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 November 1, 2022, 10:50 PM #1 tangolima Senior Member   Join Date: September 28, 2013 Posts: 3,808 Physics of shooting a rifle It is supposed to a fun discussion. Let's relax and enjoy the exchanges. No need to go after each other's throat here. Did pencil and paper doodling during lunch break. The calculation results amazed me. Caliber: .30 Barrel length: 24" Muzzle velocity: 2500fps Bullet weight: 150gr The bullet is accelerated from 0 to 2500fps in distance of 24". It must be a hell of acceleration. How much is it? What sort of force is needed for such a ride? Let's assume constant acceleration to simplify things. -TL Sent from my SM-N960U using Tapatalk
 November 2, 2022, 01:33 AM #2 JohnKSa Staff   Join Date: February 12, 2001 Location: DFW Area Posts: 24,971 Force equals mass times acceleration. Acceleration equals velocity change per second. Feet per second per second. If you assume acceleration is constant, that is the same as assuming that the velocity increases linearly and therefore the average velocity in the barrel is the average of the starting velocity and the muzzle velocity (0 fps + 2500 fps)/2 or 1250fps. 1250fps average velocity over 24" (2ft) means that the bullet is in the barrel for 2 ft /1250fps or 0.0016 seconds. The velocity changes 2500fps in 0.0016 seconds which gives us 2500fps/0.0016s= 1562500 fps/s² average acceleration. The mass being accelerated is quite small. 150gr is 0.000666 slugs. There are 7000 grs in a pound and 32.174 pounds per slug. The slug is the unit of mass in the imperial system. The average force applied to the bullet while it is in the barrel is, therefore, about 1040.6lbs force. That's 0.000666slugs x 15625000fps/s = 1040.6 We can also look at the force applied to the bullet from a pressure perspective. The maximum discharge pressure for the rifle cartridge you describe is probably around 50,000psi. The back of a .30cal bullet is about 0.0745 square inches. With 50,000psi (pounds force per square inch) acting on the bullet at the maximum pressure, the maximum force on it would be about 3725lbs force. 50,000 pounds per square inches x 0.0745 square inches We can see that the maximum force is quite a bit larger than the average force we calculated. That makes sense because the pressure drops as the bullet moves down the barrel. The gas volume must fill more space as the bullet gets farther down the barrel and that means the pressure keeps dropping as the bullet continues to move. It's not quite that simple since the combustion process isn't instantaneous and some gas is still being generated as long as it continues, but thinking of it in terms of gas filling a space that is getting larger is an easy way to get a picture of what is happening. That means, of course, although we assumed constant acceleration for simplicity, the acceleration obviously isn't constant. The force (and therefore the acceleration) varies with the pressure applied to the bullet. The pressure is zero at the instant of firing, then climbs very rapidly to the maximum pressure and then drops as the bullet goes farther down the barrel. The instantaneous force on the bullet and the instantaneous acceleration both change as the pressure changes. I went through this pretty fast. I'll check it over tomorrow, but this is someone's chance to beat me to the punch and see if they can find errors. That should be a lot more fun all the way around. __________________ Do you know about the TEXAS State Rifle Association? Last edited by Unclenick; June 29, 2023 at 10:50 AM. Reason: put the square symbol ² in.
 November 2, 2022, 07:33 AM #3 mehavey Senior Member   Join Date: June 17, 2010 Location: Virginia Posts: 6,883 Correctomundo... but I prefer a simpler presentation:
 November 2, 2022, 07:49 AM #4 JohnKSa Staff   Join Date: February 12, 2001 Location: DFW Area Posts: 24,971 Yes, I suppose it's better to express acceleration in g's. The average acceleration in the barrel is 48,564 g's, or 48,564 times the force of gravity. It's also possible to calculate the average force on the bullet in the barrel using the muzzle energy. Force is equal to energy divided by the distance over which the energy is applied. In this case the muzzle energy is about 2081ft-lbs and we know that energy was generated over a distance of 2 feet. 2081 ft-lbs / 2 ft is 1040.5 pounds force. One can also calculate the average force on the bullet in the barrel using the muzzle momentum since average force is equal to momentum divided by the time over which the momentum change occurred In this case the muzzle momentum is about 1.665 pound feet per second. We earlier calculated that the bullet is in the barrel for about 0.0016 seconds. 1.665 pound feet per second / 0.0016 seconds is 1040.6 pounds force. __________________ Do you know about the TEXAS State Rifle Association?
 November 2, 2022, 02:16 PM #5 stinkeypete Senior Member   Join Date: July 22, 2010 Location: Madison, Wisconsin Posts: 1,291 The calculations above assume that the velocity increases linearly along the length of the barrel. http://www.ballisticsbytheinch.com/ You can apply your own best-fit function to some data and get brush off your calculus to take it to the next step, needless to say, the acceleration found above is a lower bound for peak acceleration! If anyone is interested I have a pristine copy of Understanding Firearms Ballistics- Basic to Advanced Ballistics Simplified, Illustrates and Explained by Robert A. Rinker Mullberry House Publishing 1999 Which is a nice book all about ballistics with only a basic understanding of arithmetic needed, he explains all you need to work stuff out. Will trade for books on pheasant hunting, grouse hunting, or dog antics... or what have you! Let's swap stuff, it's fun. __________________ My book "The Pheasant Hunter's Action Adventure Cookbook" is now on Amazon. Tall tales, hunting tips, butchering from bird to the freezer, and recipes.
 November 2, 2022, 02:28 PM #6 tangolima Senior Member   Join Date: September 28, 2013 Posts: 3,808 Right on. I have the same results. I recall an equation from high school physics thats link speed, distance and acceleration. Spent 10 min or so to derive it. V^2=2*d*a is the simplified form for zero initial speed. The energy approach is pretty neat. I was amazed by the acceleration, which is way over 10k G. We are taking about average acceleration. Peak value is much higher. No wonder it is so hard to put anything inside the projectile for guidance. The G will destroy everything. Shift gear, shall we? What can we do if we need to lower the acceleration? The force associated with the acceleration is huge; over 1000 lbf. As per Newton's 3rd law, the rifle is also subject the same force but in opposite direction. Does it mean the same force will eventually land on the shooter's shoulder? Hard to image a human can take 1000 lbf on his shoulder without hurting bad. -TL Sent from my SM-N960U using Tapatalk
November 2, 2022, 04:01 PM   #7
tangolima
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Quote:
 Originally Posted by stinkeypete The calculations above assume that the velocity increases linearly along the length of the barrel.
It is linear with time, not quite with distance. In fact speed is proportional to square root of distance.

