Gin Breaks Pearl’s Armor
Gin breaks through Pearl’s Iron armor. This one is pretty simple, but thought it was worth looking into. Since it was fractured, that would make the energy requirement 96 J/cc.
Pearl is 2.39 m tall. He measures 737 pixels here (I didn’t include his weird helmet) compared to 426 for the iron plate’s diameter. One of his smaller plates has a thickness of 17 pixels. We’ll be using that one after this.
426 / 737 = .578 * 2.39 = 1.381 m, diameter of iron plate.
17 / 737 = .023 * 2.39 = .055 m, thickness of smaller plate.
31 / 18 = 1.722 * .055 = .095 m, thickness of iron plate.
Treating it as a cylinder, this would give the iron plate a volume of .569 m^3
.569 m^3 = 569,000 cm^3
569,000 * 96 = 54,624,000 J or 13.056 Kg of TNT
Hachi Moves a Reef
Hachi pushes a reef along the seabed. I can’t quite get the full distance he pushes it, but I can at least get something from the initial panel.
Hachi is 2.20 m tall. He measures 22 pixels here compared to 99 for distance pushed, 160 for lower diameter, 191 for height, and 19 for upper diameter.
99 / 22 = 4.5 * 2.20 = 9.9 m, distance pushed.
160 / 22 = 7.273 * 2.20 = 16.001 m, bottom diameter.
191 / 22 = 8.682 * 2.20 = 19.1 m, height.
19 / 22 = .864 * 2.20 = 1.901 m, upper diameter.
Treating this as a truncated cone, the volume would be 1,450.694 m^3. Hachi is clearly pushing rock here based on the shape and texture, so we’ll go with 2,700 kg/m^3 for the density.
1,450.694 * 2,700 = 3,916,873.8 Kg, weight of the rock.
My biggest assumption will be the acceleration of Hachi. He is clearly struggling, but also the kicked up underwater sand/dust hasn’t had a chance to settle even after he traveled some distance, so it’s not that slow. My assumption will therefore be an acceleration of 1 m/s^2.
f = ma
3,916,873.8 * 1 =3,916,873.8 N
This is not the true force Hachi exerted, however. We need to figure out the static frictional force of rock moving across soil.
fu = f * u
Where u is the friction coefficient between two objects. Unfortunately, the closest thing I can find is frictional coefficient of concrete on sand or rock, which is .3. So we’ll use that until I find something else. Concrete is quite similar to rock, so I feel that this isn’t a problem. Based on all that was kicked up from Hachi moving, I’m going to assume it’s sand, which would make the coefficient (concrete on wet sand) .4.
3,916,873.8 * .4 = 1,566,749.52 N
Finally, we multiply the two together to find the Kinetic Frictional Force.
3,916,873.8 * 1,566,749.52 = 6,136,760,146,050.576 N
Work = force x distance
6,136,760,146,050.576 * 9.9 = 60,753,925,445,900.702 J or 14.52 Kilotons (!!!)
I never would have expected such a total. Does this put it into outlier territory? Almost certainly. Does this expose a potential weakness in assuming the acceleration as I have done? Maybe…
Arlong Throws Some Water
Arlong merely throws a splash of water onto Sanji and it knocks him back as if he were punched. We can use conservation of momentum to figure out how fast he threw the water.
Arlong is 2.63 m tall. He’s 319 pixels compared to the diameter of water in his palm at 9 and the depth at 7.
9 / 319 = .028 * 2.63 = .074 m, diameter of water.
7 / 319 = .022 * 2.63 = .058 m, depth of water.
With the dome calculator, we find that the volume of water in Arlong’s palm is .000227 m^3
Density of sea water is 1,029 kg/m^3.
1,029 * .000227 = .234 Kg, weight of water thrown.
With the projectile motion calculator, and assuming that Sanji was knocked back a meter of so, the take off velocity would be 6.12 m/s. I factored in Sanji’s “Starting height” of 1.77 m here, which does lower the results a bit.
