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That's... odd. You sure you inputted the formula correctly? Sometimes, a little mistype might screw the entire result, unintentionally.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.
Definitely possible. I'll try it again when I get the chance and tell ya how it works out.That's... odd. You sure you inputted the formula correctly? Sometimes, a little mistype might screw the entire result, unintentionally.
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.Definitely possible. I'll try it again when I get the chance and tell ya how it works out.
this blog should explainQuestion, how would one go about calculating this?
Okay, I got about 1.36^39, so that seemed to fix it. Thanks!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.
Wait, do you mind if I ask where you found this equation?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))
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.Wait, do you mind if I ask where you found this equation?
Which links to thisCity 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.
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
No.A question. Can absorbing lights/photons in the surrounding area making it dark be consider as AP?
Then what about releasing them by concentrating them in single path as a ray?
Did they cause some destruction?Then what about releasing them by concentrating them in single path as a ray?
Yeah, it incinerate everything in its path.Did they cause some destruction?
you'd need to find the volume of what it incinerated.Yeah, it incinerate everything in its path.
i don't think its treated any different to normal cutting, just counts as good rangeI'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
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.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?
I keep discovering new pages everydayMount 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 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.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?
It's like a still image and narration telling that it did happenI 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.
Im confused. The calc u linked has you pixel scaling air and that air is in the vid moving from the moon.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?
You already have the mass, you can just use the speed for ke.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)
The thing is it's basically two different scenes in a way.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.
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.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
If its directly stated then yesA question. Can absorbing lights/photons in the surrounding area making it dark be consider as AP?
540 MPa = 540 million Pa = 540 million N/m^2 = 540 million J/m^3 = 540 J/cc.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
So, I just need to find energy per photon and it’s density and calculate it from there right?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
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.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
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).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.
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, 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.
one m^3 is far larger than one cm^3 (AKA one cc), 1 million times larger.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?
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 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.
I think I see what you mean here: Dividing the J/m^3 by the million to get J/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, 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?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.
why're you assuming a cut length of 1cm?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?
I had thought 1 cm width was standard for cuts.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.
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.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