Recoil and Semi-Autos (and maybe Revolvers, too)...

But, a strong grip is not a reaction, it is a pre-action. I'm guessing it can make a difference. Again, it's easy to test, but I can't think of an easy way to quantify different grip strengths. However, a qualitative difference -- ie, gripping real hard vs naormal grip vs limp grip -- should be good enough.

I suspect it'd be impossible to test objectively, because a "strong grip" will have so many other effects - anticipation/flinching being a key factor. We can work out the torques involved, but we're talking about locking the gun like a vice for the first fraction of a millisecond against a very tiny but powerful tilt motion that can't even be seen in high-speed video, and is much too fast for your brain to react to in any way before it's too late.

I don't think it's possible for the grip to matter, though you might well see a strong/wimpy grip difference in accuracy and pointing due to other factors (like flinching). That's why an objective grip test would probably be impossible.

A vice test would be interesting, though. Same gun, different bullets with different weights and velocities, is there a difference in grouping at close range? I'll bet no. Is there a difference between revolvers and pistols? I bet no again. But it's an easy bet to make since I don't have any money on the line. :D
 
Despite what anyone has said, myself included, in shooting handgun since like 1965, I've never felt the need to adjust sights for bullet weights or loads, for that matter. Whether it be 255 gr 45 ACP load with powder puff 3.2 gr of Clays or the biggest Universal 185 gr pounders...nothing made a difference in impact point nearly as much as me. Now, this is largely because I don't shoot any distance at all, and 'precision' for me is more like "Oh wow. They're all in the black!". :)
 
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45 auto said:
Front and rear sights on a semi auto are typically almost the same height, which means that the POA is very close to the barrel axis. However, the front sight on a revolver (or fixed barrel like a Contender pistol) of any power will be SIGNIFICANTLY taller than the rear sight. Therefore the barrel axis on a revolver intersects the target MUCH lower than the barrel axis of the semiauto.

You are probably right, but you could say that at least part of that high front sight is necessary because extra metal is needed in the frame top strap and the cylinder itself to keep things together. That extra metal adds height and most of it is at the rear. Here's a .450 Casull revolver. The bottom part of the sight base is there just to bring the top section of the sight up even with the rear sight. The extra 'step' isn't needed with the smaller caliber revolver.

Casull.JPG


Compare that to a less bulky .38 special

Model_36_flat_latch-1-300x205.jpg


I couldn't find images that were the same scale, but these are close. Compare the relative thicknesses of the top straps (above the cylinders)!

Semi-autos seem not to require so much reinforcement and extra strength on the top side, and the bore is always closer to the shooter's hand, too...
 
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I'm really surprised that the front sight on that 454 Casull is so close to the height of the rear sight. Biggest revolvers I have are 44 Mag and 45 Colt, all the front sights are taller than the rear sights, similar to the .38 you posted.

I wonder if the Casull is sighted in for a longer range, maybe 100 yards or more? That would require that the front sight be shorter than the front sight for the more typical shorter range of a defensive handgun.
 
Here's another. The front sight is also quite tall. The front sight might be 3/4" but the back strap is thick, too.

The front sight of the .38 isn't really that much taller than the rear -- it's just a hair above the grove in the top strap. The rear sight isn't adjustable. :)
And the sights on the bigger guns car probably cranked down as well as up.

I'm not saying your point isn't correct, but I am saying that some of the extra height is required by the design of the gun (and the need to use extra steel to cope with more powerful rounds). Hot rounds may still require a higher-than-expected front sight to cope with a round that hits higher at closer ranges. I simply don't know.

The few more powerful revolvers I've owned (a S&W 45, several S&W .357s and a Python) seemed to have sights that were close to (or actually) parallel to the bore, but I wasn't really LOOKING at that "sight" characteristic when I had them. My memory of what I saw may be way off.

Measure the distance of the tops of your front and rear sights above the bore axis. It may be just as you say.

FS3.jpg


Here's a Mateba in .454 Casull -- it has the barrel at the bottom of the cylinder. The front and rear sights also seem parallel to the barrel/bore axis. The recoil experience would be somewhat different, of course, because the barrel rise and movement induced by barrel rise would be attached to a shorter "handle" -- but a heavy round at high should still hit higher in the distance than lighter or less powerful rounds.

