How Much Can 9mm Bullet Weight Affect POI (S&W 3913)

[mavracer]

I don't see how either one of those really apply, but maybe they do. I don't claim to be a good student of physics. That may be where you need to simplify the physics for me and others.

Newton's third law states "When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body."

As the bullet's mass goes up it's momentum goes up proportionally, since barrel, slide etc mass stays the same it's velocity must go up to maintain equal momentum. Therefore as the bullets weight goes up the barrel must move further in relation to how far the bullet does.

 

[45_auto]

My assumptions:
- How far the bullet moves in the barrel is related and proportional to how far the slide and barrel move and how far a recoil spring or hammer springs are compressed.

I'm going to call the barrel and slide the "upper" since calling it barrel/slide can get confusing when also using the "/" symbol for division.

The distance the bullet moves in the barrel is NOT proportional to how far the upper moves. They are obviously related, but the squared time term in the acceleration component of the force equation (equal forces bullet and upper) is going to make it nonlinear, thus non-proportional if your bullets have different velocities. The bullet and upper will have equal momentum but different accelerations for different bullets. This is because the different masses and velocities of the bullets require different forces (thus different accelerations and positions) to achieve equal momentum to optimize the function of the gun. It will probably be as nasty getting there as the explanation above, but we will try anyway.

Summation is in the next couple of paragraphs if you want to skip everything between the "Physics Stuff" and "End of Physics Stuff" below:

As the bullet leaves the muzzle and the recoil force ends, the slide is moving at 13.7 FPS

An 8% lighter bullet moves the upper 13% less in 7% less time. When the recoil force from firing a 185 grain bullet at 1050 FPS ends, the upper has moved .061”. The recoil spring and hammer spring are compressed 13% less than when firing a 200 grain bullet at 975 FPS where the upper has moved .070”.
A 13% lighter bullet moves the upper 15% less in 15% less time. When the recoil force from firing a 200 grain bullet at 975 FPS ends, the upper has moved .070”. The recoil spring and hammer spring are compressed 15% less than when firing a 230 grain bullet at 830 FPS where the upper has moved .081”.
A 20% lighter bullet moves the upper 26% less in 21% less time. . When the recoil force from firing a 185 grain bullet at 1050 FPS ends, the upper has moved .061”. The recoil spring and hammer spring are compressed 26% less than when firing a 230 grain bullet at 830 FPS where the upper has moved .081”.

Hopefully it’s obvious from above that the upper position is NOT proportional to bullet position. The upper can be positioned at .061”, .070”, or .082” with different weight & velocity bullets located at the exact same position (the muzzle). If upper position was proportional to bullet position, at a constant bullet position (5"), the upper position would be constant.


This means that when the bullet leaves the muzzle the recoil and hammer spring are compressed different amounts by the different bullets.
The force of compressing the recoil and hammer springs is directly transmitted to the frame of the firearm, thus to the shooter’s hand and wrist.

Barrel rise is caused by the pivoting action of the firearm about the shooter’s wrist. This pivoting action is caused by the torqueing of the firearm. Since Torque = Force * Distance, and the distance from the shooter’s wrist to the bore axis remains constant, hopefully it is obvious that less force = less torque = less pivoting of the wrist while the bullet is in the barrel.

The difference in compressing the hammer and recoil spring .082” (230 grain), .070” (200 grain), and .061” (185 grain) is obviously not going to cause huge differences in the recoil forces transmitted to the shooters hand while the bullet is in the barrel. Should be easy enough to measure this with a 1911 and a scale if anyone cares. I have a hard time imagining ANY handheld shooting test using ANY amount of shots being precise enough to provide any meaningful data with the forces being so close together.

- The components listed above act as part of a fixed relationship until the bullet leaves the barrel -- when the barrel and slide go to their different stops on the frame.

It's a mechanical linkage - the physical relationships are definitely fixed.

Different weight bullets moving at different speeds will change the time it takes for the different components of this fixed relationship to go through their related motions, but not how the parts move in their relationships to each other.

True statement. Faster, lighter bullets will exit the upper at an earlier point in the upper's travel than heavier, slower bullets. In all cases all bullets exit the upper and recoil forces have ended far before the barrel unlocks. The upper will still continue rearward due to it's momentum and physical linkages.

And there was a general consistency in the results cited by others who did their own tests. I showed an image of my results. Did we all make the same errors? Or were we simply shooting our handguns at targets so close that the differences aren't significant?

Push the barrel of your 1911 against a scale and figure out how much difference in force is required to move a 1911 upper .082", .070", and .061" out of battery and we'll figure out how much difference in torque your wrist is reacting among the various weight bullets and how much the muzzle might rise while the bullet is in the barrel. My guess is that your shooting is not precise enough to measure the difference, thus my earlier comment on fixed rests and thousands of rounds.

Physics stuff

Upper momentum = bullet momentum.

Different bullet/velocity combinations can have the same momentum, but since the bullets have different velocities, how far they move in the barrel is NOT related to how far the slide and barrel move.

For example, Speer makes FMJ's in 185, 200 & 230 grains:

Momentum = mass(bullet) * velocity(bullet)

Momentum(230) = (230/7000)/32.2 * 830 = .85 sl-ft/s
Momentum(200) = (200/7000)/32.2 * 975 = .87 sl-ft/s
Momentum(185) = (185/7000)/32.2 * 1050 = .86 sl-ft/s

Notice the essentially identical momentum of each combination.

