Indentation Calcs

[Agnaa]

I've recently had some indentation calcs pointed out to me by Vapourrr. Many of these calcs had other issues, but even if those were fixed, it would still give unrealistically high results.

General Issues​

Here is a video of a real life human causing a dent in a car.

I'm going to lowball some figures for simplicity; this is fine since despite underestimating the value, I end up with a result well above what we accept humans to be able to dish out with a single strike.

Lets say that the dent was a circle with a radius of ten centimeters (despite it visually being much larger). That would give an area of 0.0314 m^2.

Multiplying that by the needed pressure of 350 MPa (since that's the accepted constant for four out of five of these calcs; one uses 275), gives a force of 11 N, Class M.

Lets say that the dent was displaced by 0.625 mm (despite some calcs using higher constants for the thickness of metal in cars).

That gives an energy needed of 6,875 Joules, Street level, only 2.18x away from 9-B.

That is ludicrous for a single kick by an ordinary human. A more accurate measurement would easily land it in 9-B. So would simply using some of the assumptions the calcs I'll link below did (displacement of 3mm would give 55,000 Joules, about 4x baseline wall level).

Due to the ludicrous inflation of values, I suggest not calculating these sorts of feats until a better method can be sorted out, and that all calcs using it should be considered invalid.

Specific Issues​

• Saitama denting a car hood
  • Application Issues: Derived area incorrectly. The area is not the width of the car multiplied by the thickness of the hood, it's the surface area of the part of the metal which was deformed. The displacement doesn't really have much reason to be half of the car hood's width; those values are just not correlated.
  • Relevant calc group members: @KLOL506
• Shinichiro Sano denting a metal street lamp
  • Application Issues: Use of 3 pixel thickness lines inflated a critical measurement (finger width in panel 2) by 25%, inflating the final result by that amount.
  • Relevant calc group members: @KLOL506, @Therefir
• Kuga makes a dent
  • Application Issues: None that I could see from a quick glance.
  • Relevant calc group members: @KLOL506, @Dalesean027, @Dark-Carioca
• Shuka dents a steel door
  • Application Issues: We can see from the scans that while the "spike" is fully embedded in the wall, past the rest of the crater, while the "hook" and the connection of it to the "spike" are sticking out into the depth of the crater. Essentially, calculating the depth of the crater as being the length of the "spike" is incorrect, since the "spike" goes deeper than the crater. On top of that, the diameter of the dent was taken from a zoomed-out panel, which showed a vastly different diameter than the zoomed-in panel immediately following it.
  • Relevant calc group members: @Psychomaster35
• Andrew crushes a car
  • Application Issues: None that I could see from a quick glance.
  • Relevant calc group members: @Flashlight237

As people who accepted this method, could you weigh in on the merits of its use?

 

[Flashlight237]

That is a good question. The 10 cm example you brought up is brought up in a concept of mechanical engineering different from what the entire calc group is used to. In fact, the most accurate way of handling depths from what I can tell is Brinell Hardness: https://en.m.wikipedia.org/wiki/Brinell_scale

Essentially it's how strong a material is against indentation, usually tested with a 10 cm steel ball. I recommend studying it before deciding on its use, however.

 

[Agnaa]

That is a good question. The 10 cm example you brought up is brought up in a concept of mechanical engineering different from what the entire calc group is used to. In fact, the most accurate way of handling depths from what I can tell is Brinell Hardness: https://en.m.wikipedia.org/wiki/Brinell_scale

Essentially it's how strong a material is against indentation, usually tested with a 10 cm steel ball. I recommend studying it before deciding on its use, however.

Yeah while that seems promising, it would in fact give a higher value. 1176 MPa for mild steel.

 

[Flashlight237]

Yeah while that seems promising, it would in fact give a higher value. 1176 MPa for mild steel.

