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

[ShootistPRS]

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.

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.

[Walt Sherrill]

Originally Posted by ShootistPRS

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.

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.
ShootistPRS is online now Report Post

Do you agree that the key relationship of the components -- barrel, slide, frame, bullet, and recoil spring are all fixed in their relationship (i.e., how they move with regard to each other) until the bullet leaves the barrel?

IF not, tell us what changes in those relationships due to different bullet weights or speeds.

IF they are fixed, the amount that the recoil spring is compressed when the slide and barrel have move that trivial 1/10th of an inch and the bullet leaves -- and the amount of forces pressed to the frame -- is the same regardless of bullet weight or speed. If they aren't the same, tell us why they're not the same. We're talking about the physical relationships and slide travel, not the TIME involved.

Nobody has claimed that the recoil has no effect in a Browning Short-Recoil Locked-Breech gun. The claim is that its effect is no different with heavier/slower bullets than with faster/lighter bullet UNTIL after the bullet is out of the barrel.

Recoil's transfer to the frame is based on slide movement. And even if it's being passed faster or slower, the slide is moving the same amount until the bullet is gone. The total recoil effect on the frame is the same -- until later.

If the slide moves the same distance to the rear, regardless of bullet weight or speed --and slide movement is what transfer recoil to the frame -- what other force is at play that causes the barrel and slide to rise when a heavier and slower load is used that isn't caused by slide movement down the frame?

[Walt Sherrill]

Originally Posted by mavracer

Ultimately yes you are because it get's to the shooter thru the frame.
One of the biggest reasons you see less difference in POI when changing bullet weights in a semi auto is bore axis above the pivot point (web of your hand).

And 45_auto is absolutely correct about the firing pin retainer and it's radius being an important factor.

Ultimately we're talking about how recoil affects barrel rise -- and why it might rise MORE BEFORE the bullet has left the barrel when the bullet is heavier/slower, or lighter/faster.

With a 1911, the firing pin retaining plate and its radius at the rear of the frame has moved only 1/10th of an inch to the rea when the front of the slide has move that same 1/10th of an inch. That is force applied to the frame, to be sure, but like the recoil spring, the same amount of slide travel occurs regardless of bullet weight or bullet speed. As noted in other responses, the distance the bullet, slide, barrel, frame and recoil spring move is a fixed relationship.

We can add the hammer-spring and hammer to the discussion, but it still doesn't change anything. If we switch to a striker-fired gun, it won't change anything, either. Ditto a linkless Browning SRLB design.

SLIDE movement (the distance traveled) when the bullet leaves the barrel is the key factor, and it's the same, regardless of bullet weight or speed. How fast all of this happens is different, but not how far things move.

As I noted in an earlier response, nobody is claiming that recoil isn't a factor with the Browning SRLB. What some of us are claiming is that with Brownging SRLB design, a light/fast bullet and a slow/heavy bullet the moves the same amount until about the bullet is gone. THEN, things change.

What would cause the slide/barrel to rise more, given the same (FIXED) amount of slide travel when a heavier, slower bullet is used than when a faster/lighter bullet is used? I've been led to believe that the bullet always leaves the barrel at the same point of slide travel. Is that not right?

[ShootistPRS]

Do you agree that the key relationship of the components -- barrel, slide, frame, bullet, and recoil spring are all fixed in their relationship (i.e., how they move with regard to each other) until the bullet leaves the barrel?

IF not, tell us what changes in those relationships due to different bullet weights or speeds.

IF they are fixed, the amount that the recoil spring is compressed when the slide and barrel have move that trivial 1/10th of an inch and the bullet leaves -- and the amount of forces pressed to the frame -- is the same regardless of bullet weight or speed. If they aren't the same, tell us why they're not the same. We're talking about the physical relationships and slide travel, not the TIME involved.

Nobody has claimed that the recoil has no effect in a Browning Short-Recoil Locked-Breech gun. The claim is that its effect is no different with heavier/slower bullets than with faster/lighter bullet UNTIL after the bullet is out of the barrel.

Recoil's transfer to the frame is based on slide movement. And even if it's being passed faster or slower, the slide is moving the same amount until the bullet is gone. The total recoil effect on the frame is the same -- until later.