If you go back to the example with the 230 grain bullet in Post 22 where we calculated the forces, you'll find that the 2 pound slide/barrel had a velocity of 13.7 ft/s. Want to bet what it's momentum will be?

Momentum(slide/barrel(Post 22) = (2/32.2) * 13.7 = .85 sl-ft/s

Notice how the momentum delivered to the slide/barrel is identical (I'm only carrying two decimal places, feel free to go out as far as you feel necessary in the calculations if you desire more resolution) for all three bullet/velocity combos from Speer, which means the slide/barrel momentum will be identical. No need to change recoil springs or do any tuning when shooting different weights of their ammo.

Since the slide/barrel mass is identical, then the slide/barrel velocities will be identical among the three bullet weights.

However, since the lighter bullets are accelerating faster, they spend a shorter amount of time in the barrel.

Assuming constant acceleration and a time in the barrel of .001 seconds from the 830 FPS, 230 grain bullet example in Post #22, then:

200 grains at 975 FPS = 830/975 * .001 = .00085 seconds in the barrel.
185 grains at 1050 FPS = 830/1050 = .00079 seconds in the barrel.

The lighter, faster bullets spend proportionately less time in the barrel. But since the upper momentum is identical, and the upper mass is identical, that means the upper velocity must be the same in each case.

However, to accelerate the upper to the same velocity but in a shorter time, that means that the acceleration of the upper must be greater. Since we cleverly know that Force = Mass * Acceleration, and the mass of the upper is constant, then greater Acceleration = greater Force.

In a nutshell, the lighter bullets are putting greater force on the slide for a shorter time to achieve momentum identical to the heavier bullet.

We also know that Power = Force * Distance. To achieve greater force in an equal distance (barrel length) would mean that the lighter bullets would need more Power. Since our power source is gunpowder, if you are a reloader it should now be obvious why lighter bullets require more gunpowder than heavier bullets. The lighter bullets have to accelerate the upper to a velocity equal to that achieved by the heavier bullets but in a shorter time.

From the example in Post #22, we calculated the force delivered to the upper by the 230 grain bullet at 830 FPS to be 851 pounds to accelerate the slide to 13.7 FPS in .001 seconds.

Since the 200 grainer at 975 FPS is only in the barrel for .00085 seconds, and we know the upper velocity must be 13.7 FPS since the upper momentum is the same and the upper mass is the same, we can calculate the slide position using the known time and velocity:

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

Xf = 0 + 0 x .00085 + (13695 x .00085**2) / 2

Xf = .0057973 feet = .070 inches

The 185 grainer at 1050 leaves the slide after it moves

Xf = 0 + 0 x .00079 + (13695 x .00079**2) / 2

Xf = .0050635 feet = .061 inches

Since we know the mass and velocity of the upper and the distance and time in which it was accelerated to that velocity, we can figure out how much force was exerted on the upper:

We know the 230 grain exerts 851 pounds of force on the slide for .001 second from Post #22.

For the 200 grain:
Velocity = Acceleration * Time
13.7 = A * 0.00085
A = 13.7/0.00085 = 16,118 ft/s^2
Force = Mass * Acceleration = 2/32.2 * 16,118 = 1001 pounds

For the 185 grain:
Velocity = Acceleration * Time
13.7 = A * 0.00079
A = 13.7/0.00079 = 17,342 ft/s^2
Force = Mass * Acceleration = 2/32.2 * 17,342 = 1077 pounds

Notice how the lighter bullets require more force? That force comes from the gunpowder. If you reload, notice the difference in powder charges on light vs heavy bullets.

End result:

185 grain bullet moves 5 inches while upper moves .061" in .00079 seconds as a result of the 1077 pounds applied for the .00079 seconds that the bullet is in the barrel.
200 grain bullet moves 5 inches while upper moves .070" in .00085 seconds as a result of 1001 pounds of force applied for the .00085 seconds that the bullet is in the barrel.
230 grain bullet moves 5 inches while upper moves .082" in .001 seconds as a result of 851 pounds of force applied for the .001 seconds that the bullet is in the barrel.

185/200 = 8% lighter bullet .061/.070 = 13% less slide movement .00079/.00085 = 7% less time in barrel
200/230 = 13% lighter bullet .070/.082 = 15% less slide movement .00085/.001 = 15% less time in barrel
185/230 = 20% lighter bullet .061/.082 = 26% less slide movement .00079/.001 = 21% less time in barrel

End of Physics Stuff

 

[Walt Sherrill]

You've created an impressive collection of data. It ought to be a sticky somewhere here on the forum (with my parts excised). I appreciate the time and effort you put into the explanation and related calculations. I wish I had your facility with numbers and a proper understanding of the physics involved. My understanding can be hazy at times.

You've shown us that different bullet weights/speeds DO affect gun behavior before the bullet leaves the barrel, with that happen at different points in slide movement. You've also shown us that bullet movement is ONLY ROUGHLY proportional to slide movement. My claims to the contrary were wrong. I was ONLY ROUGHLY correct. Close -- but no cigar.

