You can calculate it normaly but moving under water at constant speed requires constant source of power
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VS Battles Wiki Forum
Ugarik
M3X_2.0
What do you mean by constant source of power? The feat in question is about punching X object at high speeds underwater.
Ugarik
I need to know the context then. The speed can't be constant under water
M3X_2.0
Star Platinum punches and sent everything flying (?)
Star Platinum punches and sent everything flying (?)
Ugarik
I'll try to derive the formula
Ugarik
Okay let's say an object with a mass of 10 kg crossed 10 meters underwater in 2 second. You want to know its KE the moment in started to move? Is that correct?
M3X_2.0
Yes
Ugarik
Okay, well the kinetic energy is still 0.5mv^2, but finding speed is very complicated since under water the acceleration due to drag force is very high, so you can't calculate speed as D/t, 10 m / 2 s = 5 m/s.
I derived the speed formula of this particular case. It's
v=(e^(Db)-1)/(bt)
where D - distance
t - time
b - (Water density * cross-sectional area * drag coefficient / 2*mass)
Water density is 1000 kg/m^3
As I need a value for the cross-sectional area, so let the object in question be a ball with a radius of 15 cm. So the cross-sectional area is pi*0.15^2 = 0.07 m^2
Drag coefficient of a ball is 0.47
b = 1000 kg/m^3 * 0.07 m^2 * 0.47 / (2*10 kg) = 1.645 m^-1
v = (e^(10 m* 1.645)-1)/(1.645 m^-1*2 s) = 4235925.35 m/s
KE = 0.5*10*4235925.35^2 = 8.97153179*10^13 (Town level)
I honetly didn't expect it to be that high, 10 meters in 2 seconds is clearly to much for that mass, but you get the idea
I derived the speed formula of this particular case. It's
v=(e^(Db)-1)/(bt)
where D - distance
t - time
b - (Water density * cross-sectional area * drag coefficient / 2*mass)
Water density is 1000 kg/m^3
As I need a value for the cross-sectional area, so let the object in question be a ball with a radius of 15 cm. So the cross-sectional area is pi*0.15^2 = 0.07 m^2
Drag coefficient of a ball is 0.47
b = 1000 kg/m^3 * 0.07 m^2 * 0.47 / (2*10 kg) = 1.645 m^-1
v = (e^(10 m* 1.645)-1)/(1.645 m^-1*2 s) = 4235925.35 m/s
KE = 0.5*10*4235925.35^2 = 8.97153179*10^13 (Town level)
I honetly didn't expect it to be that high, 10 meters in 2 seconds is clearly to much for that mass, but you get the idea
M3X_2.0
Oof, very complicated compared to the normal formula, and why is the result that high?? It seems that the feat in question will result in Tier 7 too, well shit