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You can use the equation Kachon gave to find the initial rKE and final rKE, then take the difference in those to find the energy the character applied.
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Official Calculations Discussion Thread
- Thread starter Antvasima
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- vs battles wiki
Oliver_de_jesus
He/Him- 23,805
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OK
ALSO, I can calculate the distance from one planet to another using the time it takes for a Normal rocket to arrive?
- 18,534
- 16,111
What do you mean?
Oliver_de_jesus
He/Him- 23,805
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apparently the distance between earth and mars in a verse is different from ours, since a rocket designed by a NASA expert took 10 years to get from earth to mars
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Hey, just wanted to ask, also I'm sorry if this isn't the right thread, but how do you request a calculation evaluation? I made this calculation blog, and I want someone to evaluate it before it gets used. Could anyone help me?
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Basically follow the instructions on this thread, and post the necessary details on your post so that the calc members will know about it.
FantaRin_The_First
Username Only- 3,924
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Hey, hey, fanta here. Anybody has any idea on how I can calc this feat?
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Find the area of the split in m^2 (A)
Multiply A by 60 Mpa (60000000 N/m2)
Multiple that by the value to change tensile to shear force 0.6
Multiply your answer by diameter of the total area displaced (∆x)
The final formula would look like this: (A*60000000*.6)∆x = answer in Joules
FantaRin_The_First
Username Only- 3,924
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Question. On what you said to answer Emerald's question, does it apply to my question as well? Or just a slightly different variation of the formula you gave?
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Find the radius of the crater (length-wise)
Find the depth (just use the average thickness of a person their age's back)
Since the crater is a half-ellipsoid, use this formula to find the volume of the crater: 2/3*pi*r*r*d = answer in cm^3
Multiply the volume by 655 J/cc (average destruction value of steel) and that is the result in Joules
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FantaRin_The_First
Username Only- 3,924
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Thank you, very much!
- 32,811
- 38,029
This is yield strength, not v. frag. The steel panel's linework is also relatively intact, if it was v. fragged something that thin would be torn to tiny metal shards.
- 32,811
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We don't allow that method for splits or cutting feats like this anymore. It was rejected on the Demon Slayer cutting blog.
Splitting in general is considered incalculable simply because of its wildy upredictable formation.
But as a low-ball, just find the volume of the split (Width of the crack, depth of the crack, length of the crack) and then multiply it with the destruction value based on how severe the destruction is.
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I used pulv.
FantaRin_The_First
Username Only- 3,924
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So what should I do here? Use frag instead or some other method?
- 32,811
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Should've used yield strength instead. Pulv would've turned it to dust and would've left behind no fragments.
Metal doesn't compress as easily as other materials.
- 32,811
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Use yield strength. Best for denting metal feats.
FantaRin_The_First
Username Only- 3,924
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Yield strength is 350 MPa, right? And the only thing I have the change from what Kachon told me to do is not use 655 J/cc? Just making sure right now so I don't ask dumb questions later on.
Oliver_de_jesus
He/Him- 23,805
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do we calc the reversal of the sides of the earth (Making north into south and south into north) as rotation calc?
- 18,534
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Can you send the feat
FantaRin_The_First
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I got a hypothetical question that needs to be answered. Say you have a feat that has one character destroy something. In text, so no visuals, however, the thing in question that was destroyed was described as something that can be found in real-life and you can use low-ball height and width, but, well... can't find thickness/depth anywhere since there is no standard for it irl.
What can I do? Just assume the thickness of whatever body part was used to break the thing, or is there some other way?
What can I do? Just assume the thickness of whatever body part was used to break the thing, or is there some other way?
FantaRin_The_First
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Anybody got anything? Reason I am asking is cuz I got a calc that is riding on the answer given to my question.
- 18,534
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Can you send the text? I'm having a hard time understanding what you mean.
FantaRin_The_First
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Here you go. If you don't want to follow the link, basically, a character jumped out of a vehicle to slam forehead-first into the side window of a school bus to break through it and get inside said bus. From my own research, bus windows, are made out of tempered glass, so it should be calc it by using the frag values for ballistic glass or smth, according to a convo I had with KLOL a while back.
FantaRin_The_First
Username Only- 3,924
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Should I just assume the width and thickness is the same, for simplicity's sake?
FantaRin_The_First
Username Only- 3,924
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Can probably do the crater if the creature that made it has a canon height - or if it was standing and/or nearby a character with a canon height in another shot - but don't think the cloud feat can be calced atm due to cloud revisions
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Yeah that would work.
Recon1511
He/Him- 6,924
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The crater is caused by the shockwave of K'Sante's blow sending the creature flying, afaik he doesnt have a canon height yet but maybe average human height could do
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He looks quite big so I would go for the size of average strongmen
Oliver_de_jesus
He/Him- 23,805
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this (Yeah I already made a thread where it was accepted to use the series, if you ask), Keep in mind that he only gave the thrust, the rest of the planets were moved by the generated thrust and the connection of the planets, so it only scales to the first moments of the movement, the rest I(with help of @DarlingAurora) have already calculated as preparation/chain reaction
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Find the circumference of the Earth, divide that by two, find a timeframe, and then use Kinetic Energy formula.
Oliver_de_jesus
He/Him- 23,805
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Oliver_de_jesus
He/Him- 23,805
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Is it well done or not?
the circumference of the Earth is 40,075 km | 40,075,000 m
now divide that by two 40,075,000/2=20,037,500
Timeframe: 08:42:42-08:41:09=1.33/20,037,500=15065789.5m
Earth Mass: 5.972 × 10^24 kg
K.E. = 1/2 m v2
K.E= 1/2*5.97200e24*15065789.5^2=6.77756348e38 J or 161.987655 Ninatos (Nigh 5-A)
now divide that by two 40,075,000/2=20,037,500
Timeframe: 08:42:42-08:41:09=1.33/20,037,500=15065789.5m
Earth Mass: 5.972 × 10^24 kg
K.E. = 1/2 m v2
K.E= 1/2*5.97200e24*15065789.5^2=6.77756348e38 J or 161.987655 Ninatos (Nigh 5-A)
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Looks good
Oliver_de_jesus
He/Him- 23,805
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ok i'll put it on a blog
azontr
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How would I go about trying to calculate something like this? (Feat is from Infinite Leveling: Murim, Chapter 8)
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- Use pixel scaling to find the diameter of the crater in centimeters by using the woman as a reference
- Divide the diameter by 2 in order to find the radius (r)
- Find the depth of the crater (d) in centimeters by pixel scaling the thickness of either of these fragments still using the woman as a reference
- Plug in the radius and depth numbers into this formula (volume of a half-ellipsoid): 2/3*pi*r*r*d
- Multiple the volume (should be in cm^3) by 8, as that is the fragmentation value in J/cc of stone.
- That is your answer in joules
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