Your calculations don't show us much about barrel rise prior to bullet exit, but you do say that some force -- transmiitted through spring compression -- must have some effect on the shooter's hand. There are no calculations or data showing HOW that rise might be measured, but you wrote:

Push the barrel of your 1911 against a scale and figure out how much difference in force is required to move a 1911 upper .082", .070", and .061" out of battery and we'll figure out how much difference in torque your wrist is reacting among the various weight bullets and how much the muzzle might rise while the bullet is in the barrel. My guess is that your shooting is not precise enough to measure the difference, thus my earlier comment on fixed rests and thousands of rounds.

How the gun behaves in the shooter's hand is hard to asses or measure -- and may not be directly related to the forces transferred to the his or her hand.

This means that when the bullet leaves the muzzle the recoil and hammer spring are compressed different amounts by the different bullets.
The force of compressing the recoil and hammer springs is directly transmitted to the frame of the firearm, thus to the shooter’s hand and wrist.

I'll quibble here, as it IS about a minor point (i.e. a quibble), but I'd suggest that the forces applied to the frame thru the springs are NOT DIRECTLY transmitted to the frame -- the spring isn't a stiff piece of steel. If nothing else the transfer is a delayed process. I think, too, that some of force transmitted to the spring never makes it to the frame, but is stored in the spring. But I agree with you that the total force transmitted is arguably very small and that must, in turn, mean the amount held in the spring, if it is a factor at all, is much smaller. You show a way to measure the force actually transmitted to the frame (or a scale), but how the shooter deals with that "recoil force" will also be affected by how the shooter handles the weapon.

Barrel rise is caused by the pivoting action of the firearm about the shooter’s wrist. This pivoting action is caused by the torqueing of the firearm. Since Torque = Force * Distance, and the distance from the shooter’s wrist to the bore axis remains constant, hopefully it is obvious that less force = less torque = less pivoting of the wrist while the bullet is in the barrel.

The difference in compressing the hammer and recoil spring .082” (230 grain), .070” (200 grain), and .061” (185 grain) is obviously not going to cause huge differences in the recoil forces transmitted to the shooters hand while the bullet is in the barrel. Should be easy enough to measure this with a 1911 and a scale if anyone cares. I have a hard time imagining ANY handheld shooting test using ANY amount of shots being precise enough to provide any meaningful data with the forces being so close together.

That the forces causing different gun behavior because of different bullet weights and speeds are so similar that the very imprecise and unscientific tests done here don't show us anything definitive. Understood and agreed. But that they are so similar seems to tell us a different tale.

You've shown us some very real differences, but little that causes us to believe that there are observable-meaurable recoil-induced barrel rise except forces passed to the shooter's hand through the springs BEFORE THE BULLET LEAVES THE BARREL. It would seem that the effect of that force transfer (i.e., how much the barrel will rise) could vary by shooter, depending on their skills, grip, etc., and is not predictable or easily measured.

It would seem that IF all of the forces at play are so similar that only very sensitive measurement tools and practices will clearly show us the differences, do these differences really matter to most shooters? Are these differences really that much different than the kind of performance variances we might routinely see from one batch of hand loads or factory ammo to another?

We may all be straining to lift a gnat (or wasting our time trying to beat it to death.)

Again -- thanks for a very informative response. I'm sorry you had to do it -- but it was necessary to correct some of my incorrect assumptions -- but I'm glad you took the time to do it. It's the best analysis of 1911 behavior I've seen. I will keep a copy.

.

 

[random guy]

It would seem that IF all of the forces at play are so similar that only very sensitive measurement tools and practices will clearly show us the differences, do these differences really matter to most shooters? Are these differences really that much different than the kind of performance variances we might routinely see from one batch of hand loads or factory ammo to another?

For the purpose at hand, I think that is the bottom line. Semi-autos such as the one owned by the OP appear to show very little impact shift by bullet weight.

Compare that to fixed barrel guns which commonly shift inches at 25 yards.

 

[mavracer]

I'll quibble here, as it IS about a minor point (i.e. a quibble), but I'd suggest that the forces applied to the frame thru the springs are NOT DIRECTLY transmitted to the frame -- the spring isn't a stiff piece of steel.

Again Newton's 3rd law you can't compress a spring by exerting force on one end without putting equal force at the other end.

For the purpose at hand, I think that is the bottom line. Semi-autos such as the one owned by the OP appear to show very little impact shift by bullet weight.

Compare that to fixed barrel guns which commonly shift inches at 25 yards.

Again it has very little to do with the semi auto part and more to do with the fact that 9mm produces little recoil and semi autos have a much lower bore axis than revolvers.

 

[Walt Sherrill]

Again Newton's 3rd law you can't compress a spring by exerting force on one end without putting equal force at the other end.

I notice you didn't include the part where I said it was a DELAYED response. THE TIME REQUIRED TO COMPRESS THE SPRING (which 45_auto showed will vary depending on the bullet speed and weight) controls the time required to transfer the recoil to the frame. The transfer is still direct, but not immediate. As I read 45_auto's calculations, that's part of the reason why there is a different amount of slide travel before the bullet leaves the barrel -- it all depends on bullet speed and weight and how the springs respond.

If ALL ofa gun's recoil is transferred to the shooter's hand BEFORE the bullet leaves the barrel, it would MORE of a factor than if only PART of it is transferred. But if only part of it is transferred, but even if only part of it is transferred (as with a Browning SRLB design), the amount of recoil the shooter experiences will vary based on bullet weight and speed, AND BE DEALT WITH DIFFERENTLY, depending on the shooter.

