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Official Calculations Discussion Thread

Mr._Bambu said:
Could be speed. I don't think generating darkness can be used for any AP calc, can't imagine what you'd do for it. Suppose you might try to argue it's removing the energy of the light in the given area but that could be very tenuous depending on the means of creating darkness. If you did approach that route, it's a complicated calculation I don't have on me at the moment, but could dig it up if there's literally nothing better for your given verse and you feel like taking that shot.
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I guess depending on what the person means, it could be used like say whatever affected area where sunlight hits like watts per area.
 
CNBA3 said:
I guess depending on what the person means, it could be used like say whatever affected area where sunlight hits like watts per area.
Click to expand...
Like light absorption, sort of thing? Could be that too. Doesn't necessarily scale to AP, that, but under very specific contexts I suppose it could.
 
Question.

In this thread said that there is 1 in 12 ratio of sand and air. This is for sandstorms, what are your thoughts on this?

 
When doing an explosion speed calc where multiple separate explosives were detonated close to each other, do we use the charge weight of a single explosive or the combined weight of all the explosives?
 
Last edited:
Mr._Bambu said:
Could be speed. I don't think generating darkness can be used for any AP calc, can't imagine what you'd do for it. Suppose you might try to argue it's removing the energy of the light in the given area but that could be very tenuous depending on the means of creating darkness. If you did approach that route, it's a complicated calculation I don't have on me at the moment, but could dig it up if there's literally nothing better for your given verse and you feel like taking that shot.
Click to expand...
If we're covering something in darkness, that would more accurately fall under kinematic viscosity: https://en.m.wikipedia.org/wiki/Viscosity

I don't think we, or anyone in the vs debating community for that matter, have a good idea on how to convert kinematic viscosity into speed.
 
Flashlight237 said:
If we're covering something in darkness, that would more accurately fall under kinematic viscosity: https://en.m.wikipedia.org/wiki/Viscosity

I don't think we, or anyone in the vs debating community for that matter, have a good idea on how to convert kinematic viscosity into speed.
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Well, I think Emerald was asking about AP. And as for speed, if it's a matter of darkness spreading over an area, it'd be a simple calculation of the distance the darkness traveled over the timeframe it traveled that distance in.
 
LaserPrecision said:
Is there a method for calculating uprooting trees (or really anything with roots)? One that can be extrapolated to fictional comically large trees? Because uprooting them feels like it should be a very impressive feat, but I haven't seen any formula for it around.
Click to expand...
Bump
 
What does it mean for density values to have a negative value? like g/cm^-3 or kg/m^-3? trying to find the density values of strong ice types like Glaciers or Ice 7.
 
I'm not one that does calcs much, but i need help trying to pixel scale the size and AP of the monolith from Vita Carnis, these are some of the best images i can find for it's height and size.

I would love some help, as i'm not really that great at math.
 
Is there a formula that one can use to determine the size of an object with gravitational waves? Using the speed of light and time duration of event?
 
I think I found a method to get the size of objects different parts of the planet.

I found it difficult to comprehend the calculation of object sizes from a space view to be the same as the distance from the horizon/circumference of the planet via pixel scaling.

So I did something that I think would help, taking the diameter of the planet and equate that to the semicircumference or half the circumference of the planet, that way when you measure the distance to the object on the surface of the planet you would get the distance traveled while taking into account the curvature.

The way you can use this is with ratio calculator.

thoughts?
 
I tried this out from a satellite image of the horizon of earth and ireland. it may not be 100% (Ireland length is 486 km) accurate but should be close enough with when I did it I got 526.84 km.
 
I feel that the calculation of AP of Saitama is having a problem with the way of using formulas. Does anyone want to talk about this?
 
Can you elaborate on what you mean?
 
...Could you link to the calculation? I don't know what you're referring to.
 
I know but I think there is a mistake in this calculation here is "6.087799e+18 meters^2" this is actually the surface area of the sun we can do a simple calculation with the radius of the sun which is 695700000 meters then 4*3.14159265358979*695700000^2=6.0821044e+18 meters^2, meaning the number used in this calculation is the surface area of the sun calculated as a sphere whereas what we need to use is a circle 3.14159265358979*695700000^2=1.5205261e+18 meters^2.
 
