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December 17, 2009, 04:29 PM | #26 | |
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We also know that from the work-energy relationships that work done is equal to the change in energies. Work is also equal to force over distance (or the time rate of change of momentum). The target downrange will adsorb all the energy over a shorter distance than the shooter, and because the rate of momentum change is much higher, will see a higher force number than the shooter... Think of it this way... If you have two objects of the same momentum, and one applies a force over a short distance (the bullet) and one applies a force over a longer distance (the rifle against your shoulder), the work done must be equal in each case. To compensate, the force for the shorter distance has to be higher than the longer distance. And remember, its time rate of change of momentum... The rate of change is higher for a bullet than it is a rifle... Now, the problem is that this all uses particle dynamic maths, which is not an accurate representation of what happens when a bullet hits a body. That would require a continuum dynamic model to truly show what kind of force number you'd see...
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December 17, 2009, 04:31 PM | #27 |
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This was best described to me in a thought experiment of sorts... We all know that energy (and mass etc) cannot be created, only transferred. Thus, in a pure (as in no outside variables) environment, the bullet would exhibit the SAME force in both directions (as per equal and opposite reactions). So if it were to knock the victim down it would also have to knock the shooter over. But as many have pointed out, you have a stock, grip, shooter's stance versus victim's balance, etc etc etc. When I answer this question, I say the above: "if it were to knock the victim down it would also have to knock the shooter over". Obviously though if the shooter is a brick wall and the victim is a cucumber balanced on a board this does not hold
Also internet physics argument lol |
December 17, 2009, 05:17 PM | #28 | |
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All of the math aside...
Quote:
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December 17, 2009, 05:25 PM | #29 | |
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Then there is heat energy absorbed by the gun and expelled by the gun that never gets transfered to the shooter or to the bullet/impactee. All energy must be accounted for - I think I've addressed most of where the energy goes. Just try to visualize it this way...if you threw a punch at someone with a fist shaped like a 9mm, but one foot in length, what do you think the point of impact would look like. And, that punch is going to have quite a bit less energy than bullet fired from a gun. So, I'd expect to get a degree of penetration into human flesh/bone...but without the same level of destructive energy transfered. Last edited by Skans; December 17, 2009 at 05:34 PM. |
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December 17, 2009, 05:38 PM | #30 |
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impact
How many of y'all have been shot???
I havent but I imagine it would hurt |
December 17, 2009, 05:38 PM | #31 | |
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The OP was referring to effect, if I'm not sadly mistaken. Bah, this stuff gives me a headache. I'll leave it to the experts.
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December 17, 2009, 05:43 PM | #32 | ||
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December 17, 2009, 06:08 PM | #33 |
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....
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December 17, 2009, 06:31 PM | #34 | |
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There have been some real good explainations of KE, momentum, velocity... Feels kinda like I just sat thru a physics class. But the question could have been answered very simply.
Answer to question: Quote:
I don't understand how this is even a debate. Go to a recoil calculator like the one found here:http://www.handloads.com/calc/recoil.asp Then look up the KE produced by your bullet. Not even close. If my rifle produced 3400 ft/lbs into my shoulder...no. Dumb arguement.
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December 17, 2009, 06:58 PM | #35 | |
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But I do get you point
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December 17, 2009, 10:31 PM | #36 | |
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December 18, 2009, 04:34 AM | #37 | ||
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Wait... 1oz sabot? Are you referring to a slug, saboted bullet, or the always improperly referenced anti-tank projectiles? Or... the actual sabot hitting the target....?
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December 18, 2009, 05:32 AM | #38 |
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If your wife gets mad at you and hits you in the head with her shoe it hurts and may leave a mark.
If your wife gets mad at you and hits you in the head with her stiletto heeled shoe it will hurt more and probably leave a hole. Same mass, same velocity...different surface area. |
December 18, 2009, 07:00 AM | #39 |
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PSP: that reminds me of an example my chemistry professor in college used to explain pressure (ie. psi, atm, etc). He asked how many of us would mind resting our textbook on our heads for the remainder of class for extra credit. Naturally most of us said "sure!" Then he told us to do so but rest the book on the eraser of a pencil with the point on our head. Obviously we all declined the offer since no one wanted a pencil in the skull. A good illustration of the principle of force per unit area (pressure). In this case our 3 lb book over the area that it contacted our head would not be so bad. But take that same 3 lb book over the area of the tip of the pencil and the pressure goes through the roof (and a pencil in the head!).
