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Official Calculations Discussion Thread
- Thread starter Antvasima
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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?
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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.
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Bump
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G/cm^-3 is the same thing as G * cm^3. A negative exponent is the same thing as dividing one by its positive exponent equivalent (i.e. 2^-3 being the same as 1/2^3; both values are equal to 1/8)
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okay, thank you.
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Where does 2970.3384336 Metric Tonnes weighs around at?
StarShooter80
He/Him- 5,999
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Probably a lot
Sans2345
He/Him- 5,433
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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.
I would love some help, as i'm not really that great at math.
Mr. Bambu
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Height of a white pine tree (that first image has those, I believe) is about 27 meters.
Pixel scaling it gives us a height of 93.37 meters for the tall guy.
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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 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?
IdiosyncraticLawyer
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I would appreciate some help with sorting out the Puella Magi Verse's calcs, which are partly hosted on SpaceBattles, which isn't allowed.
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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?
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In the Punch^2 calculation, this operation uses the surface area of the sun. This is inaccurate because the contact range of the sun with the explosion is the cross-sectional area, not the surface area.
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Serious Punch Squared Recalc
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Mr. Bambu
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It uses "frontal surface area"- that is, the front.Serious Punch Squared Recalc
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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.
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No, you'd need the frontal area, which would be surface area/2. Which is pretty much what Qawsedf does in his math.
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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.
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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.
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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.
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Because you're not affecting a quarter of the star's surface? You're affecting a full half.
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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.
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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.
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.
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Yes, the result will increase because the range of each star will decrease.
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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)."
"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)."
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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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Has anyone ever had the idea of calculating Saitama's speed at I/O by assuming that he moves to create rays that cover a part of the area or fill a part of the volume of I/O?
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