Yeah, I can certainly tell you that plasticity isn't exactly too well explored. Heck, in this calc (another indentation thing but it's ballistics), I had to use compressive yield and tensile strength purely out of convenience rather than legitimate understanding of plasticity: https://vsbattles.fandom.com/wiki/U...he_Dead:_Thomas_Rogan_Shoots_and_Shoes_a_Door

 

[Arceus0x]

By this logic would this calc still be fine or not?

 

[Agnaa]

By this logic would this calc still be fine or not?

Not sure. I've seen some floating around the idea of only using this equation for particularly thick sheets of metal, but I'm not sure how it'll ultimately pan out. If we go for blanket unusability, that calc wouldn't fine.

 

[Arceus0x]

Not sure. I've seen some floating around the idea of only using this equation for particularly thick sheets of metal, but I'm not sure how it'll ultimately pan out. If we go for blanket unusability, that calc wouldn't fine.

I really hope we don't do blanket unsuability then cause otherwise it'd nuke any and all bending calcs of the sort.

 

[Flashlight237]

I'll see if @StretchSebe can help out. I've seen some of his unconventional methods before, so I think he might have a trick or two on how to handle this.

 

[Agnaa]

Vapourrr has made a concrete suggestion for a solution; only disallowing these sorts of calcs when the material in question has a thickness less than that of a hammer's head. In this case, estimated at 1.24 to 3.42 centimeters.

I'm only slightly in favour of this. While it is true that ordinary humans can't make dents in materials that thick, even with tools, meaning the inflection point where yield strength becomes applicable could be around there, that's a but uncertain.

Another possible solution I'm considering is only allowing these feats when the indent doesn't protrude very much through the other side. I think that may better indicate when yield strength becomes applicable. But again, I'm not sure.

EDIT: I now disagree with that proposed solution, since it'd make weaker feats calculable but higher feats not, which is a red flag imo.

 

[KLOL506]

I'm merely using the methods that Jasonsith and Spinosaurus used before.

 

[Agnaa]

@Jasonsith @Spinosaurus75DinosaurFan Please refer to the OP of this thread.

 

[StretchSebe]

This is one of many reasons why we need a proper Lifting Strength page

Indentation I think would imply bending, which falls under shear stress:

I've definitely used ultimate tensile strength a number of times for how much force is required to do some specific, 'unconventional' LS feats. On that, shear strength itself often doesn't have many, or even any values that are standard for materials (unlike yield and tensile strength), and is generally just considered to be at 60% of a material's ultimate tensile strength.

The (general) formula for shear stress is τ = F (Force applied) / A (Area parallel/relative to force).

There are other formulas for the types of shear stress/strength though, particularly for bending stress, though it seems basic stress formula paints a pretty broad brush across the types of shear stress shown in the image above (including bending).

As for cases in which the material is twisted (quite common in fiction, but again not really much info on how to calculate such feats on the wiki), there is Torsion. Torsion is kind of a nightmare to manually calculate, but I did find a calculator for such that streamlines said formula.

 

[Flashlight237]

This is one of many reasons why we need a proper Lifting Strength page

Indentation I think would imply bending, which falls under shear stress:

I've definitely used ultimate tensile strength a number of times for how much force is required to do some specific, 'unconventional' LS feats. On that, shear strength itself often doesn't have many, or even any values that are standard for materials (unlike yield and tensile strength), and is generally just considered to be at 60% of a material's ultimate tensile strength.

The (general) formula for shear stress is τ = F (Force applied) / A (Area parallel/relative to force).

There are other formulas for the types of shear stress/strength though, particularly for bending stress, though it seems basic stress formula paints a pretty broad brush across the types of shear stress shown in the image above (including bending).

As for cases in which the material is twisted (quite common in fiction, but again not really much info on how to calculate such feats on the wiki), there is Torsion. Torsion is kind of a nightmare to manually calculate, but I did find a calculator for such that streamlines said formula.

I knew I can count on you to chime in.

 

[KLOL506]

Speaking of bending...

Can we also discuss methods to discuss bending feats for LS and a way to make it easier to calculate them?