If the slide moves the same distance to the rear, regardless of bullet weight or speed --and slide movement is what transfer recoil to the frame -- what other force is at play that causes the barrel and slide to rise when a heavier and slower load is used that isn't caused by slide movement down the frame?

Once the trigger is pulled and the cartridge ignites there is no fixed relationship. The bullet is moving in the barrel, barrel moves (slightly) in relation to the slide and both the barrel and the slide move in relation to the frame under the recoil of the bullet and the spring is being compressed. It is, in fact a very dynamic relationship.

A heavier bullet has more mass so it imparts more movement to the parts of the gun and does so for a longer (slightly) time. (barrel time)

Clear your semi-auto and move the slide a tenth of an inch to the rear. While holding it there check the amount of vertical play in your barrel. (at both the muzzle and the breech)

After the bullet leaves the barrel it no longer has any effect on the gun at all. All the motion is imparted to the gun while the bullet is traveling down the bore.

The speed of the slides rearward movement is due to the velocity and mass of the bullet. (momentum conservation) The more momentum the bullet has, the more the slide moves in a given time.

The distance the slide moves before disengaging from the barrel is always close to the same but the time it takes to get there changes with the amount of momentum exchanged with the bullet.

The question you need to answer is, "At what point does the barrel (its locking lugs) begin to move downward away from the slide?" You will find that it is less than the time the bullet is in the barrel. You continue to bring up that the distance is the same but forget the the time is different. The time is determined by the weight and velocity of the bullet.
A car going 200 miles at 50 mph will be on the road 4 hours to get there but if you go 100 miles per hour you are on the same road for only 2 hours. The faster bullet gets out of the barrel faster and has less effect than the heavier slower bullet. The distance remains the same but the time changes.

[Walt Sherrill]

Originally Posted by ShootistPRS

The question you need to answer is, "At what point does the barrel (its locking lugs) begin to move downward away from the slide?" You will find that it is less than the time the bullet is in the barrel. You continue to bring up that the distance is the same but forget the the time is different. The time is determined by the weight and velocity of the bullet.

Does the barrel of a Browning Short Recoil Locked Breech (SRLB) weapon unlock at a different point in the bullet's travel in the barrel when you use use a heavy bullet (moving slowly) than when you use a light bullet (moving more quickly)?

If it doesn't, what else can cause the barrel and slide rise when using a heavier/slower bullet before the bullet leaves the barrel which is before the slide has traveled 1/10th of an inch. Is there some way for recoil force to be transferred to the frame rather than through slide movement?

It would seem that if the bullet is moving more slowly, the barrel rise would also be happening more slowly. WHY is such a small amount of rearward movement causing what seems to be a disproportionate amount of barrel rise?

What would happen if two different loads, one with a heavy bullet and one a lighter one, both send their bullets down the barrel at the same velocity. How does the gun behave differently, then?

I've always understood that the process that leads to the bullet leaving the barrel in a Browning SRLB gun is a fixed one. While the time required for the bullet to move down the barrel can be different, the distance it travels and how the bullet physically relates to the other affected components (barrel, slide, frame, compressing springs) are all part of a fixed relationship based on bullet and barrel/slide movement. Barrel and slide movement seem to be the only way that recoil forces can be transferred to the frame.

If the barrel does unlock differently for a heavy/slow bullet than for a fast/light one, please explain why that is so. If it doesn't unlock differently (which means there can be no different barrel rise due to the barrel unlocking), what else is causing barrel rise before the bullet has left the barrel? Do you feel that the physical movement of all of the key components are NOT part of a fixed relationship until the bullet is gone? I agree that the time it takes for the firing cycle to complete itself will be different depending on the load and how fast the slide moves, but I continue to argue that the way the components move with regard to each other doesn't change when the do it more or less quickly..

Said in a different way, the speed of the process isn't the controlling factor -- instead it's the LOCATION of the different components, which are part of a fixed, predictable relationship. Nothing about that series of physical relationships changes when things are speeded up or slowed down. The components all move in the same way and travel the same distances -- they just do it at different speeds when the loads used are different.

Does any want to say that lighter/faster bullets leave at some point before the barrel unlocks while heavier/slower bullets leave after barrel/slide separation has begun. Is that the explanation for barrel rise? I have been led to believe recoil force being applied to the frame (as is the case with a fixed-barrel gun) is what causes muzzle rise.

