I am not sure you meant pressure or kinetic energy but the answer is no. The animal will die. But the animal falling down is genereally a result of nerve reaction or sudden disruption to its system.

Not a physics expert nut I don't think the energy transfer is simply bullet energy vs weight of the animal.

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MED

The sole purpose of a rifle is to please its owner

quote:

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Originally posted by ramrod340:
Go old physics says that the rifle will give the shooter the same energy in recoil. So if it doesn't knock the shooter over the bullet at the other end won't knock over a larger animal.

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Sounds good in theory, but not in reality.

The "energy" received by the shooter is spread out over a comparatively large area, the size of the recoil pad. In addition, the weight of the rifle slows down the recoil considerably (the rifle is not recoiling at 2,000 - 3,000+ FPS), and the pad itself absorbs some of the recoil, or at least spreads the force over a longer timeline, to be absorbed by the shooter.

The animal on the receiving end however, has all of the energy contained in one small 1/2" square area, traveling at full speed, and of a hard material.

If you don't believe this to be true, attach a steel rod of bullet diameter to your buttpad, shoulder the rifle with only that contacting you, and fire a shot...

Animals fall at the shot due to massive disruption of the nervous system. A hit in the brain, spine, or with a rapidly expanding bullet of high energy in the chest, can cause an animal to loose its feet. Often, if the trauma is great enough, the animal will not regain its feet before losing consciousness, death following soon thereafter.

Unless your bullet hits the spine or brain, it will not knock down a alg (moose to North Americans).

George

[This message has been edited by GeorgeS (edited 01-15-2002).]

I'm just saying that the "equal & opposite reaction" laws don't apply in a direct fashion to this situation.

Bullets that have been fired are measured in ft/lbs of energy delivered. A shooter is certainly not receiving 3000 or so ft/lbs of energy (adjust the number to the particular round fired) when he pulls the trigger.

A portion of it is used up overcoming the inertia (or lack of it maybe) of the rifle to begin it's backward movement, some is absorbed in compressing the recoil pad, etc. Only a somewhat smaller amount is left to transmit directly to the shooter.

It's about conservation of momentum. The mass of your rifle times its recoil velocity will be equal to the mass of the bullet times its velocity (well, actually the rifle will have more due to the added effects of the powder velocity...but let's not get that picky).

To answer the question, assuming the bullet didn't exit the elk, the velocity imparted on it would be (weight of bullet times bullet velocity)/(the weight of the elk + bullet). For a 800 lb elk that comes to (750/7000*2400)/(800+750/7000) = 0.32 fps.

Hardly enough to get it moving noticably if it was floating in space--definately not enough to "knock it down" when it's standing on solid ground.

quote:

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Originally posted by Jon A:
It's about conservation of momentum. The mass of your rifle times its recoil velocity will be equal to the mass of the bullet times its velocity (well, actually the rifle will have more due to the added effects of the powder velocity...but let's not get that picky).

To answer the question, assuming the bullet didn't exit the elk, the velocity imparted on it would be (weight of bullet times bullet velocity)/(the weight of the elk + bullet). For a 800 lb elk that comes to (750/7000*2400)/(800+750/7000) = 0.32 fps.

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As they say, "right over my head with that one"!

I may not be effectively conveying what I'm trying to say, but I know enough not to argue with someone who understands this much!

I concede...

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Ray Atkinson

ray@atkinsonhunting.com
atkinsonhunting.com

Todd E

denton,

You have it backwards. In a perfectly inelastic collision, nearly all of the relative kinetic energy is lost--momentum is still conserved unless an outside force is acting on the system (which, of course, in the real world there are always some forces--friction, etc). The kinetic energy is lost doing the work of deforming the bullet, creating the wound channel, heat, sound, etc.

As I said in the other thread, they used to actually measure bullet velocity in a similar manor to shooting a bag of sand. Since a heavy bag of sand hardly will move shows just how little momentum the bullet really has.

Warren Sapp running at full click has more momentum than the biggest big bore bullet ever invented. He actually might be able to knock down an elk if he hit it unawares. But the elk would get back up because he lacks the energy to inflict as much damage to the vitals as even a lowely 243 could.

