Velocity vs. bullet weight vs. penetration
Jeff Olsen
Why would a heavier bullet going slower, penetrate further than a lighter
bullet going faster, all else (diameter, etc) being equal? For this
example, let's use loadings for 30-06 that generate equal ft/lbs at
impact. Folks talk about shooting, say, 180-grain bullets from the 30-06
for better penetration, but why exactly does this penetrate better than a
150-grain bullet going a little faster?
Thanks!
-jeff
"Daddy, sometimes I wish I wasn't made of meat"
-daughter Savvy, 6, on a walk in our woods, after I showed her a cougar
killed deer...
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Mark Yaworski
#I am confused about the physics of bullet penetration.
#
#Why would a heavier bullet going slower, penetrate further than a lighter
#bullet going faster, all else (diameter, etc) being equal? For this
#example, let's use loadings for 30-06 that generate equal ft/lbs at
#impact. Folks talk about shooting, say, 180-grain bullets from the 30-06
#for better penetration, but why exactly does this penetrate better than a
#150-grain bullet going a little faster?
It all gets very complicated because many variables are involved but
let's look at things simply.
Basic physics tells us that F=ma, that is Force equals mass times
acceleration. Now, in the most simple model, the Force or resistence
that the bullet meets is proportional to cross sectional area so we
can argue that all .30 caliber bullets are equal.
Now we reduce the mass of the bullet so for the equation to remain
balanced, the accleration must go UP because Force is the same in all
cases.
More acceleration means that the bullet loses velocity more quickly.
So, without spending a lot of time crunching number, it is very
conceivable that a lighter, faster bullet would slow down more quickly
and penetrate less.
To look at something we've all done, throw a baseball and a tennis
ball. They are about the same size but you can throw a baseball much
farther.
Trenton G. Twining
#
<SNIP>
# Why would a heavier bullet going slower, penetrate further than a lighter
# bullet going faster, all else (diameter, etc) being equal? For this
# example, let's use loadings for 30-06 that generate equal ft/lbs at
# impact. Folks talk about shooting, say, 180-grain bullets from the 30-06
# for better penetration, but why exactly does this penetrate better than a
# 150-grain bullet going a little faster?
#
The less a projectile masses, the larger proportion of its energy is
lost against any resistance. Therefor a light projectile slows at a
much greater rate than a heavy projectile. Since this includes the
resistance of air, you'll see in a ballistics table that a heavy
projectile will retain velocity much better than a light one over
distance.
--
Trenton G. Twining
Personal:
E-mail t...@usa.net
FAX 561-619-7942
URL: http://homes.acmecity.com/rosie/singing/181/
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Frank Logullo
impact energy (mass times velocity squared).
I believe this is the argument that archers use to explain why slow moving
arrows with puny impact energy can completely penetrate an animal.
Frank
Dennis Moore
Look at the book available from this web site. It thoroughly
explains all of this. It will matter whether or not the bullet
is an expandable bullet like a soft point or hollow point. But
with the same expansion or no expansion the heavier, slower
bullet will penetrate more. Within reason that is.
"Bullet Penetration" is written mainly about handgun calibers,
but a good bit applies to rifle bullets as well.
Dennis
"Jeff Olsen" <je...@efn.org> wrote in message
news:9dtt9j$c28$1...@xring.cs.umd.edu...
P. Roza
#
# Why would a heavier bullet going slower, penetrate further than a lighter
# bullet going faster, all else (diameter, etc) being equal? For this
# example, let's use loadings for 30-06 that generate equal ft/lbs at
# impact. Folks talk about shooting, say, 180-grain bullets from the 30-06
# for better penetration, but why exactly does this penetrate better than a
# 150-grain bullet going a little faster?
#
# Thanks!
#
# -jeff
#
# "Daddy, sometimes I wish I wasn't made of meat"
#
# -daughter Savvy, 6, on a walk in our woods, after I showed her a cougar
# killed deer...
Sectional Density is a bullet's weight, in pounds, divided by the square of
its' diameter, in inches. The higher the sectional density, the better the
penetration at a given velocity.
