Gun Question - Recoil. Before or after bullet is clear?


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Sean Dempsey
November 17, 2008, 11:49 AM
I have a general, non political gun question!


It's simple - has the bullet cleared the end of the barrel before there is significant recoil movement? For instance, if we had a super super slow motion camera, could we view the bullet leaving the barrel before the gun appears to recoil back at all?

Also, has the bullet left the barrel before say, the slide on a handgun moves, or the bolt in a rifle moves?

Just curiosities, thanks!

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jnyork
November 17, 2008, 12:03 PM
Recoil begins the instant the bullet starts moving. The "opposite and equal reaction" thing. You can easily observe this by firing a handgun with two different loads, one very light and the other one full power. The light load will shoot quite a bit higher due to the bullet being in the barrel longer while the barrel is moving up with the recoil.

Sean Dempsey
November 17, 2008, 12:12 PM
Yeah I realize newtons law is in effect, but I was curious as to say, what is the distance the barrel climbs in the time it takes for the bullet to clear the end of the barrel. A few nanometers?

Some really, really dirty math and I figured that a bullet from a 9mm could clear a 3" barrel in .0003 seconds, or 3 ten-thousandths of a second. So, if that math is even close (which it probably isn't), I wonder how far the recoil of a barrel is in the same .0003 seconds?

everallm
November 17, 2008, 12:21 PM
Actual or felt recoil?

In simple Newtonian physics, for every action there is an equal and opposite reaction. However.......

The recoil impulse is generated at the instantaneous moment of the round ignition and will last as long as the charge is burning. Once the charge has fully combusted, no more energy is being released, no more energy for recoil

Due to mass and inertia, the rifle will take a finite time to start to recoil and will build over time so the perceived recoil effect will still be occurring after the round has left the barrel.

If the rifle is semi-automatic and some of the pressurized, combusted gas is tapped and diverted to the reloading mechanism, there is a secondary impulse that can extend the length of time of the recoil.

The same inertia and lag affects the time it takes a slide, piston or bolt carrier to start and complete a cycle.

MAKster
November 17, 2008, 12:25 PM
The simple answer that I think you are looking for is that the bullet is moving so fast that the shooter doesn't feel the recoil until after the bullet has left the barrel. Flinching in anticipation of the recoil will throw off your aim, not the recoil itself.

misANTHrope
November 17, 2008, 12:26 PM
This seems like something that will not be consistent across the board. I'd expect a large amount of variance between different types of guns, action types, loads, weights....

The consensus with handguns is that the muzzle does begin to rise significantly before the bullet exits. As noted previously here, this is why lower-velocity loads actually shoot higher.

RKBABob
November 17, 2008, 12:31 PM
I have a general, non political gun question! Its about time someone came up with one of those!

Yes, recoil and muzzle rise start the instant the bullet starts to move... as the bullet is propelled forward, the firearm is propelled rearward (and a little upward) in proportion to its weight relative to the bullet. I'd love to see a video of this as well.

General Geoff
November 17, 2008, 01:23 PM
http://www.youtube.com/watch?v=XXZUMGRxPM4


Yes, recoil begins even before the bullet leaves the barrel, but it's so insignificant that air density probably affects shot placement more.

Dr. Tad Hussein Winslow
November 17, 2008, 01:44 PM
The overwhelming majority of recoil happens AFTER the bullet has left the muzzle - BUT, there is *some* recoil movement which occurs while the bullet is in the barrel.

BigG
November 17, 2008, 01:57 PM
Yep - the bullet cannot move unless the gun moves too. The masses are so different it seems like the bullet leaves before the gun moves, but - nope! The same amount of energy that moves the bullet causes the gun to recoil. It's as simple as that.

Sean Dempsey
November 17, 2008, 02:57 PM
Very interesting that lower velocity rounds shoot higher on target. I didn't know that.

The ultra-high speed mechanics of gun functions are fascinating to me. The video of the bullet leaving the barrel is awesome, I can clearly see the muzzle start to rise, even if it was only a fraction of a millimeter.

benEzra
November 17, 2008, 03:07 PM
The gun is already recoiling near its full recoil velocity by the time the bullet exits the barrel. However, it has not had time to actually move very far.