So everybody, next time you see somebody in a movie get blown accross the room, up against the wall, out a window, etc, after getting shot with a 9mm, 12 guage, whatever...just smile and say, "it's just a movie...."

If our 585 Nyati bullet has 10,000 ft/lbs of energy (work) that means it could lift 10,000 pounds 1 foot or 1 pound 10,000 feet or 10 pounds 1000 feet.

It does come down to how the energy is released.

HOWEVER,

If we could somehow attach the 585 Nyati bullet to a block and tackle or some other gearing system and then fired the 585 Nyati and also assuming no air resistance, then by the time the bullet stopped, the block and tackle would have lifted 1000 pounds to a height of 10 feet.

After the bullet stopped, the 1000 pound weight would start to fall from a height of 10 feet. Through the gearing system (still attached to the bullet) by the time the weight reached the ground the 585 Nyati bullet would be pulled back and would betravelling at its original muzzle velocity.

Mike

This individual is a police office (a friend of mine) who was unfortunate enough to take a hit. It knocked him through a store front. He was wearing body armor and wieghs around 220 pounds! It seems to be common practice with many people to trivialize physics problems and therefore make highly erroneous conclusions.

For those who did not understand my last post it was meant to be sarcasm.

Todd E

Stories told by those involved may not have all the details needed for a full analysis. I could tell you about the time I took 600v across my hands and it sent me flying out of my chair - unfortuantely I lost about 5 minutes of memory and have no idea what actually happened except my initial and final locations.

Electrical shock affects the signals sent to muscles. I bet your are correct though. When I played football and hit someone I now realize that it had nothing to do with the impact energy that I was tranferring to their body. They were just involuntarily jumping! Loen according to my friend he didn't even hear the shot he was shot in the back. Since the load would have been supersonic he wouldn't have had a chance to react would to the sound of the shot so how was it you feel that this was an involuntary reaction. You are only trying to justify a flawed application of physical law.

You guys totally crack me up. For the record consevation of momentum is applicable only with rigid bodies. With elastic bodies momentum is lost and transformed into energy due to deformation of said elastic bodies. That what I meant by trivializing these complex physics problems. I am not going to try and figure out this problem either.

Food for thought. An elk weighing 800 pounds is not rigid. When it is struck, say in the shoulder, only a portion of its body and therefore mass is affected by the impact. This portion is then moved, which affects the elks body as a whole.

Think about someone pushing you inthe chest. Does your entire body move back or just your upper torso.

Todd E

By the way, don't you know that Rambo can knock an armored car over on its side with a magazine full from an M-16?

Jon A is incorrect and you must be a true moron to argue with me! I will not argue with the two of you idiots! Actually, I find most of your posts to be idiotic.

Actually, I don't think Jon A is an idiot and I retract that statement. Jon A is just grossly over simplifying the problem and arriving at an erroneous conclusion! Energy is also conserved. So if the bullet does not completely exit the target all that energy has been expelled within the target. This accomplished by tearing/disrupting flesh and/or shattering bone. This energy is dispelled over a very short time.

rugerman,

You are correct. The impact energy was expelled against the armor and in a very short amount of time.

Todd E

Todd E

I agree that the "energy" of a shot will not uproot an elk. My point above about the "equal & opposite reaction" was separate from that point. But it does seem that at a certain point, energy *will* bowl something over. I'd certainly think that if you shot a rabbit with an -06, a flying quail with a 12 ga (done it & seen it happen), or even a small bird with a 22 LR, that it will move backwards at impact. Am I still on track, or do I need to go get some sleep?

(This isn't meant as an argument starter, just curious).

The momentum that is communicated to the shooter on firing the rifle is the time integral of the force exerted on the rifle by the exiting bullet and combustion gases. This quantity is further affected by what amounts to a spring - damper system in the recoil pad , clothes and shoulder muscle, all of which works to transmit the rearward forces over a longer period of time. You can safely assume that the rifle is a rigid body in front of the recoil pad if you want to model this event.

On the receiving end you have a penetration event, which is inherently non-pushy. In fact, for "perfect" penetration, no momentum would be imparted to the target. Thats purely imaginary of course, but for example a bullet that penetrates a very short distance in a very short period of time (and is fully captured) will deliver a much higher force to the target than one which penetrates slowly and deeply over a comparatively long period of time. The forces are transmitted to the non-rigid tissues on the bullet path and they are in turn accelerated away from the shotline at an angle described by the hydrodynamics.

