<blockquote><i>
    Now to Absorptivity/Emissivity in the Infra Red. 
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| 20121026.130221 |
<blockquote><i>
This website lists the emissivities for a range of materials.  http://www.engineeringtoolbox.com/emissivity-coefficients-d_447.html

They are quoted as emissivity at 300K so essentially in the temperature, and thus IR frequency  range we are concerned with when looking at radiation from the Earth's surface.

    Water - 0.95 - 0.963
    Wood - 0.885 - 0.95
    Sand - 0.76
    Marble - 0.95
    Ice - 0.966 - 0.985
    Gypsum - 0.85
    Granite - 0.45
    Basalt - 0.72

Here is another site quoting emissivities for various materials  http://www.omega.com/literature/transactions/volume1/emissivityb.html#s1
    Also the abstract from this paper discusses emissivity of land surfaces:  http://journals.ametsoc.org/doi/abs/10.1175/JCLI3720.1. Important note about the abstract -  they discuss broadband emissivity. Then there is this plot from the UCSB:  http://www.icess.ucsb.edu/modis/EMIS/images/seawat10.gif. This is emissivity of seawater over a  range from 3 to 14 microns. That doesn't cover all the IR spectrum we are interested in but it  certainly covers a large proportion of the emitted energy. They have a range of other plots:  http://www.icess.ucsb.edu/modis/EMIS/html/water.html

    Importantly in the IR range that matters here, both water and ice have a very high  emissivity, as does wood. It is mainly rocks and soils that would tend to drag any average  emissivity value for the planet down.
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The values for water of <code>( `short , 0.9 ; `long , .96 )</code> gives a temperature of  :<code>
  ( .9 % .96 ) ^ % 4 />/ 0.984
  278.68 * r />/  274.22 </code>
which I believe is not that far off from the temperature of the mass of the ocean .
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    The figure I have heard quoted for average surface emissivity is 0.97. I can't absolutely  confirm that but with most of the Earth being water, ice and forests, it is hard to imagine  exposed rocks etc being able to pull emissivity down significantly below this.
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Actually , there is quite a bit of sand and rock over large regions . I assume "Wood" is wood ,  not forest . 
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So, for an idealized grey sphere at the Earth's distance from the Sun and very importantly with  no atmosphere, the A/E ratio will be 1 as you mention . The calculation is:

341 = 5.67 10^-8 * Tsurf^4. Solve for Tsurf and we get 278.5, much in line with your calc.

But an idealized grey body is still just that, an idealization. Lets make it a bit more  realistic. Lets assume that the 2 emissivities values can be used - a 2 tone grey sphere. One  for incoming sunlight, one for outgoing IR. Now we have

341 * 0.875 = 0.97 * 5.67 10^-8 * Tsurf^4. Solve for Tsurf and we get 271.4.
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We're essentially in agreement . I see my assuption for total irradiance is 342 w%m^2 . Rather  than calling it a "two-tone gray sphere" let's call uniformly painted with a single specturm  with those integrals across the short and long spectral ranges .

By my computation in terms of <i>gray</i> temperature :
<code>  278.7 * ( .875 % .97 ) ^ % 4  />/ 271.61 </code> 
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So the emissivity difference cools the temp somewhat, now down just below freezing. This is  what a non-ideal grey body with approximately the Earth's emissivity profile would be at.  Importantly, it has no atmosphere.

Now lets slowly add more components and see what happens. This is a totally un-physical  situation. I am doing this as a thought experiment to unpack the various aspects of what  happens in the system.
</i></blockquote> This is <a href="http://climatewiki.org/wiki/Category:Essential_Physics" >the  way physics is done</i></a> .
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First lets add an atmosphere but without any absorption of EM radiation. Neither of incoming  Solar or outgoing IR. But we will allow this atmosphere to reflect sunlight, just as the  current atmosphere does.

Since the atmosphere & clouds reflect 79 W/M^2 of the incoming 341 (from K&T 09), the Albedo of  the Atmosphere is 79/341 = 23.2%
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Now we're getting somewhere . This does introduce a semitransparent layer . But you are creating a filter which is totally transparent in the long waves . Which both water vapor and clouds most assuredly are not .

I'm going to have to leave this for the next chapter because now are adding a second parameter , transparency , at each wavelength to our objects . Like having sprites with alpha channels 

| 20121115.232940 |
<blockquote><i>
    So now our energy balance eqn for the surface is

        341 * (1-0.232) * 0.875 = 0.97 * 5.67 10^-8 * Tsurf^4. Solve for Tsurf and we get 254.0 

    This result is un-physical in several ways. Firstly by artificially 'turning off' the  absorption properties of the gases in the atmosphere. Next, in such a colder world there would  be much less evaporation so much less cloud cover so actually not as much reflection off the  atmosphere. However such a world is none the less colder than now, colder than during the  depths of an Ice Age. So the figure for surface Albedo in the Solar spectrum (12.5%) will be  too low because there would be much more ice reflecting sunlight.

