Explanation and method[]
I noticed a thing that I hadn't noticed before. Stolas was able to survive the friction of the moon in the sun's atmosphere. Considering that the achievement on Large Star Level was undeservedly rejected - most likely, this feat will be able to strengthen Stolas and other Overlords. Probably.
At first I had no idea how to calculate it. Then, I found in Dante's profile a link to the calculation of the falling sword. There was the formula that we need. Probably.
I really want to sleep and I can make some mistakes. But I want to do it now.
Let's start.
Formula[]
The formula itself looks like this:
- Joules = ½*ρ*C*A*(V^2)*D
- ρ = mass density of fluid
- C = Drag coefficient
- A = Area (in meters^2)
- V = Velocity (in meters)
- D = Distance traveled over which the object caught fire (I'm not very sure here)
Apparently, only half of the moon's surface is on fire. This means that A = 18965000000000 meter^2.
(C) = 0,47.
ρ = 10^(-5) kg/m^3 density of the Sun's atmosphere (Sorry, but I did not find the source in English. Only in Russian).
Speed[]
The angular size of the moon at the beginning = 2 × atg(tg(70 / 2) × (179.2 / 306)) = 44.5927694397 degrees.
http://www.1728.org/angsize.htm
Knowing that the diameter of the moon is 3471.94 kilometers, the moon is 4233.5 kilometers.
The angular size of the moon after a second = 2 × atg(tg(70 / 2) × (186.2 / 306)) = 46.1551861076 degrees.
To the moon 4074.3 kilometers.
4233.5 − 4074.3 = 159.2 kilometers first cathetus.
The moon moved 6 pixels to the left a second later.
3471.94 / 179.2 × 6 = 116.247991071 meters of the second cathetus.
Knowing cathetus, the hypotenuse is 197.124923407 kilometers.
V = 197124.923407 m/s.
We do not know how quickly the moon lit up, because this was not shown to us. I guess it took a second. I'm going to sleep, so I won't do a few options. I'll do it after I wake up.
All the data is there. Let's find the result.
After a while, I realized that I had done a stupid thing. I took the area of the moon. And I needed to take the area of Stolas' body. Awesome.
Blitzo's head[]
Door = 558 px = 2 meters.
2/558 × 182 = 0.6523297491 meters window.
0.6523297491 / 335 × 169.2 = 0.32947520462 meters Blitzo head.
Stolas and Blitzo's height[]
(0.32947520462 / 57.5 × 265.5) + 0.32947520462 = 1.85079114943 meters Blitzo's height. In our calculation, this is not needed, but I need it to edit past calculations. Because I used to think that Blitzo was 1.5 meters. I was wrong.
0.32947520462 / 57.5 × 583.6 = 3.3440300768 meters Stolas height.
Surprisingly, he is taller than Valentino.
The body surface area of an adult male is 1.9 square meters.
The average height of a man is 175 centimeters (probably).
(3.3440300768 / 1.75) × 1.9 = 3.63066122624 m^2 body surface area of Stolas.
A = 3.63066122624 m^2.
The coefficient of friction also changes. Most of all, Stolas resembles a figure, whose coefficient is 0.04.
С = 0.04
All the data is there. Let's find the result.
Low-End:[]
0.5 × (10 ^ (−5)) × 0.47 × 18965000000000 × (197124.923407 ^ 2) × 197124.923407 = 3.413857 × 10^23 joules, or 81.5931405354 Teratons (Country Level+, 6-B).
0.5 × (10 ^ (−5)) × 0.04 × 3.63066122624 × (197124.923407 ^ 2) × 197124.923407 = 5562119760.41 joules, or 1.32937852782 Tons of TNT (Building Level+, 8-C).
The level does not change, but the fact that Stolas does not feel such damage at all only confirms that he can be at Level 8-A.
High-End:[]
After some time, I thought and made some conclusions. As the series has shown, the authors of the series strongly neglected cosmic laws (wind on the moon, in space, sounds in space). And also, the Sun's atmosphere is too dense. The photosphere is 10000 times less dense than the Earth's atmosphere. But as we can see, the moon is burning strongly. Much stronger than it should. And it is only at the edge of the Photosphere. I think it's worth using the density of the Earth's atmosphere. Apparently, the authors of the series gave the star just such parameters. Otherwise the moon would not burn in the atmosphere from friction.
Therefore, I believe that it is better to take the density of the Earth's atmosphere. Because, as practice has shown, artists and authors gave the sun exactly such density parameters without realizing it.
ρ = 1.225 kg/m^3
0.5 × (1.225) × 0.04 × 3.63066122624 × (197124.923407 ^ 2) × 197124.923407 = 6,8135967 × 10^14 joules, or 162.848869503 Kilotons (Large Town Level, High 7-C).
RESULT[]
Low-End = 81.5 Teratons (Country Level+, 6-B).High-End = 4.1 Exatons (Multi-Continent Level, High 6-A).- Low-End = 1.3 Tons (Building Level+, 8-C).
- Low-End = 162.8 Kilotons (Large Town Level, High 7-C).
Cool.
Considering that Stolas did not feel the damage at all, he should be upgraded to the Large Country Level (The difference between our result and the Large Country is less than two times, so everything is legal) and Multi-Continent Level+ (A similar situation).
Now, either Stolas remains at the Multi-City Block Level, but even more firmly, or he upgrades to the Large Town Level. I have already said my opinion about the density of the atmosphere.