Official Calculations Discussion Thread

[Emerald]

How do you calculate the energy generated by an unstable nuclear reactor?

 

[DastardlyDangerousDracoga]

Okay, I did all of that, and I got about 37465 joules, which, needless to say, I don't think is right given the size of Nixon.

That's... odd. You sure you inputted the formula correctly? Sometimes, a little mistype might screw the entire result, unintentionally.

 

[Da3ggman]

That's... odd. You sure you inputted the formula correctly? Sometimes, a little mistype might screw the entire result, unintentionally.

Definitely possible. I'll try it again when I get the chance and tell ya how it works out.

 

[koopa3144]

Question, how would one go about calculating this?

 

[DastardlyDangerousDracoga]

Definitely possible. I'll try it again when I get the chance and tell ya how it works out.

Actually, I think I realized what I did. I left the leftover r1 alone without changing it to 6371000 like I did with the other 6371000. I went ahead to edit it, so you shouldn't be stumped on that.

 

[SeijiSetto]

Question, how would one go about calculating this?

this blog should explain

 

[Da3ggman]

Actually, I think I realized what I did. I left the leftover r1 alone without changing it to 6371000 like I did with the other 6371000. I went ahead to edit it, so you shouldn't be stumped on that.

Okay, I got about 1.36^39, so that seemed to fix it. Thanks!

 

[Da3ggman]

It is, you just need to resort to this formula.

E = (G*M*m)/r1 - (G*M*m)/r2

Where

G = 6.674*10^-11

M = Earth Mass = 5.972*10^24

m = Mass (as calculated by prior formula)

r1 = 6371000m = Earth Radius

r2 = 6371000m + Height/2

E = Energy

So it would like this when you have your values ready.

E = G×M×((Height / Average Height)^3 * Average Weight) × (1/6371000 - 1/(6371000 + Height/2))

Wait, do you mind if I ask where you found this equation?

 

[DastardlyDangerousDracoga]

Wait, do you mind if I ask where you found this equation?

Funny that you mention it. I got it from a long chain of pages and links that led me to it. I was just doing my usual browsing of blog posts, and I came across this, which led me to this. Rather odd why that formula is absent from a certain page that could benefit from it.

 

[StretchSebe]

I've become increasingly aware of Lupin III having crazy shit. For example, Goemon's been shown capable of cutting clouds at a distance.

So how to calculate cutting a shooting star at a distance?

As in, how would one calculate indirect cutting? Assuming it's possible

 

[Testarossa002]

A character has this on his profile:

City level (Comparable to Jason Grace, who could create a storm powerful enough to contain Gaea; can also create storms as a side-effect of his powers as well. Caused substantial damage to Hubbard Glacier. Triggered the volcanic eruption of Mount Saint Helens, which is said to release about 24 megatons of energy.

Which links to this

How much energy in a hurricane, a volcano, and an earthquake?

How much energy in a hurricane, a volcano, and an earthquake?

web.archive.org

Now, the feat did happen. All I want to know is if it actually scales to the 24 megaton

 

[Arkenis]

A character has this on his profile:

Which links to this

How much energy in a hurricane, a volcano, and an earthquake?

How much energy in a hurricane, a volcano, and an earthquake?

web.archive.org

Now, the feat did happen. All I want to know is if it actually scales to the 24 megaton

Mount St Helens eruption

Mount St. Helens, a stratovolcano located in Skamania County, in the state of Washington, United States erupted in May 18 of 1980, at 8:32:17 am (PDT). It had a VEI of 5. It is the most disastrous volcanic eruption in United States recorded history. The eruption was so powerful that a column of...

vsbattles.fandom.com

 

[NIK_FARIS]

A question. Can absorbing lights/photons in the surrounding area making it dark be consider as AP?

 

[Arkenis]

A question. Can absorbing lights/photons in the surrounding area making it dark be consider as AP?

No.

 

[NIK_FARIS]

No.

Then what about releasing them by concentrating them in single path as a ray?

 

[Arkenis]

Then what about releasing them by concentrating them in single path as a ray?

Did they cause some destruction?

 

[NIK_FARIS]

Did they cause some destruction?

Yeah, it incinerate everything in its path.

 

[Arkenis]

Yeah, it incinerate everything in its path.

you'd need to find the volume of what it incinerated.

 

[SeijiSetto]

I've become increasingly aware of Lupin III having crazy shit. For example, Goemon's been shown capable of cutting clouds at a distance.

So how to calculate cutting a shooting star at a distance?

As in, how would one calculate indirect cutting? Assuming it's possible

i don't think its treated any different to normal cutting, just counts as good range
width of sword * area of cut to get volume destroyed, multiply by shear strength of material, i believe?

 

[StretchSebe]

i don't think its treated any different to normal cutting, just counts as good range
width of sword * area of cut to get volume destroyed, multiply by shear strength of material, i believe?

Huh. Alright then. I'll assume average katana width + the cut being 3 cm, and the three types of meteors in iron, stony-iron, and stone.

