The science of recoil

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So after all of this, we find ourselves in agreement

Yep. I started to realize that we were on the same page...but we just weren't finishing all our sentences sometimes.

Many people focus so hard on Newton 3 that they fail to consider the full impact of Newton 1 or factor it in.

Here's the video. Pay particular attention to the last 1911 in stop-action...just before the revolver. Place a pencil point at the end of the recoil spring tunnel in the frame, and watch. First you'll see a light puff of blowby gas...followed by the nose of the bullet peeking out of the muzzle, but still not out...followed by the exit gas and fireball.

You'll notice that the slide is moving even before the blowby gasses appear.

Equally interesting is the cutaway showing how much the ammunition bounces around in the magazine during recoil.

The really fascinating clip is the bolt bouncing on the AR15...three times...before it settles down.

Watch it several times. You'll always see somethin' that ya missed.

http://www.trippresearch.com/media/movement/hispeedgateway.html
 
Very cool.

Yep. Watch the 4th clip closely...the one that doesn't have the pistol anchored. Notice how little the frame moved on cartridge ignition...and that the frame doesn't really get moving in recoil until the slide has moved an inch or so, and started putting a load on the recoil spring...and how the muzzle didn't really flip all that much until the slide hit the impact abutment?

Testing has shown that in a 5-inch gun, the standard 16 pound spring's in-battery preload is about 9 pounds, on average. I'd be willing to wager that if the shooter really took a crushing, solid grip, that the frame would barely move until the slide was almost at the impact abutment.
 
One of the points made in this thread is that the first inches of travel down the barrel impart more speed to the bullet than the last few. If you find a graph of chamber pressure versus time on the web (like in the Wikipedia article on "internal ballistics") you can predict how a bullet's speed changes as it moves down the barrel.

Change in momentum is equal to force integrated over time. If you ignore friction, the force on a bullet is equal to the pressure behind it times its cross-sectional area. Since the bullet starts at rest, if you take a pressure graph, the speed of the bullet at any time T is equal to the area under the pressure curve from time 0 to time T, times the bullet's cross-section, divided by its mass. Since the pressure curve peaks as (or a bit before) the powder is completely burned, this is the point of maximum acceleration for the bullet. It continues to accelerate along the rest of the barrel, but more slowly, because with no additional gas being generated the pressure drops as the volume behind the bullet increases.
 
If you ignore friction, the force on a bullet is equal to the pressure behind it times its cross-sectional area.

But you can't ignore the friction...and that's the wild card in the game. The inertial and frictional resistance of kicking the bullet in the butt to get it moving, and swage down into the lands and grooves causes a violent pressure and force rise...and because force forward equals force backward...there is where the maximum recoil impulse is generated.

Even in a rifle caliber with a slow burn-rate propellant...slow compared to pistol powder...the pressure peaks early in the bullet's travel. Probably within a couple inches in the .308 or .30-06 class. The area under the curve represents maximum force...and how long it's sustained. Pistol powder pressure graphs show a sharp, fast peak...like the apex of a triangle... and rapid drop as the volume of the cylinder increases. As pressure drops, so goes force...so that at the point that the bullet is about to exit the muzzle...the recoil impetus is maybe 5% of the maximum, depending on...again...burn rate and barrel length. It's entirely possible to continue to increase barrel length until the bullet actually exits at a lower velocity that the maximum speed that it attained while still in the barrel if the powder is so fast that the gas plug runs out of steam before the bullet gets to the muzzle.

The other, often forgotten factor is the mass of the gasses and unburned particulate...which is roughly equal to the mass of the powder before ignition. So...in the typical .45 ACP cartridge...a 230-grain bullet actually is about 235 grains. So, that must be factored in as well.

The science of internal ballistics is a fascinating subject for discussion...and debate...as we've seen on this thread. It's at the same time more simple and more complex than one might suspect.

Good thread.
 
But you can't ignore the friction...