High speed videos of Browning SRLB guns being fired show the bullet leaving the barrel after the slide and barrel have moved a small fraction of an inch. The barrel remains level and the bullet exits horizontally. The bullet path and the slide/barrel movement seem the same regardless of bullet weight, caliber, barrel length, etc. The very high-speed videos (73,000 frames per second) show no muzzle rise until AFTER the bullet is gone and the slide has moved farther to the rear. This can clearly be seen in the links in earlier messages.

IF slide movement is the way recoil force is transferred to the frame, and all bullet weights and bullet speeds cause the same amount of slide movement UNTIL THE BULLET HAS LEFT THE BARREL, it would seem that recoil isn't playing a big role in barrel rise when it matters -- while the bullet is still in the barrel.

[JohnKSa]

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.

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.

You've posted this more than once and I've already responded to it.

Rather than repeating it again, please consider the sightline vs. boreline diagram posted earlier and explain how significant muzzle rise while the bullet is in the bore could take place given what the diagram shows.

[mavracer]

The very high-speed videos (73,000 frames per second) show no muzzle rise until AFTER the bullet is gone and the slide has moved farther to the rear.

Look again closely, keeping in mind that with a 5" barrel every .005 is nearly an inch (.9 to be exact) at 25 yards.

Let's look at your theory in reverse. Once the bullet leaves the barrel it has no more input on recoil, the gun has already reached it's maximum recoil velocity if it weren't already starting to rise it never would

[Walt Sherrill]

Originally Posted by JohnKsa

Rather than repeating it again, please consider the sightline vs. boreline diagram posted earlier and explain how significant muzzle rise while the bullet is in the bore could take place given what the diagram shows.

I suspect the boreline and sightlines for most handguns are set by the gun makers for bullet impact at 20 yards. That distance also seems to be the sweet spot used in most of the ballistic tables cited by ammo makers. I've never noticed that much difference in bullet impact points at shorter distances regardless of bullet weight or speed. My tests at much closer distances didn't show any.

Bullets that must go farther would force sight adjustments -- which changes the boreline/sightline relationship. But what we see there, arguably, is the effect of the bullet's travel over time -- gravity is constant, regardless of bullet weight. That means that a slower bullet will drop more at a given distance than a faster bullet, and you've got to adjust your point of aim to compensate for that time-related variable.

Originally Posted by ShootistPRS

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.

I guess I'm beating a dead horse, but...

You seem to be saying that a heavier/slower bullet causes a different amount of vertical movement of the barrel than a lighter/faster bullet. What causes that to happen?

As far as I can tell, the movement of the bullet, barrel and slide on the frame remain n a FIXED RELATIONSHIP until the bullet is gone and the barrel and slide are fully unlocked. The speed with which these components move can change, but not HOW they move with regard to each other.

I think that for the barrel to unlock VERTICALLY in a different way when a heavier/slower bullet is used would mean the heavier bullet has traveled farther down the barrel during the firing cycle than a faster/lighter bullet would have traveled. That seems counter-intuitive. That says there is NOT a fixed relationship between the key components prior to the bullet exit from the barrel. Is that what you believe?

.

[45_auto]

Originally Posted by Walt Sherrill

You don't need a Ransom Rest to offer evidence to prove your claims. You don't need to fire thousands of rounds. You just need the same sort of rest you'd use when sighting in a gun -- a sandbag or bag of beans or rice -- a target, a variety of bullet weights, and a slow, steady press on the trigger as each shot is fired. Firing all of the bullets from the same gun without changes to the sights, and using the same point of aim makes for an easy test. If you use three or four bullet weights -- maybe 5 or 10 rounds each --means you'd shoot less than one box of ammo.

You must be joking. How close do you think you can locate the actual center of your population from 5 or 10 rounds?

Where is the population center based on this sample of five shots? Is it the center of the five holes, or do those 5 holes just happen to be above or below or to one side of the true center of the population? The more shots you fire, the more accurately you can determine that center.



How much difference are you expecting?

Let's say the difference is .25" at 25 yards.

Let's also say your pistol/shooter combination is good for 2.5" groups at 25 yards.

