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Official Calculations Discussion Thread
- Thread starter Antvasima
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- vs battles wiki
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Question, can we use GBE for character that was able moving stars?
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Depends, if they move the star several body lengths in a very quick timeframe, then baseline 4-C it shall be (Moving a solar mass about roughly a solar diameter in 15 minutes 19 seconds or less wields the same energy level as the Sun's GBE, but since we're talking seconds, that could easily break FTL AKA violate KE rules, at which point you just go baseline 4-C, which is the GBE of our Sun)
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So if they moving more than one star, just add two sun GBE?
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Yeah.
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thank you
Vzearr
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Could you give a scan with more context ? This could mean alot of things.
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Can you get LS from a character doing this with a meteor?
In the sense that the monster just flies past by right through the asteroid and isn't slowed down by it
In the sense that the monster just flies past by right through the asteroid and isn't slowed down by it
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I think this is more power and precision than LS. We wouldn't give a sword user greater LS for cutting through a meteor I believe.
Also that bird is likely heavier than the meteor given its size compared to it.
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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.
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Yeah I don't know a formula for it, but a method I think could be used is finding the roots size if they're shown adding that to the tree's trunk mass and then just calcing how fast it was pulled to get ke.
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I think that'd still be far inferior to the result of ripping it out by accounting for tensile strength and what-not. Normally that's okay tho, but there's some cases where trees are uprooted off-screen.
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Celestial Body Feats
In fiction, some of the more impressive feats are those involving constellations and celestial bodies in general. Those feats are often hard to quantify due to going beyond the physics. The following is the agreement on how it will be done. If a constellation is created by moving stars...
vsbattles.fandom.com
About the via an explosion method on constellations, what does it mean? And how do you calculate/replicate that?
Can't you instead use the energy of one exploding star and just multiply that by how many stars exploded?
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QUESTION:
In a cloud splitting feat (in a manga), you can see the end of the split (i.e. the split is a triangular shape), so the split obviously slowed down to the point where it stopped, i.e. the speed of the split is not the same throughout the area.
How could I find that?
In a cloud splitting feat (in a manga), you can see the end of the split (i.e. the split is a triangular shape), so the split obviously slowed down to the point where it stopped, i.e. the speed of the split is not the same throughout the area.
How could I find that?
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No, you need to take into account the distance between the stars and rely on inverse-square law to find the power at the epicenter determined by how much power is left at the far edge.Celestial Body Feats
In fiction, some of the more impressive feats are those involving constellations and celestial bodies in general. Those feats are often hard to quantify due to going beyond the physics. The following is the agreement on how it will be done. If a constellation is created by moving stars...
About the via an explosion method on constellations, what does it mean? And how do you calculate/replicate that?
Can't you instead use the energy of one exploding star and just multiply that by how many stars exploded?
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Like this right?
Via using explosion method
0.6 x G(M^2)/R = 0.6 x (6.67 x 10^-11) x (1.9888x10^30)^2/(6.955x10^8) x 10 (10 stars shown destroyed) = 2.2759471^42 joules
Distance between stars = 5 light years (47303653885597800m)
Distance between 10 stars = 47303653885597800 x 10 = 473036540000000000m
473036540000000000/2 = 236518270000000000m
(2.2759471^42) x 4(π) x (2.3651827×10^17)/(6.955x10^8))^2 = 1.8330596e+49 (183305.96 Foe)
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Something somewhere the brackets got screwed up, because that's even lower than 4-A despite having a larger explosive radius.Like this right?
Via using explosion method
0.6 x G(M^2)/R = 0.6 x (6.67 x 10^-11) x (1.9888x10^30)^2/(6.955x10^8) x 10 (10 stars shown destroyed) = 2.2759471^42 joules
Distance between stars = 5 light years (47303653885597800m)
Distance between 10 stars = 47303653885597800 x 10 = 473036540000000000m
473036540000000000/2 = 236518270000000000m
(2.2759471^42) x 4(π) x (2.3651827×10^17)/(6.955x10^8))^2 = 1.8330596e+49 (183305.96 Foe)
I see why I got different results now
Also, normally speaking we just assume the target body at the edge being destroyed is the Sun for the sake of a lowball.
Distance between one star being 5 light years or 4.73e+16 meters... times ten equals 4.73e+17 meters.
The Inverse-square formula works out like this:
4*U*(Er/Br)^2
U is the GBE of the Sun, Er is the Explosive Radius and Br is the radius of the body blown up at the edge.
U= 5.693e+41 joules
Er= (4.73e+17 / 2) = 2.365e+17 meters
Br= 6.957e+8 meters
Energy: 4 * 5.693e+41 * (2.365e+17/6.957e+8)^2 = 2.6315975e+59 J or 2.6316 petaFoe (4-A)
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radius of the sun right?
Do you not need to account the other 10 stars that got destroyed?
Wait... Petafoe don't even exist in the tiering
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Yep
Nope. The AoE of the explosion takes care of that. You only need the star that is at the furthest end of this radius.
Now it does.Wait... Petafoe don't even exist in the tiering
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Probably should bring this page up since Surface Area is part of out repertoire: https://www.rxlist.com/body_surface_area/definition.htm
It's written by a doctor if that helps with the reliability. Cross-sectional area is still less, tho.
It's written by a doctor if that helps with the reliability. Cross-sectional area is still less, tho.
Da3ggman
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Bump
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Try to find the dimensions of that block if there are any other shots of it and work out the volume by pixel scaling and using a triangular prism as the shape then use the destruction value for melting concrete which is 4814.5 Joules per cm^3
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I calculated the volume, but I didn't quite understand the triangle thing.
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The volume of a triangular prism is the area of the triangular face multiplied by the depth of the prism, this would get you roughly the volume of concrete that was melted through
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Da3ggman
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Bump 2x
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You would likely use the calories calculator for that, don’t know about quickly though.
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Mob Psycho 100 has a calc of growing a giant broccoli very quickly. I'd look into that maybe?
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Someone had a similar question like this asked on this very thread. Lemme me see if could find their blog post.