V^2=2*d*a
V=sqrt(2*d*a)

-TL

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November 2, 2022, 10:31 PM   #8
JohnKSa
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 The force associated with the acceleration is huge; over 1000 lbf. As per Newton's 3rd law, the rifle is also subject the same force but in opposite direction. Does it mean the same force will eventually land on the shooter's shoulder?
The easiest way to look at it is through momentum.

The momentum of the bullet at the muzzle will equal the momentum of the rifle at the moment that the bullet exits the barrel.

Momentum is mass x velocity. The bullet weighs about 0.0214lbs and is traveling 2500fps at the muzzle.

0.0214lbs x 2500fps = 8.5lbs x rifle velocity

Solving for the rifle velocity says that it will be traveling about 6.3fps at the point that the bullet exits the muzzle.

We now know that the rifle started at 0fps at the moment of firing and accelerated to 6.3fps by the time the bullet exited the muzzle 0.0016 seconds later so we can now calculate the acceleration of the rifle.

6.3fps / 0.0016s is about 3939 fps/s (about 122 g's) which is a considerably smaller acceleration than what the bullet experienced due to the much larger mass of the rifle.

So how much force does the rifle exert on the shooter's shoulder?

Calculating that exactly is quite difficult since it depends on knowing how long (or how much distance) it takes the shooter's shoulder to decelerate the rifle's recoil velocity to zero.

