This is a serious question

Craig breaks the wall and Kyara doesnt so Craig>Kyara

W Baku not using his easy 20 metric tons with one hand and holding back against fodders to not fight by himself

Edit: Forgot he lowered the pressure difference and raised the oxygen to around 17% using the doorstop, though this is still impossible cuz rather than 20 it still needs 3 tons

Mostly 10-15m

On a note, this feat is below those of the likes of Voja or Bandai just so you understand how physically it wont result as that impressive. To note, the horizontal I'm calculation is strictly for a normal bending load on edge that makes it direct, not exactly what happens here where more torsion is present, but I cant really calculate that as well since the scene is too vague and I'm not that knowledgeable to make such assumptions with these values. This is just a low end and as well, this feat is normally impossible since the hinges will rupture before this can even happen but this is hypothetical as they might be made of tungsten/its also impossible since the door would just start rotating rather than bending as well (Its just inconsistent since its a manga btw)

Normal Section bending trough maximum stress

Using bending to calculate it https://en.wikipedia.org/wiki/Bending while the moment of inertia is a rectangle or so b^3h/12

Section Values

Leo height: 167px and should be about 190cm(Estimation)

Door Height: 224.1317cm (196px)

Doors Length: 269.2989686cm (316px)

Door Length: 134.6494843cm

Door Height: 303px

Door thickness: 7.3970858cm (10px)

Distance from neutral axis: 3.6985 cm due to horizontal bending and not vertical

We need the length of the door in order to find the force since M is a torque value

Since σ max of the door is (M c)/I by arranging the equation we can determine the torque necessary to deform the section which would be (σ I)/c

Door Yield Strength

Usual metal doors use structural steel which normally has yields ranging on an average of 357MPa (Convert to Pa) https://eurocodeapplied.com/design/en1993/steel-design-properties

Second Moment Area of Inertia: (0.073970858 * 2.241317^3)/12= 0.0000756m^4

Bending Moment: (357000000 * 0.0000756)/0.036985= 729695 N * m

Torque is equal to the force applied * the rotational axis, or so the radius. In the case the radius would be the distance between the hinges and where force is applied which is pretty much on the exact opposite edge

Force: 729695/1.346494843= 541.9kN or 54 Metric Tons

The force is applied with a single hand on each door so the actual force he applied with both hands needs x2

Total Force: 541.9 * 2= 1083kN or 108 Metric Tons

This also assumes the door is NOT hollow, which is very unlikely but I'll give it to it. Generally I made some assumptions here, but this could be higher and matching Voja pushing the containers, but for that matter I also dont want to involve the possibly of Craig using his own hands as leverage rather than the hinges resisting since the door deforms on the entire surface rather than localized area. If you dont want to use straight image scaling is also fine but results are similarly around 50 tons per door, although its probably higher but not by much

Closest there is

Kinda impossible he'll lose besides one.

Like only Cara maybe beats him

Too strong the goat

Ig Im so bored, summer lk ass cuz I aint doin shi besides ts kind of post. Manabe glaze though so W

Containers Weight

On the average based on this a container of a ship weight can vary between 2100-3800kg while being empty and can reach up to more than 26000 kg at their maximum capacity regardless what is inside.

I'll use a 10ft one for the container type and the average between his maximum gross weight and his tare/empty weight (As the one used here is probably the 12-14 ft type in japan but there isn't anything for it so I'll later use a 1.4x probably)

• Results would be 5717.5kg

Friction Coefficient

We have to determine the kinetic(or dynamic) coefficient to first find the force required to move the containers over the constant deceleration while sliding. Steel on steel has a coe of about 0.42

• Container Length: 40px or about 3m
• Distance Traveled: 5.7m to 11.4m (30+23+23px)

All Distances: 1. 2.25        2. 1.725        3. is same as 2.

