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Old December 18, 2009, 05:29 PM   #51
Hkmp5sd
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44 posts and not one old gun fighter chimed in with first hand info. Makes ya wonder what this old world is coming to when the only info appears from text bookers.
You find it odd that out of 44 posts, you are the only person to claim to have been shot? The vast majority of people I know have never even fired a gun in anger, yet I wouldn't care to stand in front of them while shooting. Amazing what you can learn without being shot.
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Old December 18, 2009, 06:42 PM   #52
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my 2 cents

I would way rather feel the recoil
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Old December 18, 2009, 08:34 PM   #53
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44 posts and not one old gun fighter chimed in with first hand info. Makes ya wonder what this old world is coming to when the only info appears from text bookers.
I'm no old gunfighter, but I'm reasonably confident that experiencing the impact of a fired bullet is undesirable. Similarly, I also don't want to get hit by a Peterbilt going 70 mph, but I don't think that it's necessary to experience it firsthand to suspect that it's probably going to be harmful.

I could be totally out of line with these observations, however.
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Old December 18, 2009, 09:55 PM   #54
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OK guys

My best pal in the world here, a very knowledgeable rifleman, reloader and hunter, took a .243 round fired by an idiot who mistook him for a deer, when he was 18. Obviously he survived, now in his late 50's, and it never phased his interest in shooting or hunting. He IS the safest, most careful man w/ a firearm I have ever met. He chooses his companions very carefully and I feel honored.

He discusses his experience pretty freely, though I have never pressed him. The time I was listening, he was joking about not being able to see the Alabama / Auburn game that afternoon!

I will ask him, discreetly, and report back, about what his wound felt like.
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Old December 19, 2009, 12:25 AM   #55
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"Time" is the key element here, and a few have noted it already. The gun has a larger surface area, and that helps, too. But the force is applied to the shooter over an interval of time as the powder burns, as the gun cycles, as the arm bends at the elbow, etc. Consider firing a revolver one-handed with your shoulder and elbow locked out; it won't kill you, but you'll get a different experience.

When the bullet hits the target, it doesn't get the advantage of any of that. The bullet hits a small surface area, very fast, where nothing bends readily.

Energy and momentum can be both conserved without contradicting that the two participants have different experiences.

BTW, the same principles apply to seat belts. Belts help spread out the area your body impacts in a car crash, but that's only part of the story. A big part is that they are spreading out the time of the impact so your body doesn't have to absorb it all at once.

It's the same momentum and energy either way; your body has to go from moving to stopped. If you have a seat belt on, the impact is over a (relatively) long period of time and a wide area. If you have a spear sticking from the steering wheel, it happens fast over a small area. But you give up your momentum and energy either way and come to a stop.
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Old December 19, 2009, 01:34 AM   #56
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Energy is conserved but it "goes in a lot of different directions" so it's hard to keep track of it all. Energy is basically the mass of a moving object times its velocity squared. Kinetic energy relates to the amount of work/damage that a moving object can do.

When a gun is fired, some of the energy moves the bullet, some is dissipated as heat, some as noise, some as light, etc. Some is expended moving gases around. As the bullet moves through the air after leaving the muzzle it gives up more energy to heat and noise due to drag.

The momentum of the recoiling mass is identical to the momentum of the ejecta at the moment that the ejecta leaves the muzzle. This is complicated a bit by the fact that not all of the gun recoils at the same time. In a recoil operated gun different parts of the gun recoil at different times and different speeds which makes it more difficult to calculate the momentum of the firearm as a whole.

Momentum is essentially mass times velocity. It relates to how hard it will be to stop a moving object--or looking at it from a different angle it tells how much an object in motion "wants" to remain in motion.

Force is not the same thing as momentum. Force is the momentum of an object divided by time taken to stop the object. This is why bracing a gun against a hard object may break the stock. By stopping the gun's recoil more rapidly, more force is applied to the stock which may result in the material failing.

So the force the shooter feels depends on the momentum of the gun and also on how fast his shoulder/body stops the gun.

On the other end, the bullet has less momentum than the recoiling gun by the time it makes impact--it has lost velocity to drag since it left the gun's muzzle. Maybe it has lost a significant amount, maybe not. Also, not all of the ejecta is bullet. Some of the ejecta is unburned powder and combustion gases--none of which has any effect on the person being shot in most cases.