CypherNinja
November 17, 2008, 03:10 PM
The ultra-high speed mechanics of gun functions are fascinating to me.

*clears throat*

Muzzle-loader, Single-shot, Bolt-action, Break-action, Lever, "Normal" Revolver:

Ignition
The bullet begins to move due to climbing pressure. Recoil begins in the same instant due to "Equal/Opposite Reactions".
As the bullet accelerates down the barrel, equal force is applied to both the bullet and the breach face (through the case, when present) by the gas pressure. If the gun was suspended from both ends by strings and was free swinging, the momentums of the bullet+gasses and the gun would MATCH during this process. (Revolvers only: When the bullet passes the cylinder gap a very small, but theoretically there, force is applied to the front face of the cylinder by the gasses escaping through the gap.)
When the bullet reaches the end of the muzzle, virtually all of the recoil has already been generated. After the bullet clears the muzzle, a (generally smaller, usually MUCH smaller) force is applied to the muzzle crown and breach face by the residual gasses still escaping the barrel.
Any "recoil" perceived after this point is simply the shooter in the act of slowing the rearward velocity of the firearm and returning it to the ready position.


All Blowbacks function similar to above. However, all recoil (except the residual mentioned in #4) is first transferred to the bolt (excepting some small rearward force applied through friction to the chamber by the case, and some small forward force applied through friction to the barrel by the bullet, the relative size of these forces could be argued forever). As the bolt travels to the rear, a growing amount of force is applied to the receiver by the recoil spring, as well as some force being applied to the receiver through cocking the hammer, if present. When the bolt reaches the rearward stop, all remaining momentum is transferred to the receiver. Further force is also applied to the receiver as the recoil spring accelerates the bolt forward, and is canceled when the bolt reaches the forward stop. (this is the source of the "lower reciprocating mass is better" arguments)

All Recoil-operated guns also function similar to above. However, all recoil (including #4) is first transferred to the barrel and slide/bolt (or the "upper frame and cylinder" in automatic revolvers ;)). As the barrel and slide/bolt travel to the rear a growing amount of force is applied to the receiver by the recoil spring, as well as some force being applied to the receiver through cocking the hammer, if present. Additionally, when the barrel reaches it's rearward stop, the barrel's momentum is transferred to the receiver. When the slide/bolt reaches the rearward stop, all remaining momentum is transferred to the receiver. Further force is also applied to the receiver as the recoil spring....etc.....etc

All Gas-operated guns function the same as above. However, the "reciprocating mass issue" mentioned at the end of the Blowback and Recoil-operated paragraphs is preceded by a force accelerating the gun forward and the bolt carrier rearward after the bullet passes the gas port. Some of this force is canceled out during the bolt unlocking process as well as some force being applied to the receiver through cocking the hammer, if present. All the remaining momentum is canceled when the bolt carrier hits the rearward stop. Further force is also applied to the receiver as the recoil ....etc.....etc


THERE! DONE! I didn't mention torque/counter-torques, chain-guns, balanced-actions, recoil boosters, bolt accelerators, elastic vs. inelastic collisions, or a few other things, but it seems relatively complete. :scrutiny:

Pardon any errors, I still need to get some caffeine in me. ;)


(i r good at teh mechanics, can i has cookie now?)


EDIT: To clarify, all/most of the various accelerations happen while the bullet is in the barrel. Most, but not all, of the movement happens after the bullet has left the barrel. You can spot all the little movements I mentioned in the video General Geoff linked.

Claude Clay
November 17, 2008, 03:21 PM
p239 with its 40 S&W bbl shoots POA
with a 9mm conv bbl at 25 feet is 2+ inches lower than the 40 group
with 357SIG conv bbl shoots 2+ inches lower than the 9mm group.
same gun and shooter; slower is higher.

rcmodel
November 17, 2008, 03:32 PM
Just look at the sights on any big-bore revolver.
The front sight is always much higher then the rear.

The barrel is pointed well below the line of sight at the moment of ignition. Recoil moves the gun up to POA before the bullet can clear the muzzle.

Another good example:
Try shooting a 30-30 Winchester carbine of a bench.

