Hey Tuner, xray of 1911 firing

[1911Tuner]

I seem to recall once hearing something along the lines of: "once a man's mind is locked onto an idea or a woman...no matter how bad or how wrong...there just ain't much way to shake him loose from either one of'em."

That's right.

SDC...Can I call ya SD? SD...I'm really not tryin' to be a condescendin' arse. I'm really not. It's not in my nature. But you're short on a few of the basics.
We've taken the thread a few inches off the original topic, but I've tried to accomodate in order to try and help you understand it...but I'm beginning to realize that I can't until you have a grasp of the basics.

I won't keep arguing the same points until you go push a bullet through a barrel with a stick. After that, we can start over.

To provide a quick reply to your question about the slide on the rail and recoil not occurring until the bullet is long gone...I'd suggest another experiment if you can beg, borrow, or steal a Ransom Rest. You know...the device that allows you to fire the gun without holding it, and rotates upward in recoil.
They have a friction screw that lets you increase or reduce the resistance to that rotation. By using that adjustment, you can change the bullet's point of impact on the target by simply adjusting the tension on the screw...and thereby changing the amount of muzzle rise as the gun rolls upward in recoil.

Yes. I've seen it. The change in point of impact is pretty dramatic when the gun is fired at the two extremes of tension adjustment. If the bullet is gone before the gun recoils...it would make no difference. You can even do it without the rest if you're careful. Fire a freehand group with the gun held loosely...then do it again with the gun gripped hard. If you do it correctly...without fudging...your point of impact will go up or down with the change in grip.

 

[Walkalong]

Yes. I've seen it. The change in point of impact is pretty dramatic when the gun is fired at the two extremes of tension adjustment.

It sure will. I have seen that myself. Just like shooting a gun in competition. A consistent grip is VERY important to keep the bullets impacting where you want. (vertical stringing) I shot Benchrest a lot of years before my dry spell of not shooting it lately and it makes a BIG difference when you are putting 5 rounds into .400 or less at 200 yards, if you want to be competitive anyway.

There is absolutely no dought there is recoil before the bullet leaves the barrel.

This is sort of like math class where you add something wrong (the easy part of the equation) several times in a row trying to find where you made a mistake. (concentrating on the hard part where you assume you made a mistake) Then you look at it the next day and it is crystal clear where your mistake is. Ya gotta get your mind off it for awhile and look at it different.

 

[SDC]

SDC...Can I call ya SD?

As long as you don't call me late for dinner

I won't keep arguing the same points until you go push a bullet through a barrel with a stick.

Yes, I KNOW it takes a lot of force to push a bullet through a barrel, but it doesn't take any more force to push that bullet through the barrel of a semi-auto than it takes to push that same bullet through the barrel of a revolver. If I'm to believe your argument, this magical "bullet dragging the barrel forward" force is not only CRUCIAL to the operation of a semi-auto, it somehow becomes inconsequential and immeasurable when the exact same ammunition is fired in a revolver.

The Ransom Rest experiment only illustrates that if you hold a firearm differently when you fire it, you're going to get different points of impact; if it takes 14 pounds of force to compress the recoil spring, and you're holding the pistol with 20 pounds of pressure the first time you fire it, and 7 pounds of pressure the second time you fire it, you're going to get two different points of impact. So?

 

[Walkalong]

Yes, I KNOW it takes a lot of force to push a bullet through a barrel, but it doesn't take any more force to push that bullet through the barrel of a semi-auto than it takes to push that same bullet through the barrel of a revolver. If I'm to believe your argument, this magical "bullet dragging the barrel forward" force is not only CRUCIAL to the operation of a semi-auto, it somehow becomes inconsequential and immeasurable when the exact same ammunition is fired in a revolver.

There is a great deal more rearward force involved ( recoil / equal and opposite reaction ) going on than a little drag forward in the barrel. We don't even notice that little bit. It's overcome by the difference. Like pissing in the ocean. Nobody notices.

 

[Canuck-IL]

The Ransom Rest experiment only illustrates that if you hold a firearm differently when you fire it, you're going to get different points of impact

He didn't reference adjusting the grip pressure on the Ransom - that's a separate adjustment. The example referred to adjusting only the spring that resists recoil - the amount the weapon flips up in the rest - point being, if recoil doesn't begin until after the bullet exits, why does that adjustment alter the POI?

