You shouldn't be using 288k in this calculation to arrive at 1.38E+17.

...

I am not using 288k "to arrive at 1.38E+17". I am using it for something else, and i believe - correctly. Namely, i am using 288k to estimate how applicable Stefan-Boltzmann law is to a planet with a thick athmosphere. This is the essense of this topic. You see, we know average actual Earth surface temperature = 288k (from measurements - satellites records, etc). Based on that, i calculate how much energy such a surface - at such a temperature, - would be radiating out according to Stefan-Boltzmann law (and using the calculator via the link provided, to simplify the process).

Then i calculate actual input (using Earth crosssection and solar constant) and compare the two figures. In case of positive greenhouse effect, the former figure will be higher than the latter. How _much_ higher - is exactly what i was interested to find; i compared such a difference for Venus and for Earth, and numbers i've found are imho quite interesting to see. In case of anti-greenhouse effect (i.e. negative change of surface temperature due to athmospheric processes - as observed on few other bodies in the Solar system), the former figure would be lower than the latter.

This is what 1st post is about before it gets to the question.

... There is just one formula and therefore just one route to the effective emission temperature of 255k.

I disagree. There are often more than one way to calculate something in physics, and this case - i think - is no exception. On top of the way you describe, there is also this path: 1. measure/calculate actual input of energy for "Earth with completely transparent athmosphere" - how much Earth would absorb from the sun; this should be quite close to the amount Earth absorbs in reality, since actual Earth athmosphere is quite transparent to visible spectrum, and most of Sun's energy comes in in exactly that - quite short-wave - form; 2. knowing that for any long-term situation, amount of absorbed energy _have to_ equal amount of radiated energy (geothermal is a microscopic share, and can be discarded for simplicity, as can be background space radiation, energy from meteorite impacts, etc), - use Stefan-Boltzman law "in reverse": calculate what temperature Earth surface (without any athmosphere) will be while knowing what energy it will be radiating out (which, again, has to be equal with how much it'd absorb, long-term).

I've done it even when writing 1st post, and as a result, i've found that Earth greenhouse effect should be - quote - "it seems that greenhouse effect is some 8...9 degrees". Namely, if you'd put T=279.3 K, surface area of Earth and emissivity = 0.694 into said calculator and hit "calculate", then the result you get - is 1.2212271426758E+17 . Which is a match to the amount of energy Earth absorbs. The difference between 279.3 K and 288 K is exactly "some 8...9 degrees kelvin" i've talked about; some ~4 times less than the "official" 33 K for greenhouse effect. See?

... output Watts for a black body earth (i.e. with completely transparent atmosphere). ...

Never was calculating anything for a "black body" earth; this would mean emissivity ~= 1, albedo ~= 0. As mentioned above, actual Earth is some ~70% water at its surface, and water has emissivity of some ~0.67. Definitely not a black body to me.

... Then SB * T^4 * surface area of Earth = output watts

5.67E-8*255^4*5.1E+14 = 1.22E+17 ...

This is correct for a black body. But not for reality of Earth. Since Earth is not a sphere covered by some thick layer of soot everywhere on its surface. Water is 0,67 emissivity, sand (Earth has vast deserts) is 0.76 emissivity, areas covered with thick layers of green plants - i don't know, but far lower than 1.0 i bet, soil surface - 0.38, clay - 0.39. I've used 0.694 in the 1st post as average emissivity for Earth surface. And got proportionally lower value as a result. From which my original question appears, of course.

... also solar constant * (1-albedo) * area of circle with Earth radius = input watts

1370*0.7*1.275E+14 = 1.22E+17 ...

This is exactly what i've done in the 1st post for both Venus and then Earth. Slightly different value for solar constant i've used (1366.1) and for albedo (0.306) are the only (cosmetic) differencies, due to which the result found in the 1st post is 1.21 (not 1.22) E+17 W. "Crosssection" and "area of circle with Earth radius" - is the same thing; crosssections for both Venus and Earth in the 1st post were calculated via "radius squared multiplied by Pi", with Pi=3.14159265 (yes i do remember it with this precision, if you wonder). Numbers for Venus and Earth crosssections in the 1st post are exactly the product of this very basic calculation.

... These match so the 255K is correctly calculated and given the current average temperature is 288k there is a 33C greenhouse effect. ...

So you calculated how much energy "black body" Earth would radiate out, defined a temperature at which the amount of radiated (by the black body Earth) energy would equal amount of energy absorbed by a NOT black-body Earth energy (since you used 1-albedo in the SECOND part, but not in 1st), - and based on that, you define "greenhouse effect". Right? =)

Excuse me, but how exactly removal of the athmosphere automatically produces a decrease of Earth albedo from the value you used to ZERO? =) And, of course, the Kirchhoff's law, while being strictly true only for an isolated frequency, - still plays a major role even if we speak about all frequences; this means, if you talk about black body emitting energy, then you _have to_ automatically assume that such a body's albedo is ZERO, and use this value when you calculate absorbed by such a body energy, as well.

Alternatively, if you use actual albedo to calculate how much energy a body actually absorbs - then you _have to_ add corresponding emissivity when you calculate how much energy such a body would radiate out. And to do it quantatively right (with precision) - is difficult, as Wipneus pointed out, since bodies often radiate in a different spectrum than spectrum they absorb at (making calculating exact effect of Kirchhoff's law a huge mess!).

... If you are using (1-albedo) then you appear to be calculating input watts. ...

Nope - not nesessarily. Again, Kirchhoff's law; plus an assumption i used - which was, that said law is directly and fully applicable to a planet as a whole (which is possibly rather big error to my calculations, but not on the order of dozens percents, i dare hope) - allowed me to find _emissivity_ of a body as (1 - albedo). The word _emissivity_ itself indicates that such a calculation is not about "input watts", but about "output watts".

... So it looks like a misguided calculation and I still can't work out what you are trying to calculate and how in order to try to explain the error ...

Hopefully, this post will allow you to see that it wasn't misguided calculation, and will allow you to work out what i was calculating.

P.S. when i am saying "Kirchhoff's law", i, naturally, mean this one:

http://en.wikipedia.org/wiki/Kirchhoff's_law_of_thermal_radiation . In particular, this corollary from it (my bold): ... the emissive power of an arbitrary opaque body of fixed size and shape at a definite temperature can be described by a dimensionless ratio, sometimes called the emissivity, the ratio of the emissive power of the body to the emissive power of a black body of the same size and shape at the same fixed temperature. With this definition, a corollary of Kirchhoff's law is that for an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium,

**the emissivity is equal to the absorptivity**. ... and, of course, absorptivity is nothing else than (1 - albedo), by the albedo's definition.