Tuner, I was going to argue with you about this. Compared to 30000 PSI chamber pressure of a 9mm round, I figured the friction of the bore was insignificant, but thought I should do an experiment first. So I bought a box of 9mm JHP bullets and tried to push one down the barrel of my Beretta 92 with a cleaning rod. I couldn't budge it, even pounding my hand on the handle of the rod. So I'm not going to suggest that the friction is negligable.

But I'm a shooting newbie and probably the only person here who didn't know this. :eek:

Does anybody know what the bore friction force is of a 9mm round? I'm curious how much of the powder's energy gets used to overcome it.
 
I couldn't budge it, even pounding my hand on the handle of the rod.

Howdy Dave. That'll give ya an idea of the magnitude of the forward drag that the bullet places onto the barrel during its break to freedom. :)

It varies. If you knew the exact construction of the bullet...jacked alloy and thickness and the makeup of the lead core...hardness...it could be closely calculated. It also tends to vary from one barrel to another because of dimensional differences and surface texture of the bore...smooth vs rough...land/groove depths...even rifling twist rates will affect the resistance of the bullet's passage. That's why we see different handloading data listed in loading manuals for different bullet manufacturers in identical bullet weights with outwardly identical bullets. The pressure figures came from a given charge of Powder X with their bullet. That can easily change with someone else's bullet.
 
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I'm enjoying this immensely, thanks to all.

I haven't had any physics classes since I finished undergrad school in '72. Which reminds me of this one old prof who refused to call it anything but The Calculus. Five days a week of The Calculus this and The Calculus that was enough to drive me up a wall. But I digress.

John
 
The parameter that Tuner is describing is called 'jerk.'

It is the rate of change of acceleration.

It is not used in many applications, but crops up in all sorts of funny places when you start slicing time into smaller and smaller periods while doing analysis.
 
Ugh..related rate calculations. Brings back horrid memories from high school.

The physics of recoil are interesting, but the fact that there is every design under the sun out there means we don't have to understand the numbers to find a gun that works for us. :D

The 40 is "snappy" compared to the 45, but with a low bore-axis gun, good two handed grip, and properly balanced stance, it is easily controllable. Try the full size M&P 40 and compare it to a Glock 40cal (or Fo-Tay as some know it). Same round, different feel.

Thanks to everyone who posted math in this thread. At least I had a cup of coffee first, so I wasn't entirely in the dark.
 
Dave,

Once the bullet is underway and engraved by the rifling the frictional force (-F) is actually quite negligible as opposed to the force (F) generated by the expanding propellant gasses, otherwise acceleration would really suffer and the bullet wouldn't achieve the speed that it does.

Looking at the frictional co-efficients for copper on steel and considering the total area in contact with the bore and making an educated guess, I would believe the magnitude of -F to be in the 150N - 200N (~40-45 pounds) range afer the bullet is engraved and in motion depending on the alloy composition of the bullet jacket and the barrel alloy.

Considering that the "average" (I know it is not a constant force and that it peaks then diminishes rapidly) force acting on a 115 gr. 9mm bullet during its trip down a 4.5" barrel that achieves a muzzle velocity of 1150 fps, is on the order of 900-1100 pounds for a little more than 6/10,000's of a second and the fluctuations that are possible due to gas blowing by the bullet that may reduce that frictional component, it is really not that significant and not something that anyone would notice or feel.
 
Once the bullet is underway and engraved by the rifling the frictional force (-F) is actually quite negligible as opposed to the force (F) generated by the expanding propellant gasses,

Well...Once engraved, it would have to drop...but I wouldn't call it insignifigant. Ever had a squib load stick a jacketed bullet halfway through the barrel? Still gotta have a hammer or an arbor press to back it out unless you soak it in penetrating oil for a couple hours.


otherwise acceleration would really suffer and the bullet wouldn't achieve the speed that it does.

See Newton 1B for the clue to that.
Once the object is in motion, it requires less force to keep it in motion...and the faster it goes, the less force it takes to keep it going and/or accelerate it to a higher speed. The coefficient of friction doesn't change...only the magnitude of force required.
 
