[LiLBucket]

I'm looking for a good rundown of bullet forces upon impact. Specifically I'm looking for various calibers in:

Force
-lb
-Newtons

or

Pressure
-psi
-N/m^2


This is completely disregarding deformity, deflection, etc etc etc. For quick FEA so just simple numbers is all that's needed.

 

[Zuki Tyler]

Wikipedia has a lot of data compiled for each caliber from manufacturers.

You have to search each round out, though.

 

[TJGreg5]

might help if you know bullet weight and velocity:

http://www.firearmexpertwitness.com/customguns/calcnrg.html

Edit, wiki did have some generic info:
http://en.wikipedia.org/wiki/Muzzle_energy

 

[LiLBucket]

All of what I'm seeing seems like muzzle energy as opposed to impact forces. I supposed I could back into the figures I'm after, but it seems like this has to have been done before no?

And not only the bullet impact forces, properties of AR500 are a bitch to find and convert

 

[TJGreg5]

All of what I'm seeing seems like muzzle energy as opposed to impact forces. I supposed I could back into the figures I'm after, but it seems like this has to have been done before no?

And not only the bullet impact forces, properties of AR500 are a bitch to find and convert

Using the link to the calculator i posted you should be able to figure it out. As long as you know the velocity at impact (tables should be easy to find) you'll have it.

 

[Weasel]

Liar no such thing as a quick fea. :flipoff2:

 

[Quick & Dirty]

I think that's going to be an impulse momentum calculation.

F ∙ t = M ∙ V

F = M ∙ V/t

t = time, which is how long it takes from the bullet first touching the target, until it comes to a complete stop. That's going to be tricky to figure out. ( High speed photography?)

The harder the bullet, the less it will deform, so time will be shorter, resulting in higher forces. A soft bullet that splatters will take longer to deform and result in much less force. I don't think you can ignore deformation and get a meaningful result. If you assume the bullet and target have no deformation, t = 0 and force becomes infinite.

 

[Al Kaholick]

You cannot calculate force without considering the object being hit. For example, a steel plate would experience a larger force than a log of jello because the force of impact would occur over a longer duration in the jello allowing for less peak force while stopping the bullet. Pressure is just force over area so the problem remains.

I'm looking for a good rundown of bullet forces upon impact. Specifically I'm looking for various calibers in:

Force
-lb
-Newtons

or

Pressure
-psi
-N/m^2


This is completely disregarding deformity, deflection, etc etc etc. For quick FEA so just simple numbers is all that's needed.

 

[crtbc]

depends also on distance from gun.... longer distance, less impact ft/lbs

 

[DavidVanVorous]

Liar no such thing as a quick fea. :flipoff2:

More specifically ones that resolve eventually and ones that dont...

D.