 

[Flashlight237]

Speaking of bending...

Can we also discuss methods to discuss bending feats for LS and a way to make it easier to calculate them?

I typically went for flexural strength when it comes to bending, though the article on it says we're calculating flexural strength wrong.

 

[Arceus0x]

This is one of many reasons why we need a proper Lifting Strength page

Indentation I think would imply bending, which falls under shear stress:

I've definitely used ultimate tensile strength a number of times for how much force is required to do some specific, 'unconventional' LS feats. On that, shear strength itself often doesn't have many, or even any values that are standard for materials (unlike yield and tensile strength), and is generally just considered to be at 60% of a material's ultimate tensile strength.

The (general) formula for shear stress is τ = F (Force applied) / A (Area parallel/relative to force).

There are other formulas for the types of shear stress/strength though, particularly for bending stress, though it seems basic stress formula paints a pretty broad brush across the types of shear stress shown in the image above (including bending).

As for cases in which the material is twisted (quite common in fiction, but again not really much info on how to calculate such feats on the wiki), there is Torsion. Torsion is kind of a nightmare to manually calculate, but I did find a calculator for such that streamlines said formula.

So how would this look in practice?

 

[StretchSebe]

Can we also discuss methods to discuss bending feats for LS and a way to make it easier to calculate them?

So how would this look in practice?

The general use of the formula of τ = F (Force applied) / A (Area parallel/relative to force) would be force needed to get the material to deform divided by the surface area that is affected in an object. Shear strength (force) again being 60% ultimate tensile strength of any given material in mPA, and the surface area (Area) that is affected in an object being in meters squared. This is because these deformations are are considered shearing, and not volumetric. They do not affect all three dimensions of the object.

I linked a torsion feat calculation from Baki in there as an example. All the shear stress examples should be based on the same principle of deformation despite how different they look. As said and suggested by surface area being used for area, the deformation of shear stress acts only on two of the three dimensions.

Using the Baki example, I got the thickness of the steel door (length), but it was the surface area of the door (height and width) that mattered in the formula/calculator. The change in area was in meters squared like it would be in bending/indentation, because the material/object is only being deformed in two of it's dimensions. It's different from say, crushing an object, or even the slicing/splitting an object. The 'shear' example in that image is a bit misleading, and should more look like the one below, because at the point of breaking, it would be ultimate tensile strength to be considered, not shear strength.

Anyways, the more simple example of this is provided on the Wikipedia page for Shear Stress in 2D:

In this, we see that the shear stress on the rectangle is dependent on the change in one dimension: The change in starting and end 'length' of what we'd probably call the 'width' side of the rectangle being the area that we'd calculate.

Height remains the same in this 2D example, like how depth (or whatever the third dimension is) would remain the same in the case of an indent or torsion in 3D. It's the change in the other dimension (or other two dimensions if this was 3D and surface area) that is what is to be considered for calculating area.

 

[DontTalkDT]

Not gonna comment on new methods for now, as this isn't really what I understand well, but I will say that I generally would assume that a denting calc must use at least:

1. Something like the area of the dent. (1 cm^2 and 1 m^2 dents should give different results)
2. Some parameter for how much it was dented. (1cm deep and 1m deep dents should give different results)
3. Some material parameter. (denting rubber vs denting steel should make a difference)
4. Some parameter for how thick the dented sheet is. (1mm thick steel plate should give different results from 1m thick steel plate)

Just by my feeling of how the world works.
At a glance, I think the calcs listed above tend to usually not use at least one of those.

 

[Arceus0x]

The general use of the formula of τ = F (Force applied) / A (Area parallel/relative to force) would be force needed to get the material to deform divided by the surface area that is affected in an object. Shear strength (force) again being 60% ultimate tensile strength of any given material in mPA, and the surface area (Area) that is affected in an object being in meters squared. This is because these deformations are are considered shearing, and not volumetric. They do not affect all three dimensions of the object.

