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Viewing blog entries in category: One Piece calcs

  • Tacocat
    Y'all know what this is.

    http://www.[Blocked Domain]/one-piece/778/8
    http://www.[Blocked Domain]/one-piece/778/10
    http://www.[Blocked Domain]/one-piece/778/11

    Zoro cut Pica up so bad that fucker was floating.

    Pretty rough scaling, but it's something.

    http://www.narutoforums.com/blog.php?b=25062

    Here I got the 2nd level of Pica's landmass to be 605.65m in height.



    First and foremost I'm gonna need the head for scaling between different panels.

    (605.65/40)23=348.25m

    Tuck that in the back pocket.

    Let's do his arms next.

    Upper arm
    (605.65/40)70.24=1063.52m

    Inner arm-elbow width
    (605.65/40)33=499.66m

    Forearm
    (605.65/40)81=1226.44m

    Shoulder width
    (605.65/40)41.2=623.82m

    Gonna calc the upper arm as a conical frustum and the forearm as a cylinder.

    vfrustum=(1/3)pi*h(r^2+r+R+R^2)
    v=(1/3)pi*1063.52((499.66/2)^2+(499.66/2)(623.82/2)+(623.82/2)^2)
    v=264649274.9m^3

    vcylinder=pi*r^2*h
    v=pi*(499.66/2)^2*1226.44
    v=240483539.36m^3

    Multiply by two because...you know.

    varmstotal=(240483539.36+264649274.9)2
    v=1010265628.52m^3

    Now for the body.

    (605.65/40)78.3=1185.56m

    (605.65/40)55.46=839.73m

    Seeing as I'm going to calc this as a trapezoidal prism because I'm lazy as fuck, I'm going to average this.

    Height of prism (l)
    (839.73+1185.56)/2=1012.645m



    Height of trapezoid (h)
    (348.25/25)108=1504.44m

    Base 1 of trapezoid (a)
    (348.25/25)91=1267.63m

    Base 2 of trapezoid (b)
    (348.25/25)70=975.1m

    vtrapezoidalprism=.5h(a+b)l
    v=.5*1504.44(1267.63+975.1)1012.645
    v=1708358808.93m^3

    Add 'em all up...

    vtotal=1708358808.93+1010265628.52
    v=2718624437.45m^3

    And find mass...

    m=v*rho
    m=2718624437.45*2700
    m=7340285981115kg



    Width I got for his waist was 975.1m, so...

    (975.1/66.91)46=670.37m

    That's our height for PE.

    PE=mgh
    PE=7340285981115*9.8*670.37
    PE=48222933628968612.99J

    or about 11.5 megatons.

    Might add in the head later. Might not. Who knows? :distracted
  • Tacocat
    The first of the year. At Black Leg's request, let us commence.

    Feat:
    http://www.[Blocked Domain]/one-piece/771/19
    http://www.[Blocked Domain]/one-piece/772/5

    Seems kind of like a...trapezoidal pyramid to me :hmm So let's go with that. We need the area of the base and the length (or height of the volume) that the attack traveled.

    They're on the 2nd Level, so let's get some dimensions.

    http://www.narutoforums.com/blog.php?b=24531



    Dastan has the red line at 139 pixels and 1683.706m. Green line is the 2nd Level of the structure Pica created.

    (1683.706/139)50=605.65m

    The 2nd Level is then 605.65m in height.

    So let's start with the base.



    (605.65/57)21=223.134m

    (605.65/57)32=340m

    The height of the base is, obviously, the height of the 2nd Level, i.e. 605.65m.

    atrapezoid=.5(a+b)h
    a=.5(223.134+340)605.65
    a=170531m^2

    Now for the height of the pyramid.



    A little awkward, but manageable nonetheless.

    (605.65/52)134=1560.7m

    Now for the volume of the pyramid itself.

    vpyramid=(1/3)b*h
    v=(1/3)170531*1560.7
    v=88715910.6m^3

    Gonna go with 69J/cc.

    88715910.6*1000000*69=6121397831400000J

    or about 1.46 megatons.

    Don Sai's kick>Sanji's confirmed :maybe

    On the real, though, that seems kind of big :hmm
  • Tacocat


    This is the one I'll be talking about. Some people seem intent on using minimum impact velocity. However, as JWL pointed out...

