Mixing
Why does shooting a handgun accelerate a bullet to deadly speed without injuring the gun user's hand?
Momentum is defined by the product of mass and velocity. Now a projectile out of a gun has to have high velocity to penetrate a human body, as its mass isn't significant. But to reach this velocity, due to inertia/the law of energy conservation, momentum on both sides is to be equal. As there is strong negative acceleration in the opposite direction of the bullet, it should result in a strong force in the opposite direction of the bullet ($F = m*a $).
Therefore I wonder why shooting a bullet with a handgun is not ripping your hand apart.
newtonian-mechanics momentum conservation-laws collision estimation
asked Dec 29 '18 at 16:37
ZurechtweiserZurechtweiser
24025
add a comment |
Momentum is defined by the product of mass and velocity. Now a projectile out of a gun has to have high velocity to penetrate a human body, as its mass isn't significant. But to reach this velocity, due to inertia/the law of energy conservation, momentum on both sides is to be equal. As there is strong negative acceleration in the opposite direction of the bullet, it should result in a strong force in the opposite direction of the bullet ($F = m*a $).
Therefore I wonder why shooting a bullet with a handgun is not ripping your hand apart.
newtonian-mechanics momentum conservation-laws collision estimation
asked Dec 29 '18 at 16:37
ZurechtweiserZurechtweiser
24025
1
Comments are not for extended discussion; this interesting conversation has been moved to chat. Please use comments only to suggest improvements to the question or request clarification from the asker.
– rob♦
Dec 31 '18 at 16:59
1
Piercing a human doesn't require as much energy as you might think. It's about pressure. Imagine a knife, or a needle. A bullet has a small area so doesn't need a lot of energy to pierce. Imagine holding a stick the diameter of a bullet in place of the gun and having someone push a leg of lamb onto it. How hard would you need to brace in order for the stick to pierce the meat?
– Richard
Jan 1 at 20:12
add a comment |
Momentum is defined by the product of mass and velocity. Now a projectile out of a gun has to have high velocity to penetrate a human body, as its mass isn't significant. But to reach this velocity, due to inertia/the law of energy conservation, momentum on both sides is to be equal. As there is strong negative acceleration in the opposite direction of the bullet, it should result in a strong force in the opposite direction of the bullet ($F = m*a $).
Therefore I wonder why shooting a bullet with a handgun is not ripping your hand apart.
newtonian-mechanics momentum conservation-laws collision estimation
asked Dec 29 '18 at 16:37
ZurechtweiserZurechtweiser
24025
Momentum is defined by the product of mass and velocity. Now a projectile out of a gun has to have high velocity to penetrate a human body, as its mass isn't significant. But to reach this velocity, due to inertia/the law of energy conservation, momentum on both sides is to be equal. As there is strong negative acceleration in the opposite direction of the bullet, it should result in a strong force in the opposite direction of the bullet ($F = m*a $).
Therefore I wonder why shooting a bullet with a handgun is not ripping your hand apart.
newtonian-mechanics momentum conservation-laws collision estimation
newtonian-mechanics momentum conservation-laws collision estimation
asked Dec 29 '18 at 16:37
ZurechtweiserZurechtweiser
24025
asked Dec 29 '18 at 16:37
ZurechtweiserZurechtweiser
24025
edited Jan 1 at 0:21
Zurechtweiser
asked Dec 29 '18 at 16:37
ZurechtweiserZurechtweiser
24025
asked Dec 29 '18 at 16:37
ZurechtweiserZurechtweiser
24025
asked Dec 29 '18 at 16:37
ZurechtweiserZurechtweiser
24025
24025
1
Comments are not for extended discussion; this interesting conversation has been moved to chat. Please use comments only to suggest improvements to the question or request clarification from the asker.
– rob♦
Dec 31 '18 at 16:59
1
Piercing a human doesn't require as much energy as you might think. It's about pressure. Imagine a knife, or a needle. A bullet has a small area so doesn't need a lot of energy to pierce. Imagine holding a stick the diameter of a bullet in place of the gun and having someone push a leg of lamb onto it. How hard would you need to brace in order for the stick to pierce the meat?
– Richard
Jan 1 at 20:12
add a comment |
1
Comments are not for extended discussion; this interesting conversation has been moved to chat. Please use comments only to suggest improvements to the question or request clarification from the asker.
– rob♦
Dec 31 '18 at 16:59
1
Piercing a human doesn't require as much energy as you might think. It's about pressure. Imagine a knife, or a needle. A bullet has a small area so doesn't need a lot of energy to pierce. Imagine holding a stick the diameter of a bullet in place of the gun and having someone push a leg of lamb onto it. How hard would you need to brace in order for the stick to pierce the meat?
– Richard
Jan 1 at 20:12
1
1
Comments are not for extended discussion; this interesting conversation has been moved to chat. Please use comments only to suggest improvements to the question or request clarification from the asker.
– rob♦
Dec 31 '18 at 16:59
Comments are not for extended discussion; this interesting conversation has been moved to chat. Please use comments only to suggest improvements to the question or request clarification from the asker.
– rob♦
Dec 31 '18 at 16:59
1
1
Piercing a human doesn't require as much energy as you might think. It's about pressure. Imagine a knife, or a needle. A bullet has a small area so doesn't need a lot of energy to pierce. Imagine holding a stick the diameter of a bullet in place of the gun and having someone push a leg of lamb onto it. How hard would you need to brace in order for the stick to pierce the meat?
– Richard
Jan 1 at 20:12
Piercing a human doesn't require as much energy as you might think. It's about pressure. Imagine a knife, or a needle. A bullet has a small area so doesn't need a lot of energy to pierce. Imagine holding a stick the diameter of a bullet in place of the gun and having someone push a leg of lamb onto it. How hard would you need to brace in order for the stick to pierce the meat?
– Richard
Jan 1 at 20:12
add a comment |
6 Answers
6
active
oldest
votes
Firstly, some guns do give quite a kick! So the effect you are thinking of is real.
However, conservation of momentum means that $m_text{bullet} cdot v_text{bullet} = m_text{gun} cdot v_text{gun}$. So the bullet's velocity is greater than that of the gun by a ratio of $frac{m_text{gun}}{m_text{bullet}}$. Then energy is distributed in the same ratio because while energy scales as velocity squared, it also scales with the mass. So, it is useful for the gun to be heavy and/or for it to have a spring-loaded mechanism to slowly distribute the kick to your hand and body.
edited Dec 30 '18 at 9:41
Wasserwaage
293316
answered Dec 29 '18 at 17:06
Paul YoungPaul Young
73417
15
Spot on about the mass of the gun. by the way, the spring loaded mechanism that dampens recoil is typically the action of a semi-automatic of automatic firearm. The purpose of that action isn't to dampen recoil, but rather to eject the spent cartridge and chamber the next cartridge. However, in using recoil energy to reload the weapon, it does indeed dampen recoil. However, there is a mechanism on firearms that is specifically designed to dampen recoil: The muzzle brake, which directs some of the propellant gasses backwards and thereby counters some of the force of recoil.
– Wayne Conrad
Dec 30 '18 at 1:34
3
@Blue - Yeah, after it overcomes the inertia of the gun and other losses due to distortions. It will still apply the same overall force to the weapon which is what then, if designed right, correctly transfers the energy to your body mercifully, instead of you ending up pointing it at your forehead after every discharge.
– Mazura
Dec 30 '18 at 21:35
4
@BlueRaja-DannyPflughoeft: The weight of the weapon plays a huge part.
– whatsisname
Dec 30 '18 at 23:01
4
@Yakk - "It may soak up energy, but it cannot soak up momentum." It doesn't need to. By extending the recoil absorption over a longer distance, and therefor longer time, it reduces the peak recoil force on the shooter, and that is an important part of perceived recoil.
– WhatRoughBeast
Dec 30 '18 at 23:55
2
@WayneConrad is correct, although I'd like to add the effect of the recoil spring in most short recoil operation handguns is huge on felt recoil. Many race shooters tune their recoil springs, and if done correctly and to the user's preference, can almost completely eliminate felt recoil. Many people use custom aftermarket guide rods and recoil springs for this purpose. They are often rated by poundage, and are tuned to reduce the amount of recoil transferred to the action to just the amount needed to cycle it, and nothing more transferred to the user.
– nostalgk
Dec 31 '18 at 13:48
|
show 13 more comments
The handgun is braced with a large surface area of the hand, and the palm and entire hand are robust; the result is that the hand, or hand and arm, or hand and upper body are sharply displaced as a whole before the motion is damped by the rest of the body.
Some details of recoil are discussed here. The recoil of rifles, which are generally more powerful, braced near the shoulder, and operated near the face, can easily cause a broken collarbone, torn rotator cuff, black eye, and/or detached retina.
