Barrel to cylinder gap and endshake.

[Mizar]

Hello,
Several times, on this forum and other places, I have read the following statement concerning correction of excessive endshake: "If you adjust your revolver's endshake to minimum spec. you will also set it's B/C gap to the largest measure (you gonna increase the gap)". Like this is something that can be avoided by simply NOT correcting the endshake. I was even explained once that the cylinder, right after ignition, is only dragged forward by the speeding bullet. Well, today I remembered a slow-motion video on YouTube that showed very clearly this exact condition - you can start watching from 0.30 min. mark.



Watch closely the lower back part of the cylinder - the gun fires, bullet drags cylinder forward, but at the instant the bullet clears the cylinder the pressure generated by escaping gasses pushes the cylinder all the way to the back where it stays until the bullet leaves the barrel - i.e., until the pressure drops. So, when it counts - firing - your revolver's B/C gap is always at it's biggest size.

Best,
Boris

 

[DDeegs]

Neat

Dan

 

[Driftwood Johnson]

Howdy

What I notice, more than the movement of the cylinder is quite a lot of gas comes out of the barrel before the bullet does, and you can plainly see the muzzle is starting to rise before the bullet exits the barrel. That's why the front sight on most revolvers is so tall.

 

[GBExpat]

bullet drags cylinder forward

That forward movement is from the hammer strike.

 

[Mizar]

That forward movement is from the hammer strike.

Not only. It's actually combined - hammer strike, bullet dragging (if said bullet is not undersized) and gases expanding and stretching the fired case

...and you can plainly see the muzzle is starting to rise before the bullet exits the barrel...

Driftwood, every firearm does this. And you cannot do nothing about it, because recoil comes from the moving bullet and not from the hot gases. Several years ago 1911Tuner did explained it in a tread discussing 1911 style pistols, but this is valid for every single firearm.

Best,
Boris

 

[GBExpat]

Not only. It's actually combined - hammer strike, bullet dragging (if said bullet is not undersized) and gases expanding and stretching the fired case

Seems to me that for that to be the case, you would first have to toggle off Newtonian mechanics.

 

[Mizar]

Would you please explain your statement in details, because I can't see how my explanation is in conflict with Mr. Newton?

 

[Ironicaintit]

Though the cylinder moves forward with the hammer strike, It looks to me like the cylinder is driven to the rear, immediately upon ignition. Which makes sense to me.

To say that the bullet is dragging the cylinder forward, forgets the fact that the gases are pushing the cylinder and bullet violently apart.

The forces driving the bullet forward, are exactly equal to those driving the cylinder to the rear.

 

[Ironicaintit]

I am surprised how much gas escaped the front of the barrel before the bullet does.
Too small a bullet for the bore, maybe?

 

[Driftwood Johnson]

Too small a bullet for the bore, maybe?

I don't think so. I have seen the same thing with other slow motion videos of revolvers. Perhaps not quite so much gas, but I recall a video of a Schofield reproduction being fired and some gas came out before the bullet.

 

[Vern Humphrey]

"Though the cylinder moves forward with the hammer strike, It looks to me like the cylinder is driven to the rear, immediately upon ignition. Which makes sense to me.

To say that the bullet is dragging the cylinder forward, forgets the fact that the gases are pushing the cylinder and bullet violently apart.

The forces driving the bullet forward, are exactly equal to those driving the cylinder to the rear".

Exactly right. Another factor is primer ignition. The primer moves backward upon ignition, and the cartridge moves slightly forward. But when the powder ignites, the case is driven back, re-seating the primer.

