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A 55 grain .22 caliber bullet at 3,600 FPS has a calculated 1,583 FPE but only 28.29 pound feet of momentum. A 360 grain .45 caliber bullet from a .454 at 1,400 FPS has a calculated energy of 1,567 FPE, yet produces 72 pounds feet of momentum. We know from expereince that the 360 grain .45 caliber bullet will out penetrate the 55 grain .22 caliber bullet at 3,600 FPS despite having less kinetic energy.

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someoldguy is right!

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Originally posted by someoldguy:

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A 55 grain .22 caliber bullet at 3,600 FPS has a calculated 1,583 FPE but only 28.29 pound feet of momentum. A 360 grain .45 caliber bullet from a .454 at 1,400 FPS has a calculated energy of 1,567 FPE, yet produces 72 pounds feet of momentum. We know from expereince that the 360 grain .45 caliber bullet will out penetrate the 55 grain .22 caliber bullet at 3,600 FPS despite having less kinetic energy.

Sorry I didn't come up with a more appropriate response earlier.
OK, then...
That .22 caliber bullet I would guess is a pointy-type expanding bullet which is likely going to lose a lot of mass. I doubt seriously that you'd get a net 17 inches of penetration out of it in ballistic gelatin. My program says 17.7", but that is straightline, with expansion, and no mass lost. (This is, after all, a short program. No, it won't do your taxes for you. )
For the .454 bullet, I would say this is one of those flatnosed deals which all the kids have to have now. I would guess that it would do around 62" in ballistic gel, maybe some better.

There are two factors which go into what I call the resisting force. The resistance on the bullet caused by the target's density and the resistance on the bullet caused by the target's strength. The resistance caused by the target's density is the one which is velocity dependent. (The square of the velocity in fact.) Calculating it looks similar to the familiar equation for drag. What I get is a value for deceleration (negative acceleration.) The other factor is a shortcut, but it also gives me another deceleration component. It involves the strength (in pounds-force per square foot) times the penetrating surface area divided by retained mass. It's also a deceleration in feet per second squared. After adding these two deceleration components together, I multiply by the retained mass and, voila, there is the resisting force.
<whew>
After I find the kinetic energy, the rest is just simple arithmetic.

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someoldguy is right!

Thanks for the support, 303guy. We're undoubtedly related, as our last name is "guy."

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Originally posted by someoldguy:

quote:

A 55 grain .22 caliber bullet at 3,600 FPS has a calculated 1,583 FPE but only 28.29 pound feet of momentum. A 360 grain .45 caliber bullet from a .454 at 1,400 FPS has a calculated energy of 1,567 FPE, yet produces 72 pounds feet of momentum. We know from expereince that the 360 grain .45 caliber bullet will out penetrate the 55 grain .22 caliber bullet at 3,600 FPS despite having less kinetic energy.

Sorry I didn't come up with a more appropriate response earlier.
OK, then...
That .22 caliber bullet I would guess is a pointy-type expanding bullet which is likely going to lose a lot of mass. I doubt seriously that you'd get a net 17 inches of penetration out of it in ballistic gelatin. My program says 17.7", but that is straightline, with expansion, and no mass lost. (This is, after all, a short program. No, it won't do your taxes for you. )
For the .454 bullet, I would say this is one of those flatnosed deals which all the kids have to have now. I would guess that it would do around 62" in ballistic gel, maybe some better.

There are two factors which go into what I call the resisting force. The resistance on the bullet caused by the target's density and the resistance on the bullet caused by the target's strength. The resistance caused by the target's density is the one which is velocity dependent. (The square of the velocity in fact.) Calculating it looks similar to the familiar equation for drag. What I get is a value for deceleration (negative acceleration.) The other factor is a shortcut, but it also gives me another deceleration component. It involves the strength (in pounds-force per square foot) times the penetrating surface area divided by retained mass. It's also a deceleration in feet per second squared. After adding these two deceleration components together, I multiply by the retained mass and, voila, there is the resisting force.
<whew>
After I find the kinetic energy, the rest is just simple arithmetic.

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someoldguy is right!

Thanks for the support, 303guy. We're undoubtedly related, as our last name is "guy."

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Duncan MacPherson's book Bullet Penetration makes it quite clear that damage is done by stress, not energy . Stresses cause damage only if they strain body tissues above their elastic limits.

