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Final Fantasy XIV - Warrior of Light blasts a Sin Eater

Published by TTGL in the blog TTGL's blog. Views: 97

Calc request for @GiveRobert20dollars and @Cipher97 .


The Warrior of Light blasts a Sin Eater hard enough to part infront and behind him. Average height of a man is 176.1cm, or 1.761m. That would mean at least 4.7km to the horizon, or 4700m. How wide is the gap though?
Spoiler:


= 2*atan(234/(673/tan(70/2)))
= 0.326554592 rad
= 18.7101999022629677 degrees
Using the angscaler, the gap in the clouds is 1548.6m across.

234 pixels = 1548.6m
1 pixel = 1548.6m/234 = 6.61794872m
6.61794872m X 90 = 595.615385m

Volume as a rectangle.

V = 4700 X 595.615385 X 1548.6
= 4.33513893e9m^3

Density of a cloud is 1.003kg/m^3.

M = 4.33513893e9 X 1.003
= 4.34814435e9kg

Next we need the timeframe.
Spoiler:


1.02 seconds.

T = 4700m/1.02s
= 4607.84314/340.29
= Mach 13.5409302

Finally for the KE.

KE = (0.5)mv^2
= (0.5) X 4.34814435e9 X 4607.84314^2
= 4.61603752e16 joules

Given it goes both ways, let's times that by two.

E = 4.61603752e16 X 2
= 9.23207504e16 joules
= 22.065188910133844757 megatons

Final Results
Warrior of Light parts clouds (speed) = Mach 13.541
Warrior of Light parts clouds (energy) = 22.0652 megatons
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