    Not only that, but OP Earth's atmosphere is larger than our own.



    Angsizing Alabasta (3226.8km) right quick...

    2atan(tan(70/2)*(431/327))=85.4

    2atan(18.97/(431/tan(85.4/2)))=4.65

    d=39,738km

    ...tells me that impact would take over an hour if the thing were moving at a meager 11km/s.

    Law and Doflamingo were certainly not standing around wondering what the hell Fujitora was doing for an entire hour. Furthermore, Block C of the competition began at the exact same time that Dofla and Fujitora arrived at the beach of Green Bit and was still in its early stages after the meteor was dropped. Even more so, the news that Doflamingo's resignation was falsified was released to the public en masse at the exact same time, and after the feat Franky says he had only just learned of the information. When everyone in the streets was raving about it, it'd be a little hard for him to have only heard it an hour later.

    Basically, 11km/s just doesn't fit the bill. Indeed, the calculated minimum impact velocity of OP Earth would leave everyone on the beach just standing there for 17 whole minutes. Also completely absurd.

    Now the question becomes this: With these circumstances, would an assumed time-frame be warranted? Also, what would that time-frame be, logically?

    Edit: Alright, well Dartg seems to advocate a minute, and that's actually what had been thinking.

    39738000/60=662300m/s

    or about mach 1948.
  • Tacocat
    http://www.[Blocked Domain]/one-piece/749/9
    http://www.[Blocked Domain]/one-piece/749/10
    http://www.[Blocked Domain]/one-piece/749/13

    Luffy Grizzly Magnums Peeka's face off.



    Colosseum is 431.838m in diameter.

    (431.838/57)9=68.185m



    I dunno if I feel comfortable including the upper half of Peeka's face considering it's still intact as Luffy's attack hits, so I decided to just use his nose down to his shoulders modeled in the volume of a conic frustum. I know the angle was declining instead of inclining as it is here, but that would only serve to increase the volume, right? :hmm

    Anyway,

    (68.185/12)23.6=134.1m (h)

    (68.185/12)53.76=305.4688m (2R)

    (68.185/12)46.23=262.683m (2r)

    v=(1/3)pi*h(R^2+Rr+r^2)
    v=(1/3)pi*134.1((305.4688/2)^2+(305.4688/2)(262.683/2)+(262.683/2)^2)
    v=8515450.89m^3

    Should at least warrant 69 J/cc. Gonna go with both 69 J/cc and 120 J/cc.

    8515450.89*1000000*69=587566111410000J
    8515450.89*1000000*120=1021854106800000J

    or about 140-244 kilotons.
  • Tacocat
    Just a really quick, really rough calculation. I'm shit tired, and I've gotta crash soon to be up for a final in a few hours. Well, here goes.

    In the new OP chapter, Pika causes all kinds of mayhem all over Dressrosa.

    http://www.[Blocked Domain]/one-piece/745/15
    http://www.[Blocked Domain]/one-piece/745/16

    I'm gonna calc the castle-raising feat.



    I know Dastan calculated the height of the Colosseum to be less than this, but he used the interior for some reason. What I need is the exterior. So I'll use his diameter.

    That happens to be 1590m.

    (1590/57)37=1032.1m

    Now that we have that...



    (1032.1/58)29=516.05m

    So about half a kilometer for the castle's base's height.



    Again, I couldn't be fucked to sit here and do a bunch of scaling, so I'm just going to calculate this as a cylinder for now, using ~the smallest diameter and go back to fix it later.

    (516.05/26.7)84.4=1631.26m

    ...for the diameter, and...

    (516.05/26.7)169.65=3279m

    ...for the height.

    vcylinder=pi*r^2*h
    v=pi*(1631.26/2)^2*3279
    v=6852951718.5m^2

    m=v*rho
    m=6852951718.5*2700
    m=18502969639950kg

    Don't really have an exact time-frame, so PE, I guess?

    PE=mgh
    PE=18502969639950*9.8*(3279/2)
    PE=297289063502040645J

    or about 71 megatons.