Your confusion stems from asking the wrong question. A 180 grain bullet of
..30 caliber, fired from a 30.06, will penetrate deeper than a 180 grain
bullet of .429 caliber, at a given velocity, because of the greater
sectional density of the former.
A 150 grain non-expanding bullet, pushed to a sufficiently higher velocity,
will equal the penetration of a 180 grain non expanding bullet, with about
the same level of energy. When an expanding bullet strikes, however, the
effect of expansion exponentially decreases sectional density, so a
fractionally lighter bullet at a fractionally higher velocity is more
greatly affected.
bevnsag
Jeff Olsen wrote:
#
# I am confused about the physics of bullet penetration.
#
# Why would a heavier bullet going slower, penetrate further than a lighter
# bullet going faster, all else (diameter, etc) being equal? For this
# example, let's use loadings for 30-06 that generate equal ft/lbs at
# impact. Folks talk about shooting, say, 180-grain bullets from the 30-06
# for better penetration, but why exactly does this penetrate better than a
# 150-grain bullet going a little faster?
#
Well, first, that isn't necessarily so. But as far as it goes,
faster/lighter bullets tend to dump their kinetic energy quicker than
larger/slower bullets. In extreme examples, very high velocity/light
bullets can generate considerable shock damage but have limited through
penetration as their "carry" below x-velocity is quite limited, while
big slow slugs still have considerable "carry" even at very low
velocity. The 150/180 gr example isn't actually much of a difference,
it's the 50/500 gr comparisons that really get dramatic.
Mike in Oregon
Jeff,
How are you doing Jeff?
Boy did you open a can of worms with this post! This is the age old question
that we've yet to fully understand. Little testing has been done to really
reduce this question to a matter of mathmatics. But one of the best
predictors of stopping power when it comes to game is the Taylor Knock Out
Formula (the revised version includes sectional density). Of course when
Pondo came up with this formula, he was refering to African game. But we can
look to his data on lesser & greater kuda and various antelope for comparison
to our game. But even the TKO formula doesn't take bullet performance into
account and this is where things get muddy real quick.
Here are my thoughts:
Bullet expansion is proportional to rate of energy transfer.
Bullet velocity squared is proportional to total amount of energy.
Amount of total energy avaliable is proportional to bullet mass & velocity
squared.
Bullet construction is proportional to penetration depth.
Bullet expansion is inversely proportional to penetration depth.
Bullet velocity is inversely proportional to penetration depth.
Therefore bullet expansion is proportional to both velocity and rate of
energy transfer, but inversely proportional to penetration.
w/ chamber pressure constant, if mass goes up then velocity must go down.
w/ equal bullet construction, if velocity goes up so does expansion, but
penetration goes down.
Velocity = expansion. Expansion = rate of energy transfer. Bullet mass =
penetration.
but Bullet mass is inversely proportional to Velocity. Since velocity equals
expansion, it goes down.
So two things cause the slower, heavier bullet to penetrate deeper. More mass
& less velocity.
The trick is to have enough total energy at target, with enough expansion to
tranfer that energy over a penetration depth equal to the animal's width
(including a few bones).
A good bullet will be constructed in a manner that allows expansion to
increase as velocity increases, so that penetration remains constant. Some
bullets are constructed so poorly that they simply go all to pieces on impact
at high velocities. That's why we have bullets of differant constructions for
differant types of game. The bullet contruction that is good for whitetail is
a poor choice for elk or moose, and vice versa. It's to bad we don't have a
EC, expansion coefficent to describe bullet construction that one could use
in conjuntion with impact velocity to predict penetration in soft tissue. But
then we'd all be arguing over the definition of "soft tissue". To bad we
couldn't develop a standard EC test using wet newsprint that we could use to
evaluate the penetration characteristics of differant bullet constructions,
bullet diameters, impact velocities, and bullet weights equally. If we could
we may be able to predict how much of a bullet's total impact energy is
imparted to the animal before the bullet exits the animal. Some bullets like
the Barnes-X and Failsafe bullets would seem to hit like the hammer of Thor,
given the light bullet weight & super high velocities, but infact impart very
little energy to the target because they don't expand quickly enough to
tranfer that energy and create that temporary shock wave that seems to spell
lights out for much of north america's game.
