38 vs 357 point of impact. Bullet weight also...

[Dman23]

So i have a question. The last few months i have been getting into this thing called handguns. Which i am loving! I have fired a few hundred rounds through my 2 357 sp101s. (One 2 inch one 3 inch)

So i am still figuring out my point of aim. At 7 yards it doesnt seam to matter what i shoot it is dead on. 38 or 357. 130 grain or 158. But go back to 15-20 yards things change. With the ammo shortage i dont have much of a chance to "decide" on what to buy, so when deals come up i get whatever i can, which has been a mixed bag. (38 in 110-158 and 357 115-148)

Generally speaking, what will impact where? To me, being an archery man, heavier will impact lower, faster higher, more power higher ect....

I have heard so much and i know each gun will be different. But lets say generally will a 38 lighter go one place, 38 heavier will go another, 357 lighter go one and 357 heavier go one. Which will do what? And more importantly the reason behind why? I know recoil and such will play a roll and just knowing whats in the chamber will play a roll. But does anyone know generally what each will do? Love 38 for practice but will also be woods/trail guns for bears and such and will have 357s in for that so i like practicing for both defense 2 leg and defense 4 leg.

 

[Webleymkv]

Generally speaking, lighter bullets will shoot lower than heavier ones. Many fixed sight revolovers in .38 Special and .357 Magnum have their sights regulated for 158gr bullets. Such has been my experience with the following revolvers: S&W M36, S&W M64, Colt Cobra, Ruger Police Service Six, Ruger GP100, and Ruger SP101. That being said, some newer revolver have their sight regulated for lighter bullets since those have become popular for CC. My dad's Ruger LCR .357 Magnum, for example, seems to have its sights regulated for 125gr bullets.

As to .38 Special and .357 Magnum ammunition shooting to different POI's, in my experience the difference is small enough to be unnoticable out to 30 yards so long as the same weight bullets are used in both.

 

[savit260]

Heavy = higher. Ligher= lower. Slower = higher. Faster=lower.

That's been my experience with re loading.

 

[Dman23]

Why slower higher faster lower? Just seams opposite to me. I dont know exactly where mine is but it just seams interesting that i have heard that

 

[James K]

Those statements seem contrary to the experience of rifle and bow shooters.

The difference in a handgun is barrel time. A gun begins to recoil the instant the bullet begins to move, which means the gun is recoiling while the bullet is still in the barrel. In a rifle, the comparative mass of the rifle and the bullet is so great that barrel time means little.

But in a handgun, barrel time affects the point of impact. The lighter/faster bullet gets out of the barrel before the gun recoils very much, so it strikes lower. The heavier/slower bullet is later getting out of the barrel and so the gun can recoil upward more and the bullet strikes higher.

Jim

 

[savit260]

But in a handgun, barrel time affects the point of impact. The lighter/faster bullet gets out of the barrel before the gun recoils very much, so it strikes lower. The heavier/slower bullet is later getting out of the barrel and so the gun can recoil upward more and the bullet strikes higher.

Yup, right on the money.

 

[JohnKSa]

The difference in a handgun is barrel time. A gun begins to recoil the instant the bullet begins to move, which means the gun is recoiling while the bullet is still in the barrel. In a rifle, the comparative mass of the rifle and the bullet is so great that barrel time means little.

In addition, rifles tend to recoil straight back without a lot of muzzle rise because the stock transfers the recoil to the shooter's shoulder which is pretty close to being right behind the muzzle.

In a handgun, the shooter's hand is almost always considerably lower than the muzzle so the recoil tends to result in muzzle rise. The more the rise before the bullet exits, the higher the point of impact.

 

[Super Sneaky Steve]

The weapon matters too.

Today I was having fun with my 6" GP100. Both the Magnum and .38 loads had the same POI.

With my SP101 which is smaller and lighter there's a noticable difference.

 

[James K]

The main factor is mass. When recoil begins, the greater mass of the rifle means greater inertia, which means it takes longer for the rifle to begin detectable recoil. Since velocity times mass of the rifle equals velocity times mass of the bullet, the higher velocity and lower relative mass of the bullet means it is usually out of the barrel before the inertia of the rifle is overcome and recoil begins. So the recoil movement of the rifle will have no effect on the bullet point of impact.

Jim

 

[Blue1]

Excellent explanations, I would have thought the opposite, but this makes perfect sense.

Thanks to all for my continuing education.

Blue1

 

[JohnKSa]

When recoil begins, the greater mass of the rifle means greater inertia, which means it takes longer for the rifle to begin detectable recoil.