 

[Mizar]

Guys, let me explain why you are wrong in this case. But first, let's get back to the video, because it's one of the best I have seen showing this exact phenomenon we are discussing. Forgive me if you already know this, but I will explain some settings that are really helpful - when you start the video (put it on full screen) there is a menu bar at the bottom of the screen. Search for the "settings" button - you can select the quality of the video and what is more important - the speed of replay. Select 0.25 - slowest possible. With pressing "space" bar you can put the video on "pause/play" frame by frame. Now start from 0.30 min. mark watching closely. What you will see, in horological order:

Hammer strikes the primer, while the cylinder is in relatively still position you can see the case head moving rearwards (to the left) and pressing against the breach face. Moving AWAY from the cylinder. The cylinder starts to move forward (to the right) while the case head stays in contact with breach face. As this cylinder movement is stopped by the front portion of the frame you can witness how the frame is stretching - look at the frame's locking latch. This indicates that the forces are quite violent and the forward movement is not simply caused by the hammer strike. As the bullet clears the cylinder fully and enters the barrel, the expanding gasses rush thru the gap and push the cylinder all the way to the back.

Why this is happening? When a cartridge is ignited, gases expand the case walls and they are gripping the chamber walls thus locking both in place. But the case head is solid and cannot expand and the bullet is effectively blocking the expanding gasses from escaping. The case head, forced by the gasses, is pushed away from the cylinder thus stretching the case. This is why one can witness a case head separation in revolvers with excessive endshake, especially on reloaded brass. No oversized chambers, no rough chamber walls, no over charged case - there is simply too much room for the case to stretch. Back on our discussion - when the case head reaches the breach face it is stopped there, can't move more, so it starts to push on the cylinder - to get out of it's way because, at the same time, a speeding bullet is still blocking the exit for the gasses and they have to expand somewhere. Oversized, or undersized bullet - drag, or no drag, the cylinder is pushed forward until it stops. Watch the video closely - everything we are discussing is there. Mr. Newton's laws are not violated, in fact this is by the book demonstration of it's third law.

Best,
Boris

 

[Driftwood Johnson]

I am surprised how much gas escaped the front of the barrel before the bullet does.
Too small a bullet for the bore, maybe?

Just watched a really good video from myth busters. Quite a lot of smoke exiting the barrel before the bullet leaves. Not a revolver, but very interesting anyway

 

[Ironicaintit]

I'd never thought about it before, but I bet it's the gas that gets passed, prior to the bullet fully engaging the barrel/rifling. Makes perfect sense, and seems obvious in hindsight.

 

[ojh]

because recoil comes from the moving bullet and not from the hot gases.

There's no interaction between the bullet and the gun that would push the gun backward. The only interaction between the bullet and the gun is friction. When the bullet is moving forward it tries to pull the gun to the same direction. Pressure of the gases pushes the case against the breechface and thus the whole gun backward. Therefore one can say that the recoil indeed comes from the hot gases.

Another way of looking things is conservation of momentum and in that context it is OK to say that recoil (gun moving backward) comes from both the bullet and hot gases moving forward.

 

[Mizar]

ojh, this was discussed in details not so long ago and it's not the purpose of this thread.

 

[Vern Humphrey]

There's no interaction between the bullet and the gun that would push the gun backward.

Yes, there is. It's called "recoil."

As soon as the bullet begins to move, a recoil impulse is transmitted to the gun. In the case of a revolver, the recoil impulse is transmitted first to the cylinder.

 

[Mizar]

On a second thought, recoil has lots to do here... A question - if a revolver cylinder is not pushed forward first, then why do break action shotguns have locking lugs? After all, the barrel will be pushed straight into the breach face, why do you need the lugs for? Make an experiment, anyone willing to - remove locking lugs and charge a shell with a slug slightly loose fitting in the barrel - mimicking a revolver cartridge minus the rifling. No over powder wads, just powder and a slug. Cylinder barrel, no chokes. Ill be some 50 yards away watching the fireworks...

And a question for ojh: if bullet weight has no effect on recoil than why you must grind of the locking lugs on a locked breach recoil operated gun, converting it to straight blowback, in order to shoot blank ammo? You must put a barrel constriction also in order to raise the pressure.You can plug the barrel entirely, you can top off the charge until the gun explodes, but it still would not reload if you do not make it a straight blowback first. Why this is happening? A hint - a gas operated gun converted for blanks - lets say an AK47, that needs only a barrel constriction (a threaded plug with about 4 mm opening) in order to shoot blank ammo, has lot less recoil than one shooting live ammo.