Deceleration due to kinetic friction is equal to

μg= 0.42 * 9.81= -4.12m/s^2

Displacement

I'll assume the first container accelerate over a distance of roughly 0.4m which is about half Voja arm length or the displacement. Though it's most likely less when the first time she accelerates she already has her arms extended

Velocity required To overcome Everything

The velocity after sliding distance d with friction deceleration a, while Vf is final velocity and Vi is initial, is equal to Vf^2= V𝑖^2 − 2𝑎𝑑 for general kinematics. To note as well, the momentum type collision is assumed to be perfectly inelastic, as the containers pretty much stick together on impact, while they should have basically the same mass which is why I won't use any type of semi elastic collisions which would rather make this higher

I'll explain different signs so it's easier to understand

• v0 = initial speed of container A (unknown)
• v1 = speed just before first collision (after sliding 𝑑1)
• v2 = speed just after first collision (containers A+B combined)
• v3 = speed just before second collision (after sliding 𝑑2)
• v4 = speed just after second collision (containers A+B+C combined)
• v5 =0 (final speed after sliding 𝑑3 and stopping)

Note how d1, d2, v1, v2 ecc is just to differentiate distances, and to not indicate things like multiplications. The calculation will be done by first starting on the last container and then reaching the first one obviously

First part with 2.25m

v1^2= v0^2 - 2ad1

Second part 1.725m

v3^2= v2^2 - 2ad2

Part 3 with with same as 2

0(Since if where it ends)= v4^2 - 2a𝑑3 ⟹  v4^2 = 2ad3

Calculating the collisions now using momentum transfer

First collision momentum

mv1= 2mv2  ⟹  v2= (v1)/2

Second collision momentum

2mv3= 3mv4  ⟹  v4= (2/3)v3

Substitute collision velocities into sliding equations

3rd part of the sliding

v4^2= 2ad3 = 2 * 4.12(deceleration) * 1.725m = 14.214

v4= Square-rot 14.214= 3.77 m/s

From second collision using the velocity from part 3

v4= (2/3)v3 ⟹ v3= (3/2)v4= 1.5 * 3.77= 5.6 m/s

2nd part of the sliding

v3^2= v2^2 - 2ad2 ⟹ v2^2= v3^2 + 2ad2

v3^2= 5.6^2= 31.9

2ad2= 2 * 4.12 * 1.725= 14.214 (same values as 3)

v2^2= 31.9 + 14.214= 46.195

v2= square-rot 46.195= 6.8 m/s

First collision

Just 2 * 6.8 since it's 1/2= 13.6 m/s

1st part of the sliding

v1^2= v0^2 - 2ad1 ⟹ v0^2= v1^2 + 2ad1

v1^2= 13.6 m/s (the same as above)= 185

2ad1= 2 * 4.12 * 2.25= 18.54

v0^2= 185 + 18.54= 203.54

v0= square-rot 203.54= 14.26 m/s

That is the velocity of the first container

Calculating now the force trough the 0.4m displacement assumed  trough kinematic

a= v0^2/2s(displacement)= (14.26)^2/(2 * 0.4)= 203.54/0.8= 254.42 m/s^2

Force Applied: 254.42 * 5717.5(Mass of the first container)= 1455717N or 150 Metric Tons

Assuming it's actually a 14ft one: About 1.4x mostly so 1455717 * 1.4x= 2038003N or 207 Metric Tons

Small notes are that 1. the gel doesn't influence the thing as it didn't get below the containers meaning friction wasn't lowered, though if you wanna somehow argue some of it went below it would still be lower only by a bit 2. I didn't use semi elastic collisions as this would have a really low coefficent and I'd rather not assume it. 3. Displacement is low balled 4. Containers weight is about that on average but using a bit less can also he correct

Is Lalo just sayin bs cuz he doesn't how strong dudes actually are or it's just inconsistent. I guess inconsistent

Some characters like Ranko, fake Bai Long or just randoms like Mogami guards werent added due to lack of actual importance or having no actual scaling to anyone at all, like for example Anoma or Fukurou. Also characters like Craig allies in chap 70 or Ron would be below Hangman Marco, Labyrinth Kadokura Is a bit below Labyrinth Marco

Most characters in each tiers are interchangeable between each others, like EOS Kadokura, prime Bai Long and Jonglyo fight Kyara or either old Bai Long, Manabe, Robert K Kyara and pre AW Kadokura

I copied the slides from my yt video, as I lost them and I ain't makin a second one. He ain't t1 cuz I only believe in 385 Kyara acceptance

Yes, Cara is stronger than Tok Marco, that's obvious, but compared to what Manabe did, Cara had to match Marco with more intensity, without even counting Eos Marco depiction. This post was kinda old from tt, even then I never really understood why Cara could ever be superior, when Marco there was clearly no match for him, yet against Cara he was matching to some extends (Obviously not referring to only strength, as in that case there was a decent difference between the two seeing the hand match)