45_auto's technical response should generally apply to a linkless design too. And if, with a swinging link gun, the link doesn't cause the barrel to separate BEFORE the bullet has left the barrel, and a linkless design doesn't also cause premature unlocking, the two systems should perform in the same way. Both designs are available in .45 acp and 9mm.

Does a barrel link ever cause too-early unlocking (and, barrel rise) if the link is the wrong length? One of the participants in this discussion talked about slower/heavier bullets possibly having that effect -- although he wasn't addressing a misfited link. I suppose that COULD happen if the link isn't properly sized. (We're getting into esoteric gun function issues here and I'm certainly not able to do more than ask the question.)

When one such difference between bullet weight group locations was noticed (with the 185 gr. groups being lower [or, perhaps, LESS HIGH] than than 200 gr. or 230 gr. groups), another participant said that the difference could have been due to the barrel link. IF, as 45_AUTO showed up, bullets leave the barrel at different points in slide travel, one wonders about the role of the link in the different group POSITIONS. (As I noted earlier, the Khunhausen book talks about 7 different link lengths.)

Again it has very little to do with the semi auto part and more to do with the fact that 9mm produces little recoil and semi autos have a much lower bore axis than revolvers.

We know that fixed-barrel guns behave differently than Browning Short-Recoil Locked-Breech guns. And it has as much to do with the fact that the barrel is DIRECTLY ATTACHED TO THE FRAME in a fixed-barrel gun as does bore axis. Recoil is immediately transferred to the frame with a fixed-barrel gun, and it starts as soon as the bullet is fired. With a BSRLB gun, begins to transfer when the slide starts to compress the springs -- And that travel is only a fraction of an inch by the time the bullet leaves the barrel. The total amount of recoil force passed to the frame via the barrel BEFORE THE BULLET EXITS IS MUCH LESS with the Browning SRLB design. That's true regardless of BSRLB gun caliber.

By the time the bullet exits the barrel of a fixed-barrel gun, nearly all of the recoil has been transferred to the frame. That barrel is going to start to rise sooner and rise more than with the Browning design. This will be true of a revolver with a low bore axis, too -- and there are some with bore axes lower than many semi-autos (like the Chiappa handguns). A low bore-axis revolver will still have more measurable recoil before the bullet leaves the gun.

Is there a reason a given 9mm bullet's effect on recoil transfer to the frame of a gun using the same Browning short-recoil locked breech design truly different than a .45's recoil transfer when both guns use the same system? A similarity in bullet impacts (group size and location) despite different bullet weights/speeds have been noticed by others when shooting .45s. and 9mm guns.

The point 45_auto made some time back -- but didn't explain in ways I fully appreciated -- was reinforced by his last technical response. I see his point: the differences we're talking when using different bullet weights and speeds ARE SO SMALL that only very sophisticated sampling methods with a large number of shots fired can show us a statistically significant difference that means something. I think this would should be true regardless of bullet weight or bullet speed OR CALIBER as long as we're dealing with Browning Short Recoil Locked Breech design guns.

A Ransom Rest test may be the only way to see the actual effect of recoil on group size and location with different bullet weights and speeds, and the barrel would probably have to probably need to be resighted with each shot -- not generally done in Ransom Rest tests. (Ransom Rest tests typically look only at group size, not size and location [i.e., higher or lower, etc.])

Those RR results wouldn't allow us to predict how a given gun and bullet weight/speed would performs in different people's hand. We all respond to recoil differently -- but they would be more meaningful than the simple and far-less-replicable tests we discussed earlier.

 

[mavracer]

If ALL of that recoil is transferred to the shooter's hand BEFORE the bullet leaves the barrel, it will be less of a factor than if only PART of it is transferred. But even then, the amount of recoil the shooter experiences will vary based on bullet weight and speed, AND BE DEALT WITH DIFFERENTLY, depending on the shooter.

I think I finally understand your disconnect you're having a real hard time differentiating speed and distance,

The recoil velocity is maximum at the time the bullet leaves the barrel after that the velocity of both the gun and bullet are slowing down.

A Ransom Rest test may be the only way to see the actual effect of recoil on the gun

Not really, iffin you were capable of shooting 3" groups off hand at 25 it's pretty easy to tell.
My 1076 Smith has a very noticeable shift in POI between my 135gr IPSC major load (1250fps) and my rip snort 180gr load (1175FPS)
Not near what my Redhawk 44 mag shifts from my Elmer Keith practice load (240@1100) to my 300gr@ 1200fps load.
But way more than either my HK P7 PSP and Smith 442 or LCR show.

 

[Walt Sherrill]

I think I finally understand your disconnect you're having a real hard time differentiating speed and distance,

The recoil velocity is maximum at the time the bullet leaves the barrel after that the velocity of both the gun and bullet are slowing down.

My comments below may show that I don't understand the point you're trying to make --at least as stated. I may certainly have a disconnect, but I don't see where that comment should apply.

If you're addressing what I said about fixed-barrel guns vs Bronwing SRLB guns, then...

Velocity may be maximum at that point of bullet departure, but the amount of recoil force transferred to the frame of a BSRLB gun as the bullet departs is much less than is the case with a fixed-barrel gun.

The forces transferred to the BSRLB semi-auto frame as the bullets leave the barrel when firing .357 SIG round and a .45 a.c.p. will be greatly different than the forces transferred to the frame when a similarly hot .357 Magnum round and a 45 a.c.p round are fired in a revolver.

