Sorry guys . . . Just felt like an ass today! So here I go. I'm pretty sure you all know which feat I am referring to, and honestly I'm to lazy to put in all the details. If you want more details as to how a certain calculation is done, or where I got a certain number, or why a certain formula is being used then just ask. Also, if you think I'm wrong then please give reasons for your answer so I can improve the post. Without further ado . . .
I might accidentally start a shit-storm of a thread!
Video Reference: https://www.youtube.com/watch?v=ewptHiX76C0
The submarine was an Akula class submarine which can weigh in from 8140 to 13800 tons. Let's high ball and say that it was 20000 tons!
Initial Vertical Velocity of the Sub as it broke the water = x m/s
Final Vertical Velocity of the Sub after it broke the water = 0 m/s
Video Time Tag --> [0:49]
Height that the sub went into the air < 30 meters. Let's High ball and say it was 50!
(V^2) - (x^2) = 2as
(x^2) = -2as
x = (-2as)^(1/2)
x = (-2*-9.81*50)^(1/2) ==> 981^(1/2) ==> 31 m/s
Acceleration = Velocity/time ==> x/t
t = [0:48] - [0:17] ==> 31 seconds :o
Acceleration = 31/31 == 1 m/s^2
F = ma
F = (20000*1000)*1/10000 ==> 2000 tons!
We also need to add in drag force
Time tag --> [0:25] Arthur begins 'bracing' the submarine. Feat turns from strength to durability from here on outwards.
Time of Acceleration = [0:25] - [0:17] ==> 8 seconds
Drag Force calculator = Co-Eff * Density * Reference Area * 0.5 * Velocity^2
Co-eff = 0.47
Density = 1000 kg/m^3
Ref-Area = 2709 m^2 [High-Balled]
Velocity = Acceleration*time ==> 1*8 ==> 8 m/s
Drag Force = 0.47 * 1000 * 2709 *0.5 * 64/ 10000 ==> 4074 tons!
Total Force = 4074 + 2000 = 6074 tons [High-Balled]
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