Mateba454casull-1.jpg


Here's a .38 Special conversion of a UBERTI black powder revolver. Notice that the front sight sets right on the barrel (which is thicker than it needs to be -- because the black powder guns depended more on the barrel and frame to keep things together -- they hadn't gotten around to using top straps, yet. The rear sight (the tip of the hammer) seems a bit higher than the front sight, but there isn't likely to be much of a difference if you measure them against the bore axis.*

*After I keyed that sentence, I realized that the hammer would be cocked when it is used as a rear sight, so it would be somewhat lower, even then.

images
 
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If John could lock his slides we might have evidence for both questions.
I can, actually and that's a good idea. I'll use a striker-fired gun and put my thumb against the back of the slide to keep it from moving for some of the testing. It won't be an exact one-to-one test because I'll be holding the gun differently for that part of the testing, but I don't think it will be enough to ruin the results.

That should also allow me to test the grip hypothesis using the same gun and ammo. Shoot some with the slide blocked and a very firm grip and some with the slide blocked and a very loose grip.
 
Walt,

We still haven't come to an agreement or satisfying conclusion as to whether semi-autos really DO reduce or delay barrel rise.

I agree with 45 auto in that I'm convinced there is a difference between revolvers and semiautos in that the mechanics of the latter absorbs recoil, thus reducing muzzle rise. I don't know about the rise being delayed, but whether the absorption is a simple matter of rise reduction, rise delay, or a combination, it has to be there, as some of the recoil energy in a semiauto is being channeled into productive work.

JeffK,

I suspect it'd be impossible to test objectively, because a "strong grip" will have so many other effects - anticipation/flinching being a key factor.

Perhaps, but it only costs 10 rounds to try. I don't know that grip strength is correlated to flinching.
 
Limnophile said:
I agree with 45 auto in that I'm convinced there is a difference between revolvers and semiautos in that the mechanics of the latter absorbs recoil, thus reducing muzzle rise.

I also suspect there is something different in how a semi-auto (that doesn't have fixed connection of the barrel to the frame) is affected by recoil than a revolver, which has the barrel and frame functioning as a single unit. But I have to remind myself that the slide is only moving a tenth of an inch +/- before the bullet is gone and barrel rise wouldn't seem to be GREATLY affected by recoil. That's the puzzle...

Saying "DELAYED" is confusing this discussion, but I haven't found a better way to describe what we're talking about or trying to explore. The bullet's forward movement IS delayed (or slowed) because the barrel/slide move a bit to the rear... recoil force isn't delayed, as the powder is doing its work at the same rate as in a revolver -- but bullet movement due to recoil is affected differently.

It would seem to me, in theory at least, that the same round shot out of semi-auto ought to be a hair slower from a unlocked slide than from a locked slide. Why? Because more of the recoil force is used to move the entire barrel/slide assembly early in the recoil cycle than with a revolver or a semi-auto with the slide locked. It's movement is being resisted by a recoil spring, and that is work being done, too: the spring is starting to be compressed. That's extra work done that isn't done with a locked slide or a revolver. Work done, it seems to me, should mean energy used or redirected. (In writing this I may be further exposing my imperfect understanding of how physics affect handguns. :()

If the recoil-induced bore-axis angle rise begins before the bullet leaves the barrel, it should be noticeable in a locked slide vs unlocked comparison. Maybe it does occur in both semi-autos and revolvers, but is less dramatic (is reduced, delayed, etc.) in a semi-auto because of barrel/slide movement, etc. A semi-auto with the barrel locked should behave much like a revolver.

It would be interesting to compare a 4" semi-auto shooting 9mm to a 4" revolver shooting 9mm. (I once had a S&W 547 that would have allowed that. Then the only differences would be bore-axis height above the hand and the revolver's cylinder gap.)

Most folks on gun forums don't seem to think there is anything to be investigated -- but they can't really offer convincing evidence for either position other then their experience. They may be right, but common wisdom is not always wise, and the reasons used to explain things don't always explain them properly.

Hopefully, JohnKSa's efforts will help us grapple with this puzzle. (JohnKSa also has had recoil-spring tests going on, too. He gives us feedback on that from time to time.)
 