I linked a torsion feat calculation from Baki in there as an example. All the shear stress examples should be based on the same principle of deformation despite how different they look. As said and suggested by surface area being used for area, the deformation of shear stress acts only on two of the three dimensions.

Using the Baki example, I got the thickness of the steel door (length), but it was the surface area of the door (height and width) that mattered in the formula/calculator. The change in area was in meters squared like it would be in bending/indentation, because the material/object is only being deformed in two of it's dimensions. It's different from say, crushing an object, or even the slicing/splitting an object. The 'shear' example in that image is a bit misleading, and should more look like the one below, because at the point of breaking, it would be ultimate tensile strength to be considered, not shear strength.

Anyways, the more simple example of this is provided on the Wikipedia page for Shear Stress in 2D:

In this, we see that the shear stress on the rectangle is dependent on the change in one dimension: The change in starting and end 'length' of what we'd probably call the 'width' side of the rectangle being the area that we'd calculate.

Height remains the same in this 2D example, like how depth (or whatever the third dimension is) would remain the same in the case of an indent or torsion in 3D. It's the change in the other dimension (or other two dimensions if this was 3D and surface area) that is what is to be considered for calculating area.

So let's say I wanna apply it to my calc here

The entire door was bent X amount of cm. I have the door height, length and width, all 3 dimensions. What do I do with it?

 

[Agnaa]

This is one of many reasons why we need a proper Lifting Strength page

Indentation I think would imply bending, which falls under shear stress:

I've definitely used ultimate tensile strength a number of times for how much force is required to do some specific, 'unconventional' LS feats. On that, shear strength itself often doesn't have many, or even any values that are standard for materials (unlike yield and tensile strength), and is generally just considered to be at 60% of a material's ultimate tensile strength.

The (general) formula for shear stress is τ = F (Force applied) / A (Area parallel/relative to force).

There are other formulas for the types of shear stress/strength though, particularly for bending stress, though it seems basic stress formula paints a pretty broad brush across the types of shear stress shown in the image above (including bending).

As for cases in which the material is twisted (quite common in fiction, but again not really much info on how to calculate such feats on the wiki), there is Torsion. Torsion is kind of a nightmare to manually calculate, but I did find a calculator for such that streamlines said formula.

If the only real difference is changing the constant from ultimate tensile strength to shear strength, which is just a 60% decrease, that sounds insufficient to accommodate the issue of humans being able to dent cars.

 

[KLOL506]

AFAIK the indentation feats used yield strength, which at times was lower than shear strength, but at this point, IDK.

 

[Flashlight237]

Not gonna comment on new methods for now, as this isn't really what I understand well, but I will say that I generally would assume that a denting calc must use at least:

1. Something like the area of the dent. (1 cm^2 and 1 m^2 dents should give different results)
2. Some parameter for how much it was dented. (1cm deep and 1m deep dents should give different results)
3. Some material parameter. (denting robber vs denting steel should make a difference)
4. Some parameter for how thick the dented sheet is. (1mm thick steel plate should give different results from 1m thick steel plate)

Just by my feeling of how the world works.
At a glance, I think the calcs listed above tend to usually not use at least one of those.

Those are all big players. Application is, of course, hard to get used to as from what I've seen, everyone's been used to fairly inaccurate volumetric calculations. Even I did it when I made that House of the Dead calc, despite ballistics (which is basically indentation but with bullets) being a completely different facet from what we're used to.

 

[StretchSebe]

So let's say I wanna apply it to my calc here

The entire door was bent X amount of cm. I have the door height, length and width, all 3 dimensions. What do I do with it?

Given the door bent, but did not break, that would be a shear.

To find the force of that door bending, you would first find the material that is being bent. I really ought to watch Murder Drones, but I'm going to assume the same steel as in the Baki calc, so the steel would have 620 Mpa ultimate tensile strength. 620 * 0.6 (to get shear strength) = 372 Mpa to bend the steel.