Hold it down by the forend and it will shoot where the sights are looking.

Let go of the forearm and let it jump, and it will shoot way high.

MrBorland
November 17, 2008, 03:54 PM
Yeah I realize newtons law is in effect, but I was curious as to say, what is the distance the barrel climbs in the time it takes for the bullet to clear the end of the barrel. A few nanometers?

Some really, really dirty math and I figured that a bullet from a 9mm could clear a 3" barrel in .0003 seconds, or 3 ten-thousandths of a second. So, if that math is even close (which it probably isn't), I wonder how far the recoil of a barrel is in the same .0003 seconds?


About 0.036". You were on the right track...sort of.

You have to solve the conservation of momentum (mass x velocity) equation, where velocity is distance per time. Time actually cancels out, since it's the same for the gun and the bullet (i.e. the amount the gun recoil while the bullet's in the barrel). The distance the bullet travels is the barrel length (L), whereas the distance the gun travels (R) is the amount of recoil you're trying to calculate. Bottom line:

R = L x (mass of bullet*/mass of gun)

(*maybe have to include mass of powder charge here as well)

For a 40 oz 4" .357 mag shooting a 158 grain bullet, the gun theoretically will move about 0.036". If this muzzle movement were all in one direction, it'd send the bullet about 8" off target at 25 yards!

Now, for some reality: This assumes the 40oz gun is hanging free in space. It doesn't generally do that, so the effective weight of the gun must include the effective weight of the shooter's body, and that of the earth they're standing on. More difficult to quantitate is how much control (via grip) the shooter has on the gun.

Many shooters can shoot much much better than 8" groups at 25 yards, so the amount the muzzle really moves is much less than what the simple physics predicts. The simple physics does say, though, that all else being equal, a longer barrel actually increases this intrinsic recoil.

wally
November 17, 2008, 04:04 PM
Recoil is not significant until the bullet leave the barrel with rifles and pistols. Watch any of the super slo-motion shots of guns firing that you can find on You-Tube if you don't believe me.

Revolvers, are different since there are two "recoils", first when the bullet leaves the cylinder and enters the barrel and a larger one when the bullet exits the muzzle. So recoil during firing (before the bullet exits) does have a noticeable effect with revolvers. But any effect with rifles or pistols are swamped by changes in barrel harmonics with different loads.

--wally.

1911Tuner
November 17, 2008, 09:27 PM
Fascinating...

Rather than become embroiled in yet another theory storm...Let's play "Just Imagine" and see if we can provoke thought with some sort of logical progression and conclusion.

Being aware that perfection only exists in the mind of God, and that there is no such thing as a perfect dimension...Let's build an imaginary cannon and assume for the sake of Isaac Newton that it is indeed perfect in all described aspects.

Let this cannon be a double-ended cannon, with a bore that is not only perfectly concentric, but also precisely the same inner diameter from one end to the other. Let's also assume that the thickness of the barrel doesn't vary even a ten millionth of an inch. The bore's surface is also precisely the same from end-to-end.

Let's assume two identical cannon balls...both perfectly round and identical in surface finish. These two projectiles are also precisely of the same mass/weight.

Let's place a flash hole that is located precisely in the center of the barrel, and load each ball precisely equidistant from its respective muzzle with a compressed powder charge.

Both balls are bearing equally hard on the powder charge. Precisely and perfectly.

Fire it.

Will the gun recoil? If so...In which direction? If not...Why not..and what WILL happen?

For every action, there must be an equal and opposite reaction. Equal being the operative word, here...it means equal in every way.

E-Q-U-A-L

I'll leave alla ya'll to wrack your brains on it and spend 3 or 4 pages on theories.

Revolvers, are different since there are two "recoils",

:scrutiny:

Popcorn, anyone? :D

moojpg2
November 17, 2008, 09:48 PM
come on guys, read a physics book or something


The cannon won't recoil (defining recoil as "move"), provided 1911Tuners premise where everything is exactly the same on both sides because both shots would produce an equal and opposite force, thus producing no net force. Without force, there is no movement.

