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Old 02-13-2010, 04:10   #382
N/Apower
Senior Member
 
Join Date: Jan 2010
Posts: 585
Quote:
Originally Posted by glock20c10mm View Post
You're right, he doesn't get it. Maybe now he will. My fingers are crossed. Now I see he's posted back as I've quoted you.



N/Apower,

Here is some info that should help you understand and answer to your curiousness -

The equation for JHP handgun bullets with 100% mass retention is -
p = (5*E)/(pi*d)

p is the peak pressure wave magnatude on the surphase of a 1" diameter cylinder centered on the wound channel (in psi). E is the impact energy (in ft-lbs) and d is the penetration depth (in feet).

If a JHP bullet fragments then generally whatever % the bullet fragments is the same % you need to add to the PBPW originally figured for nonfragmentation.

For FMJ handgun bullets the equation changes to a reasonable approximation of -
p = (3*E)/(pi*d)

For FMJ rifle bullets there is much more variation because some tumble deep and some tumble at shallow depths and some fragment. The retarding force profile (the more retarding force the greater the PBPW) is dominated by the depth at which a FMJ rifle bullet tumbles.

An FMJ rifle bullet which does not fragment and tumbles late in the penetration (10" or more) will have a peak pressure wave comparable to the formula for FMJ pistol bullets.

An FMJ rifle bullet which does not fragment and tumbles early (first 4") will have a peak pressure wave comparable to the formula for JHP handgun bullets.

You might wonder why PBPW goes up with bullet fragmentation. This involves a bunch more math which I can post if you like, but I don't see that it's necessary. What I do understand is the basic principal which I believe will be simple for you also once you simply basically understand the basic equations above for equating PBPW.

If it is necessary for you, maybe this will help, and it's about as far into as I'ld prefer to get. If kinetic energy and penetration depth are equal, bullets that fragment create a larger pressure wave than bullets that retain 100% of their mass because the average penetration depth is shorter than the maximum penetration depth. Less penetration depth with equal kinetic energy = higher PBPW.


Also, not to rush you, but I did ask you specific questions you haven't answered in your last post. I'm hoping to hear from you on them unless you're simply acknowledging you were wrong by ignoring them.


Good Shooting,
Craig

I skimmed his paper when I first saw it, saw that there was a sample size of 10, noted that he included no real data, and threw it out as unacceptable work based on lack of data/information. It would have gotten an "F" in any intermediate college class he chose to turn it during.

When you asked your question, I half-heartedly went looking for it, and didn't find it, and didn't worry about it.

The numbers above would have you believe that the answer to the problem is the shotgun loaded with bird-shot at close range. However, real-world use of this proves it's fallacy.

On the same hand, it would make one think that hard-cast solids are horrible bullets to use on nasty animals like cape buffalo, etc--yet we know that this is not the case.

Sure, BPW exists, but what does it DO? Whatever it does (or doesn't) do, it's mighty un-predictable. I would much rather choose my tools based on something predictable, like penetration and expansion.

Given the numbers/percentages of probability of BPW, there is a lot wrong with it if you review some OIS's and look at how people kept trucking after multiple shots.

This sticky from this forum shows as much.
http://www.fbi.gov/publications/leb/...eb.htm#page_15

You can talk about BPW all you want, throw all the equations out there you want, etc. but that doesn't make it any more effective.

Last edited by N/Apower; 02-13-2010 at 04:15..
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