Thus, whether injury occurs depends on the stress induced in vivo from the acceleration of the brace position vs. the relative strength of the nearby organs.
answered Dec 29 '18 at 17:12
ChemomechanicsChemomechanics
4,70131023
The hand (and wrist, forearm, &c) also has many flexible muscles and joints, which act as shock absorbers. Not just when firing a pistol: think of hitting something with your fist.
– jamesqf
Dec 29 '18 at 18:50
15
If the gun weighed as much (very important) as the projectile, and was as small (also important), it would. We have an ancient Styre rifle that was shortened at some point, vastly reducing it's weight. Its recoil went from being described as 'unpleasant' to 'grim'.
– Mazura
Dec 29 '18 at 20:39
3
The mass of the firearm also plays a significant role.
– whatsisname
Dec 30 '18 at 22:58
2
Occasionally, beginning rifle shooters attempt to "deal with" the recoil by bracing their shooting shoulder against a good stout tree. They generally only try that once.
– Jamie Hanrahan
Jan 1 at 23:06
add a comment |
Therefore I wonder why shooting a bullet with a handgun is not ripping your hand apart.
Because the mass of the handgun is greater than the mass of the bullet, and because the energy transferred from the gun to your hand is distributed across the surface of the pistol grip.
One of the more interesting things I heard on a trip to Williamsburg Virginia (which is period of ~1776; American Revolution) was the question of how much the muskets weighed. The historical figure answered (I don't recall the weight) and the questioner said in a surprised voice, "That's basically the same that guns weigh today!"
"Yes," replied this historical figure, "because the physics hasn't changed. They could make rifles lighter today, but they don't because the recoil would be worse."
There are other things put in modern pistols to reduce recoil like springs and discharge, but since you tagged it "Newtonian physics" I expect your are less interested in those things.
answered Dec 29 '18 at 22:32
J. Chris ComptonJ. Chris Compton
35124
2
This answer covers it best. If you had a 'bullet trap' of the same weight as the gun, which decelerated the bullet over an equally long trajectory as it was fired, catching a bullet using this device would be no harder than firing it. Catching a bullet with just the mass of skin in front of it, over the thickness of said skin, generally doesn't work for anything but bb's.
– Eelco Hoogendoorn
Dec 30 '18 at 3:57
1
@EelcoHoogendoorn a bulletproof vest is one example of "bullet trap"
– Eric Duminil
Dec 30 '18 at 13:05
1
Re "They could make rifles lighter today...", they do. Compare the weight of an M14 to the M16, for instance. Or for pistols, the .45 caliber Colt M1911 weighs 1100 g, the comparable Glock .45 845 g. (Other Glock models weigh even less.)
– jamesqf
Dec 30 '18 at 18:13
1
@jamesqf: M16 is lighter, but it also shoots rounds that are roughly half the mass.
– whatsisname
Dec 30 '18 at 23:04
2
@DavidRicherby There's also a tradeoff of muzzle control vs. handling as mass is moved forward or back in a rifle. More forward weight keeps the muzzle more stable, but increases the polar moment of inertia when moving the gun around, and vice versa.
– nasch
Jan 1 at 19:34
|
show 5 more comments
By holding the gun firmly and taking a proper firing pose, you arrange it so that the momentum is passed to your entire body as a whole, or a large enough part of it. Your body is much heavier than the bullet, so the speed it gets from the shot is very small and easily countered by muscles and joints.
If you don't take a proper pose or don't hold the gun properly, you may fail to steady it against the kickback, or the momentum could be passed only to a small part of the body, resulting in injury.
answered Dec 31 '18 at 14:24
ivan_pozdeevivan_pozdeev
20714
add a comment |
Lets nail down some figures with a random choice of compatible equipment with specs publicly available.
Gun https://us.glock.com/products/G19
Gun Weight with empty magazine 670 g | 23.63 oz
Barrel Length 102 mm | 4.02 inch (slightly less than ammo test, so time under acceleration is slightly less)
Ammunition https://en.wikipedia.org/wiki/9%C3%9719mm_Parabellum
Bullet mass 8.04 g (124 gr) Federal FMJ
Bullet velocity 1,150 ft/s (350 m/s) @ 118mm (4.65") barrel length, 0.00067s
Bullet energy 364 ft⋅lbf (494 J)
We could set this up as a conservation of energy problem, and that would be a reasonable approximation with a satisfying answer.
The time under acceleration is very short(0.00067s). (anything less than 0.01s feels instantaneous)
The gun weighs a lot (83 times) more than the bullet.
It is coupled to a human that weighs very much more.(180lb = 10155 times)
It is coupled over a large, soft, energy absorbing surface(hand).
The recoil is transmitted through several relatively massive energy absorbing tendons, tensed muscles, bones, flexed joints, fat, fluids, which redirect the recoil into countless vectors.
Where is the point of impact between gun and bullet? How does the gun impart force to the bullet and the bullet impart equal and opposite force to the gun? Well that doesn't really happen. It isn't a collision.
Only the timing is similar to a collision because both gun and bullet are accelerated in opposite directions by expanding gasses with same T0.
But is energy imparted equally? The center of those gasses is itself being accelerated away from the gun. Barrel friction counteracts recoil. You may be mistaken in part of your premise "momentum on both sides is to be equal".
answered Dec 30 '18 at 1:45
slomobileslomobile
514
gasses -> gases
– Faheem Mitha
Dec 30 '18 at 7:20
1
@FaheemMitha In British English, and possibly all others except US English (where it is no longer the common variation), gasses is correct.
– Andrew Morton
Dec 30 '18 at 13:52
@AndrewMorton Wow, I don't think I've come across that variant before. Or if I have, I don't remember it.
– Faheem Mitha
Dec 30 '18 at 15:34
How much the human weights is quite irrelevant: When the bullet leaves the gun, the gun has moved back by $102mm/83 = 1.2mm$ and is moving at a speed of $350frac{m}{s}/83 = 4.2frac{m}{s} = 15frac{km}{h}$. The gun subsequently basically crashes into the human at that speed: The tiny translation of $1.2mm$ is simply not enough to transmit any relevant forces.
– cmaster
Dec 31 '18 at 2:09
That said, if you half the weight of the gun, you get it crashing into the user at twice the speed, i.e. $30frac{km}{h}$, which is definitely not fun.
– cmaster
Dec 31 '18 at 2:12
|
show 3 more comments
Assuming impulse invariance, roughly the same amount of momentum is transferred to the shooter and the bullet and ultimately the target. However, if momentum was the only lethal element of a bullet, there would not be such a thing as a bullet-proof vest since a bullet-proof vest cannot keep momentum off the target (you'd need a ground-mounted shield for that). It can, however, absorb and disperse energy which, as opposed to momentum, grows with the square of the speed of moving mass. A bullet will often not transfer all of its energy to the target unless it gets stuck and will rather work by inflicting deadly injury, but there are some that are designed to stop and/or disperse on entry, like hollow point bullets. Those tend to be much more deadly than an ordinary bullet even when not hitting vital parts of the body.
So the basic point is that a pistol has more mass than the bullet, so while they share momentum equally (at least when discounting what exhaust fumes may do), they don't receive equal amounts of energy.
Note that shoulder-held rocket launchers are built in a way where the exhaust fumes (which continue accelerating the rocket via conservation of momentum) can leave the launcher at its back, thus minimizing the amount of momentum transferred to the operator (naturally, the place behind someone with a rocket launcher is a very bad place to be as opposed to being behind a gunner). Since it is the pressure of the propellant explosion rather than the momentum of something leaving the bullet, this kind of mechanism is not feasible with a gun to a good degree.
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6 Answers
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6 Answers
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oldest
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active
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Firstly, some guns do give quite a kick! So the effect you are thinking of is real.
However, conservation of momentum means that $m_text{bullet} cdot v_text{bullet} = m_text{gun} cdot v_text{gun}$. So the bullet's velocity is greater than that of the gun by a ratio of $frac{m_text{gun}}{m_text{bullet}}$. Then energy is distributed in the same ratio because while energy scales as velocity squared, it also scales with the mass. So, it is useful for the gun to be heavy and/or for it to have a spring-loaded mechanism to slowly distribute the kick to your hand and body.
edited Dec 30 '18 at 9:41
Wasserwaage
293316
answered Dec 29 '18 at 17:06
Paul YoungPaul Young
73417
15
Spot on about the mass of the gun. by the way, the spring loaded mechanism that dampens recoil is typically the action of a semi-automatic of automatic firearm. The purpose of that action isn't to dampen recoil, but rather to eject the spent cartridge and chamber the next cartridge. However, in using recoil energy to reload the weapon, it does indeed dampen recoil. However, there is a mechanism on firearms that is specifically designed to dampen recoil: The muzzle brake, which directs some of the propellant gasses backwards and thereby counters some of the force of recoil.