Once the bullet is underway and engraved by the rifling the frictional force (-F) is actually quite negligible as opposed to the force (F) generated by the expanding propellant gasses,

Originally posted by 1911Tuner:
Well...Once engraved, it would have to drop...but I wouldn't call it insignifigant. Ever had a squib load stick a jacketed bullet halfway through the barrel? Still gotta have a hammer or an arbor press to back it out unless you soak it in penetrating oil for a couple hours.

The "resistance" (-F) is of very little significance when compared against the propellant generated force (F) in that it is insufficient to overcome it and the bullet still develops enough velocity to leave the barrel at the desired velocity. That the bullet leaves the barrel is what is important here, since the internal frictional forces induced by contact with the barrel are insufficient to stop it.
A squib load is a vastly different situation than a bullet that is already in motion in the bore because the moving bullet is not static whereas the squib load is. There is considerable difference between the coefficient of friction of the two different materials (bullet jacket and barrel steel) involved in each situation (static vs. dynamic) and is why the squib load bullet requires the use of an arbor press or hammer and oil to remove it from the bore.




otherwise acceleration would really suffer and the bullet wouldn't achieve the speed that it does.

Originally posted by 1911Tuner:
See Newton 1B for the clue to that.
Once the object is in motion, it requires less force to keep it in motion...and the faster it goes, the less force it takes to keep it going and/or accelerate it to a higher speed. The coefficient of friction doesn't change...only the magnitude of force required.


Never said that the coefficient of friction could change and it doesn't in either the 'static' or the 'dynamic' regime. I said that propellant gas blowby will alter the friction component (a different concept) by reducing contact via its interposition between the two materials, thereby 'floating' the bullet, or acting as a 'gas bearing', so to speak, and would require less force as a result to overcome the resistance (-F) induced by that contact. Two different elements, both of which dictate in different ways how the friction encountered will be modified.



Originally posted by 1911Tuner:
Once the object is in motion, it requires less force to keep it in motion...and the faster it goes, the less force it takes to keep it going and/or accelerate it to a higher speed.

Incorrect.

Once in motion, the object actually requires no force to keep it in motion at a certain velocity since it tends to remain in motion unless acted upon by another impinging force (friction in the barrel or air resistance being a fine example) so there is no additional force required to maintain the initial velocity except for the exceptions noted.

However, while velocity and acceleration are related, they are two different properties and in order to accelerate (ΔV) a body increased force is required because you are now adding K.E. by changing the velocity (accelerating) of the body in motion. It does not require anymore force to accelerate a body that is already in motion because it's mass (m) remains the same regardless of its velocity (v) and is still effected in the very same way by an accelerating (or decelerating) force, because acceleration is independant of any existing velocity (even 0) the mass may have as dictated by the equation:


F = ma*

*In examining the equation above, you will note that there is no variable in the equation for velocity (v) and that there is no modification of the variable for Force (F) or for mass (m) for any 'pre-existing' velocity that the body may (or may not) possess. Once in motion, a body remains the same mass (m) and experiences the same acceleration (a) by a given force (F) regardless of whether it is in motion (v) or not.

There is no such equation for what you propose below:


Originally posted by 1911Tuner:
("the less force it takes to keep it going and/or accelerate it to a higher speed.")

If you can provide either a mathematical expression or a proven documented textual description of such a phenomena, I would love to see it since it would set the world of Newtonian mechanics on its ear.

As it exists, the only increase of mass (m) that is quantifiably related to velocity (v) is that as described by one A. Einstein in his theories of General and Special Relativity and is encountered only at the higher "relativistic velocities" as they begin to approach the speed of light (c) and I do not think that you are referring to velocities in that range as they are presently unattainable in firearms and likely to be so for the foreseeable future and are irrelevant to this discussion anyway.