I would personally convert pascals to Newton / Square Meter first for convenience, because you would then take the surface area of the section of the door that got bent (length and width likely, assuming the bend affects an area that is relatively rectangular), and divide the shear strength by the meters squared to get the force. I said earlier to convert to N/m^2 first because then you would just get the Newtons on their own by eliminating the m^2.

I'm not sure how practical it would be to determine the yield of an explosion though. Unless you're determining the character tanking it in terms of their LS resistance to it's force in newton meters. In that case though, you'd be determining how far the character moved, if at all.

 

[StretchSebe]

If the only real difference is changing the constant from ultimate tensile strength to shear strength, which is just a 60% decrease, that sounds insufficient to accommodate the issue of humans being able to dent cars.

I kind of skimmed through the calcs of the first post, but I think the issue across (most of) them is they're is finding the joules to do the damage. Given there's no direct conversion from pascals to joules (as far as I can find), and the closest is a conversion of pascals to J/m^3, I think the issue is they all have is that they're operating on finding volumetric damage when there's only deforming of two of the three dimensions.

 

[Agnaa]

I kind of skimmed through the calcs of the first post, but I think the issue across (most of) them is they're is finding the joules to do the damage. Given there's no direct conversion from pascals to joules (as far as I can find), and the closest is a conversion of pascals to J/m^3, I think the issue is they all have is that they're operating on finding volumetric damage when there's only deforming of two of the three dimensions.

I don't think that helps. Combining both of those fixes would still result in IRL humans having an LS of 6.6e6 Newtons, Class K, with their kicks.

 

[Armorchompy]

This is one of many reasons why we need a proper Lifting Strength page

It would be very nice to have standards set in stone for these various kinds of LS feats. Currently I rarely evaluate them given that I don't know what we actually accept regarding them.

 

[StretchSebe]

I don't think that helps. Combining both of those fixes would still result in IRL humans having an LS of 6.6e6 Newtons, Class K, with their kicks.

What is IRL humans having an LS of Class K in reference to? The car dent?

In any case, newtons aren't used for ap or striking strength, right? It's joules that are used. You find newtons through joules / meters. It's why I said one probably couldn't use the dent of a door to find the energy of the explosive yield for that feat Arceus0x brought up.

I'm assuming a few things in saying this, but in calculating and finding the 6.6e+6 you got (nice ), the value you found was probably in joules, not newtons, especially if it's a striking strength kick. That number in joules would seem to fall in line with the rest of the calculation blogs finding Wall level for similar dent feats.

As is said in this year old thread, converting pressure (pascals, newtons commonly, though psi was being asked about) into energy typically means dividing the pressure by surface area (what's been described in this thread with shear strength/stress). That should reduce the number upfront. Multiplying that by distance to get joules proper will likely increase the number, but again, that'd be to convert the pressure into energy, not the other way around.

It likely wouldn't result in a much higher tiering/value of joules than it should be either, as distances for these kinds of striking feats usually aren't that large.

It would be very nice to have standards set in stone for these various kinds of LS feats. Currently I rarely evaluate them given that I don't know what we actually accept regarding them.

Yes.

 

[Flashlight237]

As is said in this year old thread, converting pressure (pascals, newtons commonly, though psi was being asked about) into energy typically means dividing the pressure by surface area (what's been described in this thread with shear strength/stress).

What?: https://prnt.sc/GwJXPLjekRw9

 

[StretchSebe]

As is said in this year old thread, converting pressure (pascals, newtons commonly, though psi was being asked about) into energy typically means dividing the pressure by surface area (what's been described in this thread with shear strength/stress).

What?: https://prnt.sc/GwJXPLjekRw9

I would personally convert pascals to Newton / Square Meter first for convenience, because you would then take the surface area of the section of the door that got bent (length and width likely, assuming the bend affects an area that is relatively rectangular), and divide the shear strength by the meters squared to get the force. I said earlier to convert to N/m^2 first because then you would just get the Newtons on their own by eliminating the m^2.