– Wayne Conrad
Dec 30 '18 at 1:34
3
@Blue - Yeah, after it overcomes the inertia of the gun and other losses due to distortions. It will still apply the same overall force to the weapon which is what then, if designed right, correctly transfers the energy to your body mercifully, instead of you ending up pointing it at your forehead after every discharge.
– Mazura
Dec 30 '18 at 21:35
4
@BlueRaja-DannyPflughoeft: The weight of the weapon plays a huge part.
– whatsisname
Dec 30 '18 at 23:01
4
@Yakk - "It may soak up energy, but it cannot soak up momentum." It doesn't need to. By extending the recoil absorption over a longer distance, and therefor longer time, it reduces the peak recoil force on the shooter, and that is an important part of perceived recoil.
– WhatRoughBeast
Dec 30 '18 at 23:55
2
@WayneConrad is correct, although I'd like to add the effect of the recoil spring in most short recoil operation handguns is huge on felt recoil. Many race shooters tune their recoil springs, and if done correctly and to the user's preference, can almost completely eliminate felt recoil. Many people use custom aftermarket guide rods and recoil springs for this purpose. They are often rated by poundage, and are tuned to reduce the amount of recoil transferred to the action to just the amount needed to cycle it, and nothing more transferred to the user.
– nostalgk
Dec 31 '18 at 13:48
|
show 13 more comments
Firstly, some guns do give quite a kick! So the effect you are thinking of is real.
However, conservation of momentum means that $m_text{bullet} cdot v_text{bullet} = m_text{gun} cdot v_text{gun}$. So the bullet's velocity is greater than that of the gun by a ratio of $frac{m_text{gun}}{m_text{bullet}}$. Then energy is distributed in the same ratio because while energy scales as velocity squared, it also scales with the mass. So, it is useful for the gun to be heavy and/or for it to have a spring-loaded mechanism to slowly distribute the kick to your hand and body.
edited Dec 30 '18 at 9:41
Wasserwaage
293316
answered Dec 29 '18 at 17:06
Paul YoungPaul Young
73417
15
Spot on about the mass of the gun. by the way, the spring loaded mechanism that dampens recoil is typically the action of a semi-automatic of automatic firearm. The purpose of that action isn't to dampen recoil, but rather to eject the spent cartridge and chamber the next cartridge. However, in using recoil energy to reload the weapon, it does indeed dampen recoil. However, there is a mechanism on firearms that is specifically designed to dampen recoil: The muzzle brake, which directs some of the propellant gasses backwards and thereby counters some of the force of recoil.
– Wayne Conrad
Dec 30 '18 at 1:34
3
@Blue - Yeah, after it overcomes the inertia of the gun and other losses due to distortions. It will still apply the same overall force to the weapon which is what then, if designed right, correctly transfers the energy to your body mercifully, instead of you ending up pointing it at your forehead after every discharge.
– Mazura
Dec 30 '18 at 21:35
4
@BlueRaja-DannyPflughoeft: The weight of the weapon plays a huge part.
– whatsisname
Dec 30 '18 at 23:01
4
@Yakk - "It may soak up energy, but it cannot soak up momentum." It doesn't need to. By extending the recoil absorption over a longer distance, and therefor longer time, it reduces the peak recoil force on the shooter, and that is an important part of perceived recoil.
– WhatRoughBeast
Dec 30 '18 at 23:55
2
@WayneConrad is correct, although I'd like to add the effect of the recoil spring in most short recoil operation handguns is huge on felt recoil. Many race shooters tune their recoil springs, and if done correctly and to the user's preference, can almost completely eliminate felt recoil. Many people use custom aftermarket guide rods and recoil springs for this purpose. They are often rated by poundage, and are tuned to reduce the amount of recoil transferred to the action to just the amount needed to cycle it, and nothing more transferred to the user.
– nostalgk
Dec 31 '18 at 13:48
|
show 13 more comments
Firstly, some guns do give quite a kick! So the effect you are thinking of is real.
However, conservation of momentum means that $m_text{bullet} cdot v_text{bullet} = m_text{gun} cdot v_text{gun}$. So the bullet's velocity is greater than that of the gun by a ratio of $frac{m_text{gun}}{m_text{bullet}}$. Then energy is distributed in the same ratio because while energy scales as velocity squared, it also scales with the mass. So, it is useful for the gun to be heavy and/or for it to have a spring-loaded mechanism to slowly distribute the kick to your hand and body.
edited Dec 30 '18 at 9:41
Wasserwaage
293316
answered Dec 29 '18 at 17:06
Paul YoungPaul Young
73417
Firstly, some guns do give quite a kick! So the effect you are thinking of is real.
However, conservation of momentum means that $m_text{bullet} cdot v_text{bullet} = m_text{gun} cdot v_text{gun}$. So the bullet's velocity is greater than that of the gun by a ratio of $frac{m_text{gun}}{m_text{bullet}}$. Then energy is distributed in the same ratio because while energy scales as velocity squared, it also scales with the mass. So, it is useful for the gun to be heavy and/or for it to have a spring-loaded mechanism to slowly distribute the kick to your hand and body.
edited Dec 30 '18 at 9:41
Wasserwaage
293316
answered Dec 29 '18 at 17:06
Paul YoungPaul Young
73417
edited Dec 30 '18 at 9:41
Wasserwaage
293316
edited Dec 30 '18 at 9:41
Wasserwaage
293316
edited Dec 30 '18 at 9:41
Wasserwaage
293316
293316
answered Dec 29 '18 at 17:06
Paul YoungPaul Young
73417
answered Dec 29 '18 at 17:06
Paul YoungPaul Young
73417
answered Dec 29 '18 at 17:06
Paul YoungPaul Young
73417
73417
15
Spot on about the mass of the gun. by the way, the spring loaded mechanism that dampens recoil is typically the action of a semi-automatic of automatic firearm. The purpose of that action isn't to dampen recoil, but rather to eject the spent cartridge and chamber the next cartridge. However, in using recoil energy to reload the weapon, it does indeed dampen recoil. However, there is a mechanism on firearms that is specifically designed to dampen recoil: The muzzle brake, which directs some of the propellant gasses backwards and thereby counters some of the force of recoil.
– Wayne Conrad
Dec 30 '18 at 1:34
3
@Blue - Yeah, after it overcomes the inertia of the gun and other losses due to distortions. It will still apply the same overall force to the weapon which is what then, if designed right, correctly transfers the energy to your body mercifully, instead of you ending up pointing it at your forehead after every discharge.
– Mazura
Dec 30 '18 at 21:35
4
@BlueRaja-DannyPflughoeft: The weight of the weapon plays a huge part.
– whatsisname
Dec 30 '18 at 23:01
4
@Yakk - "It may soak up energy, but it cannot soak up momentum." It doesn't need to. By extending the recoil absorption over a longer distance, and therefor longer time, it reduces the peak recoil force on the shooter, and that is an important part of perceived recoil.
– WhatRoughBeast
Dec 30 '18 at 23:55
2
@WayneConrad is correct, although I'd like to add the effect of the recoil spring in most short recoil operation handguns is huge on felt recoil. Many race shooters tune their recoil springs, and if done correctly and to the user's preference, can almost completely eliminate felt recoil. Many people use custom aftermarket guide rods and recoil springs for this purpose. They are often rated by poundage, and are tuned to reduce the amount of recoil transferred to the action to just the amount needed to cycle it, and nothing more transferred to the user.
– nostalgk
Dec 31 '18 at 13:48
|
show 13 more comments
15
Spot on about the mass of the gun. by the way, the spring loaded mechanism that dampens recoil is typically the action of a semi-automatic of automatic firearm. The purpose of that action isn't to dampen recoil, but rather to eject the spent cartridge and chamber the next cartridge. However, in using recoil energy to reload the weapon, it does indeed dampen recoil. However, there is a mechanism on firearms that is specifically designed to dampen recoil: The muzzle brake, which directs some of the propellant gasses backwards and thereby counters some of the force of recoil.
– Wayne Conrad
Dec 30 '18 at 1:34
3
@Blue - Yeah, after it overcomes the inertia of the gun and other losses due to distortions. It will still apply the same overall force to the weapon which is what then, if designed right, correctly transfers the energy to your body mercifully, instead of you ending up pointing it at your forehead after every discharge.
– Mazura
Dec 30 '18 at 21:35
4
@BlueRaja-DannyPflughoeft: The weight of the weapon plays a huge part.
– whatsisname
Dec 30 '18 at 23:01
4
@Yakk - "It may soak up energy, but it cannot soak up momentum." It doesn't need to. By extending the recoil absorption over a longer distance, and therefor longer time, it reduces the peak recoil force on the shooter, and that is an important part of perceived recoil.