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Why? Because more of the recoil force is used to move the entire barrel/slide assembly early in the recoil cycle than with a revolver or a semi-auto with the slide locked.

The exact same amount of recoil force is being used over the exact same amount of time.

Force = Mass x Acceleration

It takes exactly the same amount of force to accelerate a bullet to a given velocity in a semiauto barrel as it does a revolver barrel of equal length.

The same amount of force that is accelerating the 1 pound slide/barrel of a semiauto .10 inches to the rear in the .001 seconds while the bullet is in the barrel is also accelerating the 3 pound revolver to the rear.

The difference is that in the semiauto only a small amount of this force is reacted through the springs into the frame thus into your wrist during this time interval, the vast majority of it going into the slide which is conserving momentum by moving linearly.

In a revolver, the exact same amount of force is accelerating the gun backward for the exact same amount of time. However, since no moving parts are being accelerated, the full effects of this force are transmitted through the frame of the gun thus into your wrist.

Bottom line is that during the .001 second that the bullet is in the barrel, the slide/barrel of the semiauto is moving linearly against nominal spring pressure with almost zero force on the frame and shooter's wrist. In a revolver, during the same time period while the bullet is in the barrel, the exact same force is applied directly to the frame. This force is transmitted through the frame directly into the shooter's wrist. Since the line of action is above the wrist, the force creates a MOMENT which torques the revolver shooter's wrist upward during the time period that the semiauto's slide is moving rearward.

We keep talking about "force", exactly how much force are we talking about to accelerate a bullet down a barrel?

From a previous post, ignore it if you don't care:

If anyone is really interested, you can use Newton’s stuff from 400 years ago to understand how your gun works. We used it to get us to the moon back in the 60’s, we use it on the rockets and guns we’re designing today. If you don’t care to know, don’t worry about it, keep believing whatever makes you happy! No big deal.

For example, how long does your bullet remain in your gun barrel, what kind of acceleration does the bullet experience, how much force is being applied to the slide or frame?

This kind of stuff is REAL important if you’re designing things like rocket mechanisms or artillery fuses or proximity detonators.

Not real easy to effectively transmit mathematical concepts in a forum format, but we can try. I’m going to ignore the contribution of the unburned powder and gases exiting the barrel with the bullet.

We’ll call the starting position of the bullet the zero of our coordinate system, so X(initial) = 0. We’ll abbreviate it as Xi cause I’m lazy. That means Xi = 0.

V(initial) is the starting velocity. We’ll call that Vi. For a bullet in a gun, it’s initial velocity is 0. So Vi = 0.

We’ll call the last position we consider, the end of the barrel, X(final). I’ll abbreviate that as Xf.

T = time in seconds.
T^2 means “T raised to the second power” or "T squared" or “T x T”.
X = position in inches.
V = velocity in feet per second.
A = acceleration in feet per second squared (fps^2).

The basic motion equation in terms of time is:

Xf = Xi + V x T + (A x T^2) / 2

If you want more on this basic motion equation, you can go here:

http://en.wikipedia.org/wiki/Equations_of_motion

Introductory calculus teaches us that the first derivative of position is velocity, and the first derivative of velocity and the second derivative of position is acceleration (I’m not going into calculus for you).

So Velocity = Vi + Acceleration x Time, or

Vf = Vi + A x T

We’ll look at a typical .45 caliber 1911. First thing we have to do is get everything into common units.

Bullet weight = 230 grains = 230/7000 pounds = .033 pounds = .001025 slugs (slug is mass unit in english system).
Barrel length = 5 inches = 5/12 feet = .417 feet
Bullet velocity when it exits the barrel at 5 inches = 830 feet per second.

Now we just use the basic motion equation:

Xf = Xi + Vi x T + (A x T**2) / 2

Put in our 1911 values that we know:

.417 = 0 + 0 x T + (A x T**2) / 2 or

.417 = (A x T**2) / 2 (NOTE – This will be EQ1)

Notice that we have 2 unknowns (A and T) but so far only one equation. From basic algebra we know that we need as many equations as unknowns if we hope to solve them.