"The unit, named after Blaise Pascal, is a SI coherent derived unit defined as one newton per square metre (N/m2)."

Probably could have been more specific, and so could the year old thing, but one pascal is equal to one newton in value. You divide the newtons by square meters.

The pascal itself has the si base units of kg⋅m^−1⋅s^−2, compared to newtons having 1 kg⋅m⋅s^−2. Doing straight pascal/surface area(square meters) is going to produce a funky result with units.

Kind of like with the 6.6e+6 newtons-for-ap/striking strength thing, the wiki doesn't use pascals (or psi) for units of Lifting Strength. It does use newtons. The value of pascals would be equal to the value of newtons in number, but to be actually equal, the division of surface area would be done for newtons when it comes to units.

You wouldn't divide pascals themselves by surface area.

 

[StretchSebe]

Also, might be obvious, but: Force (Mass * Acceleration) is in Newtons, and shares the same SI base units of kg*m*s^-2

 

[Agnaa]

What is IRL humans having an LS of Class K in reference to? The car dent?

Yeah.

In any case, newtons aren't used for ap or striking strength, right? It's joules that are used. You find newtons through joules / meters. It's why I said one probably couldn't use the dent of a door to find the energy of the explosive yield for that feat Arceus0x brought up.

I'm assuming a few things in saying this, but in calculating and finding the 6.6e+6 you got (nice ), the value you found was probably in joules, not newtons, especially if it's a striking strength kick. That number in joules would seem to fall in line with the rest of the calculation blogs finding Wall level for similar dent feats.

As is said in this year old thread, converting pressure (pascals, newtons commonly, though psi was being asked about) into energy typically means dividing the pressure by surface area (what's been described in this thread with shear strength/stress). That should reduce the number upfront. Multiplying that by distance to get joules proper will likely increase the number, but again, that'd be to convert the pressure into energy, not the other way around.

It likely wouldn't result in a much higher tiering/value of joules than it should be either, as distances for these kinds of striking feats usually aren't that large.

"The unit, named after Blaise Pascal, is a SI coherent derived unit defined as one newton per square metre (N/m2)."

Probably could have been more specific, and so could the year old thing, but one pascal is equal to one newton in value. You divide the newtons by square meters.

The pascal itself has the si base units of kg⋅m^−1⋅s^−2, compared to newtons having 1 kg⋅m⋅s^−2. Doing straight pascal/surface area(square meters) is going to produce a funky result with units.

Kind of like with the 6.6e+6 newtons-for-ap/striking strength thing, the wiki doesn't use pascals (or psi) for units of Lifting Strength. It does use newtons. The value of pascals would be equal to the value of newtons in number, but to be actually equal, the division of surface area would be done for newtons when it comes to units.

You wouldn't divide pascals themselves by surface area.

This is just incorrect. Pascals are N/m^2. You need to multiply then by the area to get the Newtons, which then are exactly the units we use for Lifting Strength. I did this in the OP. As such, they will only be equal in number when the surface area is 1 m^2.

It makes intuitive sense too, since applying the same pressure to a larger area would require more force, and so the pressure should be multiplied by the area.

Also, your claims about 6.6e6 Joules being in line is absolutely false. That was for a real feat that a real life human really performed. Humans do not land in wall level. That's where my issue comes from.

Also, might be obvious, but: Force (Mass * Acceleration) is in Newtons, and shares the same SI base units of kg*m*s^-2

That unit is not the same. In force, the meters are to the power of +1, in newtons, the meters are to the power of -1.

 

[StretchSebe]

This is just incorrect. Pascals are N/m^2. You need to multiply then by the area to get the Newtons, which then are exactly the units we use for Lifting Strength. I did this in the OP. As such, they will only be equal in number when the surface area is 1 m^2.

It makes intuitive sense too, since applying the same pressure to a larger area would require more force, and so the pressure should be multiplied by the area.

I was talking about converting pascals to newtons. I see what you mean about multiplying by area though.