Finally, the bullet may not be completely stopped by the person's body. But, if it is, it may be stopped faster than the recoiling gun was stopped by the shooter's shoulder. The result is that it's not so simple to determine if the force of the bullet on the shot is greater or less than the force of the recoiling gun on the shooter.

If the bullet were stopped in exactly the same amount of time that the shooter's shoulder stopped the recoiling gun then the person shot would feel LESS force than the shooter felt in recoil. That's because some of the bullet's velocity (and therefore momentum) was lost to drag and because only the bullet hit the person shot--the rest of the ejecta didn't and therefore affected the person shooting but not the person shot. So the force on the shooter = (mass of the ejecta) (velocity of the ejecta)/(time required to stop the recoil) and the force on the person being shot =(mass of the bullet)(remaining velocity of the bullet)/(time required to stop the recoil--since we made them the same for this particular example). Obviously the shooter feels more force than the person shot in this case.

If the bullet were stopped slower than the shooter's shoulder stopped the recoiling gun then the person shot would feel even less force. The force on the shooter = (mass of the ejecta) (velocity of the ejecta)/(time required to stop the recoil) and the force on the person being shot = (mass of the bullet)(remaining velocity of the bullet)/(time required to stop the bullet). Since the mass of the bullet is less than the mass of the ejecta, the remaining velocity of the bullet is less than the muzzle velocity and we said the time to stop the bullet is more than the time to stop the recoil, the person shot definitely feels less force than the shooter.

If the bullet were stopped faster than the shooter's shoulder stopped the recoiling gun then the person shot might feel more force or less force depending on whether the faster "stop" was enough faster to counteract the other losses. F=mv/t. Decreasing t makes F bigger but if v is enough smaller too it may counteract the effect of a smaller t. The force on the shooter = mv/t and the force on the person being shot =(smaller m)(smaller v)/(smaller t). Without knowing specific numbers you can't answer this one.

If the bullet isn't stopped at all but passes through the body then it's likely that the person shot would feel less force than the person doing the shooting. That's because the bullet exits with some of its momentum remaining. The force on the shooter = mv/t and the force on the person being shot =[(smaller m)(smaller v)-(remaining m at bullet exit)(remaining v at bullet exit)]/(time of passage). The t in this case would be the time required for the bullet to pass through the person's body. Again, without knowing specific numbers you can't answer this one for sure but if the remaining velocity at bullet exit is pretty large it's a good bet that not a lot of force was applied.
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Old December 19, 2009, 11:28 AM   #57
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peetzakilla: as someone who usually enjoys reading your posts let me say that rsgraebert is absolutely correct. Law of conservation of energy simply says that the energy released by burning powder is equal to sums of the energy of all involved components thereafter, such as kinetic energies of bullet and rifle. Law of conservation of energy by itself makes absolutely no implications as to how exactly this energy is distributed between the objects. In ideal, zero-friction environment, this last thing is controlled by the law of conservation of momentum which is completely, completely different story. As bullet exits the barrel, the momentum of bullet is equal in value to the momentum of the rifle. Which, among other things, means that the bullet (as the lighter object) carries a lot, lot more kinetic energy than the rifle.

Edit:
In case you want to pinpoint the flaw in your reasoning, let's go back to this statement:
Quote:
Conservation of energy is an absolute, indisputable law of physics. Conservation of momentum can not be violated. The total energy of the system is a constant. The energy transfered by the bullet to the target is identical to the energy transfered to the gun and shooter, discounting loses than can all be essentially traced to friction and/or noise.
This is where the problem is. You are right in that "Conservation of energy is an absolute, indisputable law of physics. Conservation of momentum can not be violated." But then you make this assumption -- "the energy transfered by the bullet to the target is identical to the energy transfered to the gun and shooter" -- as something derived from the Law of Conservation of Energy. It's not. Conservation of Energy means that combined energies of all moving parts plus energy the went into heat is exactly equal to the energy released by the powder. Once you start comparing mass/velocities of bullet and rifle, you've switched to the law of conservation of momentum.

And this:
Quote:
Momentum IS energy. It's not mathematically the same as KINETIC energy but it most certainly IS energy.
this really got me scratching my head ))

Just my 2 cents. As I mentioned earlier, I enjoy reading your posts.