So it isn't exactly correct to assert that the recoil and the terminal impact are equivalent, because they are governed by different events. But they are strongly related in cause and act over similar distances and times. Recoil is actually the stiffer of the two (as far as force or momentum is concerned). A charge of buckshot that gets caught in a few thicknesses of kevlar might well be a worse thump.

Depending on where you draw your control volume, energy is always conserved. It certainly is conserved in a cosmic sense. Momentum is only conserved in imaginary perfectly elastic collisions between rigid bodies. This never happens in the real world because resistive forces, inertia, friction and entropy basically always subtract some quantity of energy from a collision.

[This message has been edited by Harald (edited 01-17-2002).]

If you want to knock the elk over through momentum alone, you would have a better chance if you have it shoot the rifle.

I disagree, however, that the suddenness, or lack thereof, with which a projectile stops within a body has any thing to do with the amount of momentum transfered to that body, so long as the projectile comes to rest completely within the body. If you "stop" a bullet, then you will have all of its momentum exerted on you (relatively meager as that might be). It is true, however, that the momentum may not, due to hydraulic effect, projectile deflection, etc., all be directed in the same vector as the projectile (just look at the various directions prairie dog parts go when struck).

Instead of going into some in-depth explanation (that some here wouldn't believe anyway because it came from me, thus wasting my time) I'll just make a couple statements to clear up all the mis-information.

For anybody that wants to say I'm wrong, I'll show you something you apparantly haven't seen before--what the inside of a Physics book looks like! Physics for Scientists and Engineers, Tipler.

You can either read, learn and concede or you can continue to make statements that are completely incorrect (which will be obvious to anybody that does the reading). Your call.

Elastic Collision:

This is the "rigid body" type of collision that usually only happens in Physics books. In this type of collision neither object is deformed.

Kinetic energy is conserved.

Momentum is conserved.

Inelastic Collision:

In this type of collision, one or both of the object deforms. Work must be done to an object in order to deform it. Some of the kinetic energy (potential to do work) is used doing this work. Therefore, the system has less kinetic energy after the collision than before--kinetic energy is not conserved! The amount of energy consumed relative to the initial kinetic energy available can be calculated from the coefficient of restitution--a measure of the elasticity of the collision.

Momentum is conserved.

Perfectly inelastic collision:

In this particular case of the inelastic collision, both bodies stick together after the collision and share the same velocity (exactly like a bullet entering an elk but not exiting).

In this case all of the relative kinetic energy is consumed doing the work of deforming the objects, making noise, creating heat, etc.

Momentum is still conserved!

UNLESS AN OUTSIDE FORCE IS APPLIED TO EITHER OBJECT, MOMENTUM IS ALWAYS CONSERVED! NO MATTER WHAT TYPE OF COLLISION IT IS!

If you disagree with that, read this page 20 times and then tell me I'm wrong:

How about an example problem? Notice how momentum is conserved but most of the kinetic energy is lost (and note where they say it goes):

Still don't understand? Read up:

Todd, a 12 guage with the right load can have quite a bit of momentum. Enough to give somebody a nice "push" in the back if he isn't expecting it. It might cause him to lose his balance and fall forward--but it will not pick the guy up and throw him through a storefront! It is simply physically impossible.

Tell your friend to put his vest on a 200 lb bag of sand hanging from a tree, sitting on a table...whatever. Shoot it with a 12 guage. Telling his story to his friends won't be quite as dramatic after that.

Any questions?

