Force for tearing an arm off?

[Naito-desu]

Title, trying to check but we seem to only have spines in the references for common feats. Bonus as well if I can also calculate the force of tearing off armored arms.

 

[KLOL506]

I think we already have another thread like this in the Calc Group Discussion thread and a few more similar ones.

But to answer it bluntly, we actually haven't found a concrete value yet. I myself have been looking all over the internet for a few years for research papers on the matter, no dice.

There was this one 2200 lbs of force value from Quora but it's unsourced.

 

[Naito-desu]

I found this random calc on a forum as well

 

[KLOL506]

I found this random calc on a forum as well

Checked it just now but the results are simply too varied with too many variables.

 

[Flashlight237]

There's also this thread from Stackexchange: https://scifi.stackexchange.com/questions/129979/wouldnt-thor-have-ripped-this-persons-arm-off

One of the calculations featured horses, which were actually used to tear limbs off: case in point, the "Hanged, Drawn, and Quartered" execution.

I do agree that there are many variables to account for, but Class 5 seems to be the acceptable strength rating.

 

[KLOL506]

I guess that could work.

Do note that simply dislocating arms from their shoulder sockets without tearing off the limb won't qualify for this (IIRC that has its own lifting strength value but I think feats for pulling arms off sockets on their own should also be fine), you need to completely rip off the arm from the body, bone+muscle and all that sweet, sweet gore.

Now then, what are the values for ripping legs off?

 

[Flashlight237]

I guess that could work.

Do note that simply dislocating arms from their shoulder sockets without tearing off the limb won't qualify for this (IIRC that has its own lifting strength value but I think feats for pulling arms off sockets on their own should also be fine), you need to completely rip off the arm from the body, bone+muscle and all that sweet, sweet gore.

Now then, what are the values for ripping legs off?

If we just go off the horse thing, more or less the same.

 

[KLOL506]

Bump

 

[Migue79]

There's this document from the University of Leicester I found. I used it in one of my tearing arms calc.

 

[KLOL506]

There's this document from the University of Leicester I found. I used it in one of my tearing arms calc.

IDK about this, the 2006 N (204 kgf) value seems too low, this assumes that someone like Halfthor Bjornson could rip out an arm by himself, which, even for a record-breaking strongman like him, I'm pretty sure is impossible.

 

[Furudo_Erika]

IDK about this, the 2006 N (204 kgf) value seems too low, this assumes that someone like Halfthor Bjornson could rip out an arm by himself, which I'm pretty sure is impossible.

I have to agree on this.
I actually seen an article a couple years ago where it would take 30kN to 200kN to rip off a human's arm (Quite a large margin, I know)

I forgot where the article was at, but this would be around 3,059 kgf to 20,396 kgf.
Averaging around 11,726 kgf, which sounds a lot more reasonable than 204 kgf.

So, Class 5 to the high-end of Class 25
Quite low into Class 25 for the average.

 

[Migue79]

IDK about this, the 2006 N (204 kgf) value seems too low, this assumes that someone like Halfthor Bjornson could rip out an arm by himself, which, even for a record-breaking strongman like him, I'm pretty sure is impossible.

I meant the value of tensile strength value, given it's sourced. The article seemingly ****** up the conversion of cross-sectional area though: it says it's a surface area of 3.14e-4 m2, but I got 3.14e-2 m2 which then gives a force of 200,646 Newtons or 20,453.211 kgf (Class 25).

 

[Furudo_Erika]

I got 3.14e-2 m2 which then gives a force of 200,646 Newtons or 20,453.211 kgf (Class 25).

That makes a lot more sense, and is consistent to the 200kN requirement.

 

[KLOL506]

I meant the value of tensile strength value, given it's sourced. The article seemingly ****** up the conversion of cross-sectional area though: it says it's a surface area of 3.14e-4 m2, but I got 3.14e-2 m2 which then gives a force of 200,646 Newtons or 20,453.211 kgf (Class 25).

bruh

 

[KLOL506]

I meant the value of tensile strength value, given it's sourced. The article seemingly ****** up the conversion of cross-sectional area though: it says it's a surface area of 3.14e-4 m2, but I got 3.14e-2 m2 which then gives a force of 200,646 Newtons or 20,453.211 kgf (Class 25).

Wait, how did they **** up the conversion?

 

[KLOL506]

Oh yeah, right, they screwed up royally in trying to converting 10 cm to meter. 10 cm being half that of the 20 cm diameter of the arm. 10 cm is 0.1 meter.

pi*(0.1)^2= 3.141592653e-2 m^2

I knew something was wrong with the article.

 

[Migue79]

That's an embarrassing mistake for an University to make, let me tell ya.

