But I don't agree with that article you posted the pointer to. His slug f/s theory I find to not really explain all the data. Spears and arrows are effective not because of, in my opinion, his quantity, but rather because they have broad sharp edges that create wounds unlike a bullet. Sharp edges slice cleanly, especially through veins and arteries, and that causes massive bleeding which is rather a different sort of wound from that created by a bullet. Most tissue is flexible, and can stretch a good deal before tearing. Thus, some (I can't quantify how much) of the energy of the bullet is wasted in stretching tissue that's not actually damaged enough to contribute to death.

Another point his article fails to address: the different wound profiles of different bullet designs. For example, a round nose solid vs. a good truncated cone design. If both are fired at the same velocity and neither exit the animal, then they should both cause the same damage, per his theory. But this is not the case, based on what hunters report. The TC bullets allegedly cause a much larger wound channel. Yet both have the same "slugs f/s".

I'm not saying I have a better theory; I don't. But to date, bullet performance is still not on a rigorous scientific foundation. There's lots of speculation, and lots of interesting experimental results (mainly on non-living media). Joing the International Wound Ballistics Association if you're interested in this subject.

Pertinax

P.S. Todd E., I'm still curious about your test question.

Norbert,

I am completely discussing physics my friend. There is no BS in this post. Those are the forces imparted to the animal by the bullets passing and coming to rest within the animal. What you are not considering is that this phenomena happens over time. The animal's reaction to the force imparted to it by the bullet lags behind the actual event so the animal is actually moving at a greater velocity than conservation of momentum would indicate. This is because the force imparted to the animal by the bullet accelerates the animal as JonA ponted out, but JonA stopped prematurely. The bullet stopping does not suddenly end the accerleration imparted to the animal by the bullet. The animal must react to that acceleration, typically by leaning into the force. The reaction time of the animal will range from 0.1 to 0.4 seconds. You can do that math referencing JonA's post. You will find that when the animal begins to react to the bullets passing is has reached a velocity of approximately 5 - 20 fps.

Furthermore, we have not taken into account the mass of bodily fluids accelerated by the bullets passing (transformation of energy). This will increase the force acting upon the animal and the resulting acceleration further enhancing the "stopping" power of the round.

Todd E

Now it's entirely possible that the animal will compensate by leaning into the "push", a fraction of a second later, and before its nervous system shorts down, causing it to fall toward the shooter. But it's not because the bullet pushed on the buff for a long time; the bullet's stopped in just a handful of milliseconds, if that. The bullet's gone from your gun around a millisecond after the primer lights, and it'll stop if evenly decellerated over 6 feet in 5 mSec (assuming 2300 fps impact velocity).

Pertinax

An object will not continue to accelerate after the force has ceased. You asked for others to provide mathematical proof, now it's your turn. If you are successful, you've just invented the mythical perpetual motion machine--congratulations, you're a billionaire!

I suggest you start with a = f/m. If f is 0, how much is a?

The object will remain at a constant velocity until acted upon by another force--which will accelerate it (positively or negatively) changing its velocity. In this case, even before the animal reacts, all the other forces acting on the animal (various forms of friction) are decelerating the animal.

The final velocity mentioned above was a maximum theoretical possibility (if the animal was floating in space or something when shot). In this case it would continue to float along at the same velocity until hit by something else. In reality (where there is friction) the velocity will always be lower than the calculated one.

I found the problem with your physics, in the other thread. The target does _not_ continue to accelerate after the bullet stops, until some resisting force appears. If that were the case, your car would continue to accelerate even after you let off the gas. (And if you think friction is the reason it doesn't, think again. Think space probes and ice skaters.)

What Newton figured out was that _velocity_ remains the same until the object is acted on by another force. Velocity is a vector quantity, which includes "speed" and direction. Same rules apply for the direction of travel. And for objects at rest ("speed" and velocity of zero).

Acceleration does not continue after the bullet stops. No "ifs", no "ands", no "maybes", or "buts".

