” Energy & Environment Vol21 No4 2010 p277-292
    Here is an exract
    “To summarize the main features of Miskolczi’s theory:
    • The heat transfer from surface to atmosphere is only by convection, not by
    radiation.
    • We do not need to know the composition of the atmosphere.
    • We do not need to subdivide atmosphere into troposphere and stratosphere
    • We do not need to differentiate between low and high latitudes; the theory holds
    everywhere.
    • We do not need to differ between low & high clouds, only their total albedo effect
    matters.
    • The surface temperature TS is only coupled by SU = TS^4 = 1.5 F0 to the net SW
    absorption F0.
    • There is no “greenhouse gas”, no “forcing”, no “feedback”, no “climate
    sensitivity”.
    • The cloudy sky moves to that equilibrium effective optical density whereby
    the net absorbed solar heat can be reradiated out into space with the
    minimum greenhouse effect, minimum surface temperature or maximum entropy
    production.”
    (Sorry subscripts and supercripts do not appear to copy well but you should get the point)
    If you can not get hold of the paper I have a pdf of the whole issue which has number of other interesting articles including another from Van Andel, one from Miskcolzci and one from Willis Eschenbach. You can ask Jennifer for my email address or I could send it via her (if she also want a copy)
    Heat transfer from the surface only by convection fits in with Nasif’s, Wayne’s and my observation that the radiant heat absorption by CO2 is insignificant. I have also noted from my experience with heat transfer measurement and calculation that at low surface temperatures (less than 50C) convection is more important than radiation. If convection occurs before radiation and the temperature of the atmosphere just above the surface is the same as the surface then it is clear that radiation to absorbing gases (CO2 and water vapour) will be zero because there is no net heat transfer between objects at the same temperature.
    Please note Dr Van Andel is a chemical engineer (as was Prof Hoyt Hottel mentioned by Nasif) and with his heat exchanger invention certainly understands heat transfer.

  461. Mack April 28, 2011 at 5:29 pm #

    Yes Cementafriend,
    If you think of this simply and holistically, the power from the sun is CONTINUOUS and TIMELESS and all the atmosphere does (along with the oceans which in this case should be reclassified as part of the “atmosphere”) is SLOW the RATE of heat leaving the earth until it reaches the temperature of space.

  462. Mack April 28, 2011 at 5:52 pm #

    Temperature of space?
    Well as cold as the thermometer gets when you stick it outside the spacecraft 🙂
    Carry on.

  463. cohenite April 28, 2011 at 6:06 pm #

    Hi cementafriend; the reason I like Geiger is that the figures I refer to demonstrate Aa=Ed, the basis of Miskolczi, as averaged over a climate period the surface is energy neutral as these other images show:

    http://img140.imageshack.us/img140/4109/tablemountainall.png

    http://img12.imageshack.us/img12/3225/tablemountainnets.png

    I am aware of Van Andel’s paper but do not have a copy so I would appreciate if you can link to one; I do have a p/p of Van Andel’s which I find useful and nice and [relatively] simple!

    http://climategate.nl/wp-content/uploads/2010/09/KNMI_voordracht_VanAndel.pdf

  464. wayne April 28, 2011 at 11:14 pm #

    Wow, science being spoken here. Good. Cohenite and cementafriend, you both just reminded me of something. I pointed Neutrino and gavin to read Miskolczi’s paper, told them of the troubles they would have but never gave them any links to somewhat more “Plain English” overviews. You both are right on Van Andel’s summary, very good, and cohenite, I noticed you had left that link.

    If anyone does have this one, here’s one of Dr. Zagoni’s overview (~6 megs):
    http://nige.files.wordpress.com/2011/02/the-saturated-greenhouse-effect-theory-of-ferenc-miskolczi.pdf
    The latest paper (of three I think)
    http://www.friendsofscience.org/assets/documents/Miskolczi-GH%20effect%20in%20semi-transp.atm.pdf

    cementafriend, thanks for bring this up : “If convection occurs before radiation and the temperature of the atmosphere just above the surface is the same as the surface then it is clear that radiation to absorbing gases (CO2 and water vapour) will be zero because there is no net heat transfer between objects at the same temperature.”. That’s a big point that everyone must remember (hint).

