From: bercov@bevsun.bev.lbl.gov (John Bercovitz)
Newsgroups: rec.guns
Subject: Re: aluminum bullets
Date: 16 Aug 92 22:38:41 GMT
Organization: Lawrence Berkeley Laboratory, California

In article <9208141822.AA12726@dog.ee.lbl.gov> bercov@bevsun.bev.lbl.gov 
(John Bercovitz) writes:

#In article <1992Aug14.043707.16254@news.eng.convex.com> cash@convex.com 
#(Peter Cash) writes:

##In the latest issue of the Atlanta Discount Ammo flier, there's some weird
##.38 special ammo listed: a 33 grain HP _aluminum_ bullet. The manufacturer
##(American Ballistics) claims a muzzle velocity 2,750 fps. This stuff is +P.
##Weird. Can anybody figure out what the terminal ballistics of such a round
##might be like? Would you get any penetration? 

#Let's see, 33*2750*2750*2.22*10^-6 = 554 ft-lb.  That sounds like it would
#be a little hard on a standard 38 revolver.  Wonder what sort of pressures
[stuff deleted]
#As far as the ballistics go, one way of finding ballistic coefficient
#is to calculate it, which you can do if you know the form factor, i.
#The formula is C = m/(i*d*d) (Nope, they ain't no pi in there). m in lbs.
[stuff deleted]
#the form factor for a cavitied* bullet like that must be at least 1.5
#so for the 33 gn bullet, C = m/(i*d*d) = (33/7000)/(1.5*.357*.357) = 
#.025 or less.  That's pretty pathetic.  As I mentioned in an earlier
[stuff deleted]

On the way home last night I got to feeling a little guilty about maybe 
waffling a little on my answer to Peter Cash's question.  In that article,
I concluded, more or less, that the terminal characteristics of the "light-
speed spitwad" were bound to be lousy because it would be impossible for 
the most brilliant bullet designer with unlimited design and test funds to 
build a bullet which would perform properly over the enormous range of 
impact velocities which this bullet allows (from 2750 fps down to 1200 fps 
over a distance of 0 to 50 yards).  Let me try to make amends.

My reference for this is Carroll Peters' book, "Defensive Handgun Ballistics", 
a book which primarily deals with terminal ballistics.  Peters finds, through 
curve-fitting some experimental data, that the depth of penetration of a 
bullet in ballistic gelatin is as follows:

                          Pgel = (10*m*V^0.5)/(A*i)

and that the penetration of soft pine is:

                          Pwood = E/(350*A*i)

where m is the weight of the bullet in pounds, V is the velocity of the bullet 
in fps, A is the cross-sectional area of the bullet in square inches (0.1 in^2 
in our case), i is the form factor as in the previous article, and E is the 
bullet's kinetic energy in foot-pounds.  P is the penetration in inches.  From 
these two equations we see that the penetration is proportional to the 
square root of the velocity in a soft target and the square of the velocity in 
a harder target.  I wonder if soft pine is at all analogous to bone.

In the present case, Peter Cash's question is best answered by the first 
equation.  Let us assume that this aluminum hollow point expands just like 
an ordinary 9 mm 115 JHP.  In other words, our bullet expands to the same 
diameter and configures itself to have the same form factor as the 9 mm 
bullet does.  The Ai product of an expanded 9mm is .44 (from an example in 
Carroll Peters' book) and so I will assume it to be for the aluminum bullet.

A little digression: I looked up the velocity of our .025 BC aluminum bullet 
at 25 and 50 yards in Ingalls' tables to see how I did with my scaling to the 
165 gn 308 bullet yesterday.  Turns out I wasn't too far off.  The correct 
values are 1905 fps and 1287 fps at 25 and 50 yds.  I'll use these new 
values below.  I also looked up the ballistic coefficients of some pistol 
bullets and scaled them to our bullet's weight.  From that, the BC of the 
aluminum bullet must be in the range of .025 to .030.  Using .025:

at 0 yards, V = 2750 fps, and:
Pgel = (10*m*V^0.5)/(A*i) 
     = (10*(33/7000)*2750^0.5)/.44 = 5.6 inches.

at 25 yards, V = 1905 fps, and:
Pgel = (10*m*V^0.5)/(A*i)
     = (10*(33/7000)*1905^0.5)/.44 = 4.7 inches.

at 50 yards, V = 1287 fps, and:
Pgel = (10*m*V^0.5)/(A*i)
     = (10*(33/7000)*1287^0.5)/.44 = 3.8 inches.

So in fact this bullet has lousy penetration at any of its speeds if it is an 
expanding hollow point.  But what if it's a fake and it doesn't expand?  Then 
the Ai product would stay at 0.1 in^2 * 1.5 = .15.  What would the penetration
be then?  We can see that the penetration scales linearly with the Ai product
so the penetration will be .44/.15 = 2.9 times as great in all cases:

 0 yards: Pgel = 16.5 inches
25 yards: Pgel = 13.7 inches
50 yards: Pgel = 11.2 inches

This might almost be acceptable: hope that this hollow point doesn't expand 
if you have to use it.   8-)

John Bercovitz     (JHBercovitz@lbl.gov)