Pertinax

quote:

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Originally posted by Todd E:
MacD,

I guarantee you that the 2300 pound bullet going 200 MPH would kill and knock down any buffalo (except the old class 9 Chevy truck). But let me tell you that is one EXPENSIVE buffalo bullet!
Todd E

[This message has been edited by Todd E (edited 01-29-2002).]

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Yeh, I'd say that was a premium bullet for sure!

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..Mac >>>===(x)===>
also DUGABOY1
DUGABOY DESIGNS
Collector/trader of fine double rifles, and African wildlife art

Now I assure that the bullet knocked the buffalo over. Anyone that says otherwise is a fool and a liar. If I hadn't shot it that buffalo still probably wouldn't have fallen over!

Now since you guys cannot figure out how a living creature reacts to inputs have someone sneak up on you and give you a good shove. There is a damned good chance you will fall over.

While this conversation has been stimulating (actually no it hasn't) I will have to bid you all farewell. I have been given my new orders and have work to do so I will not be able to post for potentially quite sometime. I know that this will break many of your hearts.

God Bless
Todd E

Since we are not talking about nuclear reaction, of course energy remains the same.

But not as kinetic energy. To do that you need to go to my block and tackle example in 2nd post. That it is not practical to make up is a seperate isse.

The fact is that your bullet has the capacity to lift the 1300 pound buffalo 5 feet straight up.

But only if the gearing in right.

Mike

The momentum is the same at both ends of the weapon as the bullet leaves the barrel if we negate the effect of the powder.

No bullet can 'knock an animal' down but it can kill cleanly and the animal can fall wear it stands.

PaulS

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stay safe and live long!

I have a weapon that will fire a 5 grain sewing needle at 13422 FPS that gives it 2000 ft.lbs. of kinetic energy and one that launches a 12 lb. bowling ball launched at 103.5 FPS with the same 2000 ft. lbs. of energy. Which of these projectiles will make an inpenatrable object weighing 1200 lbs. move more? (the target will catch and hold either projectile) Which of these weapons will kick harder?

The answer is:

the sewing needle has 9.6 ft. lb. seconds of momentum and will accellerate the 1200 pound target to .008 FPS

the bowling ball will accellerate the target to 1.035 FPS

Needless to say the bowling ball gun would be impossible to shoot while the needle gun would have very little recoil.

So, the buffalo fell over dead - it was not knocked over by the bullet.

PaulS

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stay safe and live long!

quote:

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There is still energy to be conserved. What is the total affect of the conservation of energy?

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If you re-read the first thread you'll find that kinetic energy is not conserved because it is not an elastic collision. Most of it is turned into heat energy. Ever dig a bullet out of a penetration test a bit too quickly and burn your hands? That's where the energy went. It didn't accelerate anything (besides at the molecular level).

The only way the buff continues to accelerate after the bullet has stopped is if the bullet contains a rocket motor in its base that would apply a force--which is required for acceleration....

I think the problem here is a couple of things. We do not accurately register the precise time of impact. We see an animal react to the shot. I think it would be difficult to make any hard judgments about whether sufficient time had elapsed for the nervous system to react or not. More to the point, I think the nervous system FAILING to react is what occurs 9 times out of 10. Legs buckle and down she goes.

You can get similar force figures by using E = F * s, where E is kinetic energy of the bullet at impact (minus any remaining if it exits) and s is the depth of penetration to exit or rest. The average force applied (not the actual time integral of the force) is E / s.

But all of this misses the fact that the bullet penetrates by pushing a small region of tissue aside. The force is not applied to the whole animal, just to a tiny spot. That force is diverted radially, so the impetus to the animal is also radially applied. The net result is a rapidly accelerated cavity but not an accelerated beastie. Very, very little axial force (ie, in the direction of the bullet) is delivered to the target unless the bullet is firmly caught in almost no distance of travel and cannot translate its momentum into radial expansion. This gets into the whole rigid body thing again.