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Duncan MacPherson's book Bullet Penetration makes it quite clear that damage is done by stress, not energy . Stresses cause damage only if they strain body tissues above their elastic limits.

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Originally posted by jwp475:
... Duncan MacPhearson has the only Model that works 100% of the time perhaps you may want to read this: ...

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Originally posted by Hot Core:

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Originally posted by jwp475:
... Duncan MacPhearson has the only Model that works 100% of the time perhaps you may want to read this: ...

I believe I've seen that mentioned here at AR before. If it is about Wounds in Humans(as I seem to remember), it has NOTHING at all to do with how Energy works on Game - not even close.

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Duncan MacPhearson has the only Model that works 100% of the time perhaps you may want to read this:

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I believe I've seen that mentioned here at AR before. If it is about Wounds in Humans(as I seem to remember), it has NOTHING at all to do with how Energy works on Game - not even close.

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Originally posted by someoldguy:

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Duncan MacPhearson has the only Model that works 100% of the time perhaps you may want to read this:

What works 100 percent of the time?
Anybody?
Speak up, dammit! I can't hear you!
OK, never mind then.

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I believe I've seen that mentioned here at AR before. If it is about Wounds in Humans(as I seem to remember), it has NOTHING at all to do with how Energy works on Game - not even close.

Yes, that's the one, Hot Core. McPherson's work had to do with modeling wound effects in humans. All I'm trying to do is to predict how something penetrates in materials. Apples and oranges, really. But apparently when someone mentions "energy" and "penetration" in the same sentence, some devotees of McPherson get their feathers aruffle. No idea why this happens, although I've noticed it for years.

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Interesting indeed, your comments.....

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It seems like it would be easier to just go shoot a few animals and measure the results than try to predict those results from some formula.

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Originally posted by Hot Core:

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Originally posted by jwp475:
... Duncan MacPhearson has the only Model that works 100% of the time perhaps you may want to read this: ...

I believe I've seen that mentioned here at AR before. If it is about Wounds in Humans(as I seem to remember), it has NOTHING at all to do with how Energy works on Game - not even close.

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Originally posted by someoldguy:

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Duncan MacPherson's book Bullet Penetration makes it quite clear that damage is done by stress, not energy . Stresses cause damage only if they strain body tissues above their elastic limits.

What is stress? A unit of force per a unit of area. Right? Wrong?
What is energy? A unit of force times a unit of distance. Right? Wrong?
Now, am I alone in noticing similarities and possible relationships between the two?

Thanks for posting that article, 303guy. The original illustrations are at firearmstactical.com. Great site!

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Originally posted by someoldguy:

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It seems like it would be easier to just go shoot a few animals and measure the results than try to predict those results from some formula.

465H&H, thanks just the same, but I don't want to go out and shoot an elephant with a .223 just to see how it works.
I really don't intend to be turned into tembo toejam if I can help it.

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MOMENTUM IS ALWAYS TRANSFERED IN ALL COLLISIONS

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Elastic and Inelastic Collisions
A perfectly elastic collision is defined as one in which there is no loss of kinetic energy in the collision. An inelastic collision is one in which part of the kinetic energy is changed to some other form of energy in the collision. Any macroscopic collision between objects will convert some of the kinetic energy into internal energy and other forms of energy, so no large scale impacts are perfectly elastic. Momentum is conserved in inelastic collisions, but one cannot track the kinetic energy through the collision since some of it is converted to other forms of energy

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Inelastic Collision

Collisions between objects are governed by laws of momentum and energy. When a collision occurs in an isolated system, the total momentum of the system of objects is conserved. Provided that there are no net external forces acting upon the objects, the momentum of all objects before the collision equals the momentum of all objects after the collision. If there are only two objects involved in the collision, then the momentum change of the individual objects are equal in magnitude and opposite in direction.

Certain collisions are referred to as elastic collisions. Elastic collisions are collisions in which both momentum and kinetic energy are conserved. The total system kinetic energy before the collision equals the total system kinetic energy after the collision. If total kinetic energy is not conserved, then the collision is referred to as an inelastic collision.

The animation below portrays the inelastic collision between a 1000-kg car and a 3000-kg truck. The before- and after-collision velocities and momentum are shown in the data tables.






