Why do shorter barreled guns have more recoil?


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TargetTerror
November 5, 2007, 04:03 PM
I am aware that guns with shorter barrels have more recoil than guns with larger barrels, but does anyone know, technically, why that is? Is it simply that there is less mass because there is less barrel, or is something else going on too?

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mljdeckard
November 5, 2007, 04:45 PM
The vast majority of the time it's mass. However, if you are shooting two different model guns, the difference in bore axis (height between the barrel and the shooting hand) also makes a difference.

Yeah, not too tricky, the force of a bullet leaving a gun will push it rearward. A gun that weighs less will be pushed more easily.

Cosmoline
November 5, 2007, 04:49 PM
I believe it's mostly a question of less mass. There's also a mental factor as seen in the large number of people who will swear that an M-44 recoils more than a 91/30 even though they weigh the same and have virtually identical stock profiles. If you see a big fireball you assume there's more recoil.

buzz_knox
November 5, 2007, 04:50 PM
The increased muzzle blast will often lead people to perceive higher recoil.

TargetTerror
November 5, 2007, 05:22 PM
OK, that makes sense. For some reason I thought there was some other funky physics going on, but I guess mass always wins.

igpoobah
November 5, 2007, 06:18 PM
I am assuming that the point the pistol recoils is the point the bullet leaves the barrel. If that is wrong then the rest of my post will also be wrong.:D

The further away from the hand the end of the barrel is, the more linear and less angular the relationship is between the 'point of recoil' and the hand; thus delivering the force 'into' the arm rather than up and away from it.

I think this, plus the extra mass of the barrel go hand-in-hand in the reduced recoil of a longer pistol.

But I'm just guessing. ;) If anyone would like to tell me how big of an idiot I am, please do it respectfully. :D

AndyC
November 5, 2007, 06:41 PM
No. Strictly-speaking, recoil starts from the moment the bullet starts to move ("For every action, there is an equal and opposite reaction").

The muzzle tends to flip upward because the web of your shooting hand is below the line of the barrel - and becomes the pivot-point around which the firearm rotates. Watch what happens if you shoot a firearm held upside-down - the muzzle will recoil towards the ground, again because it's pivoting around the web of your hand.

Heavier guns are simply more difficult to move than lighter ones - inertia ;)

1911Tuner
November 7, 2007, 08:50 AM
I am assuming that the point the pistol recoils is the point the bullet leaves the barrel.

AndyC nailed it.

For every action, there must be an equal and opposite reaction. Equal being the operative word. It means that it's equal in force and in the time frame, but opposite in direction.

To simplify it...

You can't push or pull on an object without being immediately pushed or pulled, and the force that you impose on that object will be redirected onto you immediately and in equal measure. Go push on a wall and you'll see what is happening.

When the force vector created by the expanding gasses and resulting pressure pushes the bullet, it immediately begin to push on the breechblock. At the instant that the bullet moves...so does the breechblock, and hence so does the gun. The only way to get the bullet out of the barrel without having recoil is to make the gun so massive that the force imposed can't overcome the gun's inertial resistance and the resistance provided by the shooter. A .22 rifle that weighs 25 or 30 pounds would do it.

That doesn't mean that the equal and opposite reaction didn't occur. Only that the resistive forces involved didn't let you see or feel it. You can push on a brick wall without being able to move it...but that doesn't mean that you didn't push on the wall. It just means that you couldn't generate enough force to move it.

Conqueror
November 7, 2007, 11:14 AM
Y'all are forgetting the not-insignificant force of the actual hot gases themselves. The muzzle is basically a rocket nozzle once the bullet clears, ejecting hot, high-pressure gas into the surroundings. Boyle's Law tells us that a shorter barrel will allow the gas to escape at significantly increased pressure, meaning a short barrel is a "more powerful rocket". This effect is also seen in AR-15s - just compare the cyclic rate of a pistol-length vs. a rifle-length gas system.

1911Tuner
November 7, 2007, 11:37 AM
Y'all are forgetting the not-insignificant force of the actual hot gases themselves.

Not at all...but the residual recoil effect of the gasses exiting behind the bullet is fairly insignifigant in a straight-walled pistol case with the typically small powder charge. And...By the time it comes, the recoil-operated pistol's slide has already gotten all the momentum that it needs to cycle.

The mass of the gasses are roughly equal to the unburned powder charge.
The .45 ACP round, for instance...typically uses a quick-burning powder charge of about 5 grains and exits the muzzle at or just slightly above the speed of sound. We'll say a thousand fps just to keep it simple.