– WhatRoughBeast
Dec 30 '18 at 23:55
2
@WayneConrad is correct, although I'd like to add the effect of the recoil spring in most short recoil operation handguns is huge on felt recoil. Many race shooters tune their recoil springs, and if done correctly and to the user's preference, can almost completely eliminate felt recoil. Many people use custom aftermarket guide rods and recoil springs for this purpose. They are often rated by poundage, and are tuned to reduce the amount of recoil transferred to the action to just the amount needed to cycle it, and nothing more transferred to the user.
– nostalgk
Dec 31 '18 at 13:48
15
15
Spot on about the mass of the gun. by the way, the spring loaded mechanism that dampens recoil is typically the action of a semi-automatic of automatic firearm. The purpose of that action isn't to dampen recoil, but rather to eject the spent cartridge and chamber the next cartridge. However, in using recoil energy to reload the weapon, it does indeed dampen recoil. However, there is a mechanism on firearms that is specifically designed to dampen recoil: The muzzle brake, which directs some of the propellant gasses backwards and thereby counters some of the force of recoil.
– Wayne Conrad
Dec 30 '18 at 1:34
Spot on about the mass of the gun. by the way, the spring loaded mechanism that dampens recoil is typically the action of a semi-automatic of automatic firearm. The purpose of that action isn't to dampen recoil, but rather to eject the spent cartridge and chamber the next cartridge. However, in using recoil energy to reload the weapon, it does indeed dampen recoil. However, there is a mechanism on firearms that is specifically designed to dampen recoil: The muzzle brake, which directs some of the propellant gasses backwards and thereby counters some of the force of recoil.
– Wayne Conrad
Dec 30 '18 at 1:34
3
3
@Blue - Yeah, after it overcomes the inertia of the gun and other losses due to distortions. It will still apply the same overall force to the weapon which is what then, if designed right, correctly transfers the energy to your body mercifully, instead of you ending up pointing it at your forehead after every discharge.
– Mazura
Dec 30 '18 at 21:35
@Blue - Yeah, after it overcomes the inertia of the gun and other losses due to distortions. It will still apply the same overall force to the weapon which is what then, if designed right, correctly transfers the energy to your body mercifully, instead of you ending up pointing it at your forehead after every discharge.
– Mazura
Dec 30 '18 at 21:35
4
4
@BlueRaja-DannyPflughoeft: The weight of the weapon plays a huge part.
– whatsisname
Dec 30 '18 at 23:01
@BlueRaja-DannyPflughoeft: The weight of the weapon plays a huge part.
– whatsisname
Dec 30 '18 at 23:01
4
4
@Yakk - "It may soak up energy, but it cannot soak up momentum." It doesn't need to. By extending the recoil absorption over a longer distance, and therefor longer time, it reduces the peak recoil force on the shooter, and that is an important part of perceived recoil.
– WhatRoughBeast
Dec 30 '18 at 23:55
@Yakk - "It may soak up energy, but it cannot soak up momentum." It doesn't need to. By extending the recoil absorption over a longer distance, and therefor longer time, it reduces the peak recoil force on the shooter, and that is an important part of perceived recoil.
– WhatRoughBeast
Dec 30 '18 at 23:55
2
2
@WayneConrad is correct, although I'd like to add the effect of the recoil spring in most short recoil operation handguns is huge on felt recoil. Many race shooters tune their recoil springs, and if done correctly and to the user's preference, can almost completely eliminate felt recoil. Many people use custom aftermarket guide rods and recoil springs for this purpose. They are often rated by poundage, and are tuned to reduce the amount of recoil transferred to the action to just the amount needed to cycle it, and nothing more transferred to the user.
– nostalgk
Dec 31 '18 at 13:48
@WayneConrad is correct, although I'd like to add the effect of the recoil spring in most short recoil operation handguns is huge on felt recoil. Many race shooters tune their recoil springs, and if done correctly and to the user's preference, can almost completely eliminate felt recoil. Many people use custom aftermarket guide rods and recoil springs for this purpose. They are often rated by poundage, and are tuned to reduce the amount of recoil transferred to the action to just the amount needed to cycle it, and nothing more transferred to the user.
– nostalgk
Dec 31 '18 at 13:48
|
show 13 more comments
The handgun is braced with a large surface area of the hand, and the palm and entire hand are robust; the result is that the hand, or hand and arm, or hand and upper body are sharply displaced as a whole before the motion is damped by the rest of the body.
Some details of recoil are discussed here. The recoil of rifles, which are generally more powerful, braced near the shoulder, and operated near the face, can easily cause a broken collarbone, torn rotator cuff, black eye, and/or detached retina.
Thus, whether injury occurs depends on the stress induced in vivo from the acceleration of the brace position vs. the relative strength of the nearby organs.
answered Dec 29 '18 at 17:12
ChemomechanicsChemomechanics
4,70131023
The hand (and wrist, forearm, &c) also has many flexible muscles and joints, which act as shock absorbers. Not just when firing a pistol: think of hitting something with your fist.
– jamesqf
Dec 29 '18 at 18:50
15
If the gun weighed as much (very important) as the projectile, and was as small (also important), it would. We have an ancient Styre rifle that was shortened at some point, vastly reducing it's weight. Its recoil went from being described as 'unpleasant' to 'grim'.
– Mazura
Dec 29 '18 at 20:39
3
The mass of the firearm also plays a significant role.
– whatsisname
Dec 30 '18 at 22:58
2
Occasionally, beginning rifle shooters attempt to "deal with" the recoil by bracing their shooting shoulder against a good stout tree. They generally only try that once.
– Jamie Hanrahan
Jan 1 at 23:06
add a comment |
The handgun is braced with a large surface area of the hand, and the palm and entire hand are robust; the result is that the hand, or hand and arm, or hand and upper body are sharply displaced as a whole before the motion is damped by the rest of the body.
Some details of recoil are discussed here. The recoil of rifles, which are generally more powerful, braced near the shoulder, and operated near the face, can easily cause a broken collarbone, torn rotator cuff, black eye, and/or detached retina.
Thus, whether injury occurs depends on the stress induced in vivo from the acceleration of the brace position vs. the relative strength of the nearby organs.
answered Dec 29 '18 at 17:12
ChemomechanicsChemomechanics
4,70131023
The hand (and wrist, forearm, &c) also has many flexible muscles and joints, which act as shock absorbers. Not just when firing a pistol: think of hitting something with your fist.
– jamesqf
Dec 29 '18 at 18:50
15
If the gun weighed as much (very important) as the projectile, and was as small (also important), it would. We have an ancient Styre rifle that was shortened at some point, vastly reducing it's weight. Its recoil went from being described as 'unpleasant' to 'grim'.
– Mazura
Dec 29 '18 at 20:39
3
The mass of the firearm also plays a significant role.
– whatsisname
Dec 30 '18 at 22:58
2
Occasionally, beginning rifle shooters attempt to "deal with" the recoil by bracing their shooting shoulder against a good stout tree. They generally only try that once.
– Jamie Hanrahan
Jan 1 at 23:06
add a comment |
The handgun is braced with a large surface area of the hand, and the palm and entire hand are robust; the result is that the hand, or hand and arm, or hand and upper body are sharply displaced as a whole before the motion is damped by the rest of the body.
Some details of recoil are discussed here. The recoil of rifles, which are generally more powerful, braced near the shoulder, and operated near the face, can easily cause a broken collarbone, torn rotator cuff, black eye, and/or detached retina.
Thus, whether injury occurs depends on the stress induced in vivo from the acceleration of the brace position vs. the relative strength of the nearby organs.
answered Dec 29 '18 at 17:12
ChemomechanicsChemomechanics
4,70131023
The handgun is braced with a large surface area of the hand, and the palm and entire hand are robust; the result is that the hand, or hand and arm, or hand and upper body are sharply displaced as a whole before the motion is damped by the rest of the body.
Some details of recoil are discussed here. The recoil of rifles, which are generally more powerful, braced near the shoulder, and operated near the face, can easily cause a broken collarbone, torn rotator cuff, black eye, and/or detached retina.
Thus, whether injury occurs depends on the stress induced in vivo from the acceleration of the brace position vs. the relative strength of the nearby organs.
answered Dec 29 '18 at 17:12
ChemomechanicsChemomechanics
4,70131023
answered Dec 29 '18 at 17:12
ChemomechanicsChemomechanics
4,70131023
answered Dec 29 '18 at 17:12
ChemomechanicsChemomechanics
4,70131023
answered Dec 29 '18 at 17:12
ChemomechanicsChemomechanics
4,70131023
4,70131023
The hand (and wrist, forearm, &c) also has many flexible muscles and joints, which act as shock absorbers. Not just when firing a pistol: think of hitting something with your fist.