When that happens, it's often useful to make an assumption that seems reasonable and see where that takes us.

Let's say that the shooter's shoulder gives about an inch in the course of absorbing the recoil velocity. With that assumption, we can take the recoil energy of the rifle and calculate the force applied to the shoulder since force is equal to energy divided by the distance over which the energy is absorbed.

The recoil energy of the rifle is:

(1/2) x (8.5 lbs / 32.174 slugs) x (6.3 fps ^2) which is about 5.25 ft-lbs of kinetic energy.

We assumed that energy is dissipated over an inch (1/12 foot) as the shooter's shoulder absorbs it. With that assumption, the force on the shooter's shoulder is:

5.25 ft-lbs / (1/12)ft which is about 63lbs of force.

If we assume that the shoulder and recoil pad combined absorb the energy in 2 inches then the force is reduced to less than 32lbs of force.

Understanding this calculation helps explain why you don't want to put a rifle stock up against an unyielding object and then fire the gun.

What happens if the rifle with no butt pad and a stock that has no flex is placed against a 1ft thick steel plate anchored to a concrete pad and fired? That would mean that all the recoil energy was absorbed in effectively zero distance. Dividing by zero doesn't work, so let's say that it is stopped in 0.001".

The force applied to the steel plate (and also the rifle stock) will then be:

5.25 ft-lbs / (0.001/12) ft which is about 62,964 pounds of force.

Obviously that's not going to be good for the stock...

This also highlights why a relatively thin recoil pad can still increase the comfort level of firing a rifle considerably.

It also helps explain why shooting a rifle standing is more comfortable than shooting from the bench. At the bench, the shooter tends to be leaning forward which puts the stock higher on the shoulder where there's likely to be less natural "padding". Also, the forward lean means that the entire body can't "give" as easily as it can while standing. And the sitting position means that the torso also can't "give" like it can in a standing position since the shooter's seat will hold their upper body in place.
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 November 5, 2022, 01:56 PM #9 tangolima Senior Member   Join Date: September 28, 2013 Posts: 3,808 Excellent explanations. Just a few things to add. The rifle is subject to the same huge force to accelerate the bullet. Newton's 3rd applies. It lasts long as the muzzle time of about 2ms, during which the rifle accelerates rearward. That 1000lbf goes to 2 places; to accelerate the rifle and to the shooter's shoulder. Let the rifle recoil freely and there can be no force on the shooter (during muzzle time). Zero rifle recoil, like resting bare stock on solid wall, the stock would probably split after a few shots. The conservation of momentum approach assumes zero force on the shooter during muzzle time, and all the force is to accelerate the rifle rearward to its peak velocity. And then shooter's shoulder catches this "sliding hammer" and try to slow it down to zero speed. It is certainly not the best way to deal with recoil force but it does have one advantage that we seldom talk about (rabbit for later). We all learn to shoot with shoulder on the butt stock, so that the rifle's rearward motion is slowed down as soon as the primer ignites. This reduces the recoil force on the shooter's shoulder. But there is a price to pay for this benefit (same rabbit hole). Ok before diving down a new rabbit hole, let's continue with this one further a bit more. Some say how you shoulder the rifle affect the muzzle velocity. Shooting on bench has significantly higher muzzle velocity than shooting offhand. It is true theoretically. But is the difference that significant? "It is physically impossible to shoot a man flying off his feet. Newton's 3rd law says so". Heard that numerous times I bet. But is it true? Maybe you may have noticed a crack in that theory in our discussion. 1000lbf on the rifle but only 70lbf ish on the shooter's shoulder. What gives? -TL Sent from my SM-N960U using Tapatalk
November 5, 2022, 05:10 PM   #10
JohnKSa
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Quote:
 It is true theoretically. But is the difference that significant?
In this case, the rearward velocity of the rifle is less than 7fps. If the rifle didn't move at all, that velocity would be added to the velocity of the bullet. Not significant.
Quote:
 1000lbf on the rifle but only 70lbf ish on the shooter's shoulder. What gives?
The force on the shooter's shoulder is dependent on how rapidly the recoil energy of the rifle is absorbed. If it were absorbed in exactly the same distance that the rifle moves rearward during the time the bullet was in the muzzle, then the forces would be equal.