Yeah, you would multiply N/m^2 by m^2 to isolate and get the value in Newtons.

Also, your claims about 6.6e6 Joules being in line is absolutely false. That was for a real feat that a real life human really performed. Humans do not land in wall level. That's where my issue comes from.

I said it was in line with the rest of the calculation blogs. Those cover fictional feats. The IRL one was almost 7k joules, which you say yourself is only street level.

Idk, good for the guy for producing (potentially upwards of) 7k joules with a kick? Like damn in slow mo that's a good ass kick.

That unit is not the same. In force, the meters are to the power of +1, in newtons, the meters are to the power of -1.

Force: kg·m·s−2
Newton: 1 kg⋅m⋅s−2

Force finds an amount of newtons. A newton is a newton. This is just a semantics difference.

 

[Agnaa]

I said it was in line with the rest of the calculation blogs. Those cover fictional feats. The IRL one was almost 7k joules, which you say yourself is only street level.

Idk, good for the guy for producing (potentially upwards of) 7k joules with a kick? Like damn in slow mo that's a good ass kick.

That's not just good, that's unrealistic. Borderline street level+ is well above what we consider doable for humans. And as I said in the OP, that's a lowball of the feat. Using different assumptions for metal thickness, as some of our calcs do, would put it into wall level. And I expect that a more accurate measurement of the dent made, rather than just eyeballing it, would get there too.

But yeah, my bad for saying the wrong value. Still, I'd really prefer if I didn't have to spell it out again and again, risking getting it incorrect like that, when the information's already properly written up in the OP.

And as I said, your idea of just using force puts humans in Class K LS with just their kicks, which is ridiculous.

Force: kg·m·s−2
Newton: 1 kg⋅m⋅s−2

Force finds an amount of newtons. A newton is a newton. This is just a semantics difference.

Blech my bad, I mistyped. I meant to say that in pascals the meters are to the power of -1.

 

[Serlock_Holmes]

Umm so is the equation used wrong for those kind of stuff? Most cars nowadays cave easily to absord impact.

 

[Agnaa]

It gives results inconsistent with how we tend to rate things. Similar reasoning has been used to disallow the use of Potential Energy and Kinetic Energy at certain scales.

For those, the issue is that such energies are distributed across the entire body. I'm not sure where the exact issue here is. My guess would be that the experimentally-found "yield strength of steel" doesn't apply to extremely thin sheets, or that it doesn't apply when there is nothing supporting the back of the material.

The issue isn't with the actual equations themselves (except for the cases in the OP where they were applied improperly, which seems common with force > joule feats). They're true by definition of what those quantities mean.

 

[Serlock_Holmes]

It gives results inconsistent with how we tend to rate things. Similar reasoning has been used to disallow the use of Potential Energy and Kinetic Energy at certain scales.

For those, the issue is that such energies are distributed across the entire body. I'm not sure where the exact issue here is. My guess would be that the experimentally-found "yield strength of steel" doesn't apply to extremely thin sheets, or that it doesn't apply when there is nothing supporting the back of the material.

The issue isn't with the actual equations themselves (except for the cases in the OP where they were applied improperly, which seems common with force > joule feats). They're true by definition of what those quantities mean.

Probably that part not having any backing though cards do actually fold quite easily. Isn't the part basically plastic that covers the car? Sorry if I'm saying dumb stuff.

 

[Agnaa]

The frames of cars aren't made out of plastic. I wouldn't be too surprised if some had a tiny bit, but the bulk of it definitely wouldn't be.

 

[StretchSebe]

That's not just good, that's unrealistic. Borderline street level+ is well above what we consider doable for humans. And as I said in the OP, that's a lowball of the feat. Using different assumptions for metal thickness, as some of our calcs do, would put it into wall level. And I expect that a more accurate measurement of the dent made, rather than just eyeballing it, would get there too.

Ok we're getting really off topic but:

Isn't Street literally the last tier considered humanly possible?