This is where we use the rest of the stuff we know and the velocity equation:

Vf = Vi + (A x T)

We know Vf (Velocity final) is 830 fps. We know Vi (Velocity initial) is 0, so

830 = 0 + (A x T)

Since we loved algebra, we solve for A in the above equation and find

A = 830 / T (NOTE – This will be EQ2)

Now we can cleverly plug the above relationship back into our basic motion equation EQ1, replacing A with 830 / T.

.417 = (830/T) x (T**2) / 2

Simplifying this (basic algebra again) gives

.417 = 415 x T

Now we solve for T:

T = .417/415 = .0010048 seconds

In other words, T is right at 1/1000 of a second. This is the total time that the bullet is accelerating, which is the time that the bullet is in the barrel.


Since we now know T, we can easily solve for A from EQ2:

A = 830/ T = 830 / .001 = 830,000 feet per second squared.

Acceleration due to gravity = 32.2 fps**2, so our bullet is experiencing:

830,000 / 32.2 = 25776.4 G’s as it goes down the barrel.

Kind of important to know if you want to hang some electronics on your bullet. They better be able to handle more than 25,000 G’s.

How much force is the bullet and therefore the locked together barrel/slide seeing?

We know that the force from the bullet equals its mass times acceleration, and we know that for every force there is an equal and opposite reaction (that Newton stuff again).

So the force delivered to the locked together barrel/slide from the bullet is:

F = M x A

F = mass of bullet x acceleration of bullet

F = .001025 x 830,000 = 851 pounds

As you can see, you have an average of about 850 pounds of force pushing on the barrel/slide of a .45ACP semiauto or the frame of a .45ACP revolver while the bullet is in the barrel.

We can go through the same equations (F = MA) and you'll find that 850 pounds over .001 seconds will move the slide of a 1911 rearward about 0.10 inches. It's no where near fully compressing the springs or hitting the frame to transfer it's momentum to the frame. When it does, a little later in time, that force is going to react through the frame and twist your wrist upwards. But the bullet is already long gone.

In a .45ACP REVOLVER, that same 850 pounds of force is pushing rearwards on the frame along the barrel axis for .001 seconds. This means that it's going directly into twisting your revolver wrists upwards during the time that the semiauto slide is accelerating and your semiauto wrists are seeing essentially ZERO load. During the .001 second that the bullet is in the barrel the same 850 pounds that is going into accelerating the barrel/slide on the semi auto rearward is going into twisting your revolver wrists upward.

How much upward twist out of your wrists with a 5" barrel does it take to throw a bullet 6" high at 25 yards (900 inches)?

Figure wrist to grip is 3", frame is 2", barrel length is 5".

Means that the end of the barrel is 10" in front of your wrist.

6/900 = x/10

x = 0.067", or about 1/16 of an inch.

If your wrists allow the barrel to move up 1/16 of an inch when you apply 850 pounds for .001 seconds to the end of your barrel, the bullet will impact 6" higher than your bore axis. This means that your sights must be aligned to point 6" higher than your bore axis on your target at 25 yards for this particular revolver for POA to be the same as POI.

Since F=MA, it should be obvious that a lighter bullet at the same muzzle velocity will generate less recoil force. For example, a bullet with half the mass at the same velocity will generate half the force. This means half the twist in your wrists. If your revolver muzzle climbed 6" from the force accelerating the heavy bullet, it's only going to climb 3" from the force accelerating the lighter bullet. Thus the lighter bullet strikes lower than the heavier bullet.

Sorry, trying to keep it as simple as possible, no simple way to explain it on a forum especially if you don't have an engineering background.

Note: The 850 pounds is the AVERAGE force reacting over the length of the barrel to accelerate 230 grains to 830 FPS. This is equivalent to the area under the powder pressure vs barrel length curve. Peak force will be significantly higher. For example, a SAAMI .45ACP pressure of 21,000 PSI on a base diameter of .452 (typical .45 bullet) gives a force of 3,370 pounds.

JohnKSa, as you can see, you're trying to quantify time differences involving hundredths of thousandths of seconds, distances involving thousandths of inches, and forces involving thousands of pounds. I'd be very surprised if your hold is repeatable enough to draw any valid conclusions without a machine rest.
 