    Cohenite, you must own a copy of Geiger’s book, seems that is why I get blocked after the cover pages.

    It seems so many around the web don’t even understand why Miskolczi’s theory is such a big deal. They don’t realize this is a huge step forward in atmospheric physics, mainly because all of the equations and relations that he has coerced out of the radiosonde data are applicable at almost any point on the Earth and at any season, averaged over time and within tight error bars. They hold everywhere but very near the poles. And, when averaged all together you get points. To me that is a major breakthrough. Neutrino and gavin, you should read it, your missing something, maybe even like missing Newton’s breakthrough in gravitation but in atmospheric physics. If you respect science, I can’t believe you would just keep your eyes tightly shut. Then you can disagree later and argue but you would at least know what you are arguing against.

  465. Martin Mason April 29, 2011 at 4:30 am #

    Surely there can be downwelling radiation which may be adsorbed and re-emitted by the earths surface (and measured) but it can and must do this without heating it. It can’t heat the air at the surface and it can’t break the 2nd and 3rd Laws of thermodynamics which it has to do for the Greenhouse theory to work. Basic BB theory also shows that it’s impossible for lower energy level radiation to contribute to heating of the body. The science says it’s false, history says that it’s false by showing no link between CO2 and temperature and no tipping points at high CO2 levels and what we are seeing in reality (no warming with increasing CO2) shows it to be false.

    You can not heat a warmer body with a colder body at any level. You can not heat a radiation emitting source with its own emitted radiation. If you can why do we have an energy problem when patently a trace gas such as CO2 (but not water vapour?) can give it to us for free.

    Am I missing something here?

  466. wayne April 29, 2011 at 6:10 am #

    Martin Mason,

    Don’t think so, but at one place if I can inject.

    “Surely there can be downwelling radiation which may be adsorbed and re-emitted by the earths surface (and measured) but it can and must do this without heating it.”

    Of course, there is local downwelling radiation (“backradiation”) at and neat the surface. And you are right it can’t raise the temperature of the surface any higher that that same energy lowered the temperature of the surface when it was emitted upward. But that downwelling is from very low, like 5 to 50 meters from the surface (but in the window frequencies). A good way to properly look at it, but comical, is like fog. If the radiation from the surface is outside the window frequencies, and you were that radiation, you could not see but a small numbers of meters in front of you. You would soon be absorber.

    Now this is where the “backradiation” comes in. As long as you were absorbed close enough to the surface, that you, as the radiation, can still see the surface, it the random direction you are re-emitted is downward, you have a high chance to make it back to the surface. But since you are radiation, you cooled the surface as you left but you warm the surface back up BY EXACTLY THE SAME AMOUNT, no net warming. That is what any radiation meter of any type in those frequencies would read looking upward. In other words, it is very local.

    Now, that is close but not really what happens without getting into a line-be-line analysis with per line information as emissivities, Einstein Coefficients, or mean free path lengths (tau, or optical thickness at each line) taken into account. To be exact, this gets really messy. That why it is so hard for many to even imagine what is really happening, trillions upon trillions upon trillions times per second everywhere at all times.

    In the window frequencies your view would be a clear starry sky with a slight haze. You would normally zip out to space easily.

    Does that answer your “surely”?

    “It can’t heat the air at the surface and it can’t break the 2nd and 3rd Laws of thermodynamics which it has to do for the Greenhouse theory to work. …”

    You totally correct on the TD laws. Your kind of right on the first part with a few word adjusted. But, it can warm the air close to the ground, if the air is cooler, and does do that if that very energy just bounces around locally in the air maintaining equilibrium and equipartition, and this is what normally happens (really it is thermalization and re-emission from the GHGs), in your room, on your patio, in a field, on the ocean, its just it can never raise the temperature greater than the local surface itself is. The last part of that sentence I wholly agree.