In the collision between the truck and the car, total system momentum is conserved. Before the collision, the momentum of the car is +20000 kg*m/s and the momentum of the truck is -60000 kg*m/s; the total system momentum is -40000 kg*m/s. After the collision, the momentum of the car is -10000 kg*m/s and the momentum of the truck is -30 000 kg*m/s; the total system momentum is -40000 kg*m/s. The total system momentum is conserved. The momentum change of the car (-30000 kg*m/s) is equal in magnitude and opposite in direction to the momentum change of the truck (+30000 kg*m/s) .

An analysis of the kinetic energy of the two objects reveals that the total system kinetic energy before the collision is 800000 Joules (200000 J for the car plus 600000 J for the truck). After the collision, the total system kinetic energy is 200000 Joules (50000 J for the car and 150000 J for the truck). The total kinetic energy before the collision is not equal to the total kinetic energy after the collision. A large portion of the kinetic energy is converted to other forms of energy such as sound energy and thermal energy. A collision in which total system kinetic energy is not conserved is known as an inelastic collision.

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If an erg is a measure of energy whigh erg is better at becoming a non-erg?

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Yep, that would be an extremely stupid thing to do. But why not test it on a coyote?

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I guess I'm just curious and want to learn.

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Originally posted by someoldguy:
Yes, that's the one, Hot Core. McPherson's work had to do with modeling wound effects in humans. All I'm trying to do is to predict how something penetrates in materials. Apples and oranges, really. But apparently when someone mentions "energy" and "penetration" in the same sentence, some devotees of McPherson get their feathers aruffle. No idea why this happens, although I've noticed it for years.

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Originally posted by Wstrnhuntr:
The less energy your weapon has, the more relevant energy becomes..

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Originally posted by Ackley Improved User:

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Originally posted by Wstrnhuntr:
The less energy your weapon has, the more relevant energy becomes..

This is the most insightfull post on this thread! Brilliant.

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Originally posted by Hot Core:

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Originally posted by someoldguy:
Yes, that's the one, Hot Core. McPherson's work had to do with modeling wound effects in humans. All I'm trying to do is to predict how something penetrates in materials. Apples and oranges, really. But apparently when someone mentions "energy" and "penetration" in the same sentence, some devotees of McPherson get their feathers aruffle. No idea why this happens, although I've noticed it for years.

I thought so. My hero "alf" used to bring it up in nearly every thread he posted to.

For people who have first-hand experience of seeing how both people and Game react to shots taken, there is simply ZERO comparison. The Freedom Freeloaders of the world will never understand it though because they have no comparison base and are unable/unwilling to learn from those that do have it.
-----

By the way SomeOldGuy, though I've not mentioned it previously, I really enjoy your humor.

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Originally posted by Hot Core:

quote:

Originally posted by someoldguy:
Yes, that's the one, Hot Core. McPherson's work had to do with modeling wound effects in humans. All I'm trying to do is to predict how something penetrates in materials. Apples and oranges, really. But apparently when someone mentions "energy" and "penetration" in the same sentence, some devotees of McPherson get their feathers aruffle. No idea why this happens, although I've noticed it for years.

I thought so. My hero "alf" used to bring it up in nearly every thread he posted to.

For people who have first-hand experience of seeing how both people and Game react to shots taken, there is simply ZERO comparison. The Freedom Freeloaders of the world will never understand it though because they have no comparison base and are unable/unwilling to learn from those that do have it.
-----

By the way SomeOldGuy, though I've not mentioned it previously, I really enjoy your humor.

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Originally posted by vapodog:

quote:

Originally posted by Ackley Improved User:

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Originally posted by Wstrnhuntr:
The less energy your weapon has, the more relevant energy becomes..

This is the most insightfull post on this thread! Brilliant.

AMEN to that.....

The choir has sung!

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Originally posted by Ackley Improved User:

quote:

Originally posted by Wstrnhuntr:
The less energy your weapon has, the more relevant energy becomes..

This is the most insightfull post on this thread! Brilliant.

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A 9 g bullet hits a tree with impact speed of 250 m/s. The bullet makes a 8 cm long hole in the wood, then stops. Find the average force resisting bullet motion through the wood.

