Between latitude 45N and 45S 70% of the surface of the globe, 73% of the surface of the ocean and about 90% of the solar energy available for the Earth can be found. Not considering albedo, the average available energy can be as high as 612 Wm-2. With 1361 Wm-2 "average" solar energy available at the mystical top of the atmosphere (TOA), 1/4 or about 340 Wm-2 would be the global "average" peak available energy provided by the Sun. Between latitudes 45 and the poles, the amount of solar energy available, not considering the Earth's axial tilt, would decrease rapidly to zero at the poles.
The 612 Wm-2 is a "peak" value for the diurnal average it would also be the Day time average since at night the Sun doesn't provide energy. Since the Earth rotates on its axis, this "peak" is just like the "peak" voltage of a sinusoidal power supply. The root mean squared (RMS) value of a 612Wm-2 sinusoidal power supply would be 612 divided by the square root of two - 612/1.414=432 Wm-2.
Since 432 Wm-2 RMS is applied to 70% of the surface, 0.7*432=302 Wm-2 of the 340Wm-2 global "peak" is available in the 45N to 45S section of the Earth. Just like a battery or capacitor, the Earth can "charge" to any value up to the "peak" depending on the discharge resistance, with "peak" being 612Wm-2 not the RMS value.
For some odd reason, this simple analogy to a well know electric circuit function is lost in the "Back Radiation" non-sense. If I charge a battery, I do not attempt to determine the "back radiation" or back capacitance or back resistance of the surroundings of the battery case. All that is required is the capacity of the battery and the resistances, plural, of the battery and the circuit it is connected to.
Resistances is plural because a battery can charge and discharge at different rates. Internally, the battery's resistance to charging increases as it approaches its maximum capacity. The circuit or load that the battery is connected to can vary with "demand". If the demand is constant, then as the capacity of the battery reduces, the electromotive force or voltage reduces which would reduce the current or energy flow to the load or "demand". Since the Earth's oceans have the highest heat/energy capacity, the charging resistance and load resistance would determine the condition of the Earth's Charge or average heat content.
By comparing the surface average energy available to the average voltage drop of the "demand" or surface temperature, you are comparing apples to oranges. The surface temperature or "Back Radiation" if you use the energy equivalent of that temperature, is just an indication of the current flowing through the atmosphere.
Since the composition of the lower atmosphere includes much greater mass and more water vapor, the resistance to energy flow would naturally decrease with altitude. In the denser lower atmosphere, energy is "conducted" by more than radiant means, so it is easy to confuse "back radiation" an indication of energy flow, with the actual energy capacity of the atmosphere the energy is flowing through.
If you consider the atmosphere to be another battery, you can determine its "charge" by subtracting the EMF of that battery from the total "back radiation" or equivalent "voltage" drop across its charging "resistor". In the Stevens and gang Earth Energy Budget cartoon, the All-sky Long Wave absorption is 188Wm-2 and the surface EMF is 398Wm-2, 398-188=210Wm-2 is the surface energy flowing through the atmosphere and 188 is the average "charge" of the atmosphere as a result of the true Earth surface battery. Since 210 is flowing through creating the 188, 22 Wm-2 would be the voltage drop of the atmosphere or the "back radiation" felt at the "true" surface generated by energy flow from the surface to the atmosphere.
The Atmosphere has a alternate "charging" mechanism. It absorbs 75Wm-2 from the Sun. That adds to the "voltage" of the atmosphere and some portion of the "voltage" would be felt at the "true" surface. Since the resistance of the atmosphere decreases with altitude, how much of that 75 Wm-2 of energy is felt at the surface would depend on where in the atmosphere that 75 Wm-2 is absorbed. If it is absorbed perfectly in middle of the energy concentration of the atmosphere, just like a voltage divider, 1/2 of the energy would be felt at the surface. That would be 37.5 Wm-2 which would be felt in addition to the 22 Wm-2 created by the flow of energy through the atmosphere from the surface. 22+37.5=59.5Wm-2 would be the approximate "net" impact of the resistance on the atmosphere plus solar energy absorbed in the atmosphere or the "net" "back radiation" felt at the surface.
Since Stevens and gang indicate that the "net back radiation" is 53 Wm-2 (by subtracting 345.6 Wm-2 DWLR from 398 Wm-2 OLR), the atmospheric voltage divider may have slightly different resistances.
There is nothing wrong with using the gross values of radiant energies, but without considering the source of the gross energies and the methods of energy transfer, it is easy to confuse cause and effect. That is why most "Sane" people avoid gross radiant values in systems that are likely not primarily radiantly driven. If the water in one end of my pool is 10 C degrees and the water in the other end is 20 C degrees, I would not attempt to determine the average energy of the pool by converting 10 C to 283.15K, convert that to 365Wm-2 then determine that 20C has an energy equivalent of 419Wm-2 then use the gross radiant energy values to determine anything about a virtually radiantless energy transfer process in a frigging swimming pool.
I hope this helped.
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