Recoil timing

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Echo9

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Why does most of the recoil impulse happen after the projectile has left the barrel?
 
it doesn't

The nature of the human - firearm interface makes it appear that way.

In a single shot firearm, (Think bolt action, revolver, lever action, etc). The recoil energy has been changed from potential energy ( in the form of chemical energy) into kinetic energy, and no more will be transferred to the firearm (and therefore the shooter) once the projectile and propellant gasses leave the barrel.

The firearm compresses against your shoulder, in addition to compression of the recoil pad, etc. and this spreads out the recoil impulse that you feel as the shooter.

At least this is my understanding, someone who knows more than me will be along shortly to fix my mistakes, I am sure.
 
You just notice the recoil later, it doesn't actually happen after the bullet leaves the barrel.

Sir Issac Newton's Three Laws of Motion. Nothing more is relevent

A little simplistic but the laws can't be ignored. Any action causes and equal and opposite reaction.
 
I can see how someone could think it happens after the projectile leaves the barrel as the barrel points pretty skyward yet the projectile goes straight ahead, therefore it must have left before the barrel headed skyward. That is true BUT all the acceleration on the gun to push it upward (almost) completely happens while the projectile is still in the barrel being accelerated. Once the projectile has left the barrel the gun still has momentum and a human's squishy hands and grip don't slow it down instantly, so it moves a little.
 
In a Beretta 92:

Projectile weight is 7.45 grams
Firearm weight is 950 grams

The powder charge being detonated is essentially a small explosion that propels the projectile in one direction, and the firearm in the opposite direction. Since the projectile is far less massive, its inertia is overcome much more easily and achieves great velocity by the time it leaves the barrel. Conversely, the firearm is many times more massive than the projectile and thus its inertia is harder to overcome; the same explosive force which propelled the bullet, will push the handgun back at a much lower velocity.

I don't know the math off the top of my head but the handgun ends up being pushed back at somewhere around 5 meters per second. Your body acts as a shock absorber and prevents the handgun from flying away.

I can see how someone could think it happens after the projectile leaves the barrel as the barrel points pretty skyward yet the projectile goes straight ahead, therefore it must have left before the barrel headed skyward
The handgun recoils upward only because the axis upon which the explosion occurs is above the grip of the weapon, which is where you hold it. The gun would fly straight back if you were not holding it by the grip; your grip keeps the bottom of the gun relatively stationary while the upper portion (along the axis of the explosion) is forced rearward, which causes the gun to tilt from recoil. This is why firearms which have the bore axis down closer to the grip (such as the Chiappa Rhino), have less muzzle flip.
 
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Always nice to see a little physics discussion at 7:30 am.

Study Guide:

Force 'F' acts on Gun 'G' and Bullet 'B' simultaneously.

F in this case is the explosion of gunpowder. If we get really technical we could integrate the ever changing pressure of the explosion over the length of the barrel. We'd need force as a function of unburnt powder, and another function of chamber volume since the bullet starts moving immediately, blah blah too early

G has much more mass than B and therefore experiences a much slower acceleration in the opposite direction of the bullet.
 
The rifle *starts* accelerating rearward when the bullet starts moving. By the time the bullet and high-pressure gases have exited the barrel, the gun is moving rearward and/or flipping upward as fast as it is going to. However, the recoil impulse the shooter feels is the result of decelerating the moving firearm to a stop, and that occurs well after the bullet leaves the barrel.
 
Nerve signals travel ~ 100 m/s. The impulse from a rifle has a duration in hundredths of a second. In short, you're perception lags behind the forces, which peak as soon as the bullet is no longer accelerating down the barrel, so that it only seems that the majority of recoil occurs after the bullet leaves the barrel.

F=M(rifle) x A(rifle) = m(bullet) x a(bullet) --> A=(ma)/M=a(m/M)

The impulse is the integral of the force with respect to time over the time it takes the bullet to accelerate down the barrel
 
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Redneck translation:

Yup , yall got it. Equal and opposite reaction.
The bullet got gone really fast because it is light compared to the rifle that is going the other way a lot slower because it is a lot heavier than the bullet.
 
Oh Boy...formulas:)
How about some vector talk:rolleyes:
One of my favorites:D
 
This is why when you see people getting knocked down from gunshots in moves its totally made up. If a projectile struck a person with enough power to throw them say through a glass window in front of the bar in an old western, the shooter would be knocked down as well from the recoil.
 