One of the faults of textbooks is that they tend to simplify things for the sake of illustration. I'll stick by my earlier statement: "Momentum is only conserved in imaginary perfectly elastic collisions between rigid bodies. This never happens in the real world because resistive forces, inertia, friction and entropy basically always subtract some quantity of energy from a collision." That last sentence describes the sort of reasons why conservation of momentum is only really strictly correct in hypothetical terms. There are always resistive forces acting on the system, from potential fields, molecular forces, crystal lattice forces, several kinds of friction, etc. From the example of the block, if momentum were truly conserved it would continue to move with its given velocity indefinitely, and we know that isn't reality. An alternative perspective is gained from an example in Brach's "Mechanical Impact Dynamics: Rigid Body Collisions", which treats simplified mechanics more rigorously than the textbook. The example is of a very tiny particle striking a very large and massive surface (similar to a bullet impact). To quote: "With only one particle, momentum is not conserved during the collision." In reality it is never conserved, but in his treatment that assumption is generally made for simplicity's sake. This however is a case where a relaxation of that assumption (rather than its enforcement) is helpful in the analysis. The textbook muddied things by describing energy losses while preserving momentum. The correct treatment would be to analyze the problem using energy methods and then describe the particle motions on the basis of their residual kinetic energies. In practice, for some scenarios, the momentum conservation is not a bad assumption because losses are such that your answer won't be far off the mark. That is why the ballistic pendulum works reasonably well with lead bullets at modest velocities that are fully captured and do not deform the block to a large extent.

Stonecreek, the total momentum change of a system is defined as the time integral of the force(s) acting on the system. That's straight out of Newton's Principia, not just an opinion of my own invention. Its a mathematical necessity.

Thank you. That is exactly what I wanted to say.

Todd E

This has nothing to do with the text simplifying things. What you said is still incorrect by definition--the very definition you quoted, in fact. The very definition they are trying to teach that's so important they wrote an entire chaper on it.

quote:

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the total momentum change of a system is defined as the time integral of the force(s) acting on the system.

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Exactly. In the above examples there is no external force acting on the system. The bullet deforming itself and the block is not an external force acting on the system. That consumes energy, which by definition makes the collision inelastic. The very definition of an elastic or inelastic collision contains the word energy, not momentum.

The fact that ballistic pendulums did, in fact, work pretty well is a perfect example of why this is true. There is no way in the world to stop a bullet without losing well over 90% of its kinetic energy. Work needs to be performed to stop the bullet. Trying to do anything by measuring energy would be futile. In a billiard ball-type collision you could use energy because that is nearly elastic (or you could use momentum). The point is, how much the bullet and block deform makes no difference in how high the block will swing.

You are correct that in the real world there will always be external forces acting on the system. The block will have air resistance, the wires will have friction, etc. The idea in any experiment is to keep these forces tiny relative to what you are measuring. A well designed pendulum will swing back and forth on its own for quite some time before stopping--low amount of friction. And, this friction can be measured and factored into the final results.

quote:

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The textbook muddied things by describing energy losses while preserving momentum.

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That wasn't muddying things, it was explaining how things work. Using well known, mathematically provable laws of physics straight from Sir Isaac himself. If they hadn't shown momentum is conserved even in collisions with energy losses, they would have been wrong. The book would have been pulled from the shelves and somebody's career would be over (which would probably be a good thing).

Perform an experiment. Take a steel cylinder and suspend it. Now hit it with a steel bar and see how it swings. Replace the steel cylinder with an equal mass having a spring and damper system on one side. Hit the spring plate and see how it swings. The spring and damper dynamics will inhibit the motion and also cause the motion to subside much more quickly by absorbing the energy imparted by the impact. That, in a crude sense, is the reality of terminal mechanics.

The examples from that book are merely illustrations of various simplistic concepts. They are not reality. None of them. They are not special cases of reality. There are always forces acting on a system. On every system. Thats my point. It is useless to even speak of elastic collisions because there aren't any. Its just an imaginary approximation for collisions in which there is a negligible plastic deformation and negligible loss of energy to elastic deformation. And in every real world inelastic collision the loss of energy means that, by definition, momentum is not conserved.

In order to model an inelastic collision on the premise of absolute conservation of momentum it would be necessary to reduce the finite elements of the analysis to perfectly rigid and elastic bodies such that no energy was ever lost in their interactions. This would mean tracking the motions of every elementary particle in the system (atoms, and sub-atomic species) - and even this would not be correct, because those collisions are subject to energy exchanges too. The result would also differ from the experience of reality because that is not how our world behaves.

You are still confusing energy and momentum.

quote:

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And in every real world inelastic collision the loss of energy means that, by definition, momentum is not conserved.

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No, that's not what the definition says. The definitions are posted above--please read them again!