Revolvers do not have two recoils:banghead:

1911Tuner
November 17, 2008, 09:58 PM
come on guys, read a physics book or something

Very good, mooj. You didn't quite finish it, but that's close enough. I didn't figure a ringer to come along quite this early, but since you nailed it...

Let's use the same cannon, with everything the same as before...except the cannon balls.

This time, let's assume that they're identical in every way...except this time...let's make one with 10X the mass/weight of the other.

Fire it.

What will happen?

Will it recoil this time?

If so...will the recoil begin at exactly the same instant that the other projectile moves...or will it "wait" until the projectile has exited its muzzle?

If not...Why?

Hint:

You can't apply force in one direction, and you especially can't have force in one direction...and then later on force in the other direction...during the same action/reaction event within the same closed system, which is what would have to happen in order for the gun to sit still until the bullet exits.

woodfiend
November 17, 2008, 10:05 PM
Yeah, it seems that most of the recoil is caused by the action of the bolt, and not by the bullet travelling.

moojpg2
November 17, 2008, 10:10 PM
I think I'll sit back and enjoy this though, It would kind of ruin it if I kept spitting out the answers.

You kind of gave it away with the hint though.

Loomis
November 17, 2008, 10:11 PM
THe reason why recoil "feels" like it doesn't start until after the bullet exist the barrel is threefold:

1) there is friction between the bullet and the barrel. THe bullet tries to drag the barrel along with it. This counteracts some of the "felt" recoil

2) once the bullet exits the barrel, combustion gasses accelerate many times greater than the bullet ever did and add to the "felt" recoil

3) human nervous systems aren't fast enough to perceive force, impulse, momentum, in that minute a time frame.

oops. I just thought of one more item...

4) In a semi auto handgun, the slide creates it's own "felt recoil" when it reverses direction and begins it's forward motion. THis happens WAAAY after the bullet it LOOONG gone.

Conclusion: Both are correct. Recoil occurs after the bullet exits the barrel. REcoil also occurs before the bullet exits the barrel.

And now I'll read the moderators response and see if I am duplicating anything in his post. ;)

Loomis
November 17, 2008, 10:16 PM
NOpe. The mod really oversimplified the theory on this one. I guess I win.

moojpg2
November 17, 2008, 10:23 PM
No the mod didn't oversimplify anything. Once the bullet starts to move, the gun starts to move at that same instant. The movement of the gun may continue after the bullet exits the barrel due to momentum, combustion gases, and mechanical functions of the gun. The point is that recoil begins the instant the powder is ignited and the bullet begins to move.

For every action there is an equal and opposite reaction. It doesn't get any more simple.

1911Tuner
November 17, 2008, 10:56 PM
For every action there is an equal and opposite reaction. It doesn't get any more simple.


Yep. It's that "equal" thing that's hard to grasp for some.

How's this for oversimplified...

Go push on a wall, and you immediately feel a push. Immediately and instantly.

The reason that the wall doesn't move is because the magnitude of force isn't great enough to overcome its resistance...but that doesn't mean that an equal and opposite force wasn't applied.

Kuhnhausen was wrong with the "Balanced Thrust Vector" description..because it assumes that the bullet is moving. If something is moving, it's offered as proof that the forces are UN-balanced. Things only move when the forces become unbalanced. If they're balanced, nothing moves.

If the forces are great enough to move the object in question...it will move. It doesn't have a choice. It may not move very fast or very far...but it will move.

Here's another riddle for ya'll:

Assuming the rule of thumb of 35 fps per inch of barrel gained or lost is accurate...and it usually is...and:

Your muzzle velocity is 1000 fps from a 6-inch barrel, not including the chamber...then we realize 210 fps from the barrel.

Where did that other 790 fps come from?

Hmmm?

Loomis
November 17, 2008, 10:59 PM
That's too simple. Nothing in the real world is that simple.

Loomis
November 17, 2008, 11:02 PM
When you two guys get through stroking eachother, go back and read the OP question and then explain why you've made a fool of yourselves.

moojpg2
November 17, 2008, 11:30 PM
But I will entertain it.

The Op asked if the gun moves before the bullet exits the barrel, the definitive answer is yes, based on the simple laws of Newton. How significant the movement is depends on several variables, but the fact remains, based on Newton's laws that there is movement of the gun before the bullet exits.