 

[Testarossa002]

Mount St Helens eruption

Mount St. Helens, a stratovolcano located in Skamania County, in the state of Washington, United States erupted in May 18 of 1980, at 8:32:17 am (PDT). It had a VEI of 5. It is the most disastrous volcanic eruption in United States recorded history. The eruption was so powerful that a column of...

vsbattles.fandom.com

I keep discovering new pages everyday

Thanks anyways

 

[Emerald]

There's a character that through their sheer breath alone is capable of creating a bunch of air that is this heavy, is there any that I can get the AP for this? Maybe through PE or something?

 

[Arkenis]

There's a character that through their sheer breath alone is capable of creating a bunch of air that is this heavy, is there any that I can get the AP for this? Maybe through PE or something?

I think thats fine yeah. But if there's a vid of the air being blown I think you can also use that for KE and find the acceleration of it.

 

[Emerald]

I think thats fine yeah. But if there's a vid of the air being blown I think you can also use that for KE and find the acceleration of it.

It's like a still image and narration telling that it did happen
Although for the PE, what should I go for? The height of the character that blew the air?

 

[Arkenis]

It's like a still image and narration telling that it did happen
Although for the PE, what should I go for? The height of the character that blew the air?

Im confused. The calc u linked has you pixel scaling air and that air is in the vid moving from the moon.

Pixel Scaling

Moon: 1161px or 3474,8 km

2.99293712317 km/px

Using an ellipsoid volume calculator:

Semi-Axis B: 18px or 53.8728682171 km, half that would be 26.9364341085 km

Semi-Axis C and A: 2.9px or 8.67951765719 km, half that would be 4.3397588286 km

Volume: 2125004737073 m3

Density of Air: 1.293 kg m^3

So 2.7476311e+12 Kg or 2747631100 Tons (Class T)

You already have the mass, you can just use the speed for ke.

 

[Emerald]

Im confused. The calc u linked has you pixel scaling air and that air is in the vid moving from the moon.

You already have the mass, you can just use the speed for ke.

The thing is it's basically two different scenes in a way.
The scene I calced is basically the character using all of this and moving it away from the moon through some TK esc hax, of which we have a clear timeframe for it
Whereas what I want to calc, is basically the same character creating all of this air from their breath alone (which took an unknown amount of time) and apply it to their physical stats

 

[Arkenis]

Whereas what I want to calc, is basically the same character creating all of this air from their breath alone (which took an unknown amount of time) and apply it to their physical stats

Yeah idk if that really would scale to durability. Her pushing all that air out would though which you already got mass for the air you just have get speed and then do ke.

 

[StretchSebe]

Super long calc blog from that cutting feat.

Only thing I think might not be right in advance was using 3.13cm for each dimension in volume (l*w*h). Was it correct to use katana width for cut width * cut length, and for cut height to be the same as those so it's cc? Meteoroids don't often seem to be much larger than that anyways, but still worth asking to be sure.

Other than that, if anyone is crazy enough to actually go through it too, I'd be open to feedback lol

 

[Shmeatywerbenmanjenson]

A question. Can absorbing lights/photons in the surrounding area making it dark be consider as AP?

If its directly stated then yes

You could use a method similar to this Thor calc

As for light, I'm pretty sure there's a way to calculate luminosity of an attack to find a value so I'm sure you could use that to find out how much he or she absorbed

 

[SeijiSetto]

Super long calc blog from that cutting feat.

Only thing I think might not be right in advance was using 3.13cm for each dimension in volume (l*w*h). Was it correct to use katana width for cut width * cut length, and for cut height to be the same as those so it's cc? Meteoroids don't often seem to be much larger than that anyways, but still worth asking to be sure.

Other than that, if anyone is crazy enough to actually go through it too, I'd be open to feedback lol

540 MPa = 540 million Pa = 540 million N/m^2 = 540 million J/m^3 = 540 J/cc.

your math is off by a factor of like 6 or 8 orders of magnitude.

 

[NIK_FARIS]

If its directly stated then yes

You could use a method similar to this Thor calc

As for light, I'm pretty sure there's a way to calculate luminosity of an attack to find a value so I'm sure you could use that to find out how much he or she absorbed

So, I just need to find energy per photon and it’s density and calculate it from there right?

 

[SeijiSetto]

Super long calc blog from that cutting feat.

Only thing I think might not be right in advance was using 3.13cm for each dimension in volume (l*w*h). Was it correct to use katana width for cut width * cut length, and for cut height to be the same as those so it's cc? Meteoroids don't often seem to be much larger than that anyways, but still worth asking to be sure.

Other than that, if anyone is crazy enough to actually go through it too, I'd be open to feedback lol

also, you'd wanna multiply the katana thickness by the cross-sectional area of the meteorite, since that's the actual part he cut. idunno how difficult it'll be to find the size of it but that's how cutting feats work.

that AND the feat you linked has a katana blade at 0.7cm thick.

 

[Aether2734]

Is there any how to calculate feats about push back sea water pressure?

 

[StretchSebe]

540 MPa = 540 million Pa = 540 million N/m^2 = 540 million J/m^3 = 540 J/cc.

your math is off by a factor of like 6 or 8 orders of magnitude.

