User blog:ImagoDesattrolante/Sheele Bullet Times (calc)

So yeah, basically this happens.

I'm just here to calc this. Sheele cuts apart a few bullets that are fired at her - I would calculate the kinetic energy required to do shit like this but it's probably not even wall.



I first start off by measuring the length of her blade, which is 160 pixels. Her height is 217 pixels.

My images actual height is 479 pixels. Now for the wacky formula:

2*atan(217/(479/tan(70/2 deg))) = 35.19941 deg.

Now to look for the distance.

http://www.1728.org/angsize.htm Her height is 5'3, or about 1.59 meters. The angle is the number of degrees I got earlier, so 35.19941. So it equals 2.5062. That's how far away she was.

Average bullet speed is 340.29 m/s.

Now to get the timeframe.

2.5062 / 340.29 = 0.00736489464, so that's the amount of seconds.

Now I need to get Sheele moving her arms. An arm is 45% of height, so...

0.45 * 1.59 = 0.7155, that's how long her arm is. Now to get the sword.

160 pixels for the sword, 217 for her body. Rule is to divide the object you want by the measuring stick.

160/217 = 0.73732718894. Multiply that by her height and you have the swords length.

Add that onto her arm length.

0.7155 + 1.17235023041 = 1.88785023041.

It looke like she moved her arm from one side to the other completely, so 180 angles will work.

180 degrees in radians is 3.14159. Now I multiply that by her sword + arm length.

3.14159*1.88785023041 = 5.93085140535.

Velocity = distance / time

5.93085140535/0.00736489464 = 805.286659926 meters a second, or Mach 2.347774518734694.

Since she's a bit of a fodder than should scale to a decent amount of characters in the verse, so yeah - done.