But spread it out a bit with padding on the shoulder or the gun and then the force is reduced.

The same reason hitting a big lead ball with a 5lb melon and a 5lb hammer gives very different results even though the energy is the same. In one case all the energy is absorbed very rapidly, in the other case, the energy absorption is spread out over a long time.
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November 5, 2022, 08:23 PM   #11
tangolima
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Quote:
 Originally Posted by JohnKSa In this case, the rearward velocity of the rifle is less than 7fps. If the rifle didn't move at all, that velocity would be added to the velocity of the bullet. Not significant. .
I'm afraid it is not about relative velocity. By conservation of momentum, it can be derived that the ratio of kinetic energies between the bullet and rifle is same as their mass ratio.

Bullet is 150gr and rifle is 7.5lb. The ratio is less than 0.3%. Should the rifle somehow be kept stationary, the bullet's energy would increase by the same percentage. The muzzle velocity would increase by no more than 0.15%, or 4fps. Not significant indeed.

I agree with your descriptions on how the recoil force is reduced. Apparently Newton's 3rd law doesn't apply in such scenario, because the shooter's shoulder is not in "solid contact" with the butt stock. The shoulder doesn't accelerate at the same rate as the butt stock, and hence they experience different forces. Experts use Newton's 3rd law to rebutt gun myths. Sometimes the rebuttal is oversimplified or even erroneous.

While we are at it, is it physically possible to shoot a man flying off his feet? Physically means according to physics.

-TL

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Last edited by tangolima; November 5, 2022 at 08:29 PM.

November 5, 2022, 09:57 PM   #12
JohnKSa
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Quote:
 Apparently Newton's 3rd law doesn't apply in such scenario...
It does--the force on the stock is equal to the force on the shoulder and the force on the bullet is equal to the force on the rifle.
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November 5, 2022, 10:52 PM   #13
tangolima
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Quote:
 Originally Posted by JohnKSa It does--the force on the stock is equal to the force on the shoulder and the force on the bullet is equal to the force on the rifle.
But the force on the bullet doesn't equal to the force on the shoulder.

Here is the rebuttal to gun myth again. It is impossible to shoot a man flying as the shooter will be flying too. According to Newton's 3rd law, the force upon the shootee must equal to the force upon the shooter.

If we break down the big picture into smaller steps, we will see Newton's law holds in those steps when 2 objects are in solid contact to accelerate together. But if we be blindly apply the law, it may not be right.

-TL

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November 6, 2022, 08:18 AM   #14
JohnKSa
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Quote:
 According to Newton's 3rd law, the force upon the shootee must equal to the force upon the shooter.
Each force has an equal an opposing force, just as it should, just as Newton's law says it does, but that doesn't mean that all the forces in a problem are equal.

It's not possible to chain together a bunch of forces and say they are all equal just because they are all generally linked together by a common problem.

For example, the force on the rifle and the bullet are both due to the pressure of the discharge. They are equal.

The force of the bullet on the target and the target on the bullet are equal, but because they are both dependent on how fast the bullet is decelerated by the target and not on how fast the bullet is accelerated by the discharge pressure, they are not equal to the force on the bullet and the rifle created by the discharge pressure.

The force of the stock on the shoulder and the shoulder on the stock are equal, but because they are dependent on how fast the rifle is decelerated by the shoulder, not on how fast the bullet is accelerated by the discharge pressure or on how fast the bullet is decelerated by the target medium, they are not equal to the force of the bullet on the target nor to the force on the bullet created by the discharge pressure.