 

[ojh]

I understand that this is going off topic, I apologize and shut up if asked. OTOH I have PhD in physics education and I work as a university physics teacher, so I may have some science based insight about the subject matter.

Of course bullet weight has effect on recoil. The heavier the bullet is, the longer it takes the gas pressure to drop and the force caused by it to affect to the breechface, and the bigger impulse it receives. That does not to change the fact that recoil comes from the interaction between gas and the gun, not from the interaction between the bullet and the gun.

If you shoot a blank in a normal slide-operated gun, gas pressure drops very fast and impulse is too small to work the slide. If you put a barrel restriction to a locked-breech gun, forces caused by the gas push the slide backward and the barrel forward, but because they are locked together the action won't open. If the gun is converted to blowback, the slide is free to move independent of the barrel and the barrel restriction keeps the pressure up for sufficiently long that the impulse becomes big enough to work the slide.

A question to Mizar: how would you think a bullet makes the slide of a blowback gun to recoil without the bullet even touching the slide? From where comes the interaction between them? That's a crucial question in order to clear up the misconception.

 

[Vern Humphrey]

I understand that this is going off topic, I apologize and shut up if asked. OTOH I have PhD in physics education and I work as a university physics teacher, so I may have some science based insight about the subject matter.

Of course bullet weight has effect on recoil. The heavier the bullet is, the longer it takes the gas pressure to drop and the force caused by it to affect to the breechface, and the bigger impulse it receives. That does not to change the fact that recoil comes from the interaction between gas and the gun, not from the interaction between the bullet and the gun.

If you shoot a blank in a normal slide-operated gun, gas pressure drops very fast and impulse is too small to work the slide. If you put a barrel restriction to a locked-breech gun, forces caused by the gas push the slide backward and the barrel forward, but because they are locked together the action won't open. If the gun is converted to blowback, the slide is free to move independent of the barrel and the barrel restriction keeps the pressure up for sufficiently long that the impulse becomes big enough to work the slide.

A question to Mizar: how would you think a bullet makes the slide of a blowback gun to recoil without the bullet even touching the slide? From where comes the interaction between them? That's a crucial question in order to clear up the misconception.

Well, free recoil is calculated by the formula M1V1 = M2V2, where:

M1 is the mass of the ejects -- bullet and gas
V1 is the velocity of the ejecta

M2 is the mass of the gun, and .
V2, the dependent variable, is the resultant rearward velocity of the gun.

So from the standpoint of physics, the gun does not differentiate between gas and bullet -- the recoil is generated by SOMETHING that is blown out the muzzle.

A blank will not generate the recoil of a ball round because the ejecta has so much less mass.

 

[Mizar]

A question to Mizar: how would you think a bullet makes the slide of a blowback gun to recoil without the bullet even touching the slide? From where comes the interaction between them? That's a crucial question in order to clear up the misconception.

I'm not arguing about pressure generated from hot gasses and the force it applies to the breech face. It's just the simple fact that blank firing guns have lot less, and I DO mean a lot less recoil that bothers me. See, to convert a straight blowback gun to fire blanks you only need a barrel restriction and a proper blank cartridge, nothing more - no slide weight reduction, no weaker recoil spring. With the proper constriction and cartridge, although with different pressure curves, you get pretty much the same pressure generated as with a live cartridge. And still, a lot less recoil. Couple of months ago, here on the THR, there was a discussion that resulted in a test - a 1911 with completely blocked barrel that did not allowed the bullet to move was fired with live ammo. It did not blow up and more importantly, the slide did not recoil. At the same time, people have experienced normal cycle and operation with 1911 having a squib round - one that has insufficient charge to completely drive the bullet out of the barrel, only to push it slightly into the lands. And yet the gun recoiled and charged the next round despite the fact, that the pressure generated was so low. See the difference between both cases? On the first you have a non moving bullet and very high pressure, but the gun did not operate and on the second case you have a moving bullet and a lot less pressure, but the gun acted as normal. Why is that?