That doesn't have anything to do with momentum or KE though. A good lesson nonetheless.
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December 18, 2009, 07:20 AM | #40 |
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Depends on if you are the shooter or the person being hit.
You will have a difficult time convencing a person hit by a bullet that the shooter felt the same impact. I deal with physics all the time and never gave this any thought. BUT I have also been in the presence of a couple grown men that were shot as I was next to them. Neither knew they were shot until minutes later. This could be due to caliber, entry area or many other things. I have read reports from police and medical experts as to what happened after being shot. Most agreed that the caliber shooter used will have a direct bearing on how the person being shot reacts. That said, I would rather be hit by a .22 than a .50. I would rather be the shooter than the target. No matter the recoil, I could take it better than I could stand the impact. There are too many variables in such a question. If any variable changes then the results with change. |
December 18, 2009, 07:30 AM | #41 | |
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December 18, 2009, 08:43 AM | #42 |
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Apples and Newton and shells and "foot pounds"
For each force there is an equal and opposite force.
When the apple fell onto Newton’s head his head received no damage, other than muss his hair. The apple busted. Another factor in the rifle recoil is the gas expelled from the muzzle after the bullet has left the barrel and is no longer subject to the expanding gas. This means the rifle is experiencing a reactive force which is not caused by accelerating the mass of the bullet. This expelled gas also has force. The mass is small but the velocity is large. This is the principle by which rockets operate. To demonstrate this effect you might pull the bullet from the case and plug the end or use a blank, and fire the rifle. There will still be substantial recoil. Perhaps not as much as with firing a bullet, but there will be recoil. This does not take into consideration the total mass accelerated forward and the total mass accelerated back. But it does apply to the recoil force felt by the shooter and the impact force of the bullet. So, stick a modified shell in that there clip and see how many of them there “foot pounds” you get. |
December 18, 2009, 09:46 AM | #43 |
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Very entertaining thread!
I'm not a physicist nor a mathematician; strictly a layman. And my layman's understanding is that some of a bullet's energy is expended during flight due to air resistance, and it doesn't "hit" as hard because its shape and frontal mass are so narrow that it punches through its target rather than transfering all its retained energy directly, as it would if it flattened so completely that it didn't penetrate the surface. Which is why there is so much discussion of bullet expansion, and wound channel, etc. And think of the retained energy of a rifle bullet that passes completely through its target and travels on; the shot that kills the second deer after passing through the one standing in front certainly didn't transfer all its energy to the first deer! I've never been in combat, and hope never to be. But persons I know who have say that men who are shot by bullets drop; they aren't thrown back. So my layman's answer to the OP's original question: No, the impact of being hit by a bullet isn't the same as the recoil felt by the shooter.
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December 18, 2009, 10:44 AM | #44 |
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The recoil energy (what the shooter feels) is easily computed.
You conserve momentum between the ejecta (bullet and powder weight) and the gun itself. This produces a velocity for the recoiling gun. This velocity is then used with 1/2*m*v^2 to compute the kinetic energy of the recoiling gun. This is the energy the shooter must absorb. The bullet has a much higher energy than this, since while its mass is a fraction of the guns, its energy is based on v^2. You need a lot of assumptions and simplifications to determine the force delivered by the bullet. If a bullet had 2000 ft-lbf of energy, and you brought it uniformly to a stop in 1 foot, the force delivered over that foot would be 2000 pounds. If you know the bullet diameter you can obtain a pressure (force / area). The problem is that we are far from uniform, and bullets in mixed tissue do not "uniformly" come to a halt. The loss of energy is not 'smooth' since the materiel is not uniform. Conservation of momentum works all the time, but does not indicate energy. Conservation of energy is a much larger problem, since ALL sources of energy (including the energy released by the gunpowder) must be considered, and ALL loss mechanisms (friction, heating, the work done on the bullet to accelerate it, even the work done in creating the muzzle blast, work done deforming bodies). Conservation of energy is often not a useful calculation to bother with since there are so many loss mechanisms that must be accounted for. |
December 18, 2009, 11:18 AM | #45 |
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Wow. This is a better turnout than I had dreamed. Thanks everyone for pitching in. Here are some of the points, that in terms of my own argument with my potential future father in law, were very salient:
