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Official Calculations Discussion Thread

hello, I'm here again. Is there a way to measure attack-like blast that looks like a massive laser
 
Da3ggman said:
Did it impact anything?
Click to expand...
no. it was aimed up high, but it disintegrated the monsters
b597554ba13b0ff948cdc663dece2b11.png
 
Shiraito983 said:
the black dots are the monsters
Click to expand...
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?
Da3ggman said:
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.
Click to expand...
 
Hello on Vaporization and Pulverized
How much joule/cc on
Small Building
Small towns
Small country
Durability
 
Shiraito983 said:
155d03b70bd0874538b90d6df178c8b9.png
there a way to the know the distance of Mountain in this pov?
Click to expand...
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.
 
StretchSebe said:
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')
Click to expand...
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.
 
PhantomØ4 said:
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.
Click to expand...
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)
Dragonite007 said:
How I can calc this?
Click to expand...
 
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?
 
CNBA3 said:
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?
Click to expand...
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?
 
PhantomØ4 said:
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?
Click to expand...
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.
 
Last edited:
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?
 
Aether2734 said:
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?
Click to expand...
Wasn’t the whole issue with own weight scaling that LS is the additional mass you can lift atop your mass?
 
So I learned that even with the change in Temperature factor in the Omh’s Law equation would be unsuitable.

so the only thing I can think of that can get for energy from electron volts is through temperature change.

which is 1.6023-19 Joules to 1 eV corresponds to 11606 K

With the temperature I am going with is 150000273.15 Kelvin to 12926.01 eV to 2.07097503e-15 joules.

thoughts?
 
So utilizing DontTalk's method of calculating the energy of Plasma usage, I think I have what I need for calculating what I did in my previous threads.

And like in the other thread, the scenario is where the tech transcends conventional science and they have unlimited resources.

Since Fusors are fusion devices with using electricity, I will be using the material for fusion Deuterium,

Number of atoms:

6.022e+23 (Avogadro's Number) * .1967 g/cm^3 (Deuterium Density) * 308 cm^3 (Volume) / 2.014 g/mol (Deuterium Molar Mass) = 1.8114918e+25 number of atoms

----

The degree of Ionization would be likely total in this case so will be 1.8114918e+25 ions

——

1 Atomic Number of Deuterium = 13.6 eV

1.8114918e+25*13.6 = 2.4636288e+26 Watts

Thoughts?
 
vsbattles.fandom.com

Invincible (Comic) - Destruction of Viltrum

vsbattles.fandom.com vsbattles.fandom.com

I’m kinda bamboozled by this calculation (current listing for the feat on their profiles). Why were earthquake formulae used to quantify their end of the feat? Isn’t that pretty counterintuitive? It severely downgrades them IMO. A recalc should be done using the mass they dispersed out into infinity when exiting. This gets it much higher.
 
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