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Do we have a reference for the ability to break sign poles? The poles traffic signs are on? I ask because a character recently performed said feat and I'm curious what value we can get from it. Though the feat was done in Japan and the material for the poles may differ from america's.
 
Something tells me the feat's from Yakuza or JoJo.

I think there's a lifting feat for this, not a smashing one.
 
OK so the current link we use for doorbusting to get our door sizes, it's web-archived and a more accurate description of the sizes exist here

So that link states 80 inches or 203.2 cm height, 36 inches width or 91.44 cm and a standard thickness of 1.75 inches or 4.445 cm.
 
Also, fair warning: If y'all have some important blogs and wanna change your username, you might lose all the comments on those blogs.
 
I see on the tanks point; they are made of other metals and other materials. The steel calculation could have been a low-ball; I guess the calculation could be inaccurate.
 
Tank armor is mostly depleted uranium, and we found values for it which would hurl tank destruction close to High 8-C

Here are the values, shear strength of materials is usually 0.5-0.6 of tensile strength unless it's other metals we're talking about

Though we can't ignore high-strength steel either, my guess is it also uses something stronger than RHa steel alongside titanium.
 
Mtorque
This is how to calculate force F you need to apply to create M bending moment
 
There should be a generic rock crushing calc.

Average men hand surface area is 448 cm^2, 392 for women.

Assuming granite which has a crushing strength of 104 MPa.

Multiply by 2 for two hands.

Men: 44800*104*2/9.81 = 949 887.87 kg, Class K

Women: 39200*104*2/9.81 = 831 151.89 kg, Class K

Please be noted that if a character crushes a rock smaller than his hand surface area (like this feat) or that he doesn't use his full hand surface area to crush it this calc does not apply (like this feat).
 
How do we calculate punching the ground and causing a crater around?


And hitting a boulder on the ground repetely until it breaks?
 
Spinosaurus75DinosaurFan said:
There should be a generic rock crushing calc.
Average men hand surface area is 448 cm^2, 392 for women.

Assuming granite which has a crushing strength of 104 MPa.

Multiply by 2 for two hands.

Men: 44800*104*2/9.81 = 949 887.87 kg, Class K

Women: 39200*104*2/9.81 = 831 151.89 kg, Class K

Please be noted that if a character crushes a rock smaller than his hand surface area (like this feat) or that he doesn't use his full hand surface area to crush it this calc does not apply (like this feat).
This should definitely be added to the blog
 
Off topic:

People always say detonation is scarier than deflagration, but why are deflagration yields higher AP values than detonation with the same fireball radius? (Or are they?)
 
Just requesting material scientists and civil engineers help:

How much force is required to break a wood beam in the middle?

Like this?

I am quite sure this is more a lifting strength feat... or just some PSA about interior design and household woodwork safety knowledge.
 
How much force is required to break a wood beam in the middle?

Like this?

I am quite sure this is more a lifting strength feat... or just some PSA about interior design and household woodwork safety knowledge.

4 * Tensile strength of wood * Elactic section modulus / Length of the beam

But work would be much lower than just force times distanse
 
If given the size, what formulas are used?

For crater it would be volume of crater * compressive strength of the material (214.35 j/cc for rock, 40 j/cc for concrete, etc)

For smashing boulders it would be volume of rock * fragmentation/violent fragmentation/pulverization (depending on how you destroyed it)

You can find destruction values on the Calculations page.
 
Spinosaurus75DinosaurFan said:
If given the size, what formulas are used?
For crater it would be volume of crater * compressive strength of the material (214.35 j/cc for rock, 40 j/cc for concrete, etc)

For smashing boulders it would be volume of rock * fragmentation/violent fragmentation/pulverization (depending on how you destroyed it)

You can find destruction values on the Calculations page.

I have made a table too. Hope this is useful.

Now I really need to go to work irl, ttyl.
 
In the feat in question, the character lifted the rock and used it to smash someone else, doing so a few times, in the end the boulder was completely cracked, would it be different than punching the rock?
 
Assume Newton laws of motion hold.

The body of the defender would be exerting a force to the rock equal to the impact force of the rock on the defense's body. The reaction force and the kinetic (or better call striking) energy reflected causes the rock to shatter. The AP and durability of the defender would be the destruction energy of the said rock. Not much difference from punching the rock.

Imagine Vegeta is smashing against the head of Beerus but with the body of Arale.
 
So if i can find the exact value of the target's defense, then it is the energy to crack the rock?

If i use the "punching the rock" formula, should i divided it by the number of hit's took to crack it?

What if the target isn't defending? The target is k.o and the rock is being used to crush their body and it ends up broken.

Basically, the feat goes like this:

Thry punch the victim, the crater happens, the viction is KOed, they grab the boulder and start crushing the victim's body a fee times and in thr end the boulder has cracks all around it.
 
Thelastmlg said:
So if i can find the exact value of the target's defense, then it is the energy to crack the rock?
If i use the "punching the rock" formula, should i divided it by the number of hit's took to crack it?

What if the target isn't defending? The target is k.o and the rock is being used to crush their body and it ends up broken.

Basically, the feat goes like this:

Thry punch the victim, the crater happens, the viction is KOed, they grab the boulder and start crushing the victim's body a fee times and in thr end the boulder has cracks all around it.
1. Has that victim died already? At which point?

2. Looks like the attacker attacked the victim with an attack that made a crater and the victim gets KOed. - The destruction of the crater can already be taken as one yield of AP feat.

3. The subsequent boulder slamming's AP can be calculated - In this case this may just be the attacker has an AP equal to a fraction of the destruction yield to crush a boulder. If the boulder just has some "cracks" I would rather consider the pulverisation of the cracks of the rock (not the whole rock itself) be the end of the attacker's AP feat.

4. The mass of the boulder can be calculated with density of material to find the lifting strength of the attacker.

5. Generally it depends a lot on the said condition of the story in question. The general common feat calc may or may not be applicable to your case.
 
1. Is not clear, but implied that they fainted.

The crater and rock feat happened consecutively.

2. Yeah, i'll calculate the crater feat, but is there a method to calculate the volume of a irregular shape? Because the crate has spiky and curvy segments, since is a crater made by a punch.

3. Would this fraction be the cracking valur divided by the number of hits?

4. Okay
 
You would have to divide by the amount of hits, yes, since our AP system is for the power generated in a small single attack.
 
Megapascals here are changed to j/cc, but when randomly googling something, I found out that megapascals can be converted to psi. Can't we figure out the tier of sharp objects like knives with this via the surface area of the blade?
 
We don't invoke stuff like that in tiering. A few verses did it a while ago but it was rejected.
 
Thelastmlg said:
But is there a method to calculate the volume of a irregular shape? Because the crate has spiky and curvy segments, since is a crater made by a punch.
This is a little important, maybe i can google the answer but it would be faster if someone told me now.
 
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