Some food for thought,
Mike in Oregon.
NW Shooters Support
Beaverton, Oregon
"Mike in Oregon"<Mike.W....@tek.com>
Ballistics Consulting & Pressure testing
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FBC3
#Why would a heavier bullet going slower, penetrate further than a lighter
#bullet going faster,
Momentum. 150 grain Sierra softpoints from my 30-06 will dent a steel plate at
100 yards that Sierra 180 grain softpoints punch right through. Fred
fl...@alaska.net
#Why would a heavier bullet going slower, penetrate further than a lighter
#bullet going faster, all else (diameter, etc) being equal? For this
#example, let's use loadings for 30-06 that generate equal ft/lbs at
#impact....
There will be ots of answers here, but perhaps the easiest to
understand is that the bullet's momentum has more influence on
penetration than energy. Using Remington factory data for the .30-06:
150 @ 2910 fps / 2820 fpe / 1.94 lb-sec momentum
180 @ 2700 fps / 2913 fpe / 2.16 lb-sec
220 @ 2410 fps / 2837 fpe / 2.35 lb-sec
There can be other factors involved too, such as bullet
expansion/construction, bullet shape,sectional density, etc. But the
simplest one, and one which fits existing physical laws, is momentum.
Jay T
Jim Grunewald
All things being equal (except for speed and mass of the bullet), I believe
it is an issue of momentum:
M = mv2 (mass times the square of the velociy)
You don't provide velocities, but I would not expect the difference in
weight to make up for the difference in velocity. I assume that the spin
rates of the bullets are the same. I would expect that a faster spinning
bullet would penetrate a little better than a slower spinning one.
Are you really talking about penetration or stopping power? They are two
different things.
Having said all that, somebody out there might know something I'm missing
and can set us both straight. (Which would be a good thing since my head is
starting to hurt - trying to remember my engineering from 25 years ago!)
Take care,
Jim
Ron Seiden
easier to accelerate a light bullet, it also decelerates easier, while the
heavier slug tends to keep moving. Think of big heavy car vs. light car.
"Jeff Olsen" <je...@efn.org> wrote in message
news:9dtt9j$c28$1...@xring.cs.umd.edu...
# I am confused about the physics of bullet penetration.
Sue and Mark
#
# Why would a heavier bullet going slower, penetrate further than a lighter
# bullet going faster, all else (diameter, etc) being equal? For this
# example, let's use loadings for 30-06 that generate equal ft/lbs at
# impact. Folks talk about shooting, say, 180-grain bullets from the 30-06
# for better penetration, but why exactly does this penetrate better than a
# 150-grain bullet going a little faster?
Some materials seem to be penetrated by momentum and others by energy. Hard
things, like metal and wood, seem to require energy for penetration. In this
case, the bullet with the most impact energy (in a given caliber) will
penetrate farther. This usually means velocity is what is needed, because
energy is based on velocity squared (e =1/2mv^2). However, velocity is
quickly lost with distance, so you need to determine at what range you want
maximum penetration and pick the appropriate bullet weight and shape. At a
few feet from the muzzle, a really light bullet may penetrate best.
Flesh and ballistic gelatin seem to require momentum for maximum
penetration. This usually means weight is what is needed, because you don't
lose weight in flight and mass and velocity contribute equally to momentum
(p = mv).
Finally, expanding bullets further complicate things. Velocity is usually a
major factor in causing bullet expansion. As a bullet expands, its frontal
area gets larger and this causes it to penetrate less. But you get a bigger
diameter hole. So a bullet that penetrates optimally for deer will not
penetrate enough for elk. Bullet manufacturers have a much better handle
today on making expanding bullets work reliably. However, you must know the
velocity range that it was designed to work with. Shooting 30-30 bullets out
of a 300 Magnum will probably cause the bullet to fragment and penetrate
poorly within 300 yards of the muzzle. Likewise, shooting a spitzer designed
for the 300 magnum out of a 30-30 will cause a little 30 caliber hole to be
poked through the animal. Note that this gives good penetration, but no
expansion and poor terminal effects on the animal.