5 grains of mass at a velocity of 1,000 fps would generate about 1/8th the recoil impulse of a standard velocity .22 Short. In other words...not much.

Bottlenecked rifle cases are a different matter, and the impulse generated by the exiting gasses and unburned particulate can create an after effect that the bullet itself provides...and in some rounds, it can actually exceed it.

The shoulder and neck create a venturi effect, which speeds up the exiting gasses to exceed the velocity of the bullet. Coupled with the mass of the exiting gasses, the recoil impulse that is generated can be signifigant. Figure up a 39-grain powder charge exiting at nearly 5,000 fps with a 45-grain bullet fired in a .22 Swift that exits at 3800 fps, and the "aftershock" can easily exceed the primary recoil impulse generated by the bullet.

But in a straight-walled pistol case? Wouldn't disturb a fly.

TargetTerror
November 7, 2007, 12:32 PM
Aha! There IS more to it than just mass!

1911Tuner
November 7, 2007, 01:26 PM
Aha! There IS more to it than just mass!

Nope. It's still mass X velocity. The recoil impulse generated on firing the round comes from the combined mass of projectile and gases...which again have the same mass as the unburned powder charge. So, if you have a 230-grain bullet propelled by 5 grains of powder, the formula is mass X velocity. In this case...235 grains X 800 +/- fps.

Upon bullet exit, there's still a portion of the gas in the barrel, which continues to generate a recoil impulse until it's gone.

TargetTerror
November 7, 2007, 03:06 PM
Well, more than just the mass of the gun/bullet.

1911Tuner
November 7, 2007, 03:38 PM
Well, more than just the mass of the gun/bullet.


Not really. Mass of projectile and gun. Mass of projectile include the mass of the gases...so they can essentially be considered the same mass. Think of gettin' hit by a train. You don't get hit by the mass of just one car. You also get hit by the combined mass of all the cars that are behind the one that tags ya.

Funderb
November 7, 2007, 03:48 PM
Well, I know for a fact that my m44 has higher recoil.
shooting with the m91 - no bruises
shooting with the m44 - bruises.
That's a pretty clear difference to me.

longer barrel dissapates more energy before the shot leaves the rifle,
recoil is only felt after the shot leave the rifle. That's the simplest way to say it. (and yes, the longer the heavier.)

Conqueror
November 7, 2007, 03:51 PM
Some of the physics good, others not so much.

You're assuming all of the recoil energy provided by the gas comes from its mass, but you're completely neglecting its pressure and the way in which gases behave. The bolus of gas doesn't go straight out the barrel like a solid bullet. It expands in all directions, even back along the barrel, which is why your get fouling on the sides of the slide. Your average muzzle crown is quite literally shaped like a rocket nozzle. The gas expanding in all directions will produce a net rearward force on the tip of the barrel which has nothing to do with the total mass of the gas particles and far more to do with their velocity (which is based on the gas temperature, which is quite high). If the frontal area of the barrel is 0.1 square inches, and the muzzle pressure is, say, 5000psi (quite reasonable), then you are briefly exerting a force of 500lb rearward on the gun. Change it to something like 15,000psi (reasonable for, say, an AR-15) and you're at 1500lb of rearward force.

Now, that force doesn't last very long, but it's not theoretical. It's really the only satisfactory variable to explain the significant differences in recoil. Add an inch of barrel and you've added, what, a couple ounces of weight to one side of the conservation of energy equation? Not particularly significant. But cut 50% off the barrel and the gas exits at ~200% the original exit pressure.

This same principle is why Schuemann Hybrid barrel ports are shaped like rocket nozzles - they generate a helluva lot of force.

1911Tuner
November 7, 2007, 04:49 PM
The bolus of gas doesn't go straight out the barrel like a solid bullet. It expands in all directions, even back along the barrel, which is why your get fouling on the sides of the slide.

We understand that...but once the bullet exits, it follows the path of least resistance. While the event is going on, it's expanding and pushing in all directions...vectored, of course...owing to the fact that it doesn't blow the chamber.

I guess I tried to over-simplify it by stating that mass of bullet+mass of gas when it actually expands and applies force against bullet and breechblock alike...but trying to come up with a way to explain how it can do that without splitting and becoming two separate gas plugs makes my head hurt.

Of course, following the path of least resistance would seem to put the bulk of it following the bullet...but it gets tangled at that point, so I'll leave it for ya'll to carry on. Anyway...Force forward equals force backward. No such thing as force in one direction at a time within the same system. If it pushes in one direction, it pushes in the other, and recoil begins at the same instant that the bullet moves. So sayeth Sir Isaac.