As 45_auto's examples, show only a small portion of the total recoil force will be transferred to the frame and to the shooter's hand as the bullet leaves the barrel of a BSRLB gun. The amount of force that can act as a vector is small. But it IS there, and THAT's the variable I couldn't identify or understand earlier in this discussion.

(Looking back at the whole discussion, I think that the springs were the part of the "fixed" relationship that made things "less-fixed" than I thought. )

As 45_auto's examples show, the amount of force transferred to the frame in the BSRLB design can cause the slide to be in a different position when the bullet leaves based on bullet weight/speed. Recoil force is transferred by slide travel and different loads will cause the springs to be compressed slightly differently (which is why some loads might show the slide moving only .060" while another might show it moving .082" when the bullet leaves the barrel.)

I suspect, too, that the weight of the slide as it moves to the rear may also cause an additional vector force to affect barrel tilt -- but that happens long after the bullet is gone. But, if only a small amount of slide travel (and recoil travel) has taken place when the bullet leaves in a BSRLB gun, it stands to reason that only a small amount of recoil force can affect barrel rise.

If you've making a different point, you'll have to clarify.

 

[ShootistPRS]

In a fixed frame gun a 250 grain bullet imparts a given recoil to the gun.
In an auto-loading gun with the same length of travel using the same bullet at the same speed has the same amount of force (recoil) imparted before the bullet leaves the barrel - of either and both guns. They have the same amount of force (recoil) applied.

Once the bullet leaves the barrel it can't put any more force on the gun.

 

[Walt Sherrill]

In a fixed frame gun a 250 grain bullet imparts a given recoil to the gun.
In an auto-loading gun with the same length of travel using the same bullet at the same speed has the same amount of force (recoil) imparted before the bullet leaves the barrel - of either and both guns. They have the same amount of force (recoil) applied.

Once the bullet leaves the barrel it can't put any more force on the gun.

After the bullet leaves the barrel "it doesn't NEED to put any more force on the gun," it's already been put on the gun by getting the bullet out of the barrel!

And I'll agree that both types of weapons with the same ammo and same barrel length will have the same force imparted to the weapon when a bullet is fired. But, while the forces that will be imparted to the frames of the guns are the same, the manner in which they are transferred to the frames is NOT the same.

With a revolver, barrel rise BEGINS as soon as the bullet is fired -- because the barrel and frame are a single unit. As the bullet moves down the barrel, an equal and opposite force is being directly transferred to the bullet and to the barrel and frame AND to the shooter's hand. Because recoil begins immediately it starts to create a torguing (vector) force that tilts the gun in the shooter's hand. By the time the bullet leaves the barrel virtually all of the equal but opposite rearward recoil force has been transferred to the frame. The barrel/frame will, of course, continue to rise after the bullet is gone, but no additional force is added to the mix or needed for that to happen.

With a Browning Short Recoil Locked Breech semi-auto force is transferred to the bullet and barrel and slide, but the frame isn't an equal partner in that transfer process. As the bullet moves forward and the slide and barrel move to the rear, some force is transferred to the frame by slide movement via recoil spring and hammer spring compression. The bullet, we know, is gone from the gun before the slide has only moved between 060" and 082" to the rear (here using the specs from 45_auto's tables and calculations).

During that short amount of barrel/slide travel, the only connection between slide and frame that can transfer recoil forces to the frame is a recoil spring and a hammer spring, and those two elements can only transfer a small part of the recoil force still in the slide and barrel. Compressing springs 1/10th of an inch or less doesn't transfer a lot of force. The rest of the un-transferred recoil force won't be passed to the frame until AFTER the bullet is gone.

The same recoil force that moved the bullet down the revolver barrel has moved the bullet down the semi-auto barrel. The only difference in the two weapon types is how much of the rearward force has been transferred to the frame by the time the bullet leaves the barrel. The force sending the bullet forward will be the same.​

There is a very small amount of barrel rise prior to bullet exit with the SRLB semi-auto SRLB; it's so small it's hard to see or measure. (As 45_auto said, the forces are so small, that it would require sophisticated tools and techniques to properly measure the barrel rise or the center of groups generated by different bullet weights and speeds.)

As the bullet leaves the barrel, recoil force transfer continues as the slide and barrel continue their rearward movement without interruption. The springs are compressed farther, and the barrel and slide, having separated hit their respective stops on the frame. All of that "after-the-bullet-has left" activity transfers more force to the frame.

Unlike the revolver, it appears that MUCH, maybe MOST of the recoil force is transferred to the frame AFTER the bullet is out of the barrel.

You seem to be saying that because the total forces applied to the guns are the same, how the guns behave (at least with regard to how recoil is dealt with) will be the same.

Do I misunderstand your comments?

.

 

[ShootistPRS]

I think you misunderstand the what is happening when and how long it takes.
https://www.youtube.com/watch?feature=player_embedded&v=Um9Eos9bJDk

Here is a revolver being fired. In the third replay you can see that the barrel does not seem to move at all until after the bullet is out of the barrel.

This is similar to the video with the semi auto.
If you we able to measure the muzzle rise when each gun was fired you could measure the amount of rise before the bullet leaves the barrel. It is small enough that it is hard to see from the distance of the camera.
This video does show that recoil from both guns affects the point of impact.