 

[KLOL506]

Funny thing is, not even a legion of four horses is enough to reliably dismember a person, they still had to make cuts in the joints to make it easier after the horses failed from multiple attempts.

 

[KLOL506]

That's an embarrassing mistake for an University to make, let me tell ya.

Never would I ever have guessed something like this to happen to me, ever.

 

[Flashlight237]

There's this document from the University of Leicester I found. I used it in one of my tearing arms calc.

I would've commented my own opinion on the document, but I think everyone else covered it for me.

 

[KLOL506]

I would've commented my own opinion on the document, but I think everyone else covered it for me.

Pretty much, except the document royally screwed up in converting cm to m, and we had to fix it.

 

[KLOL506]

Bump

 

[KLOL506]

Bump

Should we make a blog for the whole arm-ripping thing and note down the error they made in their conversion?

 

[Migue79]

Sure.

 

[KLOL506]

Sure.

I'll let you take the honors then. You can contact a bunch of calc members to then look at it if you wish.

Once the blog is accepted, it should be golden to add to the References for Common Feats page.

 

[EliminatorVenom]

Checked it just now but the results are simply too varied with too many variables.

To be honest, most calcs are like that. In VS, most calculations are oversimplified.

 

[KLOL506]

To be honest, most calcs are like that. In VS, most calculations are oversimplified.

This was actually before we found out about the pdf that Migue provided us with.

 

[Migue79]

Alright. It's done.

@Executor_N0 @Spinosaurus75DinosaurFan @Mr._Bambu @Therefir @Ugarik @DMUA @Damage3245 @DemonGodMitchAubin @Jasonsith @Wokistan @KieranH10 @Armorchompy

Would you all be willing to evaluate/give feedback on this blog? This is to be added to the References for Common Feats page.

 

[Ugarik]

This is an extreme high end because in reality arn is only attached by few tendons and the muscles are quite thin were they attached to the bone.
Even using this oversimplified model, 20 cm is too wide

 

[Ugarik]

Funny thing is, not even a legion of four horses is enough to reliably dismember a person, they still had to make cuts in the joints to make it easier after the horses failed from multiple attempts.

It's not as impresive as it seems. If you draw a free body diagram you can see that there's only 1 horse per limb. And a single horse can only pull with as much force as the friction with the ground provides.
Assuming a horse weights 1000 kg and the friction coefficient is 0.5 you only have 500 kgf per arm at most

 

[Flashlight237]

It's not as impresive as it seems. If you draw a free body diagram you can see that there's only 1 horse per limb.

In which case you gotta account for net force. In this case, four forces are acting on one object. The idea of four forces canceling each other out like that, depending on how you look at it (whether it be one on each corner or two on each side), is impressive.

Even if we were to go off 500 kgf, the body would effectively be withstanding 2000 kgf at the center (and yes, the center does matter in physics).

 

[KLOL506]

In which case you gotta account for net force. In this case, four forces are acting on one object. The idea of four forces canceling each other out like that, depending on how you look at it (whether it be one on each corner or two on each side), is impressive.

Even if we were to go off 500 kgf, the body would effectively be withstanding 2000 kgf at the center (and yes, the center does matter in physics).

BASED

BTW, I did a rough calc using 7 cm arm diameter, that's 3.5 cm or 0.035 m radius based on my measurement (Pls don't blame me, I'm a thin and malnourished guy), and using Migue's calc's formula from the PDF above...

Area: 0.00385 cm^2

Force: 0.00385 m^2 * 6.39e6 N/m^2= 24601.5 newtons or 2508.654841 kgf (Class 5)

 

[Dark-Carioca]

So ripping an arm off requires somewhere between 2.5 and 25 tonnes of force? That's quite the gap

 

[KLOL506]

So ripping an arm off requires somewhere between 2.5 and 25 tonnes of force? That's quite the gap

2.5 tonnes for a weak-ass dude like me, and I'm not the average dude. Normal average dudes have wider arms than me and will prolly need more oomph there.

 

[Dark-Carioca]

So how would this be applied to all the characters who have done this? "At least Class 5 to Class 25"?

 

[Flashlight237]

I do find it odd how the other Calc-Groupers hadn't touched on this yet.

 

[Naito-desu]

So how would this be applied to all the characters who have done this? "At least Class 5 to Class 25"?

Well if the dude they've torn the arm off was clearly not any normal dude, Class 25 is good enough

 

[KLOL506]

bump

 

[Cropfist]

What is the energy in AP?

 

[KLOL506]

What is the energy in AP?

Couldn't say, we'd need the distance the work is being carried out on. And I haven't the foggiest idea of where that distance would be measured in the arm joints.