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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.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.
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.I guess depending on what the person means, it could be used like say whatever affected area where sunlight hits like watts per area.
If we're covering something in darkness, that would more accurately fall under kinematic viscosity: https://en.m.wikipedia.org/wiki/ViscosityCould 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.
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.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.
How can I calculate the energy generated by a hurricane? For example this one
BumpI would need to calculate the energy of the most powerful hurricane/storm
BumpIs 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.
BumpWhat kind of calculation would you use when You want to smoothen a material?
okay, thank you.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)
Probably a lotWhere does 2970.3384336 Metric Tonnes weighs around at?
Height of a white pine tree (that first image has those, I believe) is about 27 meters.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.
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.Can you elaborate on what you mean?
...Could you link to the calculation? I don't know what you're referring to.
It uses "frontal surface area"- that is, the front.Serious Punch Squared Recalc
vsbattles.fandom.com
No, you'd need the frontal area, which would be surface area/2. Which is pretty much what Qawsedf does in his math.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.
I believe for ISL we use cross-sectional area, actually.No, you'd need the frontal area, which would be surface area/2. Which is pretty much what Qawsedf does in his math.
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.I believe for ISL we use cross-sectional area, actually.
No, you'd need the frontal area, which would be surface area/2. Which is pretty much what Qawsedf does in his math.
Because you're not affecting a quarter of the star's surface? You're affecting a full half.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.
Because you're not affecting a quarter of the star's surface? You're affecting a full half.