Nerve signals travel ~ 100 m/s. The impulse from a rifle has a duration in hundredths of a second. In short, you're perception lags behind the forces, which peak as soon as the bullet is no longer accelerating down the barrel, so that it only seems that the majority of recoil occurs after the bullet leaves the barrel.

F=M(rifle) x A(rifle) = m(bullet) x a(bullet) --> A=(ma)/M=a(m/M)

The impulse is the integral of the force with respect to time over the time it takes the bullet to accelerate down the barrel
Well, the rearward acceleration of the gun ends very quickly, yes. But the peak force on your shoulder doesn't occur until significantly later, when the recoiling gun is being decelerated and your shoulder is being shoved backward. And while our perceptions do lag a little behind reality, the acceleration impulse and the peak of the deceleration impulse *are* separated in time by a significant (and probably very perceptible) amount.
 
That's all smoke and mirrors, recoil is all due to the gravitational forces generated by the spin of the bullet trying to pull the sun out of it's orbit around the earth.
 
Yeti,
"That's all smoke and mirrors, recoil is all due to the gravitational forces generated by the spin of the bullet trying to pull the sun out of it's orbit around the earth. "
Maybe in a .45 but not a 9mm. Or is it a right twist barrel in the southern hemisphere? Which is negated if the bipod hops depending on the orientation of the range or the use of saboted bullets.
 
If a projectile struck a person with enough power to throw them say through a glass window in front of the bar in an old western, the shooter would be knocked down as well from the recoil

Not entirely correct Hatter. You have to add in the speed and weight of the projectile to get Kinetic energy. Example, if you get hit with say a .45acp point blank while wearing an older style Kevlar vest, you end up with broken ribs. Now, you cant tell me that the recoil from a .45 will bust your ribs. Now granted yes you are correct in the statement that the target wouldn't be thrown back 20 feet like you see in the movies, but when you factor in the kinetic energy formula, you end up with much more power on the travel side than the recoil side. If this were not the case, a firearm would be useless as it would kill on both sides.
 
Forces are good, forces with formulas are better, now how about some vector talk.
How come the OP hasen't been back here??
 
Posted by Freedom_fighter_in_IL: [If a projectile struck a person with enough power to throw them say through a glass window in front of the bar in an old western, the shooter would be knocked down as well from the recoil [is]]Not entirely correct Hatter.

Actually, Hatterasguy's point is entirely correct. Newton's third Law of Motion, as it were.

You have to add in the speed and weight of the projectile to get Kinetic energy. Example, if you get hit with say a .45acp point blank while wearing an older style Kevlar vest, you end up with broken ribs. Now, you cant tell me that the recoil from a .45 will bust your ribs.

The reason that a bullet might break ribs through a particular type of vest is the concentration of force into a very small area. Spread it out sufficiently, and a 7.62 NATO bullet fired point blank from a rifle will not break ribs.

Now granted yes you are correct in the statement that the target wouldn't be thrown back 20 feet like you see in the movies, but when you factor in the kinetic energy formula, you end up with much more power on the travel side than the recoil side. If this were not the case, a firearm would be useless as it would kill on both sides.
In point of fact, kinetic energy in handgun projectiles contributes to killing effectiveness only insofar as it affects penetration. The bullet penetrates the body of the target and damages organs or causes bleeding or both.
 
Not entirely correct Hatter. You have to add in the speed and weight of the projectile to get Kinetic energy. Example, if you get hit with say a .45acp point blank while wearing an older style Kevlar vest, you end up with broken ribs. Now, you cant tell me that the recoil from a .45 will bust your ribs. Now granted yes you are correct in the statement that the target wouldn't be thrown back 20 feet like you see in the movies, but when you factor in the kinetic energy formula, you end up with much more power on the travel side than the recoil side. If this were not the case, a firearm would be useless as it would kill on both sides.
What would knock someone backward is momentum, not energy; basically, energy breaks things, momentum moves them. Momentum is the same on both ends of the gun (actually a little greater on the shooting side of things than the receiving side, because the propellant gases don't typically reach the target), but far more energy is delivered to the target than is delivered to the shooter.
 
What would knock someone backward is momentum, not energy; basically, energy breaks things, momentum moves them. Momentum is the same on both ends of the gun (actually a little greater on the shooting side of things than the receiving side, because the propellant gases don't typically reach the target), but far more energy is delivered to the target than is delivered to the shooter

Absolutely correct. You guys quoting Newtons 3rd law may wish to bone up a little bit on his second law which deals with momentum and velocity theory as well as "impulse"
 
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