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Just to be pedantic, your wrist and where you hold the gun won't play a role in the early part of the recoil process, though it can later (like for recovering for the next shot). The gun doesn't pivot about your wrist at least initially, it pivots about it's center of mass, which is below the barrel. If you were to lay a gun on a table and figure out how to fire it without touching it (electronic trigger, say), you would find that the recoil spins it and sends it backwards - if you were holding it, the early part of the recoil would look exactly the same. Your hand is soft and squishy, and the forces involved are large, so you can't control the recoil until your arm starts to move with the gun and soak up some of the energy, and your muscles can start to react a couple hundred ms later. The bullet is long gone by then.
 
What JeffK says is totally true. I was attempting to keep CM discussion out of it and simplify as much as possible. In reality, with the gun revolving about the CM, the muzzle jump required is proportionally less.

For example, I'm guessing that the CM of a 5" revolver is approximately halfway along the cylinder, or about 6" behind the muzzle. If we use the 6" distance to the CM instead of the 10" distance to your wrists as our center of revolution, that would mean that the muzzle climb required while the bullet is in the barrel for a 6" high POI at 25 yards would be:

6/10 x .067" which equals .040"
 
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45 auto said:
The exact same amount of recoil force is being used over the exact same amount of time.

Force = Mass x Acceleration

It takes exactly the same amount of force to accelerate a bullet to a given velocity in a semiauto barrel as it does a revolver barrel of equal length.

We sometimes seem to be adversaries in this discussion, and that's really NOT my intent. I am trying to understand. And I do appreciate you comments in your latest responses.

I wish we had a 4" Ruger or S&W (547) so that we could compare that gun to a 4" semi-auto, to see whether they do generate similar velocities. (The barrel/clyinder gap on a revolver has to have an effect when comparing a revolver to a semi-auto.) That is force lost that isn't used to power anything. It would be nice if we could create hand load rounds for each weapon so that a given bullet weight would leave the barrels at the same velocity, that would simplify things even more.

45 auto said:
Bottom line is that during the .001 second that the bullet is in the barrel, the slide/barrel of the semiauto is moving linearly against nominal spring pressure with almost zero force on the frame and shooter's wrist. In a revolver, during the same time period while the bullet is in the barrel, the exact same force is applied directly to the frame. This force is transmitted through the frame directly into the shooter's wrist. Since the line of action is above the wrist, the force creates a MOMENT which torques the revolver shooter's wrist upward during the time period that the semiauto's slide is moving rearward.

We keep talking about "force", exactly how much force are we talking about to accelerate a bullet down a barrel?

If I understand what you've written, I would agree that if the revolver is passing force the force directly to the frame without the semi-auto's seeming delay of force transfer, there should be SOME difference in how torque causes the barrel to rise BEFORE the bullet leaves the barrel. Is there any torque applied in the-semi-auto BEFORE the bullet leaves the barrel? It may be trivial.

I also wonder if the height of handle (distance above the hand) in a revolver matters greatly during that first .001 of a second. It seems as though the immediate resistance encountered would still be there, but it might be somewhat attentuated by the longer "handle" (grip) between the barrel and the hand. The difference of observed recoil after the bullet is gone would certainly be more dramatic if the "lever" or handle were longer.

You've offered a convincing case to show why semi-autos and revolvers are different, but we've still not fully addressed the key point that led to this discussion's restatement from an earlier debate:

Will there be an observable difference in the point of impact if a heavier bullet is used in a semi-auto? (We're talking the equivalent of factory loads, one with a lighter bullet and one with a heavier bullet, not a hot-rod round.)

Some folks say that a semi-auto is different, and points of impact are essentially the same (up close, at least, before "external ballistic" factors come more forcefully into play.

That also why I want to see whether there will be a difference with the same bullet in the same gun, if the if the slide is locked. It cancels out the delay effect, and makes that semi-auto act like a revolver with low bore-axis and no cylinder gap.
 
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I also wonder if the height of handle (distance above the hand) in a revolver matter greatly during that first .001 of a second. It seems as though the immediate resistance encountered would still be there, but it might be somewhat attentuated by the longer "handle" between the barrel and the hand.

A longer "handle" (higher bore axis) does NOT attenuate the effects of the recoil force on muzzle rise, it makes it worse.

Moment = Force x Distance.

Distance is how far the line of action of the force is above the center of mass.