The Op asked if the slide or bolt begins to move before the bullet exits the barrel, that is more dependent on the design of the firearm. In some cases I would say yes, in some cases no.

The friction between the bullet and barrel is obviously not greater than the force being applied to the bullet (and gun,it's the equal and opposite thing again) being fired, or the bullet wouldn't exit the barrel. It may negate a tiny amount of force but I would argue (and could prove) that it would be very negligible.

35 Whelen
November 17, 2008, 11:39 PM
rcmodel said:

Another good example:
Try shooting a 30-30 Winchester carbine of a bench.

Hold it down by the forend and it will shoot where the sights are looking.

Let go of the forearm and let it jump, and it will shoot way high.

Period. Finale. End of discussion. Nothing more to say. You can argue physics all you want, but rcmodels example says it all and mirrors my experiences to a "T".
35W

Claude Clay
November 17, 2008, 11:43 PM
even gilligan did not drift as far as this thread has

1911Tuner
November 18, 2008, 08:06 AM
go back and read the OP question

He basically asked if the gun kicks before or after the bullet exits. Rather than try to explain it...again...I simply posted a hypothetical situation to let him...and everyone else...draw their own conclusions.

When you two guys get through stroking eachother...and then explain why you've made a fool of yourselves.

How so? By trying to explain basic physics? That's all the bullet acceleration and recoil event is, after all. Basic Newtonian physics. To wit: "For every action, there must be an equal and opposite reaction."

And, we're not stroking each other at all. Simply recognizing and acknowledging that each understands that which is obvious.

That's too simple. Nothing in the real world is that simple.

But it is just that simple. Like this:

[action]<---(FORCE--->[reaction]

Or, if you prefer:

[bullet<---(BANG)--->breechblock

But, I'll go ahead and say it straight up. Yes. The gun into full or nearly full recoil well before the bullet exits.

RPCVYemen
November 18, 2008, 11:14 AM
It's simple - has the bullet cleared the end of the barrel before there is significant recoil movement?

Note that if recoil did not move the gun before the bullet left the barrel, when when I change bullet weight, etc. the point of impact would not change.

Mike

1911Tuner
November 18, 2008, 12:21 PM
Note that if recoil did not move the gun before the bullet left the barrel, when when I change bullet weight, etc. the point of impact would not change.

Precisely...and if that reply had been understood and accepted...but then the static began.

So...A couple of us tried to explain why that's true, instead of letting it appear to be more theory without any scientific or physical information to back it up. Kinda like the age-old thing about giving a man a fish or teaching him to fish.

By the way, you can also see the effect by simply changing the strength of your grip on a handgun from extra-firm to loosey-goosey. Revolvers show a greater percentage of POI shift than autopistols, due to being fixed breech with no spring to buffer the effects of recoil.

Oh...and Loomis. Your response here:

explain why you've made a fool of yourselves.

Is bordering on personal attack, which isn't allowed here. I don't wield the mod hammer when it's leveled at me whenever I'm involved in the discussion...but it included mooj...so you may consider this an unofficial/undocumented cluebat to keep it civil.

230RN
November 18, 2008, 12:49 PM
Take almost any revolver and lay it on a flat smooth table on its sights upside down.

You will note that the barrel points upward when it is upside down, meaning it points downward from the sight line when it is right-side-up.

Laying it on the table emphasizes the fact that almost every revolver manufacturer recognizes that the barrel will recoil upward before the bullet leaves the barrel, and in light, hand-held firearms, this pre-exit recoil is "significant," using the OP's terms.

(The down-ward pointing barrel is not so obvious in autos, where the barrel is usually hidden from view by the slide.)

As the weight of the gun increases, to, say, an 8-lb rifle, the effect becomes less and less, or, in the OP's words, not as "significant."

And as the bullet weight in a handgun increases, the effect is more "significant," which is why heavy bullets usually shoot higher in a handgun than light bullets, with the same sight settings.

(By the way, the accelerating powder gas while the bullet is in the barrel also contributes to this pre-exit recoil.)

