literature

# Raoh's Goten Sho (Hokuto no Ken)

The context, Raoh uses his Goten Sho to disperse the clouds, which ends up with the sky being completely clear. ### Cloud's volume

Cloud Thickness = 106 px = 2000 m
207 px = 3905.66 m
R = 1952.83 m
Total R = 384 px = 7245.283

π(7245.283 m)²(2000 m) = 329,830,321,344.063 m3
Cloud density = 1 kg/m3
M = 3.2983×10^11 kg

### Distance

207 px = 3905.66 m
panel height = 1024 px
2atan(tan(70÷2)(207÷1024)) = 10.9423032462 degrees
Distance = 13,796 m

Total distance = 13,796+2000 = 15,796 m

### Speed and Timeframe

The timeframe for the blast's travel will be the time it took to the debris around him to reach their current position.

151 px = 2.1 m
76 px = 1.05695 m

√(2d÷g) = t
√(2(1.05695 m)÷(9.81)) = 0.4642 s
(15,796 m)÷(0.4642 s) = 34,028.175269 m/s

The timeframe for the cloud's dispersion is the time it took to the blast to get through the cloud and made the initial hole that expanded across the whole cloud.

(2000 m)÷(34,028.175269 m/s) = 0.058774823633 s
(1952.83 m)÷(0.058774823633 s) = 33,225.62 m/s

### Force

KE = 0.5×(3.2983×10^11 kg)×(33,225.62 m/s)²
KE = 1.82056574×10^20 J

Raoh's power = 43,512.5655 Megatons or 43.51256 Gigatons of TNT

### Alternate calculation

The cloud's volume of the previous calculation is most likely lowballed, since pages of the manga show the clouds reaching beyond the horizon (chapters of the fight: 135 and 136).

The distace between a viewer and an elevated point behind Earth's curvature is equal to the sume of each other's view to the horizon. I previously got the base to be 13,796 m above the ground. Putting the number in this page, I get a distance of 419.7 km. Adding the horizon's view of a 2.1 m tall person like Raoh, which is 5.2 km, would result the radius of the cloud to be 424.9 km. For a low-end, I'm gonna use a lesser distance to the horizon of 20 km.

#### Low-end

π(20,000 m)²(2,000 m) = 2.51327×10^12 m
M = 2.51327×10^12 kg

KE = 0.5×(2.51327×10^12 kg)×(33,225.62 m/s)²
KE = 1.38725×10^21 J

#### High-end

π(424,900 m)²(2,000 m) = 1.134366×10^15 m³
M = 1.134366×10^15 kg

KE = 0.5×(1.134366×10^15 kg)×(33,225.62 m/s)²
KE = 6.26137×10^23 J

Raoh's toki blast speed = 34,028.175269 m/s = Mach 100.0828
Raoh's power LE = 331,561.7 Megatons or 331.56 Gigatons of TNT
Raoh's power HE = 149,650,387.77 Megatons or 149.65 Teratons of TNT

Raoh did this after giving his life force to Yuria, having a hard fight with Kenshiro, bleeding out and at the brink of death. The alternate result is actually consistent with Jagi's nuke feat.  Hey KIrito352 how fast and powerful would Raohs goten sho be, if he was at full strength and uninjured. Please do more of these for people like, say, Kirby. Your calcs are amazing.

BTW, since Raiden used 35.3 million tons of force to stop Outer Haven (could actually be more, someone calculated OH's weight to be 92 million tons which would mean Raiden used 391.317 million tons of force to stop it), does that mean that Raiden's blows have millions of tons (or megatons) of force? Like mountain level force?

As I said before, please continue these calcs! You know, I loved Wonder Woman, because not only was that movie AWESOME, but Gal Gadot is an Israeli like me! Well, I’m going to return to Israel after college (1410 on the SAT, I hope I can go to Carnegie Mellon). In terms of energy/tons of TNT, I think Raiden is below mountain level. His feats put him between 100-1000 tons (1 kiloton) of TNT worth of energy. But you said he used 35 million tons of force to stop OH (though it could've been as much as 391 million). How much is 35 million tons of force? About 343.35 billion Newtons, which would be 3.4335×10^11 J/m. Considering Raiden's arm and blade together is about 1.43 m and he can swing in a 180° angle, he gives a total 1.54×10^12 J, or 368.66 tons of TNT. Thanks. I actually multiplied my calc by 6 because it was with 1 arm and in a weaker body when raiden was already beaten up and weakened. This is 20 trillion Newton or small town lvl. 