– jamesqf
Dec 29 '18 at 18:50
15
If the gun weighed as much (very important) as the projectile, and was as small (also important), it would. We have an ancient Styre rifle that was shortened at some point, vastly reducing it's weight. Its recoil went from being described as 'unpleasant' to 'grim'.
– Mazura
Dec 29 '18 at 20:39
3
The mass of the firearm also plays a significant role.
– whatsisname
Dec 30 '18 at 22:58
2
Occasionally, beginning rifle shooters attempt to "deal with" the recoil by bracing their shooting shoulder against a good stout tree. They generally only try that once.
– Jamie Hanrahan
Jan 1 at 23:06
add a comment |
The hand (and wrist, forearm, &c) also has many flexible muscles and joints, which act as shock absorbers. Not just when firing a pistol: think of hitting something with your fist.
– jamesqf
Dec 29 '18 at 18:50
15
If the gun weighed as much (very important) as the projectile, and was as small (also important), it would. We have an ancient Styre rifle that was shortened at some point, vastly reducing it's weight. Its recoil went from being described as 'unpleasant' to 'grim'.
– Mazura
Dec 29 '18 at 20:39
3
The mass of the firearm also plays a significant role.
– whatsisname
Dec 30 '18 at 22:58
2
Occasionally, beginning rifle shooters attempt to "deal with" the recoil by bracing their shooting shoulder against a good stout tree. They generally only try that once.
– Jamie Hanrahan
Jan 1 at 23:06
The hand (and wrist, forearm, &c) also has many flexible muscles and joints, which act as shock absorbers. Not just when firing a pistol: think of hitting something with your fist.
– jamesqf
Dec 29 '18 at 18:50
The hand (and wrist, forearm, &c) also has many flexible muscles and joints, which act as shock absorbers. Not just when firing a pistol: think of hitting something with your fist.
– jamesqf
Dec 29 '18 at 18:50
15
15
If the gun weighed as much (very important) as the projectile, and was as small (also important), it would. We have an ancient Styre rifle that was shortened at some point, vastly reducing it's weight. Its recoil went from being described as 'unpleasant' to 'grim'.
– Mazura
Dec 29 '18 at 20:39
If the gun weighed as much (very important) as the projectile, and was as small (also important), it would. We have an ancient Styre rifle that was shortened at some point, vastly reducing it's weight. Its recoil went from being described as 'unpleasant' to 'grim'.
– Mazura
Dec 29 '18 at 20:39
3
3
The mass of the firearm also plays a significant role.
– whatsisname
Dec 30 '18 at 22:58
The mass of the firearm also plays a significant role.
– whatsisname
Dec 30 '18 at 22:58
2
2
Occasionally, beginning rifle shooters attempt to "deal with" the recoil by bracing their shooting shoulder against a good stout tree. They generally only try that once.
– Jamie Hanrahan
Jan 1 at 23:06
Occasionally, beginning rifle shooters attempt to "deal with" the recoil by bracing their shooting shoulder against a good stout tree. They generally only try that once.
– Jamie Hanrahan
Jan 1 at 23:06
add a comment |
Therefore I wonder why shooting a bullet with a handgun is not ripping your hand apart.
Because the mass of the handgun is greater than the mass of the bullet, and because the energy transferred from the gun to your hand is distributed across the surface of the pistol grip.
One of the more interesting things I heard on a trip to Williamsburg Virginia (which is period of ~1776; American Revolution) was the question of how much the muskets weighed. The historical figure answered (I don't recall the weight) and the questioner said in a surprised voice, "That's basically the same that guns weigh today!"
"Yes," replied this historical figure, "because the physics hasn't changed. They could make rifles lighter today, but they don't because the recoil would be worse."
There are other things put in modern pistols to reduce recoil like springs and discharge, but since you tagged it "Newtonian physics" I expect your are less interested in those things.
answered Dec 29 '18 at 22:32
J. Chris ComptonJ. Chris Compton
35124
2
This answer covers it best. If you had a 'bullet trap' of the same weight as the gun, which decelerated the bullet over an equally long trajectory as it was fired, catching a bullet using this device would be no harder than firing it. Catching a bullet with just the mass of skin in front of it, over the thickness of said skin, generally doesn't work for anything but bb's.
– Eelco Hoogendoorn
Dec 30 '18 at 3:57
1
@EelcoHoogendoorn a bulletproof vest is one example of "bullet trap"
– Eric Duminil
Dec 30 '18 at 13:05
1
Re "They could make rifles lighter today...", they do. Compare the weight of an M14 to the M16, for instance. Or for pistols, the .45 caliber Colt M1911 weighs 1100 g, the comparable Glock .45 845 g. (Other Glock models weigh even less.)
– jamesqf
Dec 30 '18 at 18:13
1
@jamesqf: M16 is lighter, but it also shoots rounds that are roughly half the mass.
– whatsisname
Dec 30 '18 at 23:04
2
@DavidRicherby There's also a tradeoff of muzzle control vs. handling as mass is moved forward or back in a rifle. More forward weight keeps the muzzle more stable, but increases the polar moment of inertia when moving the gun around, and vice versa.
– nasch
Jan 1 at 19:34
|
show 5 more comments
Therefore I wonder why shooting a bullet with a handgun is not ripping your hand apart.
Because the mass of the handgun is greater than the mass of the bullet, and because the energy transferred from the gun to your hand is distributed across the surface of the pistol grip.
One of the more interesting things I heard on a trip to Williamsburg Virginia (which is period of ~1776; American Revolution) was the question of how much the muskets weighed. The historical figure answered (I don't recall the weight) and the questioner said in a surprised voice, "That's basically the same that guns weigh today!"
"Yes," replied this historical figure, "because the physics hasn't changed. They could make rifles lighter today, but they don't because the recoil would be worse."
There are other things put in modern pistols to reduce recoil like springs and discharge, but since you tagged it "Newtonian physics" I expect your are less interested in those things.
answered Dec 29 '18 at 22:32
J. Chris ComptonJ. Chris Compton
35124
2
This answer covers it best. If you had a 'bullet trap' of the same weight as the gun, which decelerated the bullet over an equally long trajectory as it was fired, catching a bullet using this device would be no harder than firing it. Catching a bullet with just the mass of skin in front of it, over the thickness of said skin, generally doesn't work for anything but bb's.
– Eelco Hoogendoorn
Dec 30 '18 at 3:57
1
@EelcoHoogendoorn a bulletproof vest is one example of "bullet trap"
– Eric Duminil
Dec 30 '18 at 13:05
1
Re "They could make rifles lighter today...", they do. Compare the weight of an M14 to the M16, for instance. Or for pistols, the .45 caliber Colt M1911 weighs 1100 g, the comparable Glock .45 845 g. (Other Glock models weigh even less.)
– jamesqf
Dec 30 '18 at 18:13
1
@jamesqf: M16 is lighter, but it also shoots rounds that are roughly half the mass.
– whatsisname
Dec 30 '18 at 23:04
2
@DavidRicherby There's also a tradeoff of muzzle control vs. handling as mass is moved forward or back in a rifle. More forward weight keeps the muzzle more stable, but increases the polar moment of inertia when moving the gun around, and vice versa.
– nasch
Jan 1 at 19:34
|
show 5 more comments
Therefore I wonder why shooting a bullet with a handgun is not ripping your hand apart.
Because the mass of the handgun is greater than the mass of the bullet, and because the energy transferred from the gun to your hand is distributed across the surface of the pistol grip.
One of the more interesting things I heard on a trip to Williamsburg Virginia (which is period of ~1776; American Revolution) was the question of how much the muskets weighed. The historical figure answered (I don't recall the weight) and the questioner said in a surprised voice, "That's basically the same that guns weigh today!"
"Yes," replied this historical figure, "because the physics hasn't changed. They could make rifles lighter today, but they don't because the recoil would be worse."
There are other things put in modern pistols to reduce recoil like springs and discharge, but since you tagged it "Newtonian physics" I expect your are less interested in those things.
answered Dec 29 '18 at 22:32
J. Chris ComptonJ. Chris Compton
35124
Therefore I wonder why shooting a bullet with a handgun is not ripping your hand apart.
Because the mass of the handgun is greater than the mass of the bullet, and because the energy transferred from the gun to your hand is distributed across the surface of the pistol grip.
One of the more interesting things I heard on a trip to Williamsburg Virginia (which is period of ~1776; American Revolution) was the question of how much the muskets weighed. The historical figure answered (I don't recall the weight) and the questioner said in a surprised voice, "That's basically the same that guns weigh today!"
"Yes," replied this historical figure, "because the physics hasn't changed. They could make rifles lighter today, but they don't because the recoil would be worse."
