Why Recalc this now?
As I was checking through all the currently accepted calcs we have for OPM, I realized that there are two problems with the currently accepted Gouketsu cloud-split feat, which appears in chapter 75 of the One-Punch Man manga.
Number 1: the area in the current calc is only a certain fraction of the cloud deformation- the five rings outside the innermost circle do not factor in at all even though they have been visibly deformed by the shockwave.
Number 2: the height of the clouds is based on Lina Shield’s calc, whose results have been challenged by Therefir’s more recent calc which was also accepted. So it is time to revisit this feat.
___________________________________________________________________
Cloud Type, Width and Speed
Before we can calc Gouketsu’s cloud split, we will first have to determine what kind of clouds are being deformed here. Having thought about it some time, I will note that the one undeformed cloud we see outside this area seems to be relatively low, moderately thick, relatively whitish and flat-bottomed.
I think the deformed clouds are cumulus clouds. Such clouds bottom out most commonly at 200-2000 meters. The median, 1200 meters, seems like a reasonable height when compared to the high-rise buildings in the background (about 100 meters, the building on the left has 10 stories shown and is mostly hidden by the trees). We will use their average thickness to scale the clouds, circumventing the long and complicated Suiryu cloud feat debate entirely.
Cumulus cloud thickness is at least 600 meters (varies from 600 to 2 km), 10 px. Pixel ratio of 60m:1px
The deformed area goes off-screen so we will work with the shown radius that stretches from the marked center. The radius (the black line) of the total deformed area is 554 pixels, or 33,240 meters. Since the shockwaves must have traveled this far to produce the deformation, we can use this distance to determine the speed.
1 second: 33,240 m/s,
3 seconds: 11,080 m/s,
5 seconds: 6,648 m/s,
______________________________________________________
Volume and Mass
Now, Qawsedf suggested I consider the mass of the cloud rings left behind, since some clouds are being left behind by the shockwave. I will do this by finding the total mass of the rings and subtracting them from the total area. First the total volume as a cylinder and then mass....
Π * 33240^2 * 600 = 2.08*10^12 * 1.003 kg/m3 (cloud density) = 2.08624*10^12kg
There are five cloud rings. We will find their inner and outer radii and determine their area, working outwards from the innermost. Then we add their mass together and subtract it from the 2.086*10^12 total.
(Ring 1) R = 159 px or 9540m, r = 135 px or 8100m.
A = Π (R^2 - r^2), Π (9540^2 - 8100^2) = 7.98*10^7m2 * 600 = 4.788*10^10m3 *1.003 = 4.802364*10^10 kg.
(Ring 2) R=217px or 13020m, r= 190px or 11400m.
Π (14640^2 - 11400^2) = 1.24*10^8*m2 * 600= 7.45*10^10m3 * 1.003 = 7.4793306*10^10kg,
(Ring 3) R=309px or 18540m, r= 244px or 14640m.
Π (18540^2 -14640^2) = 4.07*10^8m2 * 600 *1.003 = 2.449326*10^11kg.
(Ring 4) R= 408px or 24480m, r= 343px or 20580m.
Π (24480^2 -20580^2) =5.52*108 m2 * 600 * 1.003 = 3.321936*10^11kg.
(Ring 5) R= 554px or 33240m, r= 472 px or 28320m.
Π (33240^2 -28320^2) = 9.52*10^8m2 * 600 * 1.003 = 5.729136*10^11kg.
All the rings together now:
(4.802364*10^10) + (7.4793306*10^10) + (2.449326*10^11) + (3.321936*10^11) + (5.729136*10^11) = 1.27285675*10^12kg.
If we simplify and simply subtract this left over mass from the total, we get the following:
(2.086*10^12) - (1.273*10^12) = 8.13143254*10^11kg. Now, for the results.
_____________________________________________________________
Results
1 second:
KE= 1/12 * (8.131*10^11kg) * 33,240^2
KE= 7.48700025*10^19 joules or 17.894 gigatons, or 6-C.
3 seconds:
KE= 1/12 * (8.131*10^11kg) * 11,080^2
KE= 8.31888916*10^18 joules or 1.988 gigatons, or High 7-A.
5 seconds:
KE= 1/12 * (8.131*10^11kg) * 6,648^2
KE= 2.9948001*10^18 joules or 715.8 megatons, or 7-A.