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## Pit (A Kid Icarus respect thread)

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### Re: Pit (A Kid Icarus respect thread)

After two years, I finally have resolved Pit's durability in terms of blunt trauma.

In The War’s End: Chapter 25, when Hades destroys the Great Sacred Treasure, Pit falls at least thousands of meters. [1] Using the kinematic equation for finding distance, Pit drops 2,017.32 m. This is based on the assumption that Pit was experiencing a gravitational acceleration of 9.81 m/s^2, which lasted 20.28 seconds. Then using sqrt(2(9.81 m/s^2 * 2,017.32 m.) to find instantaneous velocity, I ended up with 198.95 m/s. After this, I used the equation for drag force, which is Fd = 1/2 ρ v^2 Cd A.

To clarify, ρ is air density, v^2 is relative velocity, Cd is the drag coefficient, and A is frontal area. I used an air density of 1.225 kg/m^3, the velocity of 198.9468 m/s and squared it, and a drag coefficient of 1.4, since I suspect this is the high-end of a skydiver. The alternative was 1.0, which I presume was for vertical diving. Pit is more or less falling as if his entire body wasn’t streamlined, but horizontal. For frontal area, I used 1.33 m^2, since this is the surface area for children of ages 12 to 13. [2] I thought it would be reasonable for Pit to be 13, or at least have the appearance. I then punched in the numbers.

Fd = (0.5)(1.225 kg/m^3)(198.95 m/s)^2(1.4)(1.33 m^2)
Fd = 45,140 kg m/s^2

Knowing all that is needed to be known, I went ahead and also calculated for terminal velocity using sqrt(2(mg)/(ρ A Cd)). Here, m is mass, g is gravitational acceleration. The rest is the same. I used a chart for height and weight. [3] While the site uses 100 lbs., using Hamwi’s method gave me 112 lbs. I’ll use the latter and convert it to kilograms, which is 50.8023 kg. I filled in the equation.

sqrt(2(50.8023 kg. * 9.81 m/s^2)/(1.225 kg/m^3 * 1.33 m^2 * 1.4))
= 20.90418817864566 m/s

In order to find momentum, I need to multiply mass times velocity. Using Pit’s mass and multiplying this by the velocity, I end up with 1,061.980839108010413018 kg m/s. I attempted at dividing this by the time of the impact, but it occurred too quick for me to time it, even with the video slowed down to 25% of its speed. I could assume 1 millisecond, but I doubt that. I think 10 milliseconds would be more reasonable because 100 milliseconds is too slow. If I use 10 milliseconds, then Pit will have experienced 106,198.08 kg m/s^2, or 11.94 tons of force. This is approximately 1.2 times the maximum total thrust generated by ThrustSSC, the first car to break the Mach 1 barrier.

In hac mea interpretatione - How durable is Pit?
Mea quidem sententia