Cheers!

32winspl
November 9, 2007, 02:42 PM
Hey folks, not sure if or how much this has to do with the topic, but here goes.
A couple of years ago, I was trying to work up good accurate loads for my TC
Renegade 50cal muzzle loader using 500gr Buffalo (brand) hp bullets. At some point around 92 grains of Pyrodex, I poured in the powder and tamped it as usual, but just before shoving the bullet down the barrel, I decided to see what it felt like/sounded like. I put a cap on the nipple and pointed it downrange... it went BANG just as loud (give or take) as it had been with a bullet, but I felt no recoil. It was kinda fun so I did it a couple more times. Neat sensation... a lotta bang and no recoil that I can recall. Anyway, just my $0.02.
Sure wish I hadn't sold that smokepole.

Conqueror
November 9, 2007, 03:53 PM
Not sure about muzzleloading powders, but metallic cartridge powder doesn't explode unless it's confined in a tight space. Pour a pile of it on the groud and light it with a match, it'll burn but not detonate. Thus, simply pulling a bullet out of, say, a .45ACP cartridge and firing the primer/powder charge would not be an accurate simulation.

SoonerSP101
November 9, 2007, 03:54 PM
more barrel = more weight of the gun = less perceived recoil?

birdbustr
November 9, 2007, 04:09 PM
Sooner makes it simple. That's why big handguns (.44 mags, .50) don't come in pocket versions. There are a few exceptions. The .44 bulldog or the little 4 barreled .357 mag. Even then, people only shoot those when they have to. It's not like they are someone's favorite range gun.

Walkalong
November 9, 2007, 05:11 PM
Y'all are forgetting the not-insignificant force of the actual hot gases themselves.
High pressure, high velocity gases contribute much more to muzzle flip than recoil.

Heavy bullet .45 loads with a small charge of fast powder can give less muzzle flip than a light bullet with more powder because in the first example the powder is much closer to dying out when the bullet exits the muzzle with not much "blast" and in the second example the powder still has good pressure left and has a good "blast" leaving the muzzle, contributing greatly to muzzle flip.

When using a comp the opposite is true. We want plenty of gas and pressure to make the comp work and hold the barrel down.

Car Knocker
November 9, 2007, 05:17 PM
Think of gettin' hit by a train. You don't get hit by the mass of just one car. You also get hit by the combined mass of all the cars that are behind the one that tags ya.
Actually, that's a bad analogy. Unless all the cars and locomotives are bunched up against the lead locomotive (in a normal train), you're only getting hit by the mass of the lead loco. That's because there's a certain amount of slack between each piece of rolling stock. As the slack runs in, there would be successive jolts as the additional masses impacted. It's like one car pulling another at the end of a 20-foot chain; if you get hit by the lead car, that's the only mass involved until the towed car hits the first car's bumper.

Sorry for the digression - now back to our regularly scheduled programming. ;)

1911Tuner
November 9, 2007, 05:54 PM
High pressure, high velocity gases contribute much more to muzzle flip than recoil.

Not in a straight-walled pistol case that uses 5 or 6 grains of powder. With an autopistol, the slide smackin' the frame is the main contributor to muzzle flip. In a revolver, the bore axis and grip angle is the player.

But...I'll back out of it and just refer back to Newton 3. To wit:

"For every action, there must be an equal and opposite reaction."

Once the bullet is gone, half the action/reaction pair is missing. As far as recoil is concerned, it may as well have never been there. Whatever impulse is realized by the gas and particulate "jet" also depends on mass X velocity.
Roughly the mass of the unburned powder times its exit velocity.

.45 ACP...5 grains at about a thousand fps. Think of a .22 short and divide by eight.

1911Tuner
November 9, 2007, 06:02 PM
Actually, that's a bad analogy. Unless all the cars and locomotives are bunched up against the lead locomotive

You know what I meant...

doc2rn
November 9, 2007, 06:33 PM
I always thought the gasses expanded behind the projectile forcing it out the front thus putting a backwards inertia on the web of the hand causing the barrel to rise and thus recoil is percieved. I also thought that was why we see some cutting from those gasses on the inside of shot out S&W revo's. Or maybe I am wrong again. My .02

Funderb
November 9, 2007, 08:40 PM
I don't think bruises are psychological.
Light does not make me imagine extra force on my shoulder.
Simply put - the m44 mosin has more recoil than the m91/30

GLOOB
November 9, 2007, 09:38 PM
I always thought that the mass and velocity of the bullet had a lot more to do with recoil than the gases rocketing the gun backwards.. like on the order of 99% to 1%.

