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Official Calculations Discussion Thread

Did it impact anything?
no. it was aimed up high, but it disintegrated the monsters
b597554ba13b0ff948cdc663dece2b11.png
 
the black dots are the monsters
Oh shit, really? It that case, you'd likely need to find the size of one monster, and estimate the value for vaporization based on that. After you find it, find the dimensions of the blast and see how much bigger it is than one monster. Then, it's as simple as multiplying the result by the energy value of vaporizing one monster.
 
oh, should I use a cylinder for vol?
Oh shit, really? It that case, you'd likely need to find the size of one monster, and estimate the value for vaporization based on that. After you find it, find the dimensions of the blast and see how much bigger it is than one monster. Then, it's as simple as multiplying the result by the energy value of vaporizing one monster.
 
155d03b70bd0874538b90d6df178c8b9.png
there a way to the know the distance of Mountain in this pov?
Maybe if you use an average mountain height and use the angsizing calculation to get the degree from the pixel height of the mountain and the panel and then do the distance with the degree to the estimated mountain height itself.
 
Are avalanches calc-able? Without any visuals to go off of. Just as like, a common feat.

(I.E. a character said to trap a large group of 'savage canibal apes' in a crevasse/valley with a 'giant avalanche')
Let's ignore the question of whether or not the character simply gave the avalanche a final push for it to naturally develop without its energy.

Keeping it brief since the avalanche was described ad gigantic you can probably take a such as one with 505,850 kg of mass and an average speed of 17.9 m/s if you wanna say he creates the totality of the avalanche. And then I think just simple KE to get you about 19 kg of tnt worth of energy.

However, I would be more inclined to scaling it to something like an avalanche's initial mass more if you don't know for sure which was 84400 kg which is only 3.2 kg worth of TNT lol.

You can probably get a higher value if you use like the highest end speeds for an Avalanche of like 88 m/s but its whatever.
 
Let's ignore the question of whether or not the character simply gave the avalanche a final push for it to naturally develop without its energy.

Keeping it brief since the avalanche was described ad gigantic you can probably take a such as one with 505,850 kg of mass and an average speed of 17.9 m/s if you wanna say he creates the totality of the avalanche. And then I think just simple KE to get you about 19 kg of tnt worth of energy.

However, I would be more inclined to scaling it to something like an avalanche's initial mass more if you don't know for sure which was 84400 kg which is only 3.2 kg worth of TNT lol.

You can probably get a higher value if you use like the highest end speeds for an Avalanche of like 88 m/s but its whatever.
Damn. Well, it makes sense given it's a sort of 'chain event,' and even at a higher end is a prolonged disaster, not a simultaneous one. Thanks for the answer!
 
Am I missing something? Pixel scale the character to then pixel scale the palm trees to then pixel scale the dimensions of the island. Take it to be roughly a triangular prism (I think the caved in parts kinda balance out with the presiding parts?) But more accurate result by diving it into smaller shapes.

Then for speed it seems like the boat’s top speed is unaffected to look for a similar boat and take its top speed. (I think the shots where it’s speed is kinda shown are too cartoony stylized)
How I can calc this?
 
So I was told that Omh’s Law is not entirely reliably in situations where temperature changes and changes the resistance of the material for conduction when the calculation requires consistent temperature.

so I am using the equation with the Temperature coefficient factor to find the wattage from the Ohm’s law.

first the temperature difference would be from room temperature 20 degree celsius which would be up to 44999726.85 degrees celsius. And the Temperature coefficient for tungsten which is a common material for fusor reactors is .0045 and the Ohm's Meter for resistivity is 4.9e-8.

The equation to start is

30keV is 30000 Volts

the Ohm’s is 4.9e-8/.07 * 0.0012566^2 = 1.1053305e-12 Ohms

now we must find the resistance after temp change with current resistance:

1.1053305e-12(1+.0045(44999726.85-20)) = 2.23829073e-7 Ohms
now we can find the Amps and then the Watts
Amps = 30000 / 2.23829073e-7 Ohms = 134030845939 Ampa
Watts = 30000 * 134030845939 Amps = 4.0209254e+15 Watts

thoughts?
 
So I was told that Omh’s Law is not entirely reliably in situations where temperature changes and changes the resistance of the material for conduction when the calculation requires consistent temperature.

so I am using the equation with the Temperature coefficient factor to find the wattage from the Ohm’s law.

first the temperature difference would be from room temperature 20 degree celsius which would be up to 44999726.85 degrees celsius. And the Temperature coefficient for tungsten which is a common material for fusor reactors is .0045 and the Ohm's Meter for resistivity is 4.9e-8.

The equation to start is

30keV is 30000 Volts

the Ohm’s is 4.9e-8/.07 * 0.0012566^2 = 1.1053305e-12 Ohms

now we must find the resistance after temp change with current resistance:

1.1053305e-12(1+.0045(44999726.85-20)) = 2.23829073e-7 Ohms
now we can find the Amps and then the Watts
Amps = 30000 / 2.23829073e-7 Ohms = 134030845939 Ampa
Watts = 30000 * 134030845939 Amps = 4.0209254e+15 Watts

thoughts?
I am not sure what you are tryna ask about here. Though shouldn’t you or ask the person who told you off in particular? Or is this what you got told off for and are asking if we agree?
 
I am not sure what you are tryna ask about here. Though shouldn’t you or ask the person who told you off in particular? Or is this what you got told off for and are asking if we agree?
It was saying that on another site that Ohm’s law on it’s own would not be entirely accurate as temperature change gives varying resistivity, so I applied the temperature change factor to the equation.
 
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It was saying that on another site that Ohm’s law on it’s own would not be entirely accurate as temperature change gives varying resistivity, so I applied the temperature change factor to the equation.
Bump
 
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May i know, if there's a character who have a black hole as part their body. By default does it grant LS same as mass of the black hole?
 
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