 

[mavracer]

Do I misunderstand your comments?

To be honest I agree with Shootist. You don't seem to grasp what is going on.

I've got a simple experiment for you to try. Take a locked breach semiauto and rack the slide without holding the frame stationary.

 

[random guy]

Again it has ....more to do with the fact that 9mm produces little recoil and semi autos have a much lower bore axis than revolvers.

I suspect that those are indeed factors. Factors which could largely be isolated in practical experiments. Chamberings are easy enough to substitute. Don't know if any SAs have as high a boreline as typical revolvers but some are higher than others. And some revolvers have a rather (or even very) low boreline. I don't have a Chiappa Rhino to try but it would be interesting to see its behavior in this respect.

 

[JohnKSa]

Take a locked breach semiauto and rack the slide without holding the frame stationary.

No problem.

I loaded a mag and put in a dummy as the top round. Inserted the mag in the gun.

Held the Ruger P89 by the back of the slide without touching the frame at all, moved the gun downward and then yanked back upwards.

The dummy round chambered without my touching the frame or holding the frame stationary. I put the dummy back in the magazine and repeated the procedure with the same result to be sure it wasn't a fluke.

The point is, of course, that the coupling between the slide and frame is quite weak. An abrupt motion of the slide can overcome that coupling easily even if the frame isn't being held still.

Did a quick measurement with my dial caliper. The slide and barrel on this particular gun move 0.15" backwards BEFORE the unlocking process begins to tilt the barrel at all. During that 0.15" of movement, the barrel recoils virtually straight backwards, the only upward arc would be caused by the very loose coupling of the slide to the frame via the recoil spring. A coupling effect which the suggested experiment demonstrates is quite weak.

Go back and look at post 62 to get a feel for how far the slide/barrel moves before the bullet exits. It's actually quite a bit shorter than the 0.15" movement on this particular pistol. In this design, the bullet is easily going to be out of the bore before the barrel begins to tilt due to unlocking.

Here is a revolver being fired. In the third replay you can see that the barrel does not seem to move at all until after the bullet is out of the barrel.

But we can tell it MUST move because the boreline is pointed markedly downward with respect to the sightline. If the muzzle didn't rise, the bullet could never hit where the sights are pointing.

HOWEVER, the same is not true of the three semi-auto pistols I examined and diagrammed in my earlier post. The boreline of those pistols is actually pointed slightly upward with respect to the sightline. Clearly the muzzle isn't rising very much at all, or the sightline/boreline relationship would be similar to what is seen in revolvers, instead of being so much different.

 

[Walt Sherrill]

You say I misunderstand. That's certainly possible, but it might help to specify the things that make my "misunderstanding" most obvious. Perhaps you, as 45_auto did, can explain the physics for me and show me and everyone else still reading where I've erred. 45_auto's certainly did just that a day or so ago. His explanations and correction(s) helped improve my understanding. They were a good lesson in gun function, too.

In an earlier reply, (#13) where you tried to correct my (erroneous) claim that the barrel of a semi-auto does not rise until after the bullet leaves the barrel, you wrote the following. The underlining is mine:

Physics tells us there is an equal and opposite reaction to the action of the bullets acceleration. Even though it is transmitted through the various parts of the gun it still acts upon the device that is holding the gun. It begins as soon as the bullet starts to move and the hand and wrist are absorbing that recoil and the gun is changing the vertical angle the entire time. The longer the bullet travels in the barrel the more that gun will move.

If you reread what I wrote in responses #68 and #70, you'll see that I agree with most of your analysis above, but disagree about what I consider a key difference: I claim that recoil is NOT transferred in the same way with both a revolver and Browning SRLB gun, and that difference affects whether more or less recoil can affect barrel rise by the time the bullet leaves the barrel.

In another reply (#51) you wrote:

In a locked breech semi-auto the barrel and slide are never in "free recoil". They push on a spring that rests against the frame. That push is recoil that is felt by the shooter. Placing the gun lower in your hand (gripping the gun higher) will help to reduce the amount of angular displacement but it does not negate it until the barrel is in-line with the center of your wrist.

That is a consistent with what I wrote in responses #68 and #70, with which you disagree. But, where we differ is in our assumptions about HOW MUCH recoil force is transferred to the frame before the bullets leave a semi-auto barrel. IF most of the equal but opposite force is transferred to the frame AFTER the bullets are gone, the barrel will still rise, but it will be rising when it doesn't matter. That it's transferred to the frame differently doesn't greatly affect how the bullet moves.

You also say the following in #51, and I added the underlining:

As soon as the link or ramp begins to move the rear of the barrel down the angle of the barrel rises. Even though the barrel is not completely disengaged it is moving vertically. The entire time that the barrel and slide are moving the recoil is being absorbed by your hand. (as transfered through the recoil spring to the frame). Recoil always begins when the bullet starts to move.

Am I correct to believe that with a revolver, all of the recoil force is transferred to the frame by the time bullet leaves the barrel? I think you made the point that additional force cannot be added to the gun after the bullet is gone.

With a BSRLB semi-auto, the bullet is gone when the slide has moved as little as .060"-.082" to the rear. We also know that recoil force is transferred to the frame ONLY through slide travel. The equal/opposite force relationship causing the barrel and slide to move to the rear doesn't stop as the bullet leaves the barrel. If recoil is only transferred to the frame by slide movement, how can a set of springs that have been compressed a trivial part of their "compression potential" cause the transfer of the same amount of recoil force to be passed to the semi-auto's frame at bullet exit as the same round transfers to a revolver's frame by bullet exit? That doesn't sound like physics. It sounds like magic!