Force on the shoulder by the stock = Force on the stock by the shoulder.
Which does not equal...
Force on the rifle created by the discharge = Force on the bullet from firing.
Which does not equal...
Force on the target by the bullet impact = Force on the bullet by the target impact.
Which does not equal...
Force on the shoulder by the stock = Force on the stock by the shoulder.
Quote:
 But if we be blindly apply the law, it may not be right.
If we apply the law incorrectly then the results are nonsensical. That doesn't mean the law is wrong or invalid in some scenarios, it just means that it needs to be applied correctly to provide valid results.

For example, the Pythagorean Theorem says that it is possible to calculate the hypotenuse of a right triangle by a certain formula. If the formula is used to calculate the side of a triangle that is not a right triangle, it won't work. Not because the formula is wrong, but because it was applied incorrectly.

Newton's law does not say that all the forces in a given physics problem are equal, it only links certain forces together. Attempting to link all of the forces in a problem together and set them all equal to each other will lead to misunderstandings, but that is not a problem with Newton's law, it is a problem with the way it is being applied.

Newton's law says that for every force there is an equal and opposite force. In this problem, we see that we can group all the forces into pairs that are equal. Which is exactly what Newton's law says should happen. Newton's law does not guarantee universal equality for all the forces in a scenario, it only guarantees that the forces in a scenario will occur in equal pairs acting in opposite directions.
Quote:
 While we are at it, is it physically possible to shoot a man flying off his feet? Physically means according to physics.
There are two problems with shooting a person off their feet. First of all, humans have a lot of "give" to them. Even if the impact of the bullet imparts tremendous force, a lot of that force will be used up deforming and destroying tissue (and the bullet as well), and therefore won't be available to actually move the person around significantly.

Second, humans act a lot like water and impacts to water will apply force in all directions, not just straight forward. When a pebble is dropped into a pond, the ripples go out in all directions, and that means force had to have been applied in all directions to make those ripples. The same thing happens to the human body.

All of that means that even though tremendous force is applied by the impact, it won't move the person around much. It will poke a hole and stretch/tear/deform things, but it's not going to move the entire person very much at all in the direction of bullet travel.