The smaller, faster bullets cause more recoil. But they are also produce more kinetic energy. A 230 grain 860 fps hardball produces, what, a few hundred foot pounds (I'm guessing, here)? A 180 grain 1200 fps produced on the order of 460 ft lbs, or so (again, a guess).

A rocket has to have a specific shape to the nozzle to be effective, at all. And a muzzle is definitely NOT a good likeness of a rocket nozzle.

But the compensator guys can't be all wrong.

Now, if someone would just build a 1911 barrel with a rocket nozzle on the end, we could all find out for sure. :)

jeepmor
November 9, 2007, 11:26 PM
You have six axes of motion. 3 linear dimensions, X, Y and Z, and three rotational components around the axes of X, Y and Z.

Effectively, and more simply mathematically, KE1=KE2

KE = kinetic energy = 1/2mv^2

KE1=gun
KE2=bullet

Fast bullet, slow gun because guns weigh more. Less gun mass, less recoil, that simple.

When you get into elaborate physics calcs, we talk about all six of these axes simultaneously, which is all the energy components that make up the recoil. We have to assume the bullet left the barrel. Not entirely true, but we're talking milliseconds, so not an issue. All that energy is imparted back to the user of said firearm.

This recoil is largely determined by the form factor of the firearm.

How far is the axis of the bullet above the grip?
What's the form factor. Moment of interia in x,y,z axes.
What's the muzzle twist of the barrel?
What's the bullet mass?

It all comes into play. This is why small 9mm mouse can whomp you more than and weighty range magnum.

Funderb
December 5, 2007, 04:51 PM
I think as a side note that the exit gasses do contribute to recoil when directed in frot of the rifle. Otherwise there wouldn't be much of a market for muzzle brakes. So, 99.99% of felt recoil is after the round leaves the barrel. There isn't enough time to feel it before the round leaves. You can't process it that fast, nor has the energy reached the butt to shoulder contact enough to take effect.

1911Tuner
December 5, 2007, 05:05 PM
So, 99.99% of felt recoil is after the round leaves the barrel

Not in this universe, I'm afraid.

At no point are the forces greater than at peak pressure...which occurs very quickly after the powder ignites and the action/reaction event begins.

Pressure drops off quickly after the peak, and how quick depends on several factors...the main one being powder burn rate...but pressure in the system at the point of bullet exit is a small percentage of the peak.

In certain rifle calibers, with sharp shoulders and light bullet to powder charge ratios...the impulse after the bullet exit can nearly equal that generated by the bullet...but such isn't the norm.

Hypothetical...

Double-end, smooth-bore cannon. Bore diameter is precisely the same, with zero variation from one muzzle to the other.

Two cannon balls. Each one exactly the same in weight/mass...diameter...and each one perfectly round.

Powder charge located precisely in the center of the cannon barrel, with both balls seated precisely equidistant from its respective muzzle.

What will happen when you fire the charge?

Funderb
December 5, 2007, 05:08 PM
The problem with that is that a rifle is not an exact tuned double ended cannon of perfection.
just look at how long it takes the bullet to leave the barrel, and how long it takes a person to notice the brunt of the recoil.

1911Tuner
December 5, 2007, 05:58 PM
just look at how long it takes the bullet to leave the barrel, and how long it takes a person to notice the brunt of the recoil.

All duly noted and factored in...but the plain, simple truth is that the reaction is instantaneous. You can't push on an object without being pushed, and whatever measure of force that you impose is immediately returned to you in like measure. There is no delay, and there is no such thing as force in one direction...and then force in another direction at some later point. Can't happen. Force forward equals force backward. The movement of recoil that you experience after bullet exit is momentum. Whatever impulse is generated by the escaping gasses behind the bullet are equal to the mass of the gasses...which is roughly equivalent to the unburned powder's...and it doesn't amount to much, because momentum is Mass X Velocity.

F'rinstance:

.45 ACP. Normally about 5 grains of powder in the case. Gasses escape at roughly the speed of sound...let's say a thousand fps for simplicity's sake. The recoil impulse will be about 1/8th that of a standard velocity .22 rimfire.

Newton 3:

For every action, there must be an equal and opposite reaction. Equal means equal...both in magnitude and in the time frames involved. Whatever force is necessary to accelerate the bullet is immediately placed on the breechblock.

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