[split]Photons and MASS in calculations...Split from CO2 home experiment thread.

[Derek]

Ok, I'll try another tack to explain my "issue" with W/m2 "figures".

THE TRICK of the second 2 in P/4.
Both sides of a disc have half the surface area of a globe of the same diameter.
That is the first 2 of P/4, which all can agree with.
ie, the power of beam received by a disc on ONE SIDE, is halved, because a hemisphere (in this case a lit hemisphere) has twice the surface area of a disc.
In earth's case the globe is an illuminated yellow hemisphere, and a dark blue unlit hemisphere.
Two hemispheres, not one.

Yes, if you want you could "double" the yellow hemisphere, but that does not overwrite or replace the dark blue hemisphere.
So, there has to be a dark blue "globe" as well as a lit at half power, on average 1/4 actual solar power, illuminated yellow "globe".
If we are to look at day AND night.

But, the real trick of the second two is two fold, firstly to make divorced from reality, and therefore meaningless averages seem to apply when they do not.
And secondly to cover up the simply omitted dark blue "globe" of night.

Let us say that the lit side emits 10 units, and the unlit side emits 5 units over 12 hours. *
That would be 10 +5 = 15. But the second 2 does not do this,
it does,
the sum of 10 and 5 divided by 2 = 7.5, ie, 1/4 of the real total emitted.
A meaningless average, divorced from the physics. It is neither day or night.

The full sum of what is emitted over 24 hours would be 2 yellow hemispheres, plus two dark blue hemispheres, ie,
(10 + 10) + (5 + 5) = 30

The answer is as near as dam it zero (received - emitted), but does that make the sum 1/4 - 1/4 = 0
No.
The evidence is that the sum is
Input - [output day side + output night side] = 0
In this example that would be,
(15 + 15) - (10 + 10) - (5 + 5) = 0

Whilst we persist in a 1/4 view (7.5 - 7.5 = 0), we will never understand the physics of the full 30.

The second 2 in P/4 is misdirection from the real world physics to meaningless averages of neither day or night applied incorrectly globally.
That's all.

* = Where the lit side is illuminated by a beam supplying 15 units over 12 hours, which would be a total of 30 units over 24 hours for the whole "globe".

Please Note - I have avoided using W/m2 in this example because they confuse the issue,
please see post 21 in this thread.
What is a Watt???
W/m2 does not specify amount, it only specifies the power, anything else is just an assumption...
Which could be, but is rarely, if ever, checked....

Grey body calculations anybody????? - I thought not......
Yet, such would answer so many questions!