| 20121026.130221 |Now to Absorptivity/Emissivity in the Infra Red.

The values for water ofThis 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.

`( `short , 0.9 ; `long , .96 )`

gives a temperature of :```
( .9 % .96 ) ^ % 4 />/ 0.984
278.68 * r />/ 274.22
```

which I believe is not that far off from the temperature of the mass of the ocean .
Actually , there is quite a bit of sand and rock over large regions . I assume "Wood" is wood , not forest .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.

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 ofSo, 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.

` 278.7 * ( .875 % .97 ) ^ % 4 />/ 271.61 `

This is the way physics is done .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.

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 |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%

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