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Old December 19, 2009, 02:23 PM   #58
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I would say yes and no.
With the firearm, you have the buttstock/grip with all those cubic inches of surface area to disperse the momentum/force/whatever. With the bullet, you have a very small amount of surface area. Because of this you get penetration and subsequent terminal damage.
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Aw, now I feel that I have to try to explain what I meant by the above, so I have to get all geeky and stuff.

****GEEK ALERT****
(MAY CONTAIN GEEKY STUFF)

We know that in English units, momentum is expressed by the unit foot-lbs (mass) per second or ft-lb/sec.

But wait a minute. We also have to remember that momentum is the same thing as force multiplied by time

Momentum = lbs (force) x sec (time)

So let's say we shoot into a target at point blank range with a .458 Winchester Magnum. Conveniently, the bullet diameter of a .458 is .458", so we will be able to figure the one area which will be affected by the momentum. Since the area will involve a circular shape, we know this will be (.458/2)^2 x 3.14 or about .165 square inch.

Just for the sake of simplicity, I'll assume that our buttstock will be a rectangle of the dimensions 2"x 4". We know that the area will then be 2x4 or 8 square inches. This is the other area.


Regardless of what value we get for the momentum, when we divide by the different surface areas I think it's easy to see that what we have now is a unit of pressure (pounds per square inch) multiplied by seconds. So when you apply the momentum over a smaller surface area, you will have a significantly larger amount (over 48 times) of pressure than on the larger area, assuming the amount of time is the same.

Without making a whole series of calculations that no one cares about, this is what I was trying to illustrate.
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Old December 19, 2009, 04:11 PM   #59
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Momentum = lbs (force) x sec (time)
No, the time ends up in the denominator.

Mass * speed, and speed is length/time.

Momentum is NOT energy, and is not expressed in the units of energy.
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Old December 19, 2009, 04:24 PM   #60
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If the bullet were stopped in exactly the same amount of time that the shooter's shoulder stopped the recoiling gun then the person shot would feel LESS force than the shooter felt in recoil.
Not correct.
The bullet is carrying the energy it received from the expanding powder gases, a far larger number than any recoil energy.

If you stopped a 2000 ft-lb energy projectile in 1 foot it would exert 2000 pounds of force.
If you stop in in 6 inches, it would exert 4000 pounds of force.
Stop in in 0.5 inches and you get 48,000 pounds of force.
This is right up their with industrial punch presses, and on thinner metal has the same effect - it punches a hole though the metal.

Energy is the ability to exert force over a distance thus the units of foot-pounds(force).
Push on something with 10 pounds of force and move it for 1 foot and you have expended 10 foot-pounds of energy.

When we introduce time into the mix, we can get power, like horse power.

1 horsepower is 33,000 ft-lbf/minute.
Any combination of distance, force, or time that nets out 33,000 ft-lbf/minute is 1 horsepower.

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Old December 19, 2009, 05:27 PM   #61
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No, the time ends up in the denominator.

Mass * speed, and speed is length/time.

Momentum is NOT energy, and is not expressed in the units of energy.
<Gaah!!!>


What is force?

Our textbooks told us:
Force is mass x acceleration (feet per second squared in English units.)


Acceleration is what?
Our textbooks told us:
Velocity divided by time
Or Feet per second per second in English units

So:

Force = mass x feet per second/second

Using high school algebra, multiply both sides of the equation by the unit second to simplify:

Force x second = mass x feet per second.

Now what does mass times feet per second equal?
That's right, the momentum.

Who the heck said energy? Me?
Nope!
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Old December 19, 2009, 10:56 PM   #62
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Not correct.
The bullet is carrying the energy it received from the expanding powder gases, a far larger number than any recoil energy.

If you stopped a 2000 ft-lb energy projectile in 1 foot it would exert 2000 pounds of force.
I was very careful to speak NOT in terms of the DISTANCE required to stop the bullet but rather in terms of the TIME required to stop the bullet.

IF the bullet stops in exactly the amount of time that the rifle's recoil is arrested then you can use conservation of momentum to see that the rifle applied more force to the shooter than the bullet applied to the shootee.

Your examples are (correctly) pointing out that it is highly unlikely that the bullet will stop in the same amount of time as the rifle was stopped by the shooter's shoulder but I didn't claim that it was realistic to expect the bullet to stop in the same amount of time that the rifle was stopped by the shooter's shoulder. It was just one of several thought experiments to help people get a feel for how momentum and force are related.

