Greenhouse effect, what greenhouse effect?

I know I keep posting the same thing, but I always end up with long posts, I'm trying to whittle it down to the basics, so each subsequent post gets a bit shorter (hopefully). When it's short enough, I will email it as a question to Climatologists, or pester them in the comments. Greta says 'follow the science' and I have done.
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To quote from James Hansen et al, ca. 1980:

The effective radiating temperature of the earth, TE , is determined by the need for infrared emission from the planet to balance absorbed solar radiation... [insert complicated looking formula which is actually quite simple]... this yields TE = 255 K. The mean surface temperature TS is 288 K. The excess, TS - TE, is the greenhouse effect of gases and clouds, which cause the mean radiating level to be above the surface.

The actual definition of 'effective temperature' is correct and the part we need to focus on. And let's accept his assumptions (nobody has ever seriously challenged them) that average mean sea-level temp = 288K; the weighted average albedo of clouds and cloud-free oceans/land = 30%; and so average absorbed solar radiation (which has to be matched by outgoing LW) = 238 W/m2

The sleight of hand is the 255K result. This is completely incorrect, whether by accident or design, but is taken as Gospel by Climatologists to this day. To arrive at 255K you have to assume uniform surface; all at the same temperature; and with 100% emissivity. All of which is completely unrealistic.

And we mere Earth dwellers assume that "surface" means ocean surface or land - but as far as incoming and outgoing radiation is concerned, it's a combination of clouds and cloud-free oceans/land (there is a separate cycle between clouds and the oceans/land below them). Climatologists skip back and forth between both definitions of "surface" as it suits them.

Here's the proper way of calculating it:
* Cloud altitude 5km, temp 256K, emissivity 70%, two-thirds of surface as seen from space. Emitted LW = 170 W/m2.
* Oceans/land, temp 288K, emissivity 96%, one-third of surface as seen from space. Emitted LW = 375 W/m2.
* Weighted average LW emissions = 238 W/m2, which balances absorbed solar radiation.
* As a check, cloud temp and sea-level temp are different by exactly the amount you'd expect, given a gravity-induced lapse rate of 6.5 K/km, adjusted for latent heat of evaporation (which reduces it from hypothetical 10 K/km to 6.5 K/km).

Conclusions:
1. Earth is - by definition - at the correct effective temperature, because LW emitted by clouds and cloud-free oceans/land ("the surface") = absorbed solar.
2. No radiation goes missing on the way out or is trapped by Greenhouse Gases.
3. Therefore, using his definitions, the greenhouse effect = zero, not 33 degrees.
4. He mentions "the greenhouse effect of gases and clouds". In truth it's just clouds. There is no need to factor in "gases" (by implication C02) as there is no discrepancy left to explain.
5. And of course the "mean radiating level" is above sea-level, but it is not "above the surface". It IS the surface (as seen from space).
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Yes, I am perfectly aware that all my inputs (cloud altitude and temp, emissivity etc) are just mid-points of ranges of estimates. Nobody will ever know the exact figures, although they could be firmed up a lot by using weather balloons with IR thermometers at lots of different locations and latitudes and different times of day. The more data we have, the better.

This would cost a tiny fraction of the $ billions they spend on 'climate models' and hunting round for evidence with satellites and expeditions to the South Pole. However conscientious the measurements and however rigourous the analysis, there will always be margins of error and uncertainty. But I am sure, you could get to +/- a few W/m2 or a few degrees K of my above zero-conclusion. And Greenhouse Gas Theories that are built on what is basically margins of error and uncertainty wouldn't really be sound science, would it?