Energy and momentum are very different things, they do not go hand-in-hand. A system can have tons of kinetic energy but have zero momentum at the same time. Particals within the system can perform work on each other which will change some of the kinetic energy into other forms of energy and reduce the systems total kinetic energy. But the total momentum only changes when an outside force is applied to the system.

Yes, the real world has friction. But to say because of that, Newton's laws of motion are wrong is to misunderstand them. Doing so won't find you in agreement with anybody that knows what he's talking about. Saying they're wrong as you approach the speed of light...well, I don't think you want to go there.... Physics books also contain information about friction. There is nothing stopping you from measuring and accounting for losses due to friction that will prove Newton's laws are correct in an experiment.

Mathematical texts have many examples that would never happen exactly as described "in the real world." I guess this means that 2 + 2 might not equal 4? That math doesn't describe anything in the real world, it's all just theory?

Your proposed experiment above is flawed, mainly in the fact that your hand is introducing an external force upon the system and since the time of impact will be different the amount of work performed on the system will be different for each collision which will skew the results.

I could prove it to you mathematically but if you don't want to believe me, any Physics text, Newton or any Physics professor that ever lived, why would you believe math?. I can tell you how to perform a few simple experiments that will allow you to prove it to yourself. Would you believe it then?

I feel responsible for starting a fight between you two. So let me try and clear somethings up.

JonA,
Your text contracdicts itself by saying that mechanical energy is not conserved in several places and then it says that energy is conserved. The first law of thermodynamics states taht energy in equals energy out. Engergy is always conserved. You are also incorrect when you said that a system can have tons of kinetic energy and no momentum. Kinetic energy is a function of mass and velocity. Kinetic energy and momentum are related by the square of the velocity.

Now to the argument at hand.

1.) Momentum is a vector quantity; therefore, it acts through the center of gravity of a mass. Since the shoulder area of an elk is most likely not it center of gravity the conservation of momentum theory is in grave jeopardy of producing rather errroneous results.

2.) The elk is also not a ridig body. When the bullet strikes it is not going to move the entire elk i.e. slide it bodily across the ground. What the bullet will do is accelerate the mass adjacent to the pont of impact. In this case the shoulder. The hindquarters and head/neck will likely not move much due to Newton's third law. That would be inertia for the non scientific. The mass of the shoulder is much less than the animals total mass.

3.) Since the momentum of the bullet is not applied to the center of gravity of the elk (an assumption but one I feel is valid) we must determine the force that the bullet strike imparts to the elk. We know the mass of the bullet so that leaves the decceleration for us to calculate. It is the determination of the decceleration that causes us to consider the bullets kinetic energy. With this information coupled to the knowledge of the location of the animals center of gravity we can determine the torque being applied to the elk and the time interval over which that torque is applied. This will allow us to draw a conclusion as to whether or not there is a sufficient amount of torque applied quickly enough to knock the animal over.

4.) The energy of the bullet will provide the decceleration by a thorough understanding of the loss and/or transformation of the bullets kinetic energy as it travels through the animal by establishing the rate at which velocity is lost.

I would go on but my internet provider is screwing with me this morning. That is why there are so many edits they are like saves.

Todd E

[This message has been edited by Todd E (edited 01-19-2002).]

[This message has been edited by Todd E (edited 01-19-2002).]

I clearly understand the distinction between the internal energy of a body created by the motions of its elementary particles and the gross motion of that body. Obviously, a body at rest can (and does) have internal energy, but that is not the body's kinetic energy.

Only perfectly elastic rigid bodies obey conservation of momentum in the strictest sense. Any source claiming otherwise is incorrect, be it Newton himself. Since there are no real perfectly elastic bodies, conservation of momentum is best understood as a gross approximation of the truth. All our modifications for "elastic energy" and "inelastic energy" and friction are fudge factors. We maintain the assumption of conservation of momentum in such cases not because it is inviolate, but because it is often much simpler to do so (and sufficiently accurate for the purpose at hand) and then to approximate the energy losses due to these other effects than to first determine the energy losses and then to calculate the residual motions. However, the latter approach is the correct one.

