Discussion on answer by Green: What kind of natural armor would stop bullets?

Feeds

03:34

66

A: What kind of natural armor would stop bullets?

Mantis Shrimp Claws will do quite nicely. Carbon is extremely abundance on Earth's surface and presumably any other Earth-like planet that the author may be working on. Given the many forms that carbon can take from the ultra soft graphite to the ultra hard diamond, it should be able to satisfy...

OneSurvivor

Green, you are a legend. This is in the running for best answer.

Green

@OneSurvivor You're very kind. I wouldn't go so far as legendary though.

@OneSurvivor I am very happy with the recent addition to get light and heavy versions with just a slight modification in how resilient the armor is to oxygen degradation.

anon

I would accept this as a truly wonderful answer if it didn't smell fishy (lol). If its armor is strong enough to survive more pressure than the Earth's core how is it these guys die at all. (brb gonna do some fact checking)

Green

@anon mantis shrimp only have two small patches on their body that are this hard. Disease, old age, predators, etc are plenty to prevent mantis shrimp from living forever.

Frostfyre

@OneSurvivor Unfortunately, Green is not yet legendary, though s/he is doing well at becoming epic.

Green

03:34

@Frostfyre true. True.

Ville Niemi

Is it my imagination or is there something weird with the pressure equation part?

KalleMP

+1 for the wonderful image. It is certainly a neat way to get extra strength out of what may be more mundane materials. Nature is pretty inventive.

Green

@VilleNiemi I think there's some units weirdness. I confess I was in a rush when I wrote that part.

Ville Niemi

Okay, as long as it isn't my brain breaking down it is fine.

Vashu

Would be a perfect answer if there was estimation of bulletproof shrimp armor thickness ;)

starrise

03:34

Cool answer, I learned some stuff about mantis shrimps! I'd caution that high velocity impact behavior is not readily calculated using back-of-the-envelope type calculations, so the math may be misleading. That said, you could estimate using a static energy condition and an equivalent mass resting on the armor. I estimate the bullet mass, at a height of 0.1 in with all energy as gravitational potential energy, at 50kg, which exerts about ~500 kN on its tip only (~5% of the xsection), giving just under 400 MPa pressure.

OneSurvivor

@starrise What is the best way to calculate this kind of behavior? I would like a link to a website to learn about this kind of stuff.

Mike M

@starrise is on point ;) you can't directly equate energy and pressure..... add a realistic heuristic for that step

starrise

Your best bet for a reasonably accurate answer, as is my understanding as a materials engineer, is to use an FEA simulation with realistic material properties and boundary conditions. There isn't really a one-shot answer for this kind of problem. If you watch this video on youtube it might give you an idea why the math isn't straightforward. Basically, the interactions aren't rigid and the bullets start acting like liquids due to the extreme speed of shape changes. My estimate comes from high-school physics using PE = mgh, and so forth.

Alex Stasse

Double check your answer, you've mixed units in your pressure equation. Joules where you need Newtons. To get the answer you want you need to figure out the cross section of the bullet, you can't figure it out just from kinetic energy alone.

Scott Milner

Well, I think your calculations are a bit off. Here's some corrections: $KE=\frac{1}{2}mv^2$, $\Delta E=F\parallel d$, therefore $F=\frac{\Delta E}{d}$; Assume $E_0=0\text{ J}$, therefore $F=\frac{mv^2}{2d}$. $P=\frac{F}{A}$, $A=\pi r^2$, therefore $P=\frac{mv^2}{2d\pi r^2}$, (cont.)

(cont.) where $P$ is the pressure in Pascals, $m$ is the mass of the bullet in kilograms, $v$ is the muzzle velocity in meters per second, $d$ is the distance the bullet travels in meters, and $r$ is the radius of the bullet. With all that, we get that the pressure exerted on the armor is $\frac{.004\cdot990^2}{2\cdot d\cdot\pi\cdot0.00285^2}\approx \frac{77}{d}$ Megapascals (MPa). The farther away the gun, the less pressure exerted. With simple algebra, we can find that you would have to fire from just under 2 cm away to reach a breaking pressure of 4 GPa.

That doesn't take into account @starrise 's comment about non-rigid interactions and such, but it should be a more accurate method as far as taking into account the distance fired and getting a distance where it breaks down. (It looks like you got the correct coefficient).

OneSurvivor

03:34

@starrise Do you know if the calculations are generally skewed up from the calculations Green or Scott Milner provided (i.e. its easier to pierce armor), or down (i.e. its harder to pierce armor)?

Mark H

@ScottMilner The variable $d$ is the distance the bullet travels through the armor, not the distance to the shooter. Stopping the bullet in less distance requires greater pressure. So, the armor has to be at least 2cm thick to stop the bullet before breaking pressure is reached.

Green

@ScottMilner thank you very very much for the corrections.

Ville Niemi

@OneSurvivor If the bullet breaks or deforms and the armor does not, the energy needed for that is away from penetrating the armor. Deforming will also increase the surface area of the impact. Same would be true for some deformations of armor and ways of armor breaking, but not all. But I think it is better here to assume the armor is strong enough these effects can be ignored. I am not an expert or even well informed on this so I should be seeing it as simpler than it is, and it still seems like a "needs to be simulated" problem.

Green

@VilleNiemi I'd say "needs to be tested vigorously".

Ville Niemi

@Green Yes, but after simulating. But you are of course correct. Just simulating it wouldn't really be enough.

R..

03:34

This answer is incomplete without a citation of TheOatmeal as the reason anyone knows there's such a thing as a mantis shrimp.

starrise

@OneSurvivor I'd agree with Ville Niemi and add that it probably depends partly on a combination of strength and ductility of what's being impacted, at the appropriate strain rates. If the material is both strong and ductile, like steel, it will absorb a lot of energy and not deform much. If the material is only strong, like ceramic or glass, it will tend to shatter. If the material is only ductile, like many polymers, it will tend to deform out of the way of the bullet. I don't know enough about mantis shrimp material to know what will happen, unfortunately. It also depends on thickness...

... and a thin, flexible sheet would probably fare better than a plate. Think of a bullet-resistant vest. It's made from tightly-woven kevlar fibers that catch the bullet, rather than try to rigidly stop it like armor. If the kevlar were bound up in a rigid matrix, the bullet might just shear the whole material, fibers and all, and punch a hole through the entire thing.

Scott Milner

@MarkH Whoops. Thanks. You're completely right.

Monica Cellio

Are any of these comments obsolete now?

Green

@MonicaCellio many of them.

fectin

@MarkH not even that, unfortunately. You're looking at it moving 2 cm while stopping. To be undamaged, you'd need to look at stress/strain curves for mantis material and know the specific shape and specific deformation, or have enough margin that it didn't matter. For ablative armor it's even harder.

Mark H

03:34

@fectin Yeah, the kinetic energy to calculation is an order of magnitude estimate at best.

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Discussion on answer by Green: What k…