Given that the momentum (the quantity that is proportional to recoil) of a typical rifle loading far exceeds the momentum of a typical pistol loading, and given that the typical rifle barrel length far exceeds the typical pistol barrel length, my gut feel is that it all pretty much cancels out.

I'll run some numbers and post the results.

Ok, here are some rough estimates.

Firearm, Recoil velocity, Barrel Dwell Time
Beretta 92FS, 9.2fps, 0.6ms
1911, 11.7fps, 0.8ms
AR-15, 3.9fps, 0.9ms
Remington 700, 8.0fps, 1.1ms

The .223 comes in with a pretty low recoil velocity, but the rest of the firearms have roughly comparable recoil velocity (8, 9.2 and 11.7fps).

The barrel dwell time is pretty similar across the board (0.6, 0.8, 0.9 and 1.1ms). So the rifles have a little less recoil velocity but a little longer barrel dwell time. The two effects should help to cancel each other somewhat.

So given that barrel dwell time and recoil velocity are pretty similar, at least between the very limited sample in my quick & dirty calculations, it would seem that muzzle rise is the best candidate for why POI changes more significantly with bullet weight in handguns than it does in rifles.

That makes sense when one notes that the POI change also tends to be more pronounced in revolvers than in autopistols. Again, it's muzzle rise that makes the difference. While they both begin recoiling as soon as the bullet begins to move, in the autopistol, that early recoil motion, while the bullet is still in the bore, is slide/barrel movement. Muzzle rise begins primarily when the slide bottoms out against the frame at it's rearward travel, long after the bullet has left the barrel.

However, recoil in a revolver tends to result in immediate muzzle rise.

Therefore, there tends to be less difference on the target between different bullet weights when shooting autopistols than when shooting revolvers.

 

[James K]

I agree except on the difference between the revolver and the auto pistol. If you lay a ruler across the sights of, say, a Model 10 revolver, you will see that the barrel is pointing down from the line of sight. If the shooter aims at a target at his own eye level, he hits the target because as the bullet moves down the barrel, the recoil causes the muzzle to rise until at bullet exit the barrel is horizontal, pointing at the target.

The same is true of the auto pistol. The slide masks it, but the barrel of an M1911 pistol is pointing downward just like the revolver. When the bullet begins to move, the barrel and slide are locked together and recoil together. As the bullet moves down the barrel, the barrel and slide recoil upward until at bullet exit the barrel is pointing at the target. Not because the barrel has begun to drop in relation to the slide but because the barrel and slide have moved to a horizontal position.

In fact, the barrel will not unlock from the slide until long after (relatively speaking) the bullet has left the barrel. That is the whole idea of a locked breech pistol, that the barrel and slide are locked together until the bullet exits and the pressure drops.

Here is an x-ray photo of an M1911 pistol being fired. The bullet is almost at the exit point, yet the barrel is still locked to the slide and the unlocking movement has just begun. So the slide impacting the frame has no effect at all on the position of the gun at bullet exit.


http://www.google.com/imgres?imgurl...yGqnI0QHLvYHQDw&sqi=2&ved=0CDoQ9QEwAg&dur=364

Jim

 

[JohnKSa]

So the slide impacting the frame has no effect at all on the position of the gun at bullet exit.

I agree 100%. That's actually what I was trying to say, but I must have botched my explanation.

The recoil motion of an autopistol prior to bullet exit is nearly exclusively slide/barrel motion and therefore has very little muzzle rise component as a result.

Ok, since I apparently screwed up the explanation, here's a picture to make it clear.

I lined up the sights of a 4"bbl GP100 revolver with a ruler and drew a line parallel to the sights. Then I put a rod down the bore and drew a line that matched the line of the bore.

I repeated the exercise with a 9mm Caracal F autopistol.

It can be easily seen that what Jim said is correct. The boreline of the revolver is obviously angled down compared to the line of the sights. That's to compensate for the recoil/muzzle flip that will take place before the bullet exits the muzzle.

However, the slide/barrel initially move virtually horizontally in recoil in the autopistol and there's very little muzzle rise until the slide hits the frame. So while the recoil starts as soon as the bullet begins to move, the recoil is primarily in a straight line backwards because only the slide and barrel move appreciably at first. Significant muzzle rise in a floating barrel, locked breech autopistol happens long after the bullet is gone. That means that the boreline and the sight line of the autopistol can be essentially parallel since there's no need to compensate for muzzle rise before bullet exit like there is in the revolver.

In fact, the bore of the autopistol actually angles slightly upward with respect to the sights although it's not really possible to see it in the picture. That upward angle is what brings the bullet up to the sight line. That's necessary because the bullet starts out below the sights since the muzzle is below the sights.

