Calc Shiro Inukai Kinetic Energy calc

Calculations
Well, in chapter 27 and 28 of Orient Shiro Inukai one of the obsidian eight cut the floor behind him deep enough to reach the Asthenosphere and lift the stone pillar of the rock cut it really fast and destroy it in the process.


Now, the hole was deep probably was the distance between the crust to the Asthenosphere(410km of deepness).

Now the dimensions of the rectangle pillar, Length: 32.7 meter, Width: 37.2 meters and Height: 410000 meters. As reference i used one of the standard size 7.6 meters(25 feets).
Screenshot_20240106-231652_ImageMeter.jpg

Volume = Length × Width × Height
= 32.7×37.2×410000=498740400 meters^3 or 4.987404e+14cc

The rocks had a density of 2.7g/cc or 2700kg/m^3

After knowing all that, we can get the mass.

Mass=Volume×Density
Mass= 498740400 meters^3×2700kg/m^3
Mass=1346599080000kg
The last part is get the Kinetic Energy of moving the rectangle pillar and the pulverization of the pillar.

For pulverization is needed the value of joules/cc of the rock, the value is 214 joules/cc.

The value of pulverize a pillar that had 4.987404e+14cc is:
4.987404e+14cc*214 J/cc=1.0673e17 joules or 25 Megatons of TNT | City level

The Kinetic Energy of moving a rectangle pillar of 1346599080000kg at high speed and if the movement was the same one as the height of the object the energy generated by that was something like this:
KE=(1/2)mv^2
KE=(1/2)1346599080000*410000^2
KE=1.13181652674e+23 joules or 27 Teratons of TNT | Country level

Any advice or correction is welcomed.

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