Mal H
November 18, 2008, 01:28 PM
Tuner - I didn't see the definitive answer to your second thought problem in which one cannonball is 10 times heavier than the other. I know the answer, but what is your answer to the problem? Does the cannon now recoil or not?

1911Tuner
November 18, 2008, 05:20 PM
Take almost any revolver and lay it on a flat smooth table on its sights upside down.

You will note that the barrel points upward when it is upside down, meaning it points downward from the sight line when it is right-side-up.

Yup...but you don't get much muzzle flip with an autopistol during the powder ignition and the resulting recoil impulse because the opposite side of the Equal/Opposite thing is on a sliding rail. The muzzle flips mainly because of the slide impacting the frame. You'll get a little from the spring compressing rapidly, and setting up a separate action and reaction between slide and frame...but most of it cmes from impact.

(The spring is a vectored force between slide and frame. As it compresses, it pushes forward on the slide and backward on the frame.)

one cannonball is 10 times heavier...Does the cannon now recoil or not?

Yes. The light ball becomes the projectile and the heavier one is the breechblock...even though it's also moving...like an autopistol.

It moves at 1/10th the lighter one's rate of acceleration, so the lighter one is gone while the heavier one is still moving via its conserved momentum. Its frictional contact with the bore causes the barrel to be dragged backward with it. Recoil.

In the first example, each one is the projectile AND the breechblock...each one providing a resistance for the force to propel the other. Because their mass is identical, their rate of acceleration is likewise. They exit at the same instant at equal speeds and opposite velocities, and their "drag" effect on the barrel is removed at the same instant.

For all who are still unconvinced...and there are a few, I promise...think of this:

A gun firing is simply Newston's 3rd law of motion and conservation of momentum. It requires an action/reaction pair of objects and a vectored force.

If the bullet exits before recoil starts...recoli can't start BECAUSE the action side of the equation is missing. If there is no action, there can be no reaction.

While we understand that the powder gasses have mass and velocity...and thus momentum...it's not enough to provide a detectable recoil impulse unless the mass of the unburned powder and its exit velocity is substantial.

But, recoil as we have come to understand it...action and reaction between bullet and bolt...can't occur without the bullet.

Mal H
November 18, 2008, 08:19 PM
Ah! Ok, you're correct if you add in some friction between the cannonballs and the cannon bore. In your first description you didn't mention friction and therefore the implication was no friction between the ball and the bore in that most perfect of all cannons. :) (Well, perfect in most respects except that it would kill the firing team!)

If there is no friction, then the cannon itself would still not move since the forces on it are evenly dispersed and all in a radial direction. The heavier cannonball would accelerate considerably slower than the lighter one, but as each ball moves along the bore, no force is impending on the cannon barrel in a longitudinal direction. Even if/when the lighter ball leaves the bore, the cannon still remains unmoved even though it would seem, intuitively, like the escaping gases would act like a rocket and propel the barrel backward since there is still no force acting on the bore itself (other than radially).

CypherNinja
November 18, 2008, 08:57 PM
Where did that other 790 fps come from?

A non-linear acceleration curve. :)

1911Tuner
November 18, 2008, 10:08 PM
Ah! Ok, you're correct if you add in some friction between the cannonballs and the cannon bore.

My bad. I assumed that it was accepted that there would be some press-fitting of the balls in the bores. BUT...Recoil would occur anyway. The balls themselves are the objects in the action/reaction system. The barrel is just a pressure vessel.

Recoil would occur in the other one, too...even with equal mass on both ends. Each ball would be both a projectile and a breechblock...and each one would be both an action and a reaction object. The difference is that the barrel wouldn't move in either direction...friction or no friction.

A non-linear acceleration curve.

Nope. The initial PUNCH delivered by the rapidly spiking pressure curve and peak force applied during that rise.
i.e. It requires more force to get the object moving than it does to keep it moving...and the faster it's moving, the less force it requires to accelerate it to a higher velocity.

Refer to Newton 1A and 1B, to wit:

Objects at rest tend to remain at rest. (Inertia)

Objects in motion tend to remain in motion. (Momentum)

Pressures in an internal ballistic event spike fast...within the first half inch or less of bullet travel for a pistol caliber, depending on powder burn rate. That's also where the greatest percentage of the velocity and recoil impulse occur. Whatever is gained after that is pretty much irrelevant because you won't have time to feel much extra push after the initial punch has stung your hand.

