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This feat is way to complicated, but I figured out a good way.

The Eye Of The Storm when brought down, pushes a large amount of clouds out of the way, I can calc the KE of that.

I probably would fail badly though

Eye Of The Storm KE

So a reasonably strong waterspout is 1095 meters, so I'll use that in lack of a better number.

The waterspout is 285 pixels tall

1095/285 = 3.8421052631578947 meters per pixel

The length of the Eye Of The Storm is 516 pixels

516*3.8421052631578947 = 1982.5263157894736652 meters in diameter or 991.2631578947368326 meters in radius

The storm seems fairly really rough, so I guess Cumulonimbus clouds? I think their 12000 meters thick according to this

Volume = pi x radius^2 x height

Volume = pi x 991.2631578947368326^2 x 12000 = 37043247131.7750662262886305601 meters cubed

Density of clouds is 1.003 kg/m^3

37043247131.7750662262886305601 x 1.003 = 37154376873.1703914249674964517803 kilograms

Timeframe: Starts at frame 7389 and ends at Frame 8101

8101-7389 = 712 frames

712/25 = 28.48 seconds

Speed: 991.2631578947368326/28.48 = 34.8055884092253101334 m/s, which is Subsonic... kek

KE time

0.5*37154376873.1703914249674964517803*34.8055884092253101334^2 = 2.2504945e+13 joules or 5.37881094646 kilotons, Small Town level+.

I probably butchered this a bit but Don't hurt me

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