Have fun!
On this page:
Another units
Balls
Albedo
Another units: TOA Net CRE,
WIN(clear) and Blocks
If you are boring with LWCRE = 26.68 Wm-2 as 1
unit, have fun with trying TOA Net CRE a lighter
"quantum" with = 20.01 Wm-2 = 1
(green) unit.
With this:
TOA Net CRE = 20.01 Wm-2 = 1
TSI =
51 = 1360.68 Wm-2 = 68,
OLR(all-sky) = 9
= 240.12 Wm-2 = 12,
G(all-sky) = 6
= 160.08 Wm-2 = 8,
Absorbed solar in the atmosphere = 3
= 80.04 Wm-2 = 4,
Absorbed solar at the surface = 6
= 160.08 = 8,
and
ULW = 15
= 400.20 Wm-2 = 20.
* * *
Or, if you want even more substantial, "heavier" units, here are your nucleus: the BLOCKS.
3 quarks = 1 nucleon
4 TOA Net CRE = 3 LWCRE = 1 BLOCK = 80.04 Wm-2
TSI =
51 = 1360.68 Wm-2 = 17,
OLR(all-sky) = 9
= 240.12 Wm-2 = 3,
G(all-sky) = 6
= 160.08 Wm-2 = 2,
Absorbed solar in the atmosphere = 3
= 80.04 Wm-2 = 1,
Absorbed solar at the surface = 6
= 160.08 = 2,
and
ULW = 15
= 400.20 Wm-2 = 5.
* * *
Costa
and Shine (2012) computed the clear-sky global mean window radiation
(called surface transmitted irradiance) and found that WIN
(clear) = 65 Wm-2. Notice that their model-OLR was 259 Wm-2. Assuming
proportionality, with our OLR(clear) = 266.80 Wm-2, WIN (clear) would
be 266.8/259 × 65 = 66.96 Wm-2. Let us recognize that an integer
position in our system is at 66.70 Wm-2, being equal to 10/4
(that is, 10
units on the disk, where OLR is 40
units). This way, the clear-sky OLR (on the disk) is added up as 30
units atmospheric upward emission and 10
units surface transmitted (window) radiation; on the sphere: 266.8 Wm-2
= 200.10 Wm-2 (atmosphere up) + 66.70 Wm-2 (window), maintaining the
one/two ratio for ATM up/ULW up.
Using now WIN(clear-sky) as the clear-sky UNIT:
WIN (clear) = 66.70 Wm-2 = 1 (blue)
G(clear) = 133.40 Wm-2 = 2
OLR(clear) = 266.80 Wm-2 = 4
ULW = 400.20 Wm-2 = 6.
Just recall here the cloudy unit, LWCRE (tot) = 1 (black underlined) = OLR(clear) – OLR(cloudy) = LWCRE / βeff = 1/0.6 = (5/3) × 26.68 = 44.47 Wm-2,
G(cloudy) = 4 = 177.87 Wm-2
OLR(cloudy) = 5 = 222.33 Wm-2
ULW = 9 = 400.20 Wm-2
This way
both the clear-sky part and the cloudy part of the atmosphere have
their own units, and the all-sky value is added up as the area-weighted
sum of the clear and cloudy units:
G(clear) = 2 = 3 = 5
OLR(clear) = 4 = 6 = 10
ULW = 6 = 9 = 15
Thus TSI in our atomic system:
TSI = 1360.68 Wm-2 = 17 = 20 + 1 = 30 + 1 = 50 + 1 = 68.
What
happens is something like this: Sun provides us here, at 1 astronomical
unit, with 1 TSI = 17 energy blocks, from which 5 bounce right back
from the disk, 12 are absorbed. Going down to the sphere, 3 energy
blocks arrive, building up 5 blocks energy at the surface by the help
of 2 blocks of greenhouse effect. This is simply the play of
the fine-tuned two-plate (single-layer atmosphere) geometry.
Then the blocks decay into three quarks per each, and build up the
clear-sky and the cloudy-sky parts, having 20 + 1 energy for the clear-sky system on the disk and 30 + 1
for the clouds.Their interplay is conducted by the help of our
"leptons", the NetCRE units (click on the figure to enlarge or
download).
The
whole global mean energy flow system seems to exhibit a well-organized
arithmetic integer ratio system, an "atomic" structure, with
"prescribed" (controlled, constrained clear-sky, cloudy and all-sky greenhouse
effects.
Have
fun: Balls. Let TSI be ONE foot-ball, falling on the intercepting disk
(a). It breaks into 17 tennis balls (b), from which 5 is reflected
(albedo 5/17), 12 yellow tennis balls are transmitted (c). 12 are
absorbed at the surface, then reemitted as brown tennis balls which are
completely absorbed by the atmosphere, and emitted upward and
downward equally. Altogether, 24 brown balls are moving there (d). Fine
tuning: Let 4 yellow be absorbed in the atmosphere, decreasing the
number of the emitted brown balls to 20 (e). Divide by 4, our incoming
12 yellow balls become 3 (f). After division by 4, from our 3 incoming
balls 1 will be absorbed in the atmosphere and 2 at the surface,
emitted up 5, atmosphere emnits 3 up and down (g). Then the
three
tennis balls decay into 9 tabletennis-balls (h), and the final
distribution we can experience in global mean energy budgets is there
in (i).
and (i) is the final all-sky result, to be compared to:

Have fun: Albedo.

The
consideration here is incomplete and yet unproven, maybe even wrong.
The
plane-parallel model albedo, with 51/4
units incoming solar and 15/4
units reflected solar is 15/51 = 5/17, identical to the IPCC AR6 albedo
of 100/340 = 0.294.
We have read in AMS Meteorological Monograph (2019) Chapter 4 that "The CERES flying on the
Terra and Aqua satellites confirm that Earth’s albedo is
29.4% (±0.3%) and that the planet emits at an equivalent
blackbody temperature of 255 K (±1 K; Wielicki et al.
1996; L’Ecuyer et al. 2015)."
Recent CERES datasets propose only 99 Wm-2
reflection from the same incoming energy, hence their suggested albedo
is 99/340 = 0.291 (at the lower boundary of the unceertainty range given above).
We
tried to use a real spherical geometry instead of
the stratified, and we found a geometric albedo between the
two, α = 1 – sin 45° =
0.293. But we were unable to derive this result from angular
distribution models or geometric
principles.

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* *

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