Oh shoot, really? I thought that Pascals were equal to J/m^3? Of which, the 540 J/m^3 to J/cc conversion seemed to return 5.4e+8 cc (the 540 million).

I could be misunderstanding, or getting the conversion wrong, but if the 540 J/m^3 does convert to only 540 J/cc: How does the conversion from J/m^3 to J/cc make the value lower when the value is initially at the larger unit of measurement of m^3, and then is being converted to the smaller unit of measurement of cc?

also, you'd wanna multiply the katana thickness by the cross-sectional area of the meteorite, since that's the actual part he cut. idunno how difficult it'll be to find the size of it but that's how cutting feats work.

that AND the feat you linked has a katana blade at 0.7cm thick.

I can at least fix this right away: For the future, katana thickness is "Thickness at the Shinogi" at 0.67cm then, and not the Hamachi?

Also will use the (seemingly) standard 1cm cut length instead of the 3.13cm, and the in-between (4.75cm) of the average lengths found for ordinary chondrites.

 

[SeijiSetto]

How does the conversion from J/m^3 to J/cc make the value lower when the value is initially at the larger unit of measurement of m^3, and then is being converted to the smaller unit of measurement of cc?

one m^3 is far larger than one cm^3 (AKA one cc), 1 million times larger.
in a rather self-evident way, it takes more energy to break something of larger volume than it does of smaller volume.
the value is 540 million J/m^3, or 540 MEGAjoules per metre cubed. this then becomes 540 joules per cubic centimetre.

Also will use the (seemingly) standard 1cm cut length instead of the 3.13cm, and the in-between (4.75cm) of the average lengths found for ordinary chondrites.

when you use said length, you get the area of a circle that size, then multiply this by katana thickness. you're actually calculating the volume of a very thin slice of the meteor (where the katana must have passed through) and then multiplying that by the shear strength to get your energy value.

 

[StretchSebe]

one m^3 is far larger than one cm^3 (AKA one cc), 1 million times larger.
in a rather self-evident way, it takes more energy to break something of larger volume than it does of smaller volume.
the value is 540 million J/m^3, or 540 MEGAjoules per metre cubed. this then becomes 540 joules per cubic centimetre.

I think I see what you mean here: Dividing the J/m^3 by the million to get J/cc.

That said, I am sorry if I am still misunderstanding, but in looking up converting m^3 to cc further, I'm not sure that is the correct way to convert:

The formula always seems to be multiplying a given value for m^3 by one million to find cc. Likewise, it is dividing cc by one million to find m^3. That's why I was confused on how converting the larger unit of measurement to the smaller one would actually result in a lower value. The conversion ratio is 1 m^3 = 1000000 cc.

Edit: Nvm, I found what you say to be the case with converting J/m^3 to J/cc. Was too focused on the conversion of m^3 and cc. My bad.

when you use said length, you get the area of a circle that size, then multiply this by katana thickness. you're actually calculating the volume of a very thin slice of the meteor (where the katana must have passed through) and then multiplying that by the shear strength to get your energy value.

Also, it would still be cc, right? Just low values for the length and width (the katana length and likely 1 cm for the width), then the meteoroids height?

 

[StretchSebe]

The fixed calc blog

 

[SeijiSetto]

I think I see what you mean here: Dividing the J/m^3 by the million to get J/cc.

That said, I am sorry if I am still misunderstanding, but in looking up converting m^3 to cc further, I'm not sure that is the correct way to convert:

The formula always seems to be multiplying a given value for m^3 by one million to find cc. Likewise, it is dividing cc by one million to find m^3. That's why I was confused on how converting the larger unit of measurement to the smaller one would actually result in a lower value. The conversion ratio is 1 m^3 = 1000000 cc.

Edit: Nvm, I found what you say to be the case with converting J/m^3 to J/cc. Was too focused on the conversion of m^3 and cc. My bad.

Also, it would still be cc, right? Just low values for the length and width (the katana length and likely 1 cm for the width), then the meteoroids height?

why're you assuming a cut length of 1cm?
the way i see it, it's like this

you're multiplying the AREA of the shooting star (assuming it's spherical, this is just the area of the great circle) by the thickness of the katana, this is the volume.
multiply said volume by your destruction value.

 

[StretchSebe]

why're you assuming a cut length of 1cm?
the way i see it, it's like this

you're multiplying the AREA of the shooting star (assuming it's spherical, this is just the area of the great circle) by the thickness of the katana, this is the volume.
multiply said volume by your destruction value.

I had thought 1 cm width was standard for cuts.

So area would be length of shooting star multiplied by height of shooting star multiplied by katana thickness (width of cut) then to get volume cc?

This is easily my second most messy calculation

 

[SeijiSetto]

I had thought 1 cm width was standard for cuts.

So area would be length of shooting star multiplied by height of shooting star multiplied by katana thickness (width of cut) then to get volume cc?

This is easily my second most messy calculation

specifically the area of the cut multiplied by the thickness of the blade. if it's a sphere then it'll be pi * r^2 (cut area) * blade thickness.