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Official Calculations Discussion Thread

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.
 
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.
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)."
 
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)."
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.
 
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?
 
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)."
That is what I was saying earlier, yes.
[...] because the area of any sphere is just 4x the area of a circle of equal radius, hence why [...]
 
Can a calc member please answer this
 
How do you decide which end is useable?
 
Based on what has been discussed, can Saitama's calculation be changed from /2 to /4?
Sure.
You don't even need to calculate the full thing then divide by 4, it's literally just the area of a circle with equal radius to the sphere. The area of the sphere's Great Circle.
 
How do you decide which end is useable?
Self-explanatory. Frag is large chunks. V. Frag is tiny chunks compared to the original volume (This one needs a bit of eyeballing). Pulverization is straightforward, entire thing turns to dust with no solid remains.
 
for feats like this, do I still just use spherical cap or is there some other shape to account for the part in the middle not destroyed?
 
I'd one question regarding ground explosion calcs, are we supposed to calculate volume of non-spherical explosions (such as parallel or kind of conical) then equate it with spherical volume to get its radius then apply explosion formula? Or we just take radius as whatever it extends to on ground disregarding it's shape/volume?
 
I'd one question regarding ground explosion calcs, are we supposed to calculate volume of non-spherical explosions (such as parallel or kind of conical) then equate it with spherical volume to get its radius then apply explosion formula? Or we just take radius as whatever it extends to on ground disregarding it's shape/volume?
Bump, I've seen this method used in some calcs, not sure if we still do use it.
 
I'd one question regarding ground explosion calcs, are we supposed to calculate volume of non-spherical explosions (such as parallel or kind of conical) then equate it with spherical volume to get its radius then apply explosion formula? Or we just take radius as whatever it extends to on ground disregarding it's shape/volume?
Pretty much the latter, in the lack of better options.
 
4.75 Kilojoules would equate to 9-C.
 
Without further knowledge, I can at least say that it could be as simple as (Baseline Star Level * 8) / 6, which wwould equate to 7.591e41 Joules (Star level). If they were done one at a time, then just baseline Star level is correct.
 
Honestly didn't expect anyone to respond on this thread as soon as I sent it. Granted, it was in hour intervals but still \^o^/

If you already know the dimensions of the elongated part, why not just use that to find the dimensions of the hole at the edge?
Well, here's the situation about this whole thing. I have considered that, but just needed some clarification as to... what volume should I use to calculate. Back then like about... two years ago, I just assumed it was a semi-ellipsoid and called it a day. But the more and more I look back on what I did wrong and what I concluded was completely off the mark. The feat itself on that hill(?) is a lot like this but on a more spherical incline than a pyramid. It's like you grabbed a straw and poked through a cone very near at the edge. I visualize it like this patch on the cone, but more rounded around the edges. And I genuinely have done a lot of searching around to understand what... that is supposed to be or if anyone has a name for something like that.
 
Honestly didn't expect anyone to respond on this thread as soon as I sent it. Granted, it was in hour intervals but still \^o^/


Well, here's the situation about this whole thing. I have considered that, but just needed some clarification as to... what volume should I use to calculate. Back then like about... two years ago, I just assumed it was a semi-ellipsoid and called it a day. But the more and more I look back on what I did wrong and what I concluded was completely off the mark. The feat itself on that hill(?) is a lot like this but on a more spherical incline than a pyramid. It's like you grabbed a straw and poked through a cone very near at the edge. I visualize it like this patch on the cone, but more rounded around the edges. And I genuinely have done a lot of searching around to understand what... that is supposed to be or if anyone has a name for something like that.
It's due to perception that it appears to be a cone. Use cylindrical volume. Imo.
 
Would anybody here know how to get the length of the red line in this image here? I'm trying to calculate how strong the explosion is and I've seen other blogs get the height of a character through a curved line before I just don't know how to get the exact measurement.
latest
 
Would anybody here know how to get the length of the red line in this image here? I'm trying to calculate how strong the explosion is and I've seen other blogs get the height of a character through a curved line before I just don't know how to get the exact measurement.
latest
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