Sanji’s weight is never given, so the best I can do is use the average Japanese male’s weight of 62.5 Kg.
Sanji m*v = water m*v
6.12 * 62.5 = 382.5 / .234 = 1,634.615 m/s, or Mach 4.81
Luffy’s Gomu Gomu Battle Axe
Luffy knocks Arlong through five floors of Arlong park, and four of them appear to be stone tile. The floor and building in general are then cracked, and Arlong Park collapses. Unfortunately I don’t have a way to calculate the full extent of the attack, but I can still do the 4 holes made through stone, and the fact that the entirety of each floor was at least fractured. I can’t account for the one wooden floor because I am unable to find a figure in J/cc for fracturing that material. If you come across anything like that, though, please do let me know!
As previously established, Arlong is 2.63 m tall. He’s 353 pixels here compared to the tile length of 66.
66 / 353 = .187 * 2.63 = .492 m, tile length.
213 / 75 = 2.84 * .492 = 1.397 m, hole diameter… how the hell did Arlong fit through this? lol
41 / 75 = .547 * .492 = .269 m, depth of floor.
This would give the destroyed floor a volume of .413 m^3
Assuming that all holes here are a similar size, the total volume of destroyed tile would be 1.652 m^3 . For this portion of the feat, I’ll be using 20 J/cc, explosive fragmentation.
1.652 m^3 = 1,652,000 cm^3
1,652,000 * 20 = 33,040,000 J
Alright, now let’s figure out what it would take to fracture (8 J/cc) those four floors. The fractures trvael all the way to the walls too and that’s probably what made the structure collapse, but again I must say that I unfortunately have no way to calculate that, so the floors is the best I can do.
Hachi is 2.20 m tall as previously established. He measures 64 pixels compared to the column diameter at 14.
14 / 64 = .219 * 2.20 = .482 m, diameter of column.
The column measures at 10 pixels compared to the third floor (Same size as second floor) at 317, 210 for the fourth floor, and 121 for the sixth. I also measured the height at 428, just out of curiosity.
317 / 10 = 31.7 * .482 = 15.279 m, width of second and third floor.
210 / 10 = 21 * .482 = 10.122 m, width of fourth floor.
121 / 10 = 12.1 * .482 = 5.832 m, width of sixth floor.
428 / 10 = 42.8 * .482 = 20.63 m, height of Arlong Park.
As a reminder, we are ignoring the fifth floor here because the floor is wooden and I don’t have the figures for fracturing that material. Thankfully stone tile is 8 J/cc, as has been often used in this blog.
My main assumption from here will be that the width dimension will be the same as length for each floor. Based on scans I’ve seen, I feel that this is pretty accurate.
With that said, treating these as rectangular prisms…
15.279 * 15.279 * .269 = 62.798 m^3, volume for the second and third floor each.
10.122 * 10.122 * .269 = 27.56 m^3, volume for fourth floor.
5.832 * 5.832 * .269 = 9.149 m^3, volume for sixth floor.
Total fractured volume is 162.305 m^3 = 162,305,000 cm^3
162,305,000 * 8 = 1,298,440,000 J
Now combining the two totals…
1,298,440,000 + 33,040,000 = 1,331,480,000 J, or 318.231 Kg of TNT.
Now like I said, this doesn’t account for the fractures of the fifth floor or the walls of the building, which ultimately caused the collapse. Someday, if I get figures for wood, I’ll come back to this, but for now, I feel that this is a good minimum at least.
RESULTS
Gin Breaks Pearl’s Armor – 13.056 Kg of TNT
Hachi Moves a Reef (Probable Outlier) – 14.52 Kilotons
Arlong Throws Some Water – Mach 4.81
Luffy’s Gomu Gomu Battle Axe – 318.231 Kg of TNT
Cable's Calculations 