The slide and barrel of the semi-auto will continue to move after the bullet has gone – and recoil force will continue to be transferred to the frame after bullets exit through slide movement, too. The mechanism for recoil transfer hasn't changed and it's working in the same way both before and after bullet exit. Except afterwards the springs may be fully compressed (the recoil spring may be stacked), and the barrel and slide have separated and hit their respective frame stops. All of that action caused recoil force transfer to the frame, too.

In both a revolver and a semi-auto, the force moving the bullet forward and frame to the rear are the same – but the amount of force transferred to the respective frames, before and after bullet exit, are not the same. The total force is the same, but it doesn't affect the frame and related barrel rise in the same way. The recoil force applied to the frame for a given amount of bullet travel is not he same.. (Bore axis can be greatly different when comparing semi-s and revolvers affects the torquing/vector force, but they do make revolvers with very low bore axes.)

I would note, too, that the recoil spring in a semi-auto isn't primarily there to dampen recoil or to delicately transfer recoil to the frame. That recoil transfer is an inevitable side effect of its use as a force storage mechanism – used to store part of the force of recoil for later use, when it will be used to strip the next round from the magazine and chamber it. You need a moving slide (or bolt, or some similar mechanism) to make a self-loading gun self-load.​

45_auto's calculations show us that slide movement, which does transfer recoil to the frame, varies with bullet weight and speed. It also shows us that the amount transferred -- via spring compression is so slight that's it's difficult to see or measure. Barrel rise will be different, but hard to measure or see. As I noted previously, that's probably why it doesn't show up in high speed videos -- it's too subtle.

If the bullet's gone and some recoil has already been transferred to the frame, what about the transfer mechanism (i.e., slide travel) changes after the bullets leave the barrel? I don't think the moving slide and barrel know that the bullets are gone -- or care. (That's a joke.) The equal but opposite force continues just as it started -- to the rear.

More importantly, if, as you claim (at least in the video you offered), the barrel doesn't rise until after the bullet leaves the barrel, what is causing the force moving the bullet forward NOT to having an equal and opposite effect on a revolver's barrel and frame and shooter's hand until then? Doesn't recoil directly affect the frame from the instant bullet travel starts? It may be more of the trick photography I was complaining about in an earlier reply.

With a semi-auto, the force pushing the bullet out of the barrel IS also offset by an equal and opposite force pushing the slide and barrel to the rear. That push continues even though the bullet is gone. (That force was transferred to the slide and barrel rather than the frame, and it moves to the frame in a different manner.) As the slide goes more to the rear the frame will receive more and more force, and the slide will shift also weight to the rear, adding to the torquing caused by recoil transfer,etc, etc. There will be MORE (continued) barrel rise, but it won't matter, but that "continued" barrel rise won't affect a bullet that's already out of the barrel.

 

[45_auto]

45 ACP 1911, 5" barrel: 230 grain bullet at 830 FPS exerts an average of 851 pounds of force for .001 seconds along the bore axis. See Post #22.

45 ACP Revolver, 5" barrel: 230 grain bullet at 830 FPS exerts an average of 851 pounds of force for .001 seconds along the bore axis. See Post #22. It's the same force as a 45 ACP fired from a semiautomatic.

I know a revolver barrel is technically longer than a semi-auto because the semi includes the chamber, and also a revolver has a flash gap, yada, yada. Feel free to do your own analysis. End result is going to be that to accelerate a 230 grain bullet to 830 FPS in about 5 inches is going to be about 850 pounds. Feel free to quibble the decimal points if you find it necessary.

Bore axis above shooter's wrist: 2.5 inches, = 2.5/12 = .21 feet

Force component transmitted to shooter's wrist through frame of gun during the .001 seconds bullet is in barrel:

Revolver: = 851 pounds. 100% of recoil force is transmitted to shooter's wrist while the bullet is in the barrel.

1911: = 8 pounds, due to compression of recoil and hammer spring during the .001 seconds the bullet is in the barrel. See pics below. First pic is weight of gun on scale, second pic is force required to compress hammer and recoil spring 0.082 inch. 9.8 lb - 1.8 lb = 8 pounds.

1.8 pounds weight of gun supported by scale:

The 9.8 pounds shown below minus the weight of the gun supported by the scale is the force required for .082" slide travel (rear of slide is .082" out of battery). This is how far your slide has traveled when the bullet leaves the barrel. The only force reacted by your hand while the bullet is in the barrel is equal to the force required to compress the springs this far. The remaining recoil force has no effect on the bullet, because the remaining force is a component of the slide momentum and will be delivered to the frame and thus your hand as the springs compress and slide and barrel hit their stops:

Torque transmitted to shooters wrist while bullet is in barrel = Force applied while bullet is in barrel x Bore Axis Distance above shooter's wrist.

Revolver = 851 x .21 = 177 ft-lb
1911 = 8 x .21 = 1.7 ft-lb

Lighter bullets in the semi auto will exert even less torque because the recoil & hammer springs are compressed corresponding less. If I had 3 hands, I could measure the scale force at .070 inches of slide movement and .061 inches of slide movement. It will be less than the 8 pounds measured at 0.082". Less torque = less muzzle flip while the bullet is in the barrel = lower impact point.