    Close to what you meant?

    Sorry to be so verbose, but I usually get trashed over missing words if not.

  467. cohenite April 29, 2011 at 8:59 am #

    Thanks cementafriend, Jennifer has sent that edition of E&E focused on Miskolczi; very good!

  468. Graeme M April 29, 2011 at 10:11 am #

    I see SoD is doing a thorough series on Miskolczi at the moment…

  469. Mack April 29, 2011 at 11:26 am #

    Wayne,
    Further to that conversation I’m having with myself. Taking everything into consideration,if the AGWers think that some “backradiation” from the CO2 we have here is going to have any effect whatsoever, especially net effect, on the RATE of heat loss from the earth,they must be totally barmey.

  470. Neutrino April 29, 2011 at 12:34 pm #

    L.J.,

    No evasion on my part,

    Yes, if the atmosphere is radiating 202W/m^2 and the solar input is 240W/m^2 then the surface will be absorbing 442W/m^2.

    One of the points of going to the SURFRAD site was to show that the the solar input is not even close to emitted from the surface. Without some additional source then the measured temperatures and emission from the SURFRAD site would be much much lower.

    For the human wrapped in an IR reflective blanket.
    If the human is producing 524W/m^2 before being wrapped by the blanket then yes the 472W/m^2 is added to what the person is producing.

    To look at it in another way.
    The person is still producing the same power after the blanket is wrapped around them. But because of the blanket he or she is severely impaired from losing heat via radiation. The emissivity of the system of blanket+person is significantly lower(0.1) than the system of just the person(near 1.0).
    So to be in thermal equilibrium the temperature of the blanket+person system must be much much higher than the temperature of the system containing just the person.

    For a quick calculation assume that the person is floating in space. Just the person has a temperature of about 37C( (524/(1.0*σ))^.25 ), but with the blanket it would be raised to an extreme 278C( (524/(0.1*σ))^.25 ). So yes in that case the person would be in danger. But since the person is not in a void they would be losing a great deal of that heat by other means(conduction, convection, and evaporation) and the temperature would be much lower.

    The raised temperature of the person is not a creation of energy. The power generated in both cases is still the same 524W/m^2.

  471. Martin Mason April 29, 2011 at 12:43 pm #

    Thanks Wayne

  472. wayne April 29, 2011 at 2:20 pm #

    Nasif, or I feel more like addressing you Dr. Nahle,

    I must admit I wandered onto Dr. Marohasy’s site a bit blind. Saw a link somewhere that there was a discussion on carbon dioxide’s emissivity at jennifermarohasy’s site and that was something I needed to know how to calculate.

    I just searched Nasif and up pops all of these previous articles you have written here. I had no idea. I feel a bit foolish, should have acclimated myself before jumping in with both feet, but, when I hear wrong science, I can’t help it. Also it would help if I read some older comments to get a idea who thinks what first.

    Seems you are the very one from whom I came across the mean free path figure I am so grateful for. That one single figure in the last couple of month’s has helped me so much untangle this question on our atmosphere. Just knowing it was not kilometers was all I needed and I thank you.

    And more, seems by skimming through your older posts, other specifics on the atmosphere you also might be the original source or the comments underneath. Seems some of what I was saying has already been discussed here before. Bravo to you and please excuse my ignorance, much I say in my own words seems to trace back to here.

    So, I have some reading on previous posts here before I engage again. This time hopefully a bit wiser.

    .

  473. mkelly April 30, 2011 at 1:48 am #

    Comment from: Neutrino April 29th, 2011 at 12:34 pm

    For a quick calculation assume that the person is floating in space. Just the person has a temperature of about 37C( (524/(1.0*σ))^.25 ), but with the blanket it would be raised to an extreme 278C( (524/(0.1*σ))^.25 ).

    Please use proper temperature scale.

  474. Neutrino April 30, 2011 at 3:07 am #

    Is Celsius not a proper scale?

    The calculation results are in Kelvin and Celsius is just 273 less than that. Since I was using L.J.’s nu