A quick estimation of how many shots you would need to locate the group center within a given error at a 95% confidence level can be approximated by 1/B^^2, where B is the error bound. Quick explanation here:

The sample size is an important feature of any empirical study in which the goal is to make inferences about a population from a sample.

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

If we want to locate your group center within plus or minus .25", which is 10% of the 2.5" group you are shooting, you would need 1/(.10)^^2 = 1/.01 = 100 shots.

So if the bullet weight caused a .25" difference, and 100 shots only allows you to locate your group within plus or minus .25", that means that given 200 shots (100 shots of each bullet weight) you can't tell any meaningful difference between them.

If you want to locate the center within .10", which is 4% of your 2.5" group, then it will take 1/(.04)^^2 = 1/.0016 = 625 shots per group.

At that resolution of .10" on group center, even if your upper group center was off .10" inch low, and your lower group center was off .10" high, you would still have .050" between the centers of your groups if there was a .25" difference, and it will only take 1,250 shots.

You must have some BIG boxes of ammo to do it in less than a box!

I'll take another shot at simplifying the physics later if I get a chance.

[Walt Sherrill]

Originally Posted by 45_Auto

I'll take another shot at simplifying the physics later if I get a chance.

I wish you WOULD simplify the physics. But you may have to do some other explaining, as well.

Re: your graphic. It shows us that it's VERY POSSIBLE to misinterpret results, but I'm not sure it's a meaningful proof for this discussion. Another participant here who teaches statistics and analysis seemed to have no problem coming to a conclusion about bullet rise vs bullet weight using far fewer rounds. Maybe he needs to re-examine his findings and how he teaches statistical analysis to his students?

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?

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.
  
• 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.
  
• 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.
  

This basic set of assumptions seem to be confirmed by the comments of many people considered experts on the subject. As I interpret them, these assumptions also seem to be confirm by observations, by simple tests (at least one of which was analyzed by a person familiar with statistical practices), and by high-speed vidoes. It appears that barrel tilt is the same UNTIL AFTER the bullet leaves the barrel. The point is that a very, very minor amount of barrel rise can have a big impact farther down range. Perhaps so, but farther down range many other factors can also affect bullet placement -- not the least of them being gravity. Can it have a large effect at typical handgun (self-defense) ranges?

It seems to be generally accepted that a .45, bullet exits the barrel before the barrel and slide have moved a bit less than 1/10th of an inch. In the 1911s I've owned, it always seemed that the barrel had to move a bit more than that before the link began to pull the barrel down. But that's hard to measure or verify if your not a gun builder or gunsmith.

But, even IF the link can come into play before the barrel has traveled that far, even before the bullet leaves the barrel, I would expect the link to pull the barrel down the same way/distance regardless of bullet weight or bullet speed -- because the link's action is dependent on slide/barrel movement with regard to the frame (to which the bottom of the link is attached). It's not about slide or bullet speed, but about the physical slide movement, and what I consider a FIXED relationship between the various parts as they move.

For a heavier/slower bullet to cause more barrel rise, it would seem to mean that the supposedly FIXED relationship of barrel/slide, bullet, and frame has to become become one that isn't fixed. It would seem to mean that the heavier/slower bullet has traveled less far when the the barrel begins to unlock from the slide than would be case with a lighter/faster bullet.

ShootistPRS mentioned different bullet momentum due to a heavier bullet, and also said that recoil was transferred in a subtly different manner when the barrel unlocks as a heavier/slower bullets is used.

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.

If any part of my assumptions above are wrong, show me why -- correct my misunderstandings about Browning SRLB function -- and we'll go from there.

.

[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]

Originally Posted by Walt Sherrill

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.

Originally Posted by Walt Sherrill

- 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.

Originally Posted by Walt Sherrill

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.

Originally Posted by Walt Sherrill

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:

Originally Posted by 45_auto

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.

Originally Posted by 45_auto

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.

Originally Posted by 45_auto

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.

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[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]

Originally Posted by mavracer

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.)

Originally Posted by mavracer

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]

Originally Posted by mavracer

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]

Originally Posted by 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.

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?

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[ShootistPRS]

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

Originally Posted by ShootistPRS

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:

Originally Posted by ShootistPRS

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:

Originally Posted by ShootistPRS

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 [U]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.