KamitoSam119 said:
I know but I think there is a mistake in this calculation here is "6.087799e+18 meters^2" this is actually the surface area of the sun we can do a simple calculation with the radius of the sun which is 695700000 meters then 4*3.14159265358979*695700000^2=6.0821044e+18 meters^2, meaning the number used in this calculation is the surface area of the sun calculated as a sphere whereas what we need to use is a circle 3.14159265358979*695700000^2=1.5205261e+18 meters^2.
Click to expand...
No, you'd need the frontal area, which would be surface area/2. Which is pretty much what Qawsedf does in his math.
 
KLOL506 said:
No, you'd need the frontal area, which would be surface area/2. Which is pretty much what Qawsedf does in his math.
Click to expand...
I believe for ISL we use cross-sectional area, actually.

the 4*U*(Er/Tr)^2 thingy we use for cosmic ISL simplifies nicely the way it does because the area of any sphere is just 4x the area of a circle of equal radius, hence why you can take the radii squared into their own bracket.
 
SeijiSetto said:
I believe for ISL we use cross-sectional area, actually.
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More or less the same thing really, usually it's surface area/2, for humans it varies a bit from 40-50% that of the total surface area depending on your bulk.
 
Why is it /2 instead of /4 this is not reasonable because in isl using the cross-sectional area here is the area of a circle.
KLOL506 said:
No, you'd need the frontal area, which would be surface area/2. Which is pretty much what Qawsedf does in his math.
Click to expand...
 
KamitoSam119 said:
Why is it /2 instead of /4 this is not reasonable because in isl using the cross-sectional area here is the area of a circle.
Click to expand...
Because you're not affecting a quarter of the star's surface? You're affecting a full half.
 
I think that the calculation on isl uses density to calculate the total energy needed for the explosion with a star when it comes into contact with the surface of the explosion, then the range that the explosion compresses it is calculated by the cross-sectional area of the star here is the area of a circle so within a range on the surface of the explosion with an area equivalent to the cross-sectional area of the star will come into contact and destroy the entire star so we need the Surface area of the explosion/Surface area of the sun's cross section * GBE of the sun.
KLOL506 said:
Because you're not affecting a quarter of the star's surface? You're affecting a full half.
Click to expand...
 
So, based on what you're saying is, it'd be like this?

6.087799e+18/4 = 1.5219498e+18 m^2

Energy (Low-end): 5.693e+41 * (9.6488373e+37 / 1.5219498e+18) = 3.6092406e+61 J (Individual yield of Saitama and Garou being 1.8046203e+61 J)

Energy (Mid-end): 5.693e+41 * (6.0305233e+40 / 1.5219498e+18) = 2.2557754e+64 J (Individual yield of Saitama and Garou being 1.1278877e+64 J)

Energy (High-end): 5.693e+41 * (3.8905547e+43 / 1.5219498e+18) = 1.4552995e+67 J (Individual yield of Saitama and Garou being 7.2764975e+66 J)

Results seem to be higher.
 
KLOL506 said:
So, based on what you're saying is, it'd be like this?

6.087799e+18/4 = 1.5219498e+18 m^2

Energy (Low-end): 5.693e+41 * (9.6488373e+37 / 1.5219498e+18) = 3.6092406e+61 J (Individual yield of Saitama and Garou being 1.8046203e+61 J)

Energy (Mid-end): 5.693e+41 * (6.0305233e+40 / 1.5219498e+18) = 2.2557754e+64 J (Individual yield of Saitama and Garou being 1.1278877e+64 J)

Energy (High-end): 5.693e+41 * (3.8905547e+43 / 1.5219498e+18) = 1.4552995e+67 J (Individual yield of Saitama and Garou being 7.2764975e+66 J)

Results seem to be higher.
Click to expand...
Yes, the result will increase because the range of each star will decrease.
 
I did find on the explosion yield calcluations regarding this statement at least for spheres (Not for human bodies):

"For a sphere for instance this would not be 4*pi*r^2 / 2, but instead pi*r^2 as that is the area of the largest cross-section which, in this case, is a circle with the same radius as the sphere laid through its center (in such a way that it is orthogonal to the expansion of the explosion)."
 
KLOL506 said:
I did find on the explosion yield calcluations regarding this statement at least for spheres (Not for human bodies):

"For a sphere for instance this would not be 4*pi*r^2 / 2, but instead pi*r^2 as that is the area of the largest cross-section which, in this case, is a circle with the same radius as the sphere laid through its center (in such a way that it is orthogonal to the expansion of the explosion)."
Click to expand...
That's what I wanted to say from the beginning so the calculation has to change from /2 to /4 right? It will raise the result higher than before.
 
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