Doubt it?

It will require a highly illegal act to see, so this is purely hypothetical, and I don't advise doing it. Your call.

Go find a cheap, beater single-shot 12 gauge shotgun in a pawn shop. Lop the barrel off a quarter-inch past the end of a fired shell. Add weight to the stock to bring it back to the pre-lop weight...and fire it. It'll belt you just as hard as it would have with a 30-inch barrel.

If you actually do it...I recommend immediately destroying the barrel with a large hammer or torch. No sense in spending time at Club Fed for the sake of science.

In case you're wondering, the answer is yes. I did it. Many years ago. They wouldn't have sent a 12 year-old to prison, although I did have to deal with my ol' man when he found my toy.

CypherNinja
November 19, 2008, 05:19 PM
Nope. The initial PUNCH delivered by the rapidly spiking pressure curve and peak force applied during that rise.
i.e. It requires more force to get the object moving than it does to keep it moving...and the faster it's moving, the less force it requires to accelerate it to a higher velocity.

Refer to Newton 1A and 1B, to wit:

Objects at rest tend to remain at rest. (Inertia)

Objects in motion tend to remain in motion. (Momentum)

Pressures in an internal ballistic event spike fast...within the first half inch or less of bullet travel for a pistol caliber, depending on powder burn rate. That's also where the greatest percentage of the velocity and recoil impulse occur. Whatever is gained after that is pretty much irrelevant because you won't have time to feel much extra push after the initial punch has stung your hand.


That's actually about what I meant. :)

The majority of acceleration (and thus, recoil) happens in the early stages (first few inches of the barrel, if that, depending on caliber/load). Using your example of an ultra-short barrel shotgun, if you started at 18" and lopped off 1" after firing a shell and recording the velocity each time, you'd see a "non-linear acceleration curve" become apparent in the final velocities. Meaning, every time you shortened the barrel, you'd see the velocities become lower. First in small steps, but they'd be rapidly growing in size. By the time you got really close to the chamber, the velocity reductions would be pretty big. The length/size of the forcing cone would be an issue clouding the results (in shotguns; there would be other little issues in rifles/pistols), but for the most part the shape of the acceleration curve is defined by the caliber and load. I've little reason to doubt that a huge portion of the acceleration happens while the load is still in the case in shotguns. If you think about it, those cases actually represent 1.5-2 inches of barrel. ;)

Pistols may have a little bit longer acceleration curve than shotguns, and most rifles probably have a longer one still. But the same principle still applies.

We actually pretty much agree on this. :)

230RN
November 23, 2008, 12:40 PM
Its frictional contact with the bore causes the barrel to be dragged backward with it.

Don't you mean "dragged forward?"

1911Tuner
November 23, 2008, 04:06 PM
Don't you mean "dragged forward?"

No. I was talking about the heavier ball dragging the barrel in its direction of travel. Since the lighter one is now defined as the projectile, and the heavier one the breechblock...direction has been established. The projectile exits forward and the gun recoils backward.

Pistols may have a little bit longer acceleration curve than shotguns,

Why? The burn rate of the powders used in typical autopistol ammunition is on a par with that used for shotshells, and in many cases is interchangeable. Unique...Red Dot...Green Dot, etc.

Walkalong
November 23, 2008, 04:19 PM
I love it when Tuner gets on a roll. I didn't think he could sit idle in this thread. :D

James T Thomas
November 23, 2008, 06:10 PM
May I add that the equasion is Force = Mass X Acceleration.

For example: A 180 grain bullet is accelerated to 2700 fps from, from zero velocity, in a distance of 24 inches. Accel. = the change in velocity with respect to time. Enormous acceleration! I'm not going to do the mathematics, but it is many times the accel. of gravity, for instance.

So, there is an exchange here. The gun weighs perhaps 45000 grains, so the acceleration of the gun will be in proportion to it's own weight versus the bullet weight; that is a small in proportion acceleration of the bullet.