There are other things put in modern pistols to reduce recoil like springs and discharge, but since you tagged it "Newtonian physics" I expect your are less interested in those things.
answered Dec 29 '18 at 22:32
J. Chris ComptonJ. Chris Compton
35124
answered Dec 29 '18 at 22:32
J. Chris ComptonJ. Chris Compton
35124
answered Dec 29 '18 at 22:32
J. Chris ComptonJ. Chris Compton
35124
answered Dec 29 '18 at 22:32
J. Chris ComptonJ. Chris Compton
35124
35124
2
This answer covers it best. If you had a 'bullet trap' of the same weight as the gun, which decelerated the bullet over an equally long trajectory as it was fired, catching a bullet using this device would be no harder than firing it. Catching a bullet with just the mass of skin in front of it, over the thickness of said skin, generally doesn't work for anything but bb's.
– Eelco Hoogendoorn
Dec 30 '18 at 3:57
1
@EelcoHoogendoorn a bulletproof vest is one example of "bullet trap"
– Eric Duminil
Dec 30 '18 at 13:05
1
Re "They could make rifles lighter today...", they do. Compare the weight of an M14 to the M16, for instance. Or for pistols, the .45 caliber Colt M1911 weighs 1100 g, the comparable Glock .45 845 g. (Other Glock models weigh even less.)
– jamesqf
Dec 30 '18 at 18:13
1
@jamesqf: M16 is lighter, but it also shoots rounds that are roughly half the mass.
– whatsisname
Dec 30 '18 at 23:04
2
@DavidRicherby There's also a tradeoff of muzzle control vs. handling as mass is moved forward or back in a rifle. More forward weight keeps the muzzle more stable, but increases the polar moment of inertia when moving the gun around, and vice versa.
– nasch
Jan 1 at 19:34
|
show 5 more comments
2
This answer covers it best. If you had a 'bullet trap' of the same weight as the gun, which decelerated the bullet over an equally long trajectory as it was fired, catching a bullet using this device would be no harder than firing it. Catching a bullet with just the mass of skin in front of it, over the thickness of said skin, generally doesn't work for anything but bb's.
– Eelco Hoogendoorn
Dec 30 '18 at 3:57
1
@EelcoHoogendoorn a bulletproof vest is one example of "bullet trap"
– Eric Duminil
Dec 30 '18 at 13:05
1
Re "They could make rifles lighter today...", they do. Compare the weight of an M14 to the M16, for instance. Or for pistols, the .45 caliber Colt M1911 weighs 1100 g, the comparable Glock .45 845 g. (Other Glock models weigh even less.)
– jamesqf
Dec 30 '18 at 18:13
1
@jamesqf: M16 is lighter, but it also shoots rounds that are roughly half the mass.
– whatsisname
Dec 30 '18 at 23:04
2
@DavidRicherby There's also a tradeoff of muzzle control vs. handling as mass is moved forward or back in a rifle. More forward weight keeps the muzzle more stable, but increases the polar moment of inertia when moving the gun around, and vice versa.
– nasch
Jan 1 at 19:34
2
2
This answer covers it best. If you had a 'bullet trap' of the same weight as the gun, which decelerated the bullet over an equally long trajectory as it was fired, catching a bullet using this device would be no harder than firing it. Catching a bullet with just the mass of skin in front of it, over the thickness of said skin, generally doesn't work for anything but bb's.
– Eelco Hoogendoorn
Dec 30 '18 at 3:57
This answer covers it best. If you had a 'bullet trap' of the same weight as the gun, which decelerated the bullet over an equally long trajectory as it was fired, catching a bullet using this device would be no harder than firing it. Catching a bullet with just the mass of skin in front of it, over the thickness of said skin, generally doesn't work for anything but bb's.
– Eelco Hoogendoorn
Dec 30 '18 at 3:57
1
1
@EelcoHoogendoorn a bulletproof vest is one example of "bullet trap"
– Eric Duminil
Dec 30 '18 at 13:05
@EelcoHoogendoorn a bulletproof vest is one example of "bullet trap"
– Eric Duminil
Dec 30 '18 at 13:05
1
1
Re "They could make rifles lighter today...", they do. Compare the weight of an M14 to the M16, for instance. Or for pistols, the .45 caliber Colt M1911 weighs 1100 g, the comparable Glock .45 845 g. (Other Glock models weigh even less.)
– jamesqf
Dec 30 '18 at 18:13
Re "They could make rifles lighter today...", they do. Compare the weight of an M14 to the M16, for instance. Or for pistols, the .45 caliber Colt M1911 weighs 1100 g, the comparable Glock .45 845 g. (Other Glock models weigh even less.)
– jamesqf
Dec 30 '18 at 18:13
1
1
@jamesqf: M16 is lighter, but it also shoots rounds that are roughly half the mass.
– whatsisname
Dec 30 '18 at 23:04
@jamesqf: M16 is lighter, but it also shoots rounds that are roughly half the mass.
– whatsisname
Dec 30 '18 at 23:04
2
2
@DavidRicherby There's also a tradeoff of muzzle control vs. handling as mass is moved forward or back in a rifle. More forward weight keeps the muzzle more stable, but increases the polar moment of inertia when moving the gun around, and vice versa.
– nasch
Jan 1 at 19:34
@DavidRicherby There's also a tradeoff of muzzle control vs. handling as mass is moved forward or back in a rifle. More forward weight keeps the muzzle more stable, but increases the polar moment of inertia when moving the gun around, and vice versa.
– nasch
Jan 1 at 19:34
|
show 5 more comments
By holding the gun firmly and taking a proper firing pose, you arrange it so that the momentum is passed to your entire body as a whole, or a large enough part of it. Your body is much heavier than the bullet, so the speed it gets from the shot is very small and easily countered by muscles and joints.
If you don't take a proper pose or don't hold the gun properly, you may fail to steady it against the kickback, or the momentum could be passed only to a small part of the body, resulting in injury.
answered Dec 31 '18 at 14:24
ivan_pozdeevivan_pozdeev
20714
add a comment |
By holding the gun firmly and taking a proper firing pose, you arrange it so that the momentum is passed to your entire body as a whole, or a large enough part of it. Your body is much heavier than the bullet, so the speed it gets from the shot is very small and easily countered by muscles and joints.
If you don't take a proper pose or don't hold the gun properly, you may fail to steady it against the kickback, or the momentum could be passed only to a small part of the body, resulting in injury.
answered Dec 31 '18 at 14:24
ivan_pozdeevivan_pozdeev
20714
add a comment |
By holding the gun firmly and taking a proper firing pose, you arrange it so that the momentum is passed to your entire body as a whole, or a large enough part of it. Your body is much heavier than the bullet, so the speed it gets from the shot is very small and easily countered by muscles and joints.
If you don't take a proper pose or don't hold the gun properly, you may fail to steady it against the kickback, or the momentum could be passed only to a small part of the body, resulting in injury.
answered Dec 31 '18 at 14:24
ivan_pozdeevivan_pozdeev
20714
By holding the gun firmly and taking a proper firing pose, you arrange it so that the momentum is passed to your entire body as a whole, or a large enough part of it. Your body is much heavier than the bullet, so the speed it gets from the shot is very small and easily countered by muscles and joints.
If you don't take a proper pose or don't hold the gun properly, you may fail to steady it against the kickback, or the momentum could be passed only to a small part of the body, resulting in injury.
answered Dec 31 '18 at 14:24
ivan_pozdeevivan_pozdeev
20714
answered Dec 31 '18 at 14:24
ivan_pozdeevivan_pozdeev
20714
answered Dec 31 '18 at 14:24
ivan_pozdeevivan_pozdeev
20714
answered Dec 31 '18 at 14:24
ivan_pozdeevivan_pozdeev
20714
20714
add a comment |
add a comment |
Lets nail down some figures with a random choice of compatible equipment with specs publicly available.
Gun https://us.glock.com/products/G19
Gun Weight with empty magazine 670 g | 23.63 oz
Barrel Length 102 mm | 4.02 inch (slightly less than ammo test, so time under acceleration is slightly less)
Ammunition https://en.wikipedia.org/wiki/9%C3%9719mm_Parabellum
Bullet mass 8.04 g (124 gr) Federal FMJ
Bullet velocity 1,150 ft/s (350 m/s) @ 118mm (4.65") barrel length, 0.00067s
Bullet energy 364 ft⋅lbf (494 J)
We could set this up as a conservation of energy problem, and that would be a reasonable approximation with a satisfying answer.
The time under acceleration is very short(0.00067s). (anything less than 0.01s feels instantaneous)
The gun weighs a lot (83 times) more than the bullet.
It is coupled to a human that weighs very much more.(180lb = 10155 times)
It is coupled over a large, soft, energy absorbing surface(hand).
The recoil is transmitted through several relatively massive energy absorbing tendons, tensed muscles, bones, flexed joints, fat, fluids, which redirect the recoil into countless vectors.
Where is the point of impact between gun and bullet? How does the gun impart force to the bullet and the bullet impart equal and opposite force to the gun? Well that doesn't really happen. It isn't a collision.