I return (one last time) to my earliest remark about textbooks. Books often make statements based on conventional methods while failing to make clear that these methods are based on convenient assumptions and not based on the reality. In other words, they are wrong. In many cases the authors are blissfully unaware of their mistake. In general it does not matter to them because their aim is to to illustrate a method, not to debate the niceties of the Third Law of Thermodynamics and the motions of complex systems with inherent resistive losses.

Todd, your analysis is (from the standpoint of the original discussion) exactly why a penetrating projectile will have little effect on the gross motion of a body in this case. Hitting a steel ram (a nearly rigid body) is one thing, but a living animal is a loose linkage of members, all attached with multiple degrees of freedom and most capable of exerting force in several (if not all) of those degrees. Its always a lot harder to drag a dead deer into the back of a vehicle than to just lift the dead weight too!

As I recall all I ever said was that the conservation of momentum approach was a gross simplication of the problem. On this I believe we agree. Actually, I agree with everything you have said.

No to the original question asked. The animal may well get knocked over by the impact if it generates a yaw rate greater than the animal is capable of compensating for. It is my experience that with a "soft" lead core bullet in calibers of this size that an animal that is unaware of your presence and hence the shot is knocked over.

Todd E

quote:

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Your text contracdicts itself by saying that mechanical energy is not conserved in several places and then it says that energy is conserved.

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I've tried to be very clear that it's kinetic energy that is lost in a collision, not total energy of all types. If you have 1000 Joules of kinetic energy before and 100 Joules of kinetic energy after as well as 900 Joules of thermal energy, then kinetic energy was not conserved.

quote:

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You are also incorrect when you said that a system can have tons of kinetic energy and no momentum. Kinetic energy is a function of mass and velocity. Kinetic energy and momentum are related by the square of the velocity.

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Consider a system that consists of two balls of the same mass that are heading straight for each other with the same speed. As you correctly pointed out, momentum is a vector quantity. Add the momentum of each ball to find the total momentum of the system and you'll find it's zero.

A joule is not a vector quantity. Total kinetic energy of the system is the sum of each ball's energy--not zero.

Let's say they are billiard balls. After they collide the will bounce away from each other at nearly the same speed of approach. The total kinetic energy will be nearly the same (collision is close to elastic) but slightly lower. The momentum of the system is still zero.

Say they are oranges. They will bounce away with a much slower speed of departure. Much of the kinetic energy is lost (inelastic collision). The momentum of the system is still zero.

Say they are balls of clay that stick together, stopping at the point of collision. Remaining kinetic energy is zero (perfectly inelastic collision).

Guess what, the final momentum of the system is still zero. It's zero before the collision and zero after the collision in all three cases.

Your point about a bullet acting on a portion of an animal and causing a torque about the center of mass is a good one. But, if you try to say the animal is picked up and thrown for some distance by the bullet, you need to consider all the animal's mass.

Harald,

I don't know what to say. If you want to believe that 2 + 2 = 5, the world is flat, etc, I guess I won't be able to convince you otherwise.

If you want to say that Newton's Laws of Motion are wrong, that every Scientist or Engineer for the last 300 years has been wrong, so be it. Just don't expect anybody that studied Physics beyond the high school level to agree with you.

In the other thread you mention that torque and energy are measured in the same units and try to convert one to the other...egad! !!!

This shows a fundamental lack of understanding of work, force and Physics in general at the most basic level. It certainly shows that you are not qualified to disprove anything in a Physics book, much less Newton's Laws of Motion.

I was too polite to point that out before, but I can only take somebody repeating "2+2=5" so many times before it starts to irritate me.

I have no desire to argue this further. In my last post I stated what I believe is the reason that an animal is knocked over. that being that a yaw rate is imposed upon the animal by the impact of the bullet that the animals brain cannot comprehend and compensate for in the time allowed. The animal then loses it's balance and falls over. The bullet is then "knocking the animal off it's feet".

I do not know how many BIG guns you have shot game animals with. I have shot many although not with a 585 (I've used 500's) they always knock the animal down. That doesn't mean they don't get back up if the hit wasn't good though.

I hope if nothing else you understand that this is a much more complex problem than knocking over a steel silluette target. That was all I was really trying to say. I do have considerable experience with impact analysis albeit on structures. It has become a considerably more enjoyable field of endeavor since we have developed proprietary CAE programs to work out all the details!

Todd E