 

[DaleA]

Dman23 and other new folks be a little wary. This topic has and can spark controversy and arguements.

My experience is nowhere near that of the staff here but it matches just what they have said both from a practical shooting point of view and a (limited) knowledge of physics.

In other words IMhO you can take what the John's here have said to the bank and I'll add my thanks for once again answering a question and providing even more and clearer details about what is going on.

 

[James K]

Recoil on any gun is around its center of gravity (and partly around the CG of the shooter/gun system). That is what causes the muzzle to rise. Recoil comes straight back, in line with the barrel. But the barrel is higher than the CG of the gun, so the straight-back tangential force is translated into a circular motion around the CG. If the barrel is on the CG, as in some guns, there is no muzzle rise, only a straight back push.

I don't have a Caracal pistol, but it will obey the laws of physics just like any other pistol. While the bullet is moving out the barrel, the recoil, which started at the same time as the bullet, is moving the whole gun, barrel, slide and all. That is independent of what the barrel and slide are doing at the time.

For any pistol of significant recoil, the bore line and sight line cannot be parallel. There is an exception for pistols of small caliber and low recoil. Since the barrel is lower than the sights, the designer depends on the muzzle rise to bring the barrel in line with the target as the gun recoils. Because recoil is low, there is no need to start with the barrel pointing downward.

Jim

 

[Dman23]

Okay that is what i heard, and what i thought. I just never really saw a good subject on the matter. And while all that makes sense, i wondered why that in a pistol it would be that way and not lets say a high powered rifle.

I really enjoy all this. Very interesting! And hopefully someday ammo will be stocked in huge numbers and i will be able to find out what works for me and stick with it. In the mean time, i guess its a good way to force me to experiment with different loads.

Btw, my 2 sp101s i LOVE!!!!!! Just getting used to where each shoot and with what loads. But they sure are built like a tank

 

[JohnKSa]

I don't have a Caracal pistol, but it will obey the laws of physics just like any other pistol. While the bullet is moving out the barrel, the recoil, which started at the same time as the bullet, is moving the whole gun, barrel, slide and all.

The Caracal is quite similar to a Glock. It's a conventional 9mm locked breech autopistol--just happened to be the first one I put my hands on when I opened the safe.

The recoil acts directly on the breech. In a locked breech autopistol, the breech is part of the slide, not part of the frame. The slide is coupled to the frame (in terms of the direction of recoil) only via the recoil spring until the slide bottoms out at the end of travel. That means that only minimal recoil force is transferred to the frame (via the recoil spring) before the end of slide travel when the slide hits the frame.

As a consequence, there is no significant muzzle rise until the end of slide travel. Muzzle rise is the effect of the recoil force being applied above the point of resistance and since there is essentially no effective resistance applied to the recoil force until the slide bottoms out, there is no significant muzzle rise until that point.

To confirm this, look at the bore & sight lines in the diagram from the autopistol. There is clearly no significant downward offset to the bore to compensate for muzzle flip as would be absolutely necessary if there were any significant muzzle flip prior to bullet exit.

I did the same measurements and diagrams for 2 additional locked-breech autopistols which are more common, the Ruger P89 and the CZ-75B to eliminate the possibility that the Caracal is somehow unique.

Note that in both cases that the bore line actually angles slightly upward in relation to the sight line. A clear departure from what is plain in the diagram made using the revolver.

 

[James K]

"Muzzle rise is the effect of the recoil force being applied above the point of resistance and since there is essentially no effective resistance applied to the recoil force until the slide bottoms out, there is no significant muzzle rise until that point."

That is what is not being understood. Recoil is the "equal and opposite" force resulting from the movement of the bullet in the opposite direction. It doesn't depend on the slide hitting or bottoming out on anything. That would be like saying that recoil of a rifle depends on the butt striking the shooter's shoulder.

The amount of muzzle rise before bullet exit depends on the caliber, bullet mass, barrel time, and other factors, but it is there if the barrel is above the center of gravity of the gun. Even a muzzle brake won't negate it because the brake depends on using the gas that escapes after the bullet exits to slow recoil. Will more of the recoil be seen after bullet exit? Of course, since the motion of the gun has begun and is not easily stopped.

Jim

 

[JohnKSa]

That is what is not being understood. Recoil is the "equal and opposite" force resulting from the movement of the bullet in the opposite direction. It doesn't depend on the slide hitting or bottoming out on anything.

That is 100% correct and I have not said anything that contradicts those statements. RECOIL doesn't require that there be any hitting/bottom out.