Only the timing is similar to a collision because both gun and bullet are accelerated in opposite directions by expanding gasses with same T0.
But is energy imparted equally? The center of those gasses is itself being accelerated away from the gun. Barrel friction counteracts recoil. You may be mistaken in part of your premise "momentum on both sides is to be equal".
answered Dec 30 '18 at 1:45
slomobileslomobile
514
gasses -> gases
– Faheem Mitha
Dec 30 '18 at 7:20
1
@FaheemMitha In British English, and possibly all others except US English (where it is no longer the common variation), gasses is correct.
– Andrew Morton
Dec 30 '18 at 13:52
@AndrewMorton Wow, I don't think I've come across that variant before. Or if I have, I don't remember it.
– Faheem Mitha
Dec 30 '18 at 15:34
How much the human weights is quite irrelevant: When the bullet leaves the gun, the gun has moved back by $102mm/83 = 1.2mm$ and is moving at a speed of $350frac{m}{s}/83 = 4.2frac{m}{s} = 15frac{km}{h}$. The gun subsequently basically crashes into the human at that speed: The tiny translation of $1.2mm$ is simply not enough to transmit any relevant forces.
– cmaster
Dec 31 '18 at 2:09
That said, if you half the weight of the gun, you get it crashing into the user at twice the speed, i.e. $30frac{km}{h}$, which is definitely not fun.
– cmaster
Dec 31 '18 at 2:12
|
show 3 more comments
Lets nail down some figures with a random choice of compatible equipment with specs publicly available.
Gun https://us.glock.com/products/G19
Gun Weight with empty magazine 670 g | 23.63 oz
Barrel Length 102 mm | 4.02 inch (slightly less than ammo test, so time under acceleration is slightly less)
Ammunition https://en.wikipedia.org/wiki/9%C3%9719mm_Parabellum
Bullet mass 8.04 g (124 gr) Federal FMJ
Bullet velocity 1,150 ft/s (350 m/s) @ 118mm (4.65") barrel length, 0.00067s
Bullet energy 364 ft⋅lbf (494 J)
We could set this up as a conservation of energy problem, and that would be a reasonable approximation with a satisfying answer.
The time under acceleration is very short(0.00067s). (anything less than 0.01s feels instantaneous)
The gun weighs a lot (83 times) more than the bullet.
It is coupled to a human that weighs very much more.(180lb = 10155 times)
It is coupled over a large, soft, energy absorbing surface(hand).
The recoil is transmitted through several relatively massive energy absorbing tendons, tensed muscles, bones, flexed joints, fat, fluids, which redirect the recoil into countless vectors.
Where is the point of impact between gun and bullet? How does the gun impart force to the bullet and the bullet impart equal and opposite force to the gun? Well that doesn't really happen. It isn't a collision.
Only the timing is similar to a collision because both gun and bullet are accelerated in opposite directions by expanding gasses with same T0.
But is energy imparted equally? The center of those gasses is itself being accelerated away from the gun. Barrel friction counteracts recoil. You may be mistaken in part of your premise "momentum on both sides is to be equal".
answered Dec 30 '18 at 1:45
slomobileslomobile
514
gasses -> gases
– Faheem Mitha
Dec 30 '18 at 7:20
1
@FaheemMitha In British English, and possibly all others except US English (where it is no longer the common variation), gasses is correct.
– Andrew Morton
Dec 30 '18 at 13:52
@AndrewMorton Wow, I don't think I've come across that variant before. Or if I have, I don't remember it.
– Faheem Mitha
Dec 30 '18 at 15:34
How much the human weights is quite irrelevant: When the bullet leaves the gun, the gun has moved back by $102mm/83 = 1.2mm$ and is moving at a speed of $350frac{m}{s}/83 = 4.2frac{m}{s} = 15frac{km}{h}$. The gun subsequently basically crashes into the human at that speed: The tiny translation of $1.2mm$ is simply not enough to transmit any relevant forces.
– cmaster
Dec 31 '18 at 2:09
That said, if you half the weight of the gun, you get it crashing into the user at twice the speed, i.e. $30frac{km}{h}$, which is definitely not fun.
– cmaster
Dec 31 '18 at 2:12
|
show 3 more comments
Lets nail down some figures with a random choice of compatible equipment with specs publicly available.
Gun https://us.glock.com/products/G19
Gun Weight with empty magazine 670 g | 23.63 oz
Barrel Length 102 mm | 4.02 inch (slightly less than ammo test, so time under acceleration is slightly less)
Ammunition https://en.wikipedia.org/wiki/9%C3%9719mm_Parabellum
Bullet mass 8.04 g (124 gr) Federal FMJ
Bullet velocity 1,150 ft/s (350 m/s) @ 118mm (4.65") barrel length, 0.00067s
Bullet energy 364 ft⋅lbf (494 J)
We could set this up as a conservation of energy problem, and that would be a reasonable approximation with a satisfying answer.
The time under acceleration is very short(0.00067s). (anything less than 0.01s feels instantaneous)
The gun weighs a lot (83 times) more than the bullet.
It is coupled to a human that weighs very much more.(180lb = 10155 times)
It is coupled over a large, soft, energy absorbing surface(hand).
The recoil is transmitted through several relatively massive energy absorbing tendons, tensed muscles, bones, flexed joints, fat, fluids, which redirect the recoil into countless vectors.
Where is the point of impact between gun and bullet? How does the gun impart force to the bullet and the bullet impart equal and opposite force to the gun? Well that doesn't really happen. It isn't a collision.
Only the timing is similar to a collision because both gun and bullet are accelerated in opposite directions by expanding gasses with same T0.
But is energy imparted equally? The center of those gasses is itself being accelerated away from the gun. Barrel friction counteracts recoil. You may be mistaken in part of your premise "momentum on both sides is to be equal".
answered Dec 30 '18 at 1:45
slomobileslomobile
514
Lets nail down some figures with a random choice of compatible equipment with specs publicly available.
Gun https://us.glock.com/products/G19
Gun Weight with empty magazine 670 g | 23.63 oz
Barrel Length 102 mm | 4.02 inch (slightly less than ammo test, so time under acceleration is slightly less)
Ammunition https://en.wikipedia.org/wiki/9%C3%9719mm_Parabellum
Bullet mass 8.04 g (124 gr) Federal FMJ
Bullet velocity 1,150 ft/s (350 m/s) @ 118mm (4.65") barrel length, 0.00067s
Bullet energy 364 ft⋅lbf (494 J)
We could set this up as a conservation of energy problem, and that would be a reasonable approximation with a satisfying answer.
The time under acceleration is very short(0.00067s). (anything less than 0.01s feels instantaneous)
The gun weighs a lot (83 times) more than the bullet.
It is coupled to a human that weighs very much more.(180lb = 10155 times)
It is coupled over a large, soft, energy absorbing surface(hand).
The recoil is transmitted through several relatively massive energy absorbing tendons, tensed muscles, bones, flexed joints, fat, fluids, which redirect the recoil into countless vectors.
Where is the point of impact between gun and bullet? How does the gun impart force to the bullet and the bullet impart equal and opposite force to the gun? Well that doesn't really happen. It isn't a collision.
Only the timing is similar to a collision because both gun and bullet are accelerated in opposite directions by expanding gasses with same T0.
But is energy imparted equally? The center of those gasses is itself being accelerated away from the gun. Barrel friction counteracts recoil. You may be mistaken in part of your premise "momentum on both sides is to be equal".
answered Dec 30 '18 at 1:45
slomobileslomobile
514
edited Dec 30 '18 at 1:55
answered Dec 30 '18 at 1:45
slomobileslomobile
514
answered Dec 30 '18 at 1:45
slomobileslomobile
514
answered Dec 30 '18 at 1:45
slomobileslomobile
514
514
gasses -> gases
– Faheem Mitha
Dec 30 '18 at 7:20
1
@FaheemMitha In British English, and possibly all others except US English (where it is no longer the common variation), gasses is correct.
– Andrew Morton
Dec 30 '18 at 13:52
@AndrewMorton Wow, I don't think I've come across that variant before. Or if I have, I don't remember it.
– Faheem Mitha
Dec 30 '18 at 15:34
How much the human weights is quite irrelevant: When the bullet leaves the gun, the gun has moved back by $102mm/83 = 1.2mm$ and is moving at a speed of $350frac{m}{s}/83 = 4.2frac{m}{s} = 15frac{km}{h}$. The gun subsequently basically crashes into the human at that speed: The tiny translation of $1.2mm$ is simply not enough to transmit any relevant forces.
– cmaster
Dec 31 '18 at 2:09
That said, if you half the weight of the gun, you get it crashing into the user at twice the speed, i.e. $30frac{km}{h}$, which is definitely not fun.