:)
 
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Once in motion, the object actually requires no force to keep it in motion at a certain velocity since it tends to remain in motion unless acted upon by another impinging force (friction in the barrel or air resistance being a fine example) so there is no additional force required to maintain the initial velocity except for the exceptions noted.

Agreed...but the frictional resistance is constituted as an outside force that is working to stop the bullet and restore equilibrium.
A bullet fired in outer space would continue to travel at the exit velocity unless and until it struck something or entered the gravitational field of something large enough to alter its constant state of motion.

"Objects tend to remain in motion *unless* acted upon by an outside force."

The "resistance" (-F) is of very little significance when compared against the propellant generated force (F) in that it is insufficient to overcome it

Agreed again...but the point wasn't compulsive force vs resistive force. It was simply that because the coefficient of friction doesn't change throughout the trip through the bore...that the forward drag on the barrel also remains constant as long as the bullet is still there and moving, and that it requires less compulsive force to accelerate it to a higher velocity than it took to initiate its acceleration. If it didn't...the bullet would slow down after the initial punch. (Which can happen in some barrels with certain powder charges and burn rates.)
 
Just got back to this post. Hmmm....

The original poster asked about being snappy, which I took as torquey because that is what many .40 shooters complain about, vs a push in the .45, and I said "velocity my friend"

That was totally dismissed, but I stand by it.

During the ride through the bore is when the sideways(rotational) torque happens as the bullet gets up to its rotational speed there.

To get up to its velocity with the same rate of twist in that length creates more sideways torque in the .40 than the .45. It gets up to speed much faster.

Am I wrong?
 
velocity my friend

It's not all about the velocity. It's determined by mass times velocity.
As noted...you can have extremely high velocity and light recoil. The .22-250 with a 50-grain bullet at 3800 fps is an example.

Conversely, you can have fairly low velocity and heavy recoil. My 7.5 pound Sharps carbine in .45-70 will demonstrate that to you clearly with a 405-grain lead bullet at 1400 fps. :eek:
 
But we are comparing something much closer together than the .22-250 and .45-70.

I still think the velocity difference in those two calibers has a lot to do with the torque felt by most shooters in the .40 vs the push in the .45.
 
I still think the velocity difference in those two calibers has a lot to do with the torque felt by most shooters in the .40 vs the push in the .45.

Sure. But it's still mass times velocity, no matter how close the two are.

Be aware too, that only a tiny fraction of the felt recoil in an autopistol comes from the explosion of the cartridge itself the way that it does in a fixed-breech weapon...like a revolver or a single-shot pistol.

The recoil spring's resistance and the slide's speed in compressing the spring...and slide to frame impact determines what is transferred to your hand.

All else being equal...the stiffer the spring, the sharper the recoil. The faster the slide moves...the sharper the recoil with a given spring. The slide's speed and its mass when it hits the frame's impact abutment "flips" the muzzle upward the greatest amount.

But the actual internal event? Nada.

Hypothetical pistol with 30-foot rails with the slide mounted on the front. Weight of the gun is identical to a normal pistol. No recoil spring. Fire the gun, and you probably won't feel a thing...and since the slide has to move 30 feet to reach the impact abutment...you probably never will, aside from a slight disturbance from the rail to rail friction.
 
If you want to talk about the actual physical profile of the recoil energy, the impulse curve, then just think of it in terms of delivered impulse to the projectile. That is the easiest way to look at it, because we can take a slice of the curve during any given time frame to determine how much impulse and thus total energy has been delivered in recoil. The recoil curve is essentially directly relative to the impulse curve delivered by the propellant energy release into the projectile.

The true recoil curve at hand is entirely dependent upon the burn time and energy release of the propellant. The impulse curve shapes and peak pressure times can vary widely between calibers and powders. The problem trying to quantify felt recoil this way is that it truly means diddily squat to a human perception of it. The reason is quite simple. We feel the total impulse curve of energy delivered to us in only a couple of milliseconds, making it impossible for us to distinguish the actual impulse profile itself when referring to two like total energy amounts with different pressure curves. However, what we can account for ourselves is the total impulse energy delivered by that event, as it hangs around much longer as it dissipates and decelerates in our hand and arms. Because of this human limitation, we must speak in terms of total impulse delivered as the basic means of our perception and comparison.