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impulse = change in momentum

FΔt = p1 - p2
F = (p1-p2)/Δt

p1 = mv = (.009 kg)(250 m/s) = 2.25 kg*m/s
p2 = 0

v^2 = v0^2 +2a(x-x0)
a = (v^2 - v0^2) / 2d
a = (0-(250 m/s)^2) / 2(.08 m)
a = -390625 m/s^2

v = v0 + at
t = (v-v0)/a = (0-250 m/s) / -390625 m/s^2
t = .00064 sec

F = (p1-p2)/Δt = (2.25 kg*m/s - 0) / .00064 sec
F = 3515.63 N

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Originally posted by someoldguy:
OK. This isn't the physics problem I was looking for, but it's quite similar, so I'll use it. With it I will prove that kinetic energy, resisting force, and penetration depth definitely have a relationship. (But momentum also, yes indeed.)

Here is the problem itself.
It's located on this page:
http://answers.yahoo.com/quest...0100302215610AAmDFEy

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A 9 g bullet hits a tree with impact speed of 250 m/s. The bullet makes a 8 cm long hole in the wood, then stops. Find the average force resisting bullet motion through the wood.

For my purposes, I need to convert these values to English units because this is what I'm most comfortable using. I have also calculated the kinetic energy, in ft-lbs.

9 g bullet = 139 grains
250 m/s = 820 feet per second
8 cm penetration = 3.15 ins (0.2625 ft) [Must be some pretty strong wood!]
KE = 820^2 x 139 / 7000 / 32.2 / 2 = 207 ft-lbs
(What kind of bullet is this? Maybe a .38 Special handgun bullet?)

In the answer, the poster used momentum to arrive at the force, expressed in Newtons. Of course, the answer is in metric units.

quote:

impulse = change in momentum

FΔt = p1 - p2
F = (p1-p2)/Δt

p1 = mv = (.009 kg)(250 m/s) = 2.25 kg*m/s
p2 = 0

v^2 = v0^2 +2a(x-x0)
a = (v^2 - v0^2) / 2d
a = (0-(250 m/s)^2) / 2(.08 m)
a = -390625 m/s^2

v = v0 + at
t = (v-v0)/a = (0-250 m/s) / -390625 m/s^2
t = .00064 sec

F = (p1-p2)/Δt = (2.25 kg*m/s - 0) / .00064 sec
F = 3515.63 N

So they've arrived at the answer:
Force = 3515.63 Newtons

Converting to English units, this quantity is
3515.63 x 0.2248 = 790 pounds-force

But guess what? If you're using English units, there's a quick and easy shortcut to arrive at this same answer.
It's simply the kinetic energy divided by the penetration depth in feet.
Resisting force (English units) = 207 ft-lbs / 0.2625 ft = 789 pounds-force

(Likely the 1 pound-force discrepancy is due to rounding.)

So this shows pretty conclusively that, yes, you can most certainly calculate the penetration depth from kinetic energy as long as you have an idea of the resisting force.

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The matmaticl expression for the drag that occurs when a projectile passes through a fluid is:
where CDis the proportionality constant known as the coefficient of drag, is the density of the target me-dium,A is the projectile’s cross-sectional area, and Vits velocity.
The drag equationindicates if remain the same, changing the density of the medium from that of air (0.001 g/cm3) to that of water (1.00 will result in a thousandfold increase in drag. Similarly, increasing the cross-sectional area of the projectile (for example, as when a deforming bullet such as a dumdum hits a solid target, or when a bullet tumbles) will increase drag. Because velocity is raised to the second power in the equation, doubling the projectile’s velocity increases drag fourfold.
All other factors, such as the coefficient of drag and the shape, being the same, drag is much greater with high-than with low-velocity projectiles. For example, high-velocity rifle bullets slow more rapidly than do similarly shaped pistol bullets.
The drag equation omits a factor for the splitting of the medium, a phenomenon that depends upon the medium’s material strength. Because of this omission, the increase in drag that occurs when a projectile passes from air into human soft tissue will be much greater than a prediction based simply on the change in density (from 0.001 in air to 1.05 in muscle). This factor is unknown and is one of the areas that requires the attention of sophisticated ballistics researchers.
When drag is treated mathematically as the force of retardation, it is usually normalized for projectile mass (that is, and can be expressed:
or, by transposition: A
However, a projectile’s sectional density /A) is an important determinant of drag: the higher the sectional density, the lower the drag.