 

[AZAndy]

Does a bullet really drag against the end of the chamber? I'm surprised by the idea that it would, since it's smaller. I've never had streaks of lead there that I noticed.

 

[ojh]

There seems to be some confusion between force and impulse. Similar pressure does produce similar force, but force alone does not define how much recoil velocity the slide (or in case of a revolver the whole gun) gets. It's impulse, force multiplied by time (or to be exact, integral of force over time) acting on a body that defines how much the motion (momentum) of the body changes.

I'm not arguing about pressure generated from hot gasses and the force it applies to the breech face. It's just the simple fact that blank firing guns have lot less, and I DO mean a lot less recoil that bothers me. See, to convert a straight blowback gun to fire blanks you only need a barrel restriction and a proper blank cartridge, nothing more - no slide weight reduction, no weaker recoil spring. With the proper constriction and cartridge, although with different pressure curves, you get pretty much the same pressure generated as with a live cartridge. And still, a lot less recoil.

If the slide works the same as with a live round, then the impulse acting on the slide must be roughly the same. But as you say, to accomplish this a restriction in the barrel is needed. That means that gasses act against the restriction and push the barrel and thus the whole gun forward. So there is an impulse forward and an impulse backward, and they partially, or mostly, cancel each other out. What's left causes the small felt recoil.

Couple of months ago, here on the THR, there was a discussion that resulted in a test - a 1911 with completely blocked barrel that did not allowed the bullet to move was fired with live ammo. It did not blow up and more importantly, the slide did not recoil.

Since nothing moved inside the gun, it was a solid, sealed, pressurized vessel. Forces caused by the gases in every direction canceled each other out. No net force, no impulse, no recoil.

At the same time, people have experienced normal cycle and operation with 1911 having a squib round - one that has insufficient charge to completely drive the bullet out of the barrel, only to push it slightly into the lands. And yet the gun recoiled and charged the next round despite the fact, that the pressure generated was so low. See the difference between both cases? On the first you have a non moving bullet and very high pressure, but the gun did not operate and on the second case you have a moving bullet and a lot less pressure, but the gun acted as normal. Why is that?

Must say this is an interesting one. Here's my explanation.

When the squib fires the bullet starts to move forward. Forces acting on the locked slide-barrel do not cancel each other out - force acting backward on the slide caused by gases wins, so the slide-barrel combo starts to move backward. It moves enough to unlock. After that the bullet sticks in the barrel. The slide has gained some momentum backward but still not enough to cycle. But now the gun works just like a blowback blank gun with a total barrel restriction. Gases act on the slide pushing it back, and since the gas has no way out except backward and that can happen only after the case comes out of the chamber, in spite of the reduced charge pressure stays high enough long enough and the slide receives impulse big enough to cycle.

 

[Mizar]

ojh, I'm not arguing that pressure from burning powder does the work. But in order for pressure to do it - i.e. "to do" the recoil you need a moving object - a projectile with considerable weight (or a very, very heavy powder charge - think rocket engine). You need to make the bullet to move in order for the recoil to happen. So, all I'm saying is that the bullet is the one "responsible" for the recoil - it happens as a consequence of it's movement. Pressurized gases from burned powder can be seen as a medium with which a force is applied - it can be a spring, it can be a hammer blow... But because of reaction to the forces that made the bullet move you get a recoil.

...The slide has gained some momentum backward but still not enough to cycle. But now the gun works just like a blowback blank gun with a total barrel restriction. Gases act on the slide pushing it back, and since the gas has no way out except backward and that can happen only after the case comes out of the chamber, in spite of the reduced charge pressure stays high enough long enough and the slide receives impulse big enough to cycle...

Nope, sorry - just a quick push on the slide is all it takes. After that inertia does the job - see Pedersen devise for example (Remington 51 pistol).

But, as I said earlier - this was covered here, on THR, in details not so long ago.

Best,
Boris