We can see this is true when shooting small varmints with large caliber rifles. It doesn't send them flying backwards nearly as much as it just splatters them around in all directions. If there is a prevailing direction to the movement of the largest part of the carcass, it tends to be the result of jet effects from the fluid being ejected by the impact more than the result of the direction of the bullet travel.
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 November 13, 2022, 10:59 AM #15 Picher Senior Member   Join Date: December 14, 2004 Location: Maine Posts: 3,694 Correction of typo: Acceleration of a bullet takes place in 24 inches, not 24'.
 November 13, 2022, 03:07 PM #16 stagpanther Senior Member   Join Date: March 2, 2014 Posts: 11,762 Interesting discussion--unfortunately physics was never my strong suit so I don't get most of the techie stuff. But I'd like to interject something on the discussion of deceleration of the stock on the shoulder, at least in terms of how I feel them. I shoot my share big magnums and have noticed for the most part pulling the stock in firmly to the shoulder prior to the shot makes a big difference in the consequential felt recoil--how is that accounted for? __________________ "Everyone speaks gun."--Robert O'Neill I am NOT an expert--I do not have any formal experience or certification in firearms use or testing; use any information I post at your own risk!
 November 13, 2022, 03:34 PM #17 JohnKSa Staff   Join Date: February 12, 2001 Location: DFW Area Posts: 24,971 By holding the stock against your shoulder, you are doing a couple of things. First of all, you are actually keeping it from accelerating to full recoil velocity by starting to absorb the recoil momentum immediately. With the gun unrestrained, it will accelerate to full recoil velocity and then all of that momentum must be absorbed. The calculations above assumes that the rifle achieves full recoil velocity, but that's not going to happen if the rifle is being significantly restrained right from the moment of firing. Second, impacts are bad. By holding the gun firmly against you, the force can be absorbed over a longer distance/time by the buttpad, the give of your flesh and the give of your shoulder/torso/body without any impact. If the recoil is more of an impact against the shoulder, then it will apply more force and cause more pain. Your shoulder won't be able to give much, your torso and body won't be able to help absorb the impact as much because it's applied as a blow instead of as a push. So the flesh of your shoulder will suffer more than it otherwise would. __________________ Do you know about the TEXAS State Rifle Association?
 November 13, 2022, 03:52 PM #18 stagpanther Senior Member   Join Date: March 2, 2014 Posts: 11,762 I kinda figured that was the case--but couldn't put into words like that. __________________ "Everyone speaks gun."--Robert O'Neill I am NOT an expert--I do not have any formal experience or certification in firearms use or testing; use any information I post at your own risk!
 November 13, 2022, 09:54 PM #19 tangolima Senior Member   Join Date: September 28, 2013 Posts: 3,808 Too busy to reply lately, sorry. JohnKSa. Even my clumsy English doesn't sound like so(my mother tongue is Cantonese), I'm in agreement with you. I kept mentioning the 2 cases not to argue they were right. They are the examples of applying Newton's law blindly. Unfortunately we have quite a few of those floating around. People just repeat what they heard without understanding it. Shoulder on butt stock. That leads to the next rabbit hole; muzzle rise. Here we are only interested in what happens during barrel time, which affect the exit angle of the projectile. With the shoulder pressing on the butt stock, force is applied to the butt stock to decelerate rifle as soon as the primer is ignited. The center of the butt stock is not inline with the barrel. 5" for regular rifle designs, 2.5" for AR type designs. The force on the butt stock produces a torque to rotate the muzzle upward. It lasts as long as force is asserted on the butt stock, but what affect the POI is the portion during the barrel time of 2ms or so. If there is small clearance (1/4" or so) there won't be such torque before the bullet exits the muzzle, and hence no effects on POI. Of course the felt recoil would be much more uncomfortable. Some say the change in POI is due to the change of MV caused by shoulder on the butt stock. We have already concluded that the change in MV is insignificant. The muzzle rise is difficult to keep consistent. Quite likely it doesn't help the group size. What can we do to minimize such effects? -TL Sent from my SM-N960U using Tapatalk
November 14, 2022, 01:29 PM   #20
JohnKSa
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 JohnKSa. Even my clumsy English doesn't sound like so(my mother tongue is Cantonese), I'm in agreement with you. I kept mentioning the 2 cases not to argue they were right. They are the examples of applying Newton's law blindly. Unfortunately we have quite a few of those floating around. People just repeat what they heard without understanding it.
Quote:
 The muzzle rise is difficult to keep consistent. Quite likely it doesn't help the group size. What can we do to minimize such effects?
Minimize the time in the barrel. In open-sight competitions, "bloop tubes" are sometimes used. These extend the sight radius without extending the barrel.

The main thing is to be consistent and understand that different shooting positions will affect the muzzle rise. The effect is very small, but then we are concerned with very small effects if we are trying for tiny groups.