Theoretically, and at least initially, the answer is yes. Equal forces pushes in both directions. Good points were made about energy loss during bullet flight from friction with the atmosphere, indicating that, to some extent, the bullet has less energy when it reaches the target. At first, I thought this might not be so relevant. How much energy could a bullet in flight really lose? The answer is related to the weight of the bullet, but the answer is potentially a LOT. The counter point is that the mechanical parts in the firearm will also absorb some of this force – especially if the firearm is a semi auto . But even if not, stocks, grips, and even bolt housings and threading will absorb a small amount of force as the all move a tiny amount. There are several comments (and a mention of a thought experiment) that I think give me some serious “ammunition” for the next time we get into this: rsgraebert: “The firearm weighs an order of magnitude more than the bullet.” wpcexpert: “Go to a recoil calculator like the one found here:http://www.handloads.com/calc/recoil.asp” I looked at the calculator, realized an important aspect of the calculation was the weight of the firearm, was thinking about rsgraebert’s magnitude comment, and I came up with the following thought experiment: If a ruger 10/22 weighs five pounds, that’s actually 35,000 grains. That would be THREE orders of magnitude greater than the weight of a standard 22lr round. So the thought experiment is this: Imagine if you took your bullet and made it a THOUSAND times heavier before you shot it. Imagine, if instead of a bullet, what you were firing was the rifle itself. If the force of the propellant was used to accelerate something the same size and weight as the rifle, the effect when it hit your target would probably NOT be particularly brutal. It might just be (ignoring atmospheric friction and mechanical force absorption on the part of the firing rifle) about the same impact as the recoil experienced by the shooter. Theoretically, the same concept could be extended to ANY firearm. The subjective experience of recoil is less for rifles because of their greater weight than firearms (assuming the same caliber cartridge). If you were to take a .40 pistol, and use it to shoot a round that was the same size and weight as the pistol (physically impossible – I know, this is just theoretical), then I would wager that the impact of that pistol sized round hitting a target would be very comparable to the force felt when firing said pistol (although the feeling of being struck by something, compared to holding something that jerks, would of course be different). Anyway – just my two cents, given all the great feedback. |
December 18, 2009, 11:34 AM | #46 |
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Wow!! 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. As for me-- 3 AK rounds in the legs, soft tissue wounds,,no bone, don't remember "felt" impact, only the burn and the pain, lots of blood, screaming--one minute I'm up--next second I'm down--world of ****. Lots of noise, smoke, and not once did I wonder if the "slope"that lit me up had recoil that was equal to the impact of the rounds that took my legs out from under me. Could be I'm just having a bad day, or maybe I think if ya don't have first hand experience ya should just go back to watching Rambo on tv. I'm done !!!
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December 18, 2009, 12:37 PM | #47 | |
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2.Recoil is absorbed initially by a much larger surface area then transferred through most of your body allowing you to absorb the force like a spring. 3.The energy is delivered much slower as it accelerates the gun and the bullet. The bullet hits a very small area. You are not braced for it and not standing such that the energy is absorbed throughout your body. The bullet has negative acceleration over a few inches in the body at a much faster rate as opposed to 16"-30" out of the gun. |
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December 18, 2009, 01:41 PM | #48 | |||
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December 18, 2009, 02:17 PM | #49 |
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Law enforcement students in this area see a film titled "Fire Power" I have viewed it several times and replayed many segments over and over. One was all about what this thread is about.
The shooter with a 308Win. The shootiee, is protected with a plated vest designed to withstand a 458 win Mag. and is standing 8-10 feet in front of the rifle. he is shot in the center of mass and there is no discernible reaction as the flinch could be caused by many things, but the recipient stood his ground. The next shot the recipient was standing one legged and again he stood his ground. Everything that was discussed in conjunction to this exercise was supported in the previous post by Frankenmauser.
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December 18, 2009, 05:04 PM | #50 | |
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Hit the last 3 at 30 yards or less, they were running right at me. I stepped away from the tree I was standing against, they turned I shot, they all went right down to the ground. One had to be finished off as it was hit in the lower spine right before the hips. Lot of power in a 12 ga slug at that range. When I was shot in the left shoulder from behind it hit the bone and hurt like a fire was in my shoulder a very intense hot fire. not so much blood tho. |
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force , hit , physics , recoil , shot |
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