    For that, the time-frame would have to be about 10 seconds, which makes sense because some civilian ^ (use bro) was still surprised by the castle moving when it was already at this height. Also, Pika raised it quickly relative to the descent of Dofla's Birdcage.

    But again, I'll work out the kinks later. At least, for this not to be city level, the time-frame would have to be over half a minute which is...dumb.

    Anyway, feel free to comment, I'm going to sleep. I'll address whatever tomorrow afternoon or something. Or you guys can work it out and I don't have to do anything but update :noworries
  • Tacocat
    Someone's going to kill me for making that comparison :sanji

    Anyway :maybe

    I was asked to re-calc Luffy's Elephant Gatling considering the destruction of the entirety of Noah, as supposed...



    Luckily, I get to be lazy because Chaos already has everything here.

    Get ready for parenthesis and half ellipsoid volume formulas.

    ((4/3)pi(4515/2)*(2171.715/2)*(2670/2))/2=6,853,942,769 m^3

    That's the volume of the bottom portion of Noah. But to get how hollow it is, we'll subtract the thickness of the hull from each dimension twice.

    ((4/3)pi((4515-(260*2))/2)*((2171.715-(260*2))/2)*((2670-(260*2))/2))/2=3,714,146,098.8 m^3

    So that's the volume of the empty space. Really, it should be lower considering all the shit inside, but whatever.

    Subtract these two for the volume of the material that would have been destroyed.

    6,853,942,769-3714146098.8=3,139,796,670.2 m^3

    Chaos used 8 J/cc, so I'll use the same.

    3139796670.2*1000000*8=25118373361600000J

    or about 6 megatons.

    I guess you'd add that to the energy of the destruction of the superstructure?
  • Tacocat
    Disclaimer: Most of the scaling in this blog was done by GM and Chaos. I'm simply posting their work here to make it easier for the viewrs of this blog.


    http://www.narutoforums.com/blog.php?b=14333

    GM's scaling in the above calc has been purposed for many things, but I noticed that said scaling depended on the beanstalk which was an entire half an island away from the object of reference. So I looked around a little and found the following:



    Well, according to GM's scaling of Merry, the proportion would then be...

    (23.83/25)132=125.8224m

    I scaled kinda lazily, but Waka got 134px for Arc Maxim and 28px for Merry.

    (23.83/28)134=114m



    The lines were made thicker for ease of the viewer, but the pixel sizes were measured with the thinnest possible lines.

    (114/7)13=211.7m

    A little disappointing, what with the previous value coming from scaling off of Merry at kilometers away, but more accurate nonetheless.



    Having a different value obviously changes some dimensions of the island.

    (211.7/9)106=2493.4m

    (211.7/9)521=12255m

    Enel's Raigou's are great and all, but I'll leave those for when I have more time. Right now what I want to look at is Chaos's calc for Wiper's Reject Dial.



    0.359*211.7=76m

    0.513*211.7=108.6021m

    I suppose he calc'd it as a cylinder, so...

    vcylinder=pi*r^2*h
    v=pi*(76/2)^2*108.6021
    v=492669m^2

    492669*1000000*107.175=52801800075000J

    or about 12.6 kilotons.

    More to come, I suppose.

  • Tacocat
    Edited 'cause the bridge is clearly not to scale.

    http://www.narutoforums.com/blog.php?b=20063

    Darth calc'd the size of Fujitora's crater on the beach to be 842.87m in width, and it spans pretty much the entirety of the beach.



    I think that's the biggest stretch of shore.

    (842.87/3.6)35.36=8278.86m

    (842.87/3.6)27.8=6508.83m

    Aellipse=pi*a*b
    A=pi*4139.43*3254.415
    A=42321723.8m^2
  • Tacocat
    In a recent thread, Dartg asked for proof that OP Earth was bigger than IRL Earth. Some people offered that it has 6 moons, but that's not impossible for a small celestial body. Thus began my search.

    The crew sans Luffy is traveling from Rainbase to Alubarna to intercept the rebel forces when Vivi states...



    Well then ;33



    Dartg-sama insists on 3px, so...

    (50/3)193.61=3226.8km

    Now we have to find the size of OP's planet.



    An island on the side of the globe, which means it is at a distance relative to perspective similar to that of the diameter, measures 2 px.