Position shooters often make sight adjustments for each position (prone/standing/kneeling/sitting) and some of that is to compensate for muzzle rise.
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 November 14, 2022, 04:39 PM #21 tangolima Senior Member   Join Date: September 28, 2013 Posts: 3,808 Backing up a bit to shooting a man flying off his feet. Theoretically I do think it is possible (not impossible). A rifle bullet can easily have kinetic energy of 1,000 ft-lb. It is enough to send a 1000 lb mass 1 foot straight up, or to send 200 lb mass moving at 18 fps (about 12mph). But in practice, it is rather difficult to get that energy fully transferred to the target. One story account is the closest that I have heard. It was on a documentary. An army medic was hit by a sniper round. The chest plate in his body armor saved him. It was slightly nicked and they couldn't recovered the projectile. It was believed that the bullet bounced back upon impact. Basically his body absorbed the substantial portion of the energy. He remembered it was like being ramped by a fast moving truck. When he regained his bearings, he was on the ground flat on his back meters away from where he was standing. His teammates' quick reaction and fierce covering fire kept the enemy from follow-up shots. They managed to pull him to safety. Nobody remembered seeing him flying in air, but I would think it must be quite close to. -TL Sent from my SM-N960U using Tapatalk
 November 14, 2022, 06:02 PM #22 JohnKSa Staff   Join Date: February 12, 2001 Location: DFW Area Posts: 24,971 Mythbusters did a test with a crash test dummy. They set it up with an AR500 plate in its chest. They shot it with a .50BMG at arm's length distance. The bullet did stop in the dummy. It went through the plate but stopped in the metal spine. The dummy was not moved backwards by an appreciable amount. Here's more on that test, including some screen captures from the test. https://thefiringline.com/forums/sho...5&postcount=15 The problem is that there's all kinds of energy used up doing things other than moving the target. I've seen slow motion footage where a bullet impact on steel causes the steel to splash like water and/or the bullet to deform like a marshmallow. That's energy that's no longer available to push the target backwards because it was used to do something else. Here's a post with a link to a guy with a ballistics vest getting shot with a .308 rifle at close range. He stands on one foot and the rifle still doesn't knock him down. There's another link showing rifle testing using a live person wearing a bulletproof vest. https://thefiringline.com/forums/sho...0&postcount=71 Even with the bullet stopping in the target, even when the bullet is from a .50BMG there's no significant push backwards. Remember, momentum transfer requires inelastic collision. Once things start deforming appreciably there's no way to get an accurate picture of what's going on from simple momentum calculations. __________________ Do you know about the TEXAS State Rifle Association?
 November 14, 2022, 11:48 PM #23 tangolima Senior Member   Join Date: September 28, 2013 Posts: 3,808 It is indeed very difficult to set up just right to make it happen. But proving impossibility itself is an impossibility. The usual proof is a quote to Newton's 3rd law, claiming it is physically impossible. Both elastic and inelastic collisions follow principle of conservation of momentum. They transfer momentum and energy alright, just by different amount. A bug smashed on windshield is inelastic collision. A baseball bat landing on a skull is elastic collision. Both hurt like hell. -TL Sent from my SM-N960U using Tapatalk
November 15, 2022, 05:41 AM   #24
JohnKSa
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Quote:
 But proving impossibility itself is an impossibility.
Proving a negative, depending on the specific negative can be impossible.

For example proving that a particular person never said a particular quote is impossible because it's not possible to get an exhaustive list of everything any person ever said. On the other hand, proving that the person never wrote the quote in one of their published books could be easy if one had access to all the books that the person published.

Proving a .308 rifle bullet can't send a person flying is certainly possible.

Set up a scenario involving a cartridge with tremendously more energy/momentum than the .308 cartridge can provide. Make sure that in that scenario, the target is similar to a human in shape and weight but is designed so that it absorbs all of the momentum/energy of the impact.

If that scenario doesn't result in the target being sent flying then it's pretty clear that using less energy/momentum can't possibly impart more energy/momentum to the target.
Quote:
 Both elastic and inelastic collisions follow principle of conservation of momentum.
The way I stated it was incorrect. What I was getting at was this:

Shoot a water balloon with a bullet. If conservation of momentum is followed in the conventional manner to determine the outcome, then the water balloon will be driven in the direction of bullet travel with a velocity that can be calculated by the momentum transfer equation.

Clearly that is not the case. In fact, most of the mass that once made up the water balloon will travel in directions other than the direction of the bullet impact. Some of it will actually travel in a direction almost directly opposite the direction of the bullet impact, splashing the shooter if the distance is short enough.

Momentum is conserved, but not in any sense that is useful for this discussion.

Check out the egg being shot at 4:33 in this video.

Even if the balloon doesn't burst, say it's made up of a gel that is self-healing and stretches to any length without breaking, the bullet impact still won't drive it downrange to any significant extent because the impact energy will be used expanding the ball out in all directions. It will move overall a little bit in the process of stretching and snapping back, but not anything like the momentum transfer equation will predict.
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 November 15, 2022, 07:03 AM #25 mehavey Senior Member   Join Date: June 17, 2010 Location: Virginia Posts: 6,883 I think the endless YouTube videos of gel blocks -- even "skeletonized" gel blocks -- not going anywhere in particular is fairly conclusive.

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