Think of two rocket motors.

One burns over the course of 1 second, and generates a steady 10lbs of thrust for that time duration.

The second rocket motor burns for 2 seconds, and generates a steady 5lbs of thrust for that duration.

At this point, we know that both rocket motors deliver the same total impulse, they just burn for different durations.

Now, if you had to hold each one in your hand, the motor burning for only 1 second is going to seem like it gives you more of a kick. If gunshots took 1 second for the pressure event to complete, this motor would represent a high pressure load with a very fast pressure rise time.

The second rocket motor is going to feel like a much more modest push, as its duration is longer. It would similarly represent a low-pressure load, with longer pressure event.

At these amounts of time for the events to occur, we ourselves can easily distinguish a difference in feel between the pressure events of the rocket motors.

OK, lets try something interesting. Speed the rocket engines' burnrates up, but keep them at the same total impulse. The 1 second rocket motor now burns for 1/20 of a second, and the 2 second motor is increased to 1/10 of a second, still twice as slow as the first motor. What do you think you will feel at this point? The rocket engines are getting much closer together in feel, both lasting for just a very quick burst of energy, instead of the large difference they had before.

Now, push those times even faster, down to the level of firearms. Down to the point where both of these events each occur over only a few thousandths of a second. Our "high-pressure" rocket motor #1 now lasts for a duration of 0.002 seconds, and our slower rocket #2 now lasts for a duration of 0.004 seconds.

Now imagine holding the two in your hand at these speeds, and trying to tell the 2/1000s of a second difference in pressure profile between them. Do you think you could distinguish the difference?

This is why the true pressure profiles themselves are far less relevant than the total impulse delivered by the energy involved.

I get what 1911Tuner was saying about "muzzle velocity", and I agree. The simple fact is that recoil calculators assume that the bullet is still accelerating, and has not lost any acceleration before muzzle exit. Otherwise, it's not really about muzzle velocity, it is rather about total impulse delivered during any given time frame. But, from muzzle velocity we can determine energy involved which had to occur at some point before the muzzle exit to achieve that projectile speed. Then, from that total impulse value we can then approximate feel, hence my example of why true profile curves are irrelevant.

Here is a little graph I drew to attempt and illustrate what the recoil on the bolt face or breech of the gun would look like over time and, as you would imagine, bullet travel:

recoilcn6.png


Also do notice the shape of the impulse curve in the graph. Since this profile is directly proportional to the shape of the acceleration curve on the projectile, each 1" of barrel removed will never do the same as any other inch of barrel. Think of printing that graph out on a piece of paper, and then taking scissors and cutting vertical slices off from the right side, making it shorter. Each slice, you would have a different amount of blue area, or pressure. This is the way you would be removing acceleration impulse as you cut the barrel shorter, closer to its maximum pressure point.
 
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Excellent illustration...and it identifies clearly the point of peak pressure and force delivered...to bullet and breechblock alike.

Again...The majority of the bullet's total velocity and the total recoil energy is delivered early in the event.
 
Re:1911 tuner

I learned a good bit from reading your posts. But I can't resist the urge to correct you on one point.

"Peak force occurs with peak pressure...With some extremely quick pistol powders, it may peak before the bullet has even left the case."

Until the bullet is moving, there is obviously no recoil going on.

Force of recoil is directly related with acceleration. That first quick jolt of greatest acceleration (during which time the peak pressure may theoretically still be building, but in all probability is actually being reduced), would be coincident with the highest peak force of recoil, assuming the rest of your argument to be sound.

Take a bottle of compresssed air, for example. You take a hammer and break off the valve, and the bottle shoots across the room. The peak pressure is there before you break the valve. The acceleration and force occur afterward, during the time the pressure is reduced.
 
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