To revisit that argument I have another thought that occurred to me after it was over as a way to make the point I failed to make effectively before. The classic demonstration of rigid body mechanics and conservation of momentum from freshman physics is the billiard balls on strings. Smack the outside one into its neighbor and the far opposite ball gets knocked off. But here is the cute little thing that the professor fails to call attention to: the balls cease all motion in a very few seconds. Now how is that? I thought momentum was always conserved. Where did it go? Given that the velocity of the outside balls decays over time, there is a number of balls in line such that impacting the outside ball on one side will produce no movement of any sort on the far ball.

The wires on the two outside balls behave like springs but that's not the trick. Were they perfect springs (similar to perfectly rigid balls) they would restore the velocity as the ball comes back down.

You can imagine a system of connected bodies in a vacuum far away from any potential field in which the two outside bodies are connected to the others by a spring. Apply an initial velocity to the outside body and you have a closed system with a momentum input just like the suspended balls. Unless those bodies (or the balls) are perfect rigid bodies and those springs (or wires) are perfect springs, then that system will lose all of its momentum according to a transient dynamic response that can be calculated very precisely. And there are no perfect bodies or springs in the real cosmos.

All bodies in motion lose energy due to resistive forces between elementary particles. Some of these losses are due to linear translations and some are rate dependent. There is a world of difference in the time constant that will characterize the decay of motion in that system if the outside springs are replaced with spring and damper systems in which the damping is velocity dependent, and there is a difference depending on whether the spring and damper arrangement is in series or in parallel. Tissue behaves like a parallel non-ideal spring and damper system.

While I don't have a clue what your talking about I don't believe that I agree with you about the bullet and radial pressure. I am thinking of a wedge and splitting firewood. If the bullet is pointed I would agree that it would split its way through. If the bullet flattens out it would punch it way through. A flat bullet must then push forward more than outward wouldn't you agree.

Kent

This is the same behavior that I see in shaped charge penetration in my job. Somebody finally thought to weigh the steel plates before and after the test shot. Guess what? The plates weighed the same with a hole down the middle! The jet simply pushed the material aside. It didn't shove it out the end or back out the entrance of the crater as some had theorized. Odd, but true.

[This message has been edited by Roger Rothschild (edited 01-31-2002).]

I am still confused. I agree with what you say about pressure on the sides, but I still think that there must be pressure in fron to. I am thinking of this as a hydraulic cyclinder. If what you say is true then there is no need for seals at the end because there would be no pressure in front of the piston. I can tell you that if you lost a seal on the no pressure side oil will spray out with of the leak at the same speed of as the piston is travelling. Obviously, there is pressure in front of the piston. It is not as great as the pressure on the pressure side of the cyclinder.

If also if I plug the exit port on that cyclinder I will blow the seals. Don't ask how I know this. I apologize for sounding like I am arguing but I have a hard time understanding how a shaped charge acting against armor plate has much to do with what appears to me to be a hydraulics problem. I am still thinking of a flattened out bullet to.
If you could lower your explanation a couple degrees maybe my I can finally understand what you are saying.
Thanks,
Kent

Hydrodynamic penetration is the case in which the projectile and/or target experience pressures (stresses) exceeding their yield strengths and begin to flow like a fluid. SHaped charges make this happen to all materials because of the incredible velocities involved.

Would this statement be true then with regard to the shape charge? Since the explosion is radial someone directly in front of the point of impact but on the opposite side would not be injured by the explosion.

Also, I still don't believe that there is no pressure in fron to the bullet. I have shot animals and seen blood 15 feet or so beyond the animal. It seems to me that some of that spray is from the pressure bursting out of the animal. I just reread you last post again and I am confused it sounds like you are agreeing with me and saying that the maximum pressure is directly in front of the bullet. Is that what you are saying?

Kent

There is not really a safe place to be near a shaped charge when it detonates, but if I had to pick a place it would be at an angle from the axis of the jet in front of the charge. There won't be as many case fragments there and most of the explosive force has been focused down onto the jet. To the sides and back it is definitely bad and directly in front is the worst.

Getting back to bullets. The cookie cutter kind of effect that you described in your first post is seen in non-hydrodynamic penetration cases, especially when a reltively thin and hard, brittle material is struck. Then you often see the projectile shear and push out a plug of material from the back of the target.