--
Mark
Renton, WA
John Bercovitz
# Why would a heavier bullet going slower, penetrate further than a
# lighter bullet going faster, all else (diameter, etc) being equal?
# For this example, let's use loadings for 30-06 that generate equal
# ft/lbs at impact. Folks talk about shooting, say, 180-grain bullets
# from the 30-06 for better penetration, but why exactly does this
# penetrate better than a 150-grain bullet going a little faster?
See L. Prandtl, "Ergebnisse der aerodynamischen Versuchsanstall zu
Goettingen". That'll give you drag coefficients and how to use 'em.
You know when you bat a balloon it zips across the room and suddenly
decelerates? That's because when the Reynolds' number of a smooth
sphere falls below about 350,000, the drag coefficient jumps due to
a change from turbulent to laminar flow. Above 350,000, the drag
coefficient, Cd, is pretty constant at 0.2.
So let's try a very simple example. We choose a sphere because it is
well characterized. Then we change the density (but not the diameter)
of the sphere to simulate the different masses of two 30 cal bullets.
I'll pick 150 and 200 grain to make the differences more apparent. If
the 150 gn bullet hits at 2000 fps (24,000 inches/sec), at what velocity
does the 200 gn hit? You stipulated kinetic energies are equal.
KE = 2.22E-6*w*V^2
KE(150@2000) = 2.22E-6*150*2000^2 = 1332 ft-lb
V(200@1332) = (1332/(200*2.22E6-))^0.5 = 1732 fps 20785 in/sec
First let's make sure we're above the dip in the drag curve.
Reynolds' number, Nr = D*V/nu
so
V = nu*Nr/D
Let's use 100 degree F water for the medium we're traversing.
kinematic viscosity, nu = 1.05E-3 inches^2/sec
density, rho = 9.29E-5 lb-sec^2/inch^4
D = 0.3" (approximately)
So what V do we have to stay above to keep the drag coefficient
low and constant?
V = nu*Nr/D = 1.05E-3 * 350,000/0.3 = 1226 in/sec = 100 fps
Yup, I think your bullet will be going faster than that during
the period of interest!
Drag force, Fd = Cd * rho * V^2/2 * A
Fd(@2000fps) = 0.2 * 9.29E-5 * 24000^2/2 * (.7854*.3^2) = 378 lb
Fd(@1732fps) = 0.2 * 9.29E-5 * 20785^2/2 * (.7854*.3^2) = 284 lb
What's the deceleration like for the two cases? a = F/m
For the 150 gn, deceleration in Gs = 378*7000/150 = 17640 G
For the 200 gn, deceleration in Gs = 284*7000/200 = 9940 G
Now, this is all very rough, but it shows the trend.
Left as an exercise for the student: Integrate this puppy from impact
down to some arbitrarily low velocity to find depth of penetration.
John B
WVanhou237
writes:
#
#Are you really talking about penetration or stopping power? They are two
#different things.
#Having said all that, somebody out there might know something I'm missing
#and can set us both straight. (Which would be a good thing since my head is
#starting to hurt - trying to remember my engineering from 25 years ago!)
#Take care,
#Jim
#
It's really very simple. For the same diameter bullet, the heavier bullet
has more behind pushing the front.
Bill Van Houten (USA Ret)
"No matter how hard you try, you can't throw a potato chip very far."
"Linus"
rosignol
Jim Grunewald <jgrun...@nc.rr.com> wrote:
# Jeff,
# All things being equal (except for speed and mass of the bullet), I believe
# it is an issue of momentum:
#
# M = mv2 (mass times the square of the velociy)
#
# You don't provide velocities, but I would not expect the difference in
# weight to make up for the difference in velocity. I assume that the spin
# rates of the bullets are the same. I would expect that a faster spinning
# bullet would penetrate a little better than a slower spinning one.
#
# Are you really talking about penetration or stopping power? They are two
# different things.
Stopping power is so very dependent on bullet placement that I'd
appreciate it not being introduced into a relatively straightforward
question of physics/ballistics, thanks.
[zap]
Eugene A. Pallat
# Jeff,
# All things being equal (except for speed and mass of the bullet), I believe
# it is an issue of momentum:
#
# M = mv2 (mass times the square of the velociy)
Oops, mv2 = energy. Momentum is just mv. Just check any freshman physics
text. My 45 year ols Sears and Zemansky even lists it that way.