Peak humans like Bruce Lee are 9-C. Beyond what the wiki considers examples of 'peak humans,' professional boxers punches average around like one thousand or so joules, which should scale to their durability (debatably, mainly through body shots). While very dangerous and damaging for them, normal people can survive attacks from weapons considered 9-C on the irl verse wiki like knives and baseball bats. Many examples are on the Peak Human Physical Characteristics page already.

I just don't see why it's so outrageous. Not everything in life falls into neat little boxes all the time. Let the one-off-car-dent man have his 9-C tiering.

But yeah, my bad for saying the wrong value. Still, I'd really prefer if I didn't have to spell it out again and again, risking getting it incorrect like that, when the information's already properly written up in the OP.

And as I said, your idea of just using force puts humans in Class K LS with just their kicks, which is ridiculous.

Just don't use Newtons for striking strength or ap? Like the wiki already doesn't do? Idk man it feels like you're asking me to explain why the apples aren't oranges.

Blech my bad, I mistyped. I meant to say that in pascals the meters are to the power of -1.

Fair enough.

 

[Agnaa]

Ok we're getting really off topic but:

Isn't Street literally the last tier considered humanly possible?

Peak humans like Bruce Lee are 9-C. Beyond what the wiki considers examples of 'peak humans,' professional boxers punches average around like one thousand or so joules, which should scale to their durability (debatably, mainly through body shots). While very dangerous and damaging for them, normal people can survive attacks from weapons considered 9-C on the irl verse wiki like knives and baseball bats. Many examples are on the Peak Human Physical Characteristics page already.

I just don't see why it's so outrageous. Not everything in life falls into neat little boxes all the time. Let the one-off-car-dent man have his 9-C tiering.

Those are all low-end in street tier. Not 2.1x away from 9-B.

And, I keep mentioning this but you're not addressing it; the calc I did in the OP was a lowball. Using values used in some of these calcs it would land in 9-B, and I am pretty confident that actually measuring the dent would land it in 9-B. Would I need to actually go through measuring that for you to respond to it?

EDIT: I just measured the dent in the OP, and it looks like my guess may have just been accurate. Assuming that car has the smallest tire size on this website since I cbf looking up the car model, it's an ellipse with radii of 8cm and 15.7cm, and so it ends up with an area of 0.0394 m^2, only 30% higher. It'd land in 9-C+, not 9-B.

Just don't use Newtons for striking strength or ap? Like the wiki already doesn't do? Idk man it feels like you're asking me to explain why the apples aren't oranges.

Our wiki already converts from Newtons to Joules a bunch. Changing that would require massive revisions.

And even if we don't do that, there is still the issue of those kicks resulting in Class K lifting strength.

 

[StretchSebe]

Those are all low-end in street tier. Not 2.1x away from 9-B.

And, I keep mentioning this but you're not addressing it; the calc I did in the OP was a lowball. Using values used in some of these calcs it would land in 9-B, and I am pretty confident that actually measuring the dent would land it in 9-B. Would I need to actually go through measuring that for you to respond to it?

EDIT: I just measured the dent in the OP, and it looks like my guess may have just been accurate. Assuming that car has the smallest tire size on this website since I cbf looking up the car model, it's an ellipse with radii of 8cm and 15.7cm, and so it ends up with an area of 0.0394 m^2, only 30% higher. It'd land in 9-C+, not 9-B.

The fictional feats all seem to be at 9-B, and converting the newtons to joules seems consistent in that.

I'm sorry the irl guy is too strong, I guess.

Our wiki already converts from Newtons to Joules a bunch. Changing that would require massive revisions.

And even if we don't do that, there is still the issue of those kicks resulting in Class K lifting strength.

We've already discussed converting newtons to joules earlier in the thread, and at this point multiple times. Yes:

Convert the Newtons to Joules if it's a striking/ap feat. Use Newtons if it's slow-twich fiber muscles and/or LS feat.

With all due respect, I would assume as an admin and calc group member you'd not need this repeated to you...