    The important point is that adding an atmosphere with some level of additional reflection  will have a net cooling effect. Firstly due to whatever level of cloud cover does apply (and  lets not forget that the atmosphere itself reflects some sunlight due to reflection off  aerosols and scattering). Secondly due to increased ice cover in a colder world.

    Lets now add back the next feature of the atmosphere. Absorption of incoming solar by the  atmosphere. This is a little more complicated since the atmosphere can't just absorb this  energy. It has to go somewhere. As a simplifying assumption, I will assume that once the  atmosphere has absorbed energy, 50% of this eventually gets radiated to space without reaching  the surface, the other 50% does reach the surface. But this radiation reaching the surface from  the atmosphere is IR, so we need to apply the IR band absorptivity

    So our energy balance at the surface for this example is:

        (((341 * (1-0.232)) -78) * 0.875) + (0.97 * 78/2) = 0.97 * 5.67 10^-8 * Tsurf^4. Solve  for Tsurf and we get 245.2 K

    So the fact that some incoming Solar gets absorbed by the atmosphere and that a certain  percentage never reaches the surface (here I have assumed the split is 50/50) causes more  cooling. And again we would need to reduce reflection from the atmosphere due to fewer clouds,  but increase reflection from the surface due to more ice.

    Now lets add back another feature. Evapo-transpiration (ET) & Convection (C) from the  surface. Radiation isn't the only method available for removing heat from the surface although  these processes can't ship heat all the way out to space; they only add it to the atmosphere.  Lets make the same simplifying assumption we used with absorption of incoming Solar. 50% of the  heat added to the atmosphere by these processes goes to space. 50% is radiated back to the  surface. K&T 09 put the value for ET and C combined at 97 W/M^2. So lets run the calc again,  taking that into account.

        (((341 * (1-0.232)) -78) * 0.875) + (0.97 * (78+97)/2) = 97 + (0.97 * 5.67 10^-8 *  Tsurf^4). Solve for Tsurf and we get 236.1 K

    Again this is assuming that the ET & C values that apply in current climate would have the  same magnitude in this hypothetical world. Obviously they would be much smaller. But still a  net cooling effect on surface temperatures of some amount. And this would increase ice sheet  expansion and drive temps lower still due to even more reflection.

    Importantly, so far I haven't introduced any effect from GH gases into this thought  experiment. And the Earth is frigid.

    Now lets start adding the effect of the GH gases. K&T 09 put the absorption of outgoing IR  radiation by GH gases at 356/396 = 89.9%. So lets factor that in but retain our 50/50 split of  where radiation goes when it is emitted by the atmosphere, up or down.

        (((341 * (1-0.232)) -78) * 0.875) + (0.97 * (78+97)/2) + (0.899 * 0.5 * (0.97 * 5.67  10^-8 * Tsurf^4)) = 97 + (0.97 * 5.67 10^-8 * Tsurf^4).

    Solve for Tsurf and we get 264.8 K a huge increase. 

    Now lets add back the final component. The distribution of radiation up and down isn't  50/50. According to K&T09, the atmosphere radiates 199 W/M^2 up vs 333 W/M^2 down. 333/ (199+333) = 62.6% down. Plug this into my eqn and we get:

        (((341 * (1-0.232)) -78) * 0.875) + (0.97 * (78+97)/2) + (0.899 * 0.626 * (0.97 * 5.67  10^-8 * Tsurf^4)) = 97 + (0.97 * 5.67 10^-8 * Tsurf^4).

    Solve for Tsurf and we get 285.0 K - another increase. And within the ball park of actual  accepted surface temps.

    My point Bob is that,starting with an idealized grey body with no atmosphere, then  progressively adding more and more real world features results in continually lower and lower  temperatures, until we start adding in GH effect features. The exact values of some of these  stages may be questionable, but they all have a cooling impact. So too does the impact on  Albedo through expanded ice extent. Starting with an idealized grey body, every factor in the  system makes the Earth colder. Only the effect of the GH gases has a warming impact.

    The -33 C figure often quoted is referenced as an indicative number, an 'all else being  equal' number. It is not intended as anything precise. More along the lines of the thought  experiment I performed here. Some effects would certainly not have as much cooling effect -  cloud cover, ET & C particularly. But they all still have some net cooling effect compared to  your airless grey ball. And equally, ice cover certainly wouldn't be 'equal' either. Consider  that the first temp, for the airless grey ball, is already down to the sorts of temps we see at  the bottom of an Ice Age, when Ice sheets reached down to Kansas City. Cool the Earth 10-20 C  colder than that due to the absence of the GH Effect and how much further would the ice stretch  - Galveston, Mexico City? Cover the Earth with enough ice and it most certainly could get that  cold.

    And we know that in the very distant past, 100's of millions to billions of years ago,  there were instances of such Snowball Earth's. Turn off the heating system that keeps the ice  at bay and the Earth freezes over.

    Best Regards
    Glenn 
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