Digimon Feats List

Introduction

In order to help with the next Digimon revisions, I’m doing this blog to collect old Digimon calculations (That don’t need recalc) and also add new calculations (Or recalc old ones). This will mostly be done for Baby Level up to Adult Level Digimon in order to establish the reasoning for their tiers. Since most of the Ultimate level Digimon deals with metaphysical feats with universal/multiversal destruction, they will have a blog of their own.

Fladramon’s Fire Rocket

Feat and Context

During the various searches for the Digimon Kaiser’s base, Koushiro and Iori found out that the base was in a place full of Dark Towers.

Iori would return to there with help, but the base wasn’t there anymore, in the place were the Base was we can only see a big hole. There they were attacked by a Dokugumon controlled by an Evil Ring, in order to defeat it Fladramon used “Fire Rocket” in the hole were the Base was before, you can see the result in the gif above.

The Hole

So, the Kaiser’s Base was inside the hole, so the first thing that we need to do is to find the size of the Kaiser’s Base.

To do that, I’ll use the size of a Devidramon that was close do the base, and to find its size I’ll use Digimon Kaiser’s own size.

The official Digimon Adventure 02 Height chart puts the Digimon Kaiser’s height at about 170 cm with its hair.

This is the scene that I’ll be using (I choose this scene because of the body configuration of Devidramon in that scene, the closest to its full height as it’s in the scene with the Base scaling);

Devidramon Height = (Digimon Kaiser’s Height/12) * 202 = 2861.6 cm = 28.616 meters

Now, the Digimon Kaiser’s base itself. I’ll be doing a lot of scaling (The full height for future proposes, the total part that was underground, the maximum diameter and the minimum diameter since the undergrounded part is more similar to a cone section, etc).

Full Height = Maximum Height = (28.616 m/11.2 px) * 420px = 1073.1 meters

Undergrounded Depth = (28.616 m/11.2 px) * 216.2px = 552.391 meters

Maximum Diameter = (28.616 m/11.2 px) * 504px = 1287.72 meters

Minimum Diameter = (28.616 m/11.2 px) * 202.4px = 517.132 meters

The Fire

So, the volume of the fire is equal to the total volume of the hole (That I’m considering to be a circular truncated cone) and the total volume from outside of the fire, that is also of a similar shape.

For the hole itself, I’ll simply use the minimum and maximum diameter and the depth of the hole in the formula

V=1/3 * π * (Rmax² + Rmax*Rmin + Rmin²) * h

V = 1/3 * pi * (643.86² + 643.86*258.566 +258.566²) * 552.391 = 374781448 m³

Now for the upper region of the fire, I’ll use this scene. Also, I think that we can clearly see that there’s even more fire above of what we can really see, so this calculation is a low-end for the total volume itself.

The minimum radius is that of the hole itself, for the height and the maximum radius

Maximum Diameter = (1287.72 m/470) * 565 = 1548.00383 meters

Height = (1287.72 m/470) * 399 = 1093.19209 meters

V=1/3 * π * (Rmax² + Rmax*Rmin + Rmin²) * h

V = 1/3 * pi * (643.86² + 643.86*774.001915 +774.001915²) * 1093.19209 = 1.73090073e9 m³

Total Volume = 1.73090073e9 m³ + 374781448 m³ = 2105682178 m³

Energy

Now, doing the same that is accepted in other feats of similar nature, I’ll be using air density as 0.3kg/m³, change in temperature as 1184 Kelvin and heat capacity as 929j/K

Mass = 2105682178 m³ * 0.3kg/m³ = 631704653 kilograms

Heat = 631704653 kg * 1184 K * 929j/K = 6.94834689e14 joules, Hight 7-C