The problem is that one can not equate muzzle rise and recoil. They are not the same thing. Muzzle rise is an effect of recoil under certain circumstances, namely "recoil force being applied above the point of resistance".

The recoil begins as soon as the bullet moves--that is an immutable law of physics. Recoil takes the form of force applied to the breechface which is part of the slide. The force is applied along the axis of the bore--perpendicular to the breechface. The slide moves in the direction that the force is applied (straight back) because there is initially nothing to prevent it from moving straight back.

What we're talking about is muzzle rise, not just recoil. THAT is why it makes a difference when the slide hits something. Until the slide is somehow prevented from moving straight backwards, there is no muzzle rise. The recoil just moves the slide straight back since the recoil force was applied straight back and there is no significant force acting on the slide to change its direction or to cause it to torque upward.

That is another immutable law of physics. An object in motion stays in motion and continues along a straight path until something acts on it to slow it or change its direction.

Straight back motion of the slide does not cause muzzle rise.

BUT, when the slide hits the frame, the frame rotates upwards due to the fact that the force of the slide impact is applied to the frame above the shooter's hand. THAT is when muzzle rise occurs in a semi-auto.

Since that happens LONG after the bullet has left the barrel, there is no need for the sights and the boreline to be offset to account for muzzle rise as there is in a revolver. That is exactly what we should expect since the diagrams created from the 3 autopistols show that there is clearly no compensation in the sights for muzzle rise. In other words, the diagrams show exactly what the physics tell us should be true.

In a revolver, the recoil is also applied straight back into the breechface. However, as soon as the breechface begins to move, the gun is IMMEDIATELY affected by the shooter's hand. There is no slide, and the breechface, instead of being attached to a movable part, is fixed to the frame of the gun. The frame of the gun is being held in place by the shooter's hand. Since that obstruction is below the bore, the gun torques upward and the muzzle rises. So, in a revolver, muzzle rise begins at the same time that the recoil force begins moving the gun.

In an autopistol, muzzle rise begins when the slide/breechface can no longer move straight back--when it hits the frame--long after the bullet has exited. Then the shooter's hand obstructs the motion of the frame--the gun torques upward and the muzzle rises.

It's not really possible to argue that there IS significant muzzle flip in a locked-breech autopistol before the bullet leaves the bore given the diagrams provided which clearly show that there is no sight compensation for muzzle rise in any of the 3 different locked breech autopistols tested.

If there were muzzle rise before bullet exit, the bores would have to be angled down with respect to the sights, just as the revolver bore is. In fact, they are actually angled slightly UPWARD. That's not easy to see in the Caracal diagram, but it is plainly visible in the other two diagrams.

That alone provides hard evidence that there is clearly something very different going on in an autopistol.

The amount of muzzle rise before bullet exit depends on the caliber, bullet mass, barrel time, and other factors, but it is there if the barrel is above the center of gravity of the gun.

That's only true if muzzle rise begins with the bullet still in the bore. As I've explained, and as the diagrams show, significant muzzle rise in a locked breech autopistol does not occur until well after the bullet leaves the bore.

Here's a cleaned up version of the diagram with all 4 handguns on the same page.

 

[James K]

John, the problem is that recoil is not "... the form of force applied to the breechface which is part of the slide". Recoil is a force along the barrel because that is the direction the bullet is moving. The gun does not recoil because there is a "push" on the breechface, it recoils because the bullet is moving and there is an equal and opposite force in the other direction. No matter how much pressure is exerted on the breech face, if the bullet doesn't move, there is no recoil and no movement of the gun or the slide.

If the barrel is blocked so the bullet cannot move, the gun will not recoil and the slide won't move.

As to how much the gun moves in recoil while the bullet is still in the barrel, that depends on bullet velocity, bullet mass, gun mass, powder mass, and maybe other factors. If the gun will move very little, the maker may not feel a need to adjust the sights to compensate. The M1911 may not be a good example because of the high recoil due to bullet mass, but it does show a case of the barrel pointing down. (The barrel is not pointed down to compensate for recoil - the barrel is that way because the design dictates it. The sights compensate for that part of the recoil that takes place before the bullet leaves the barrel.)

But muzzle flip is not caused by the barrel or slide being stopped by the frame, and if it were it would have no effect on the bullet which is long gone by the time the slide stops.

I confess your drawings are puzzling, so there is some other factor involved. If we agree that the revolver is sighted to compensate for recoil while the bullet is in the barrel and we agree that that recoil causes the muzzle to rise, then those diagrams would show that the muzzle moves down in recoil, which I seriously doubt.