If you work through what I said the way I said it you will see that it's correct.

What you said is also correct but your post is looking at it from the perspective of energy and distance instead of from the perspective of momentum and time as my post was.
Quote:
Force = mass x feet per second/second

Using high school algebra, multiply both sides of the equation by the unit second to simplify:

Force x second = mass x feet per second.

Now what does mass times feet per second equal?
That's right, the momentum.
I'm not sure where you're going with this. You are agreeing with brickeye's correct definition of momentum.

Force x time = momentum = mass x velocity = mass x distance / time
Force = momentum / time = mass x velocity / time = mass x distance / ( time x time) = mass x acceleration
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Old December 20, 2009, 01:45 AM   #63
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no they are not the same. Shoot a .357 mag out of a 2" barrel. Now shoot the same round out of a airlight model of the same size. In both cases the bullet imparts the same energy on the target, but the shooter feels 2 very different recoils.

p.s. I'm gonna mythbuster it up. So we know that small weapons fire will most likely not knock someone through a window. My question is what is the smallest caliber that will knock a person through a window.
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Old December 20, 2009, 03:56 AM   #64
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I'm not sure where you're going with this. You are agreeing with brickeye's correct definition of momentum.

Force x time = momentum = mass x velocity = mass x distance / time
Is that anything different than what I pointed out? That should be exactly what I pointed out. But, hey, give Brickeye the gold star. No biggy.

I was disagreeing with this, or at least what I was understanding about this because the statement was rather vague:

Quote:
No, the time ends up in the denominator.
It ends up in the denominator of what?

If you want to see "where I was going", please look at my post #58. That was all.
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Old December 20, 2009, 04:30 AM   #65
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Is that anything different than what I pointed out? That should be exactly what I pointed out. But, hey, give Brickeye the gold star. No biggy.
I said that you were both saying the same thing. What I didn't understand was why you seemed to be saying it as if you disagreed with him.
Quote:
If you want to see "where I was going", please look at my post #58. That was all.
Ok, I get it.

In post #58 you expressed Momentum in terms of lbs and time. The confusion arose because at first glance it's easy to confuse lbs with mass which made it look like you were saying Momentum was mass x time.

Normally Momentum is expressed in terms of mass and velocity which would put time in the denominator--> mass x distance / time.

Anyway, you're both right. Time can be in either the numerator or the denominator depending on what units you choose to express momentum.

Force x time = momentum = mass x distance / time

Of course, if you break it down to the base units and simplify then time ends up in the denominator because the time in the numerator cancels with one of the times in the denominator in the F x t expression since F = mass x distance / (time x time).
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Old December 20, 2009, 04:45 AM   #66
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Quote:
What I didn't understand was why you seemed to be saying it as if you disagreed with him.
On the contrary, he seemed to be disagreeing with me.

I said this:

Quote:
Momentum = lbs (force) x sec (time)
Then he said:

Quote:
No, the time ends up in the denominator.

Mass * speed, and speed is length/time.
He seemed to have been talking about oranges when I was talking about apples.

Dang, now this is why I get headaches during these discussions!

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Old December 20, 2009, 04:54 AM   #67
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On the contrary, he seemed to be disagreeing with me.
Understandable. People get used to thinking about things in base units--it's confusing to see it expressed other ways.

For example, if I were to start referring to a day as being 86.4Kohm-farads I would expect to be corrected a few times even though I would be technically correct to do so.
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Old December 20, 2009, 08:43 AM   #68
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Not correct.
The bullet is carrying the energy it received from the expanding powder gases, a far larger number than any recoil energy
Nope. Remember the law - For every action there is an EQUAL and opposite reaction.
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Old December 20, 2009, 02:19 PM   #69
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Nope. Remember the law - For every action there is an EQUAL and opposite reaction.
You think the bullet energy is the same as the recoil energy?


The momentum of the ejecta (bullet and powder gasses) and the rifle ARE equal.

The energy is FAR higher for the bullet since the potential chemical energy of the powder has been used to accelerate it.

Recoil energy and bullet energy are both easily computed numbers, and are NOT the same.

0.5 * m * v^2 allows fast moving objects to have very significant energy.

Momentum is mass * v, and IS conserved.

Energy is only conserved in the 'big picture.'

The powder energy is lost as heat, barrel friction, air friction acting on the bullet, barrel pressure at the moment of bullet exit (it cannot do significant work in the bullet once clear of the barrel).