– cmaster
Dec 31 '18 at 2:12
|
show 3 more comments
gasses -> gases
– Faheem Mitha
Dec 30 '18 at 7:20
1
@FaheemMitha In British English, and possibly all others except US English (where it is no longer the common variation), gasses is correct.
– Andrew Morton
Dec 30 '18 at 13:52
@AndrewMorton Wow, I don't think I've come across that variant before. Or if I have, I don't remember it.
– Faheem Mitha
Dec 30 '18 at 15:34
How much the human weights is quite irrelevant: When the bullet leaves the gun, the gun has moved back by $102mm/83 = 1.2mm$ and is moving at a speed of $350frac{m}{s}/83 = 4.2frac{m}{s} = 15frac{km}{h}$. The gun subsequently basically crashes into the human at that speed: The tiny translation of $1.2mm$ is simply not enough to transmit any relevant forces.
– cmaster
Dec 31 '18 at 2:09
That said, if you half the weight of the gun, you get it crashing into the user at twice the speed, i.e. $30frac{km}{h}$, which is definitely not fun.
– cmaster
Dec 31 '18 at 2:12
gasses -> gases
– Faheem Mitha
Dec 30 '18 at 7:20
gasses -> gases
– Faheem Mitha
Dec 30 '18 at 7:20
1
1
@FaheemMitha In British English, and possibly all others except US English (where it is no longer the common variation), gasses is correct.
– Andrew Morton
Dec 30 '18 at 13:52
@FaheemMitha In British English, and possibly all others except US English (where it is no longer the common variation), gasses is correct.
– Andrew Morton
Dec 30 '18 at 13:52
@AndrewMorton Wow, I don't think I've come across that variant before. Or if I have, I don't remember it.
– Faheem Mitha
Dec 30 '18 at 15:34
@AndrewMorton Wow, I don't think I've come across that variant before. Or if I have, I don't remember it.
– Faheem Mitha
Dec 30 '18 at 15:34
How much the human weights is quite irrelevant: When the bullet leaves the gun, the gun has moved back by $102mm/83 = 1.2mm$ and is moving at a speed of $350frac{m}{s}/83 = 4.2frac{m}{s} = 15frac{km}{h}$. The gun subsequently basically crashes into the human at that speed: The tiny translation of $1.2mm$ is simply not enough to transmit any relevant forces.
– cmaster
Dec 31 '18 at 2:09
How much the human weights is quite irrelevant: When the bullet leaves the gun, the gun has moved back by $102mm/83 = 1.2mm$ and is moving at a speed of $350frac{m}{s}/83 = 4.2frac{m}{s} = 15frac{km}{h}$. The gun subsequently basically crashes into the human at that speed: The tiny translation of $1.2mm$ is simply not enough to transmit any relevant forces.
– cmaster
Dec 31 '18 at 2:09
That said, if you half the weight of the gun, you get it crashing into the user at twice the speed, i.e. $30frac{km}{h}$, which is definitely not fun.
– cmaster
Dec 31 '18 at 2:12
That said, if you half the weight of the gun, you get it crashing into the user at twice the speed, i.e. $30frac{km}{h}$, which is definitely not fun.
– cmaster
Dec 31 '18 at 2:12
|
show 3 more comments
Assuming impulse invariance, roughly the same amount of momentum is transferred to the shooter and the bullet and ultimately the target. However, if momentum was the only lethal element of a bullet, there would not be such a thing as a bullet-proof vest since a bullet-proof vest cannot keep momentum off the target (you'd need a ground-mounted shield for that). It can, however, absorb and disperse energy which, as opposed to momentum, grows with the square of the speed of moving mass. A bullet will often not transfer all of its energy to the target unless it gets stuck and will rather work by inflicting deadly injury, but there are some that are designed to stop and/or disperse on entry, like hollow point bullets. Those tend to be much more deadly than an ordinary bullet even when not hitting vital parts of the body.
So the basic point is that a pistol has more mass than the bullet, so while they share momentum equally (at least when discounting what exhaust fumes may do), they don't receive equal amounts of energy.
Note that shoulder-held rocket launchers are built in a way where the exhaust fumes (which continue accelerating the rocket via conservation of momentum) can leave the launcher at its back, thus minimizing the amount of momentum transferred to the operator (naturally, the place behind someone with a rocket launcher is a very bad place to be as opposed to being behind a gunner). Since it is the pressure of the propellant explosion rather than the momentum of something leaving the bullet, this kind of mechanism is not feasible with a gun to a good degree.
add a comment |
Assuming impulse invariance, roughly the same amount of momentum is transferred to the shooter and the bullet and ultimately the target. However, if momentum was the only lethal element of a bullet, there would not be such a thing as a bullet-proof vest since a bullet-proof vest cannot keep momentum off the target (you'd need a ground-mounted shield for that). It can, however, absorb and disperse energy which, as opposed to momentum, grows with the square of the speed of moving mass. A bullet will often not transfer all of its energy to the target unless it gets stuck and will rather work by inflicting deadly injury, but there are some that are designed to stop and/or disperse on entry, like hollow point bullets. Those tend to be much more deadly than an ordinary bullet even when not hitting vital parts of the body.
So the basic point is that a pistol has more mass than the bullet, so while they share momentum equally (at least when discounting what exhaust fumes may do), they don't receive equal amounts of energy.
Note that shoulder-held rocket launchers are built in a way where the exhaust fumes (which continue accelerating the rocket via conservation of momentum) can leave the launcher at its back, thus minimizing the amount of momentum transferred to the operator (naturally, the place behind someone with a rocket launcher is a very bad place to be as opposed to being behind a gunner). Since it is the pressure of the propellant explosion rather than the momentum of something leaving the bullet, this kind of mechanism is not feasible with a gun to a good degree.
add a comment |
Assuming impulse invariance, roughly the same amount of momentum is transferred to the shooter and the bullet and ultimately the target. However, if momentum was the only lethal element of a bullet, there would not be such a thing as a bullet-proof vest since a bullet-proof vest cannot keep momentum off the target (you'd need a ground-mounted shield for that). It can, however, absorb and disperse energy which, as opposed to momentum, grows with the square of the speed of moving mass. A bullet will often not transfer all of its energy to the target unless it gets stuck and will rather work by inflicting deadly injury, but there are some that are designed to stop and/or disperse on entry, like hollow point bullets. Those tend to be much more deadly than an ordinary bullet even when not hitting vital parts of the body.
So the basic point is that a pistol has more mass than the bullet, so while they share momentum equally (at least when discounting what exhaust fumes may do), they don't receive equal amounts of energy.
Note that shoulder-held rocket launchers are built in a way where the exhaust fumes (which continue accelerating the rocket via conservation of momentum) can leave the launcher at its back, thus minimizing the amount of momentum transferred to the operator (naturally, the place behind someone with a rocket launcher is a very bad place to be as opposed to being behind a gunner). Since it is the pressure of the propellant explosion rather than the momentum of something leaving the bullet, this kind of mechanism is not feasible with a gun to a good degree.
Assuming impulse invariance, roughly the same amount of momentum is transferred to the shooter and the bullet and ultimately the target. However, if momentum was the only lethal element of a bullet, there would not be such a thing as a bullet-proof vest since a bullet-proof vest cannot keep momentum off the target (you'd need a ground-mounted shield for that). It can, however, absorb and disperse energy which, as opposed to momentum, grows with the square of the speed of moving mass. A bullet will often not transfer all of its energy to the target unless it gets stuck and will rather work by inflicting deadly injury, but there are some that are designed to stop and/or disperse on entry, like hollow point bullets. Those tend to be much more deadly than an ordinary bullet even when not hitting vital parts of the body.
So the basic point is that a pistol has more mass than the bullet, so while they share momentum equally (at least when discounting what exhaust fumes may do), they don't receive equal amounts of energy.
Note that shoulder-held rocket launchers are built in a way where the exhaust fumes (which continue accelerating the rocket via conservation of momentum) can leave the launcher at its back, thus minimizing the amount of momentum transferred to the operator (naturally, the place behind someone with a rocket launcher is a very bad place to be as opposed to being behind a gunner). Since it is the pressure of the propellant explosion rather than the momentum of something leaving the bullet, this kind of mechanism is not feasible with a gun to a good degree.
answered Dec 30 '18 at 17:58
user218452
add a comment |
add a comment |
protected by rob♦ Dec 31 '18 at 16:54
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– rob♦
Dec 31 '18 at 16:59
1
Piercing a human doesn't require as much energy as you might think. It's about pressure. Imagine a knife, or a needle. A bullet has a small area so doesn't need a lot of energy to pierce. Imagine holding a stick the diameter of a bullet in place of the gun and having someone push a leg of lamb onto it. How hard would you need to brace in order for the stick to pierce the meat?
– Richard
Jan 1 at 20:12