"Recoil" is the reaction mentioned. It's direction is opposite of the bullet direction. In line with the barrel, and combined with the torque; twisting reaction to the rifling.

Rise of the barrel, side ward turn of the barrel, is all the shooter's "reaction" to the gun's reaction. We just have our body masses at the head and shoulders and arms, flexible spines, and feet planted on the ground to transfer that reaction to the ground and hopefully keep us from being knocked onto our "rear" body masses.

The answer to the question as to how much?
Well, how much body mass is the shooter adding to the "reaction" coming through the gun? That is, are you gripping it hard, and are you a full back, or
are you afraid and lightly holding it and a thin anorexic?

Machinegunners are taught to really grip that gun, in the prone position, and almost do a push up with the body weight of the gunner bearing down through the gun. With tracer ammo, it can be seen in the bullet pattern what difference it makes by having added all that weight to the gun.

For a quantataive number to give you as to the typical rise or movement; sorry, I can't help. However, one thing may give you some insight into that.
The mechanical pistol rest. Consider the groups given from an example pistol to the groups obtained by a marksman with the same gun. The difference can mostly be accounted for by the movement you speak of.

And even then, with the mechanical pistol rest, all you have done, is add the rigidity of a braced frame to resist the "reaction;" in place of the flexiblilty of the human frame.

1911Tuner
November 23, 2008, 06:17 PM
Well, how much body mass is the shooter adding to the "reaction" coming through the gun?

Which makes up "outside" force...or a resistive force that lies outside of the closed system that is made up of the barrel...bullet...breechblock...and compulsive force generated by the burning powder gasses.

What does that mean? It means that the gun will move a varying distance according to the mass of the outside resistive force.

But...To answer the OP's original question again:

Does the gun move before bullet exit?

Yes. It must. The only way that it won't is for the gun to be so massive that the compulsive force can't overcome its inertial resistance to acceleration. A .223 caliber rifle that weighs 300 pounds is a good example.

That doesn't mean that recoil didn't occur. it only means that the object was too massive for the applied force to set into motion.

You can push on a house without moving it...but that doesn't mean that you didn't apply the force. It just wasn't a great enough force.

Blakenzy
November 23, 2008, 07:01 PM
Great, now I want a high speed camera for Christmas...

In a hand held .45 government model loaded with standard 230gr., how much should we expect the bore's longitudinal axis to rise before the bullet completely exits the barrel?

Jubjub
November 23, 2008, 10:16 PM
There's a method of operation known a API (advance primer ignition) blowback that was originally developed for use in submachine guns but found its niche in automatic cannons. The gun fires from an open bolt, and the round is fired while it and the bolt are still moving forward. The recoil impulse thus has to overcome the inertia of the forward motion of the bolt before pushing it rearward. It makes it possible to use a lighter bolt and get a higher rate of fire with less recoil impulse. The downside is that cases have to be lubed. Probably the most common example would be the Oerlikon 20mm antiaircraft cannon used in WWII.

camies
November 24, 2008, 03:22 AM
Don't forget the humble airgun ESPECIALLY spring powered.
They subject you to a number of jerks during itīs firing cycle which cause the springier to thrash round in your hands like a mad thing.
The trigger releases and the sear lets a couple of hundred weight of spring pressure loose. This is the start of the game. The piston flies forward. It's never an equal and opposite force and generally it jerks the rifle back into your shoulder.
The piston compresses enough air to form a cushion, which bounces the piston back the way it came from. The spring overcomes this bounce forcing the compressed air onto the base of the pellet. The pellet flies out the barrel hopefully before the piston stops with a controlled crash an the end of it's travel. This is followed by a jet of compressed air giving a jet effect push back into your shoulder.
Net result, little true recoil, and the gun appears to want to fly down, back up and forward out of your hands.
Just about the time you get the idea that the rifle has fired, the pellet is still only just starting up the barrel.
Lock time is so much slower on airguns compared to firearms so you have to adjust and follow through far longer than a firearm. :)

Hud
November 24, 2008, 09:49 AM
Blakenzy said: "Great, now I want a high speed camera for Christmas...
"
A picture for everyones' consideration.

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