All energy eventually degrades to heat.

You cannot get more energy out of the bullet than the powder contained, but that is about the limit of 'conserving energy' in a gun.
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Old December 20, 2009, 11:54 PM   #70
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The energy is FAR higher for the bullet since the potential chemical energy of the powder has been used to accelerate it.
This is misleading. The chemical energy of the powder simultaneously accelerates the pistol backward and the bullet forward. It pushes with exactly the same force in both directions.
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Old December 21, 2009, 12:21 AM   #71
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The chemical energy of the powder simultaneously accelerates the pistol backward and the bullet forward. It pushes with exactly the same force in both directions.
That's true, but it doesn't mean that the recoiling gun and the bullet have the same energy, only that they have the same momentum.

You have two force equations that are equal. One is the force on the gun and the other is the force on the bullet.

The force on the gun will accelerate the mass of the gun and the identical force on the bullet will accelerate the mass of the bullet. HOWEVER, since the two masses are not the same they will not be accelerated to the same velocity. Think about two cars being driven by the same force for the same amount of time. The heavier car will not achieve the same velocity as the lighter car.

So why is the momentum the same while the energy is different? Because momentum is linearly dependent on velocity and mass while energy is linearly dependent on mass but is dependent on the SQUARE of the velocity.

Ok, let's simplify the problem by saying that nothing but the bullet comes out of the muzzle of the gun. (Ignore any unburned powder or gases that exit the muzzle.)

Let's say we have a 7lb gun that fires a 150gr bullet at 2700fps.

The momentum of the gun and the bullet at the instant that the bullet exits the muzzle must be equal according to conservation of momentum. So the product of the mass of the bullet and the velocity of the bullet at the instant it leaves the muzzle must equal the product of the mass of the gun and the recoil velocity of the gun.

So Gun Mass x Gun Velocity = Bullet Mass x Bullet Velocity

That means the recoil velocity of the gun is 8.27 fps.

Now that we know the mass and velocity of both the bullet and the gun we can calculate the energy of each.

The energy of the gun is 7.4 ftlbs and the energy of the bullet is 2427.7 ftlbs.

Clearly not the same.

If you want to work this problem from a conservation of energy standpoint you can do it but it will be MUCH more complicated.

First you start with the amount of energy released from the powder. Then you calculate and total everything done by that energy and if you've accounted for everything and done your calculations correctly the sum of everything done by firing the gun and the amount of energy released by the powder will be equal. Some of the things you'd need to consider would be: the heat generated, the noise generated, the bullet energy, the energy of the gun, etc.
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Old December 21, 2009, 04:43 AM   #72
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This thread is too geeky for even me!
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Old December 21, 2009, 04:52 AM   #73
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Quote:
This thread is too geeky for even me!


Me too. And I even contributed!

I don't understand what my own calculation means, if anything. I just noticed it.

See, it was late one night, nothing much to do...

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Old December 21, 2009, 05:27 AM   #74
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A Pearl From Da Sarge

Why don't you ask someone who's been shot? I've been told by the few people that I have met who have survived a shooting that at first they don't realize that they've been shot, and then the body reacts in shock, and down they go! Just remember something about firing a round; when a bullet hits a SOLID target, depending on just how heavy or muscular the victim is can attenuate the effect of the round. And also one must take into account the bullet itself. If you hit with a high-speed round in 9mm, and a solid ball or FMJ, then it may not tumble well or "crumple" and make a nasty wound channel. But a .38sp with a 125 grain hollow point will expand up to .63 caliber and will "stick in" the wound channel itself, sending all of the force of the round into the body of the victim and causing a LOT of damage due to shock and massive wound trauma.
You're right about the amount of force of the round (foot pounds of energy) hitting the target, but as the round loses a bit of it's energy traveling from the weapon to the target, it lands with something less than what it started out with. Not much less, maybe, but something less.
And when it hits the target all those foot pounds don't have an "immediate" effect on the target. As you can now tell, there will always be different variables generating a differing range of effects, and remember that the diameter of the round is a major one. And the tendency of the round to make an effective wound channel is dependent upon the characteristics of the bullet itself: hollow-point vs full-point, full metal jacket vs semi-jacket, solid head vs partition-nosed, and so many more. And all this has to do with the effects of the impact.
Almost too much to want to trouble oneself with, ennit?
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