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Official Calculations Discussion Thread

SeijiSetto said:
you can find this out using steps on the Large Size Calculations page.
Click to expand...
Wanna make sure I'm doing this right.

So I have 8766.62857144m for the characters height, and 8766.62857144m / 1.71m = 5126.683375x for the x he's taller than avg male

If the character is x-times larger than its normal-sized counterpart, and made of roughly the same materials, then its mass would be about x^3-times larger.

What do I do now?
 
Arkenis said:
Wanna make sure I'm doing this right.

So I have 8766.62857144m for the characters height, and 8766.62857144m / 1.71m = 5126.683375x for the x he's taller than avg male

If the character is x-times larger than its normal-sized counterpart, and made of roughly the same materials, then its mass would be about x^3-times larger.

What do I do now?
Click to expand...
the paragraph you quoted tells you what to do. you have your scale factor (the x times taller), so [if the guy is made of the same materials and isn't like stone or smth], he's x^3 times heavier.
 
Arkenis said:
Wanna make sure I'm doing this right.

So I have 8766.62857144m for the characters height, and 8766.62857144m / 1.71m = 5126.683375x for the x he's taller than avg male

If the character is x-times larger than its normal-sized counterpart, and made of roughly the same materials, then its mass would be about x^3-times larger.

What do I do now?
Click to expand...
This is the whole process of a Large Size Calc:
  • 8766.62857144/1.71 = 5126.68338
  • 5126.68338^3 = 1.34744017e+11
  • Average human mass: 62 kg
  • Mass of big dude: 62*1.34744017e+11 = 8.35412905e+12 kg
 
can Sound intensity (W/m^2) be used to find energy? Is Sound intensity unilateral across the whole area or does it take into account of being projected from a source and spread outward in all directions covering the area m^2?
 
If it is spread equally across the area, would i have I have to use Inverse Square law equation to find how much sound energy is weakened with distance?
 
I found that for sound intensity, the decibel reduces by 6 dB for each doubling of distance from the source, if the source distance is around 3502 km, what would the doubling be? like divided by 2, then divide the result by 2, etc.?
www.acoustical.co.uk

Distance Attenuation: How Sound Reduces with Distance

The farther you are from a point of sound, the less sound you hear. This is called distance attenuation. But what is happening to the sound as it travels?
www.acoustical.co.uk www.acoustical.co.uk
 
Question, is there a formula that can be used to calculate the KE of the movement of the clouds if part of them move slower than another part? (not omnidirectional)
 
GFArrow said:
How can I calculate this feat?
Click to expand...
The fall is simply weight*height*9.81 for gravitational potential energy.
Denting the car roof would require you to find the yield stress of a car roof and multiply it by its volume in m^3 to get force which can then be converted to work done by multiplying the force by how far it was dented in meters
 
SunDaGamer said:
The fall is simply weight*height*9.81 for gravitational potential energy.
Denting the car roof would require you to find the yield stress of a car roof and multiply it by its volume in m^3 to get force which can then be converted to work done by multiplying the force by how far it was dented in meters
Click to expand...
I couldn't find the yield stress for the car roof. Is there an average value I can use for yield stress?
 
GFArrow said:
Thank you. I calculated the gravitational potential energy, now I don't quite understand what I should do. Could you please explain in more detail? And is there a sample calculation blog I can look at?
Click to expand...
Make a blog with a link to the feat, the verse it's from and the episode. Then type out all of your calculations in a coherent way on a blog and present your yields in Joules, you can see some of my calcs to get a sense of formatting
 
how can i get the mass of a meteor? There is no other detail besides its a big burning brown meteor at several thousand kilometers
 
Shiraito983 said:
can anyone check if I did the density and mass right because i am still unsure the process
vsbattles.fandom.com

TODAG another ap calc

vsbattles.fandom.com vsbattles.fandom.com
Click to expand...
I could be wrong but you should've gotten this.
Cov6Kjp.png

Divide the length by half and width by half.
 
Rutæhh said:
I think i did something wrong here
could someone check it out please
Click to expand...
For explosions that happen on the ground, we use this formula:

  • W = R^3*((27136*P+8649)^(1/2)/13568-93/13568)^2, where W is the yield in tons of TNT, R is the radius in meters, and P is the shockwave pressure in bars, where we generally use 1.37895 bars or 20 psi of pressure. For this specific formula, there is no need to divide the result by 2
Click to expand...
W = 2667^3*((27136*20+8649)^(1/2)/13568-93/13568)^2 = 43476261.7 Tons of TNT
43.5 Megatons for blowing up a warehouse is pretty excessive because warehouses aren't 17500 feet wide, that statistic is talking about the area some warehouses can cover.
 
Rutæhh said:
I think i did something wrong here
could someone check it out please
Click to expand...
The square root of 180,000 is 424.26 feet. Discerning the length of the warehouse from your previous calc should be done like so- admittedly, most warehouses are longer than they are wide, but this is not a hard and fast rule and as you're working with a general "warehouse" rather than a specific building, it's best to treat it as a square.

424.26 feet is 129.32 meters. Half of that would be 64.66 meters.

I will note that @SunDaGamer also made an error by using 20 psi for the equation- that is correct, but it should be listed in bars, as mentioned. So instead of 20, the value used in its place should be 1.37895.

Your calculation should look like this for blowing up a standard warehouse: W = 64.66^3*((27136*1.37895+8649)^(1/2)/13568-93/13568)^2, which equals 21.73 Tons of TNT equivalent. That's low-end City Block level. This, of course, assumes a ground-based explosion, rather than something elevated from the ground.
 
Stupid question but- Is there a formula for or way of calcing destroying the moon? Or is it just the same as blowing up a planet's KE?
 
I was wondering about calculating something like this.

All we can see are individual, firey dots on the planet's surface from ground-detonated nukes.

What would be the best method to use here for the explosion itself? I.e, shockwave formula, 5 psi, mushroom cloud stuff, etc.

It's also worth noting that these are neutron bombs.
 
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