bercov@bevsun.bev.lbl.gov (John Bercovitz) (06/18/91)
[I put a Z in the subject as I said I might earlier. Z means it's a tech-
nical snoozer-topic which is fascinating only to masochists and the like.]
In article <35742@mimsy.umd.edu> DCROWE@GTRI01.gatech.edu writes:
#This discussion seems to be based on the assumption that supersonic shock
#waves are the mechanism for tranferring energy to the target.
I'm not sure if you're referring to one of my posts as I was trying to prove
that the opposite was true. Hope my explanation wasn't _that_ bad!
#Hydraulic pressure is the more common mode, and it accounts for the large
#damage channels that are observed to vastly exceed the bullet size.
That's an interesting statement. Could you elaborate on the subject?
I have an idea that your statement is similar to the following statement
I made in an earlier post:
"I see a bullet traversing the euphemism as primarily a momentum transfer
problem. This is because the forces involved in the momentum transfer
are much higher than the forces involved in destroying the structural
integrity of the euphemism. So I would say the wound channel is the
result of all the little dMs (that's dee masses) becoming sub-miniature
secondary projectiles. But I'm just guessing." .....and the secondary
projectiles incite tertiary projectiles, ad infinitum, ad absurdum.
For comparison with your line of thought, I'll expand on the above statement.
As is known from high-speed X-ray photos of ballistic gelatin, the nose of
a bullet is in contact with the leading edge of an evacuated cavity. So
surface drag is a minor character in this problem. That leaves momentum
transfer and structural breakage to account for.
Assuming it's an impulse-momentum kind of problem, we start with:
F*dt = m*dv
from which: F = mdot*dv
Where mdot is the rate at which euphemistic mass is being encountered by the
bullet's nose; this would be mathematically equal to the volume swept by the
bullet per unit time times the density of the material swept. Volume swept
per unit time is just bullet velocity times bullet cross sectional area. Now
dv would be equal to the velocity of the bullet because the mass encountered
by the bullet would be accelerated from 0 velocity up to the velocity of the
bullet regardless of what happens to that mass later. From this:
mdot = r*A*V where: r is density of the euphemism
A is nose area of bullet
V is the velocity of the bullet
dv = V
Combining these two into F = mdot*dv:
F = r*A*V^2
Well, I screwed up because there's supposed to be a 1/2 on the right side of
the equation. So sue me or find an elementary fluid dynamics text and look
up a _good_ explanation. Also, there's a form coefficient, C, which is
supposed to be on the right side. How it sposed to was:
F = 1/2 * C * r * A * V^2
But Carroll Peters, in his book, "Defensive Handgun Effectiveness", says
that the drag is proportional to V^1.5 or thereabouts over the range of
velocities of interest. I think that change in exponent somehow represents
the structural damage being done. In other words, F should really be
what we have above _plus_ a linear number for structural damage. (Well,
maybe a little non-linear to account for explosive deformation effects.)
I think that v^1.5 just approximates this truer analysis. V^1.5 would
then be just a curve fit but a very handy curve fit.
Anyone still awake? Aha! You sir - any comments?
JHBercovitz@lbl.gov (John Bercovitz)