Gene Pallat
Orion Forensics
Orion Data Systems
Jim Grunewald
I was hoping nobody would notice! Please remember that I did say it was 25 years
ago! My Sears and Zmansky and yours do indeed agree. Would you believe that I
wasn't sure which was the controlling factor, momentum or kinetic energy, so I
"blended" the equations? Oh, come on! Cut me a little slack!
Take care,
Jim
"Eugene A. Pallat" wrote:
> ...
Jim Grunewald
#
# It's really very simple. For the same diameter bullet, the heavier bullet
# has more behind pushing the front.
Actually, Bill, what I was thinking is that a bullet can completely penetrate a
body but not stop as well as, say, a hollow point of similar caliber.
WVanhou237
<jgrun...@nc.rr.com> writes:
#
#Actually, Bill, what I was thinking is that a bullet can completely penetrate
#a body but not stop as well as, say, a hollow point of similar caliber.
#
Then what you will have is less behind trying to push the same weight
with increasing caliber. 8>}
I will have to go along with the group that contends that it is practicaly
impossible to quantify "stopping power". Or , I think, reliably predict it.
Too much depends on a lot of conditions. Whether an animal is
unsuspecting or in a panic. Whether the bullet disrupts the bone structure
or the central nerve system. etc. If it does neither, will it disrupt the
oxygen
to the subjects brain quickly.
Bill Van Houten (USA Ret)
"No matter how hard you try, you can't throw a potato chip very far."
"Linus"
Dr. Bob
Kinetic energy = (1/2)mv2
Dr. Bob
---------------------------------------------------------------
In article <9e5kr4$5k$1...@xring.cs.umd.edu>, "Eugene A. Pallat"
<eapa...@apk.net> wrote:
> ...
Adam Kippes
# Momentum = mv
# Kinetic energy = (1/2)mv2
ke = .5mv^2/7000/32.18
ke = kinetic energy
m = mass, in grains
v = velocity in feet per second
7000 is one pound in grains
32.18 (or 32.175) is gravitational acceleration
-- AK
--
adam....@pobox.com
PGP keys available from servers
Eugene A. Pallat
> ...
Right! I forgot the 1/2
Gene
Michael Ballai
it relates to an animal or human body, a bullet crushes through
tissue. Just as a lightweight tin hammer isn't much good at squashing
something and a heavy steel or iron hammer is effective without much
help from us, so the heavier bullet crushes through more tissue and
penetrates deeper. If the heavier bullet expands or deforms as it
impacts, the effectiveness can be devasting because the crush cavity
increases.
I don't know if this comes across with all the hallmarks of good
science, but I believe it to be accurate.
John Bercovitz
ke = kinetic energy
w = weight, in grains
v = velocity in feet per second
7000 = 1 pound in grains
32.175 = gravitational acceleration
ke = wv^2/2/7000/32.175
ke = (2.220000/1,000,000) * wv^2
= 2.22E-6 * w * v^2
;-)
John B
PS: But isn't it 32.174 fps^2 ???
Gene Paul
"John Bercovitz" <ber...@csg.lbl.gov> wrote in message
news:9f5glb$a46$1...@xring.cs.umd.edu...
# ke = kinetic energy
[snip]
# 32.175 = gravitational acceleration
Yes, I haven't seen it mentioned, but it's being assumed here that the units
for ke are foot-pounds (work done by a force of one pound-force(lbf) acting
through a distance of one foot.)
If the units were foot-poundals (energy unit for the lb-ft-sec "absolute"
unit system) we wouldn't need that pesky "g" of 32.174 f/s^2, which converts
to foot-pounds.
Adam Kippes
# If the units were foot-poundals (energy unit for the lb-ft-sec "absolute"
# unit system) we wouldn't need that pesky "g" of 32.174 f/s^2, which converts
# to foot-pounds.
Quite true, but since virtually all literature, certainly popular
literature, various tables, etc. use pounds and not pounders, you
would confuse the heck out of everyone.
-- AK
--
adam....@pobox.com
PGP keys available from servers
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