    (3226.8/2)93=150046.2km

    Good enough. Note that in the scan containing the aerial view of Alabasta, using a diagonal and scaling from the island gives us 92,000km already, so...stop bitching :maybe
  • Tacocat
    Milkin' dem Usopp feats :tomato

    During Usopp's fight with Mr. 4 and Ms. Merry Christmas, Usopp launches Mr. 4 with one of his hammers.



    I had a hard time finding a way to scale Mr. 4, but here we see him towering over Robin. Perspective isn't really an issue, considering he's at the corner of the table and she's at the edge of it. Robin is 1.88m tall, so 2.5m for Mr. 4 seems pretty solid.



    (2.5/191.9)118.23=1.54m

    So 1.54m for this Lasso or whatever its name is.



    Now I have to angsize this thing.

    2atan(32/(661/tan(70/2)))=3.882948809586

    d=22.715m

    Adding to that distance...

    (1.54/54)296=8.44m

    x=31.155m

    So we'll consider the range to be 31.155m, even though it should be larger.

    Now to find a projection angle...

    (1.54/54)46=1.312m

    tanA=opposite/adjacent
    tanA=1.312/22.715
    tanA=0.0577591899625798
    A=3.305685019187 degrees

    Input that shit in here and we get 51.496130331668m/s.

    To find Mr. 4's mass, I'll use the BMI formula, which, when using the metric system, is...

    BMI=m/(h^2)

    Where m is mass and h is height.

    Mr. 4 is quite overweight, and the minimum BMI to be considered overweight is 30.

    30=m/2.5^2
    m=30*2.5^2
    m=187.5kg

    mv=mv

    187.5*51.496130331668/3=3218.5m/s

    or about mach 9.5.
  • Tacocat
    Spoiler:



    Usopp fights Chu and hits him with his signature move. We see in the first scan that Chu is pretty much centered in the road. The second scan shows that Usopp's attack landed Chu in the water. So, it's sensible to take half of the road's width for a range.



    Nami is 1.69m tall according to the wiki.

    (169/106)120=1.9132m

    Chu is then 1.9132m in height.



    (1.9132/163.03)623.35=7.315m

    Halve the width of the road for Chu's projectile range and we get 3.6575m.

    The angle of the shot is a little weird, so 45 degrees will serve as a low-end for Chu's projection angle.

    http://www.ajdesigner.com/phpprojectilemotion/range_equation_initial_velocity.php

    So Chu's velocity is 5.9869441286854m/s.

    Now, Usopp presumably shoots lead ball bullets.

    http://www.speer-bullets.com/products/components/lead_round_balls.aspx

    The most massive ball in the link above consists of approximately 278 grains. I grain is roughly equivalent to 6.479891e-5 kilograms.

    278(6.479891e-5)=0.01801409698kg

    We'll peg Chu at the average of 70kg.

    mv=mv

    70*5.9869441286854/0.01801409698=23264.34m/s

    or about mach 68...

    Spoiler:
    Spoiler:
    Spoiler:
    Spoiler:
    Spoiler:

    GURARARARARARARARARARARA!!!
  • Tacocat
    Don't take this too seriously. I was kinda bored 'cause Bleach was on Toonami and fuck if I'm watching that shit. I was kinda mourning the absence of OP calculations on my blog.



    The Wiki says Luffy's 1.72m tall in chapter 1.

    1.72m/275px=x/124.1px
    x=.78m



    .78m/22px=x/76.9px
    x=2.73m

    2.73m for the distance between the marines and about where Luffy landed.

    Now, I know nothing about guns, so I had no idea what to look for when I was checking for the speed these guys could fire at. I looked up minie ball speed and got 950 feet per second, or 289.56m/s, http://www.historynet.com/minie-ball. Anyone wanna point me in the right direction, feel free.

    2.73/289.56=0.00943s

    Now to angsize Luffy's distance (the sixth panel is quite obviously from Luffy's perspective).

    2*atan(tan(70/2)x(280/350)=58.51213528192

    2*atan(22/(350/tan(58.51213528192/2)))=4.033152595952

    Gives me a distance of 11.076m.

    11.076m/0.00943s=1174.549310710498m/s

    or about mach 3.5.
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