Revolver simulation: Get a torque wrench and have a buddy apply 177 ft-lb of torque for .001 seconds and see how far your wrist moves.

1911 Simulation: Have the same buddy apply 1.7 ft-lb of torque to your wrist for .001 seconds and see how far your wrist moves.

Since the torque is 100 times higher, my guess is that your wrist will twist about 100 times further in the revolver simulation than in the semi simulation. That means 100 times more muzzle flip in the revolver than in the semi.

Looks like the sight line difference above the bore is about .200 inches between a .45 ACP semi and a .45 ACP revolver. (Rule 1: Do NOT laugh at the chaos of my wife's vanity.) I'm going to call the rear sight heights equal and the distance between the sights 6" in both cases.

With a 6" sight radius, that .2" would make a difference of 30" at 25 yards. In other words, if you set your 45 ACP revolver sights at the same relationship to the bore as the semi sights, your revolver would shoot 30" high at 25 yards. That 30" at 25 yards is the difference between resisting a 177 ft-lb moment and a 1.7 ft-lb moment (remember the torque wrench experiment) for .001 seconds.

Given the 100 to 1 relationship between the reaction torques on the revolver and the semi, I would expect the sight compensation for the torque to be similar. In other words, if a 0.2 height difference is required for 177 ft-lb of torque, I would expect 1/100 of that torque to require 1/100 of the sight height difference. If the 177 ft-lb of the revolver requires 0.2", then 1.7 ft-lb of the semi would require .2/100 = .002". .002" on the front sight is .3" at 25 yards. I would imagine that is less than the manufacturing tolerances on the sights, so the manufacturers probably install the front and rear sights on their semis nominally parallel to the bore.

Back to the original question of does the bullet weight make a difference. By my estimates, a 1911 sight is required to be aligned .30 inches above the bore axis (front sight .002" higher in relation to rear than bore axis) at 25 yards to compensate for the muzzle flip of a 230 grain bullet at 830 FPS. That means if a 230 grain bullet requires .002" of sight compensation for the recoil which makes a .3 difference at 25 yards, then lighter bullets are going to make less than a 0.3" difference.

Walt, I would think that this is why you're having a hard time seeing any significant differences in your test. The maximum difference possible between zero recoil and 230 grain recoil (assuming equal momentum, like the Speer in Post #62) is .002" on the sight height, .30" at 25 yards.

If we estimate a spring compression of 7 pounds for the 200 grain at .070" slide travel, and 6 pounds for the 185 grain at .061" of slide travel, we can estimate the impact differences at 25 yards. That would mean that the 200 grain would require 7/8 x .002" = .00175" of sight compensation at 25 yards, which is .26" at 25 yards. The 185 grain would require 6/8 x .002 = .0015" of sight compensation which is .22" at 25 yards.

Summary:

If your 1911 is sighted in to be exactly on target at 25 yards with the 230 grain bullets, that means that the bore axis is aligned .30" below the Line Of Sight at 25 yards to compensate for the recoil effects. Firing the different weight bullets without changing the sights will have the following effects:

230 grain: Will hit .30" above initial bore axis alignment due to recoil effects. So they are right on target.
200 grain: Will hit .26" above initial bore axis alignment due to recoil effects. Since the sights are set for .30" above the bore axis, this will result in the 200 grain hitting .040" lower than the Line of Sight.
185 grain: Will hit .22" above initial bore axis alignment due to recoil effects. Since the sights are set for .30" above the bore axis, this will result in the 200 grain hitting .080" lower than the Line of Sight.

1/16 of an inch is .0625", easy to see why it's hard to tell any difference. I doubt that there is a gun or ammo made, much less a shooter, capable of discerning the difference.

9mm will obviously be even less of a difference, I think the answer to the OP's question is "NOT MUCH".

 

[Walt Sherrill]

(Rule 1: Do NOT laugh at the chaos of my wife's vanity.)

Rule 2: you also do Not SEE what appears to be Kilroy peeking over the edge of the mirror.

 

[JohnKSa]

Since the torque is 100 times higher, my guess is that your wrist will twist about 100 times further in the revolver simulation than in the semi simulation. That means 100 times more muzzle flip in the revolver than in the semi.

...

1/16 of an inch is .0625", easy to see why it's hard to tell any difference. I doubt that there is a gun or ammo made, much less a shooter, capable of discerning the difference.

9mm will obviously be even less of a difference, I think the answer to the OP's question is "NOT MUCH".

Very thorough analysis. Thanks for working out the math/physics.

 

[lwestatbus]

My Goodness

Well, this thread has taken on a (very interesting) life of its own.

Anyone have any more ideas about what I should do to improve my shooting with this pistol?

 

[Walt Sherrill]

Have you tried any of the things suggested in response #37? That's a start.

I suspect that a lighter hammer spring might make a big difference in results. (It did with a couple of 40XX S&Ws Iv'e owned.)

(I've had a number of 3rd Gen S&Ws, and they all had heavy (from the factory) triggers. But one of them that had been worked on by Cylinder & Slide was VERY NICE.)

Here's a link to Wolff Springs. Scroll down to find the 3913 springs.

https://www.gunsprings.com/SMITH%20&%20WESSON/NUMBERED%20SERIES/cID1/mID58/dID261#886

.