For example, double the distance ("the height of the handle") and the exact same force will produce twice the moment. Same reason that putting a cheater bar on a wrench makes it easier to remove bolts.

If a given force raises the muzzle 1/8" on a gun with a bore height of 1" while the bullet is in the barrel, the exact same force will raise the muzzle 1/4" on a gun with a bore height of 2" while the bullet is in the barrel (assuming that the guns have an essentially equal rotational moment of inertia).

Minimizing this moment is why Olympic Free pistols have the bore axis aligned as closely as possible with the CG. Less "height of handle" equals less moment equals less muzzle rise equals less variables to compensate for.

This is also one of the selling points of the Mateba revolver (barrel is aligned with BOTTOM of cylinder instead of top):

The Mateba Autorevolver's barrel alignment is different from most other revolvers. The barrel is aligned with the bottom of the cylinder instead of the top. This lowers the bore axis (line of the barrel) which directs the recoil in line with the shooter's hand thereby reducing the twisting motion or muzzle flip of normal revolvers.

https://en.wikipedia.org/wiki/Mateba_Autorevolver

This is also why the CG of rockets without active controls is so important. If the line of action of the propulsive force (usually called thrust) does not go through the CG, the rocket will not go where it's pointed. Get it far enough off axis, and you have a pinwheel.
 
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It seems that some folks think the barrel axis of, say, a 1911, is the same as the line of the slide, in other words, that the bore is parallel to the top of the slide. Nope.

Jim
 
A longer "handle" (higher bore axis) does NOT attenuate the effects of the recoil force on muzzle rise, it makes it worse.

Moment = Force x Distance.

Distance is how far the line of action of the force is above the center of mass.

That all makes sense.

As I first thought about it, I realized that the LENGTH of the handle wasn't the only factor at play. That handle has weight/mass and while the lever effect offsets that somewhat, that longer lever is heavier, and there's more stuff to be moved. That was why I wondered whether it might attentuate things a bit. Perhaps it does (as might be the case with lead vs aluminum) -- but not enough to greatly affect the power of the "lever."
 
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Is there any torque applied in the-semi-auto BEFORE the bullet leaves the barrel? It may be trivial.

I maintain that there is. The "equal and opposite" reaction is instantaneous. As the bullet moves forward, the gun or its parts move backward. There is no delay in these movements or forces in physics. In a revolver that force is applied direction to the frame because all the intervening parts (cylinder and attachments) are, for all practical purposes, rigid. In a pistol that force moves the slide back. The force transmitted to the frame is reduced because some of it is stored momentarily in the compression of the spring. The part that some folks are missing is that the reason the spring compresses instead of flying out the back of the pistol is because the back end of the spring is in direct contact with a part of the frame. The spring necessarily transmits some of the force of recoil to the frame. It is reduced by the amount of mechanical force absorbed by the compression of the spring, but not to zero, and since that force is applied above the center of resistance (the hand of the shooter) torque does exist. It should be noticeably smaller than a revolver using this logic, and it may produce a very small amount of rotary movement in the amount of time that the barrel is in the revolver, but it is not zero. In this model, muzzle flip continues after the bullet leaves the barrel because the unequal forces have produced a rotary motion that has its own momentum, and has to be countered by the work of the shooter in controlling recoil.
 
Recoil of any firearm is around the center of gravity of the gun-hand(s) system. The higher the barrel is off the CG the more the gun will "flip" upward. The more in line the barrel is with the CG, the more the recoil will be straight back.

Jim
 
Is there any torque applied in the-semi-auto BEFORE the bullet leaves the barrel?

Almost 100% of it is, since by the time the bullet leaves the barrel the pressure behind it is small (ideally very small, since whatever is left is doing no good and is just creating more recoil, flash and bang - the reason .357 snubbies are such beasts to fire, even though the bullet doesn't wind up much faster than a .38 out of the same gun), and the only thing left to eject is gas and some powder. The whole recoil process is set off in that first millisecond, the torque and the backward force.
 
Is there any torque applied in the-semi-auto BEFORE the bullet leaves the barrel? It may be trivial.
A: there is,the bullet tries to rotate the barrel while exiting and and a correct seating of the barrel lug on the slide stop cross pin controls it.
No it is not trivial,John Browning took it into account with his barrel lug design.
 
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