Ergo, recoil begins when there is an action - which must have an equal and opposite....etc.etc.
/Bryan

 

[SDC]

He didn't reference adjusting the grip pressure on the Ransom - that's a separate adjustment. The example referred to adjusting only the spring that resists recoil - the amount the weapon flips up in the rest - point being, if recoil doesn't begin until after the bullet exits, why does that adjustment alter the POI?

For exactly the same reason that you get a different POI when you fire a pistol holding it only with your thumb and middle finger (pressing the trigger with your index finger) than you do when you have a proper grip on the pistol. If you're only resisting the motion of the compressing recoil spring with the force of a few fingers, of course you're going to get a different result than when you're resisting the motion of the compressing recoil spring with the force of your full grip, hand, arm, shoulder, and body.

 

[SDC]

There is a great deal more rearward force involved ( recoil / equal and opposite reaction ) going on than a little drag forward in the barrel.

Exactly; yet, Tuner is asking me to believe that it's not that rearward force that's responsible for the way that a 1911 operates, it's this relatively tiny (so small we can't even NOTICE it) force that's going the other way.

 

[brickeyee]

It is the forward pull that holds everything locked even as the slide and barrel recoil together.
No use is made of the friction between bullet and barrel in a revolver, but that does not mean the force is not present.
If you ground the threads off the revolver shank and pulled the trigger what do you think would happen?
When the bullet hit the forcing cone both bullet and barrel would become the projectile.

It is the delay in unlocking that allows a 1911 to have far less mass than a blow back action would.
I have fired a 9mm blow back gun (a guy I know purchased it for all of $100).
It is heavy and extremely loud with deformed brass that is scrap.
It stretches some and bulges some since the action is moving and extracting the shell while the pressures are still high.

While the dynamic friction will be less than the static friction, it still takes a hammer to drive a squib, let alone a bullet that has not engraved on the rifling.

That force has been used to help ensure locking until the bullet has exited allowing pressures to drop considerably.

 

[1911Tuner]

Late for Dinner

It is the forward pull that holds everything locked even as the slide and barrel recoil together.

YES! YES! YES! By GOD, Brickeyee! I knew that you understood!
If you ever get to North Cackalackey, turbocoffee and ammo is on me!

Brickeyee...He just ain't thought about it hard enough yet. I have hope for him, though. He seems to have evolved a bit...but just a bit.

SD...In a revolver, that forward force on the barrel and the rearward force on the...RECOIL SHIELD...at the instant of the recoil punch...is what stretches the topstrap and creates a little problem known as excessive endshake.

The change in point of impact with a change in grip is caused by the bullet exiting the muzzle at a different point in its upward rotation...and it happens because...wait for it...because the gun is in recoil before the bullet exits.

Now then...There is a way to almost get the bullet out before the gun punches your hand. It will negate much of the bullet's change in impact, though it won't completely do so.

If you can figure it out...I'll call ya in time for dinner.

The Ransom Rest experiment only illustrates that if you hold a firearm differently when you fire it, you're going to get different points of impact;

Do you understand why that happens? Think about it. Think long and hard.

Exactly; yet, Tuner is asking me to believe that it's not that rearward force that's responsible for the way that a 1911 operates,

Nosir. I never asked any such thing. The rearward force is what drives the slide. The slide, in turn pulls the barrel back with it. Because the bullet is pushing the barrel forward and resisting the slide...the lugs engage horizontally and achieve lockup. During this phase, the barrel's mass is added to the slide's...slowing the slide and delaying the breech opening...just like using a more massive slide in a blowback pistol. Once the slide has pulled the barrel rearward to the linkdown point, the link draws the barrel down and away from the slide, disengaging the lugs vertically. The barrel stops against the vertical impact surface...drops to bed...and the slide continues on momentum.

 

[presspuller]

Can you tell me what happens to these "barrel pulling forward" forces when fired in a revolver? If they amounted to anything, why wouldn't we see them to an even GREATER extent in something like a bolt-action rifle?

You can.
Watch somebody that has not shot a big bore revolver before, espescially one handed. Stand behind them and watch that revolver twist in their hand. That the force of the bullet pulling on the barrel.
Same thing with a big bore rifle shot from a rest. Watch it and see if it does not rotate. Again the bullet pulling on the barrel.

Most of the time we don't pay attention to it but it has always been there.

Keep on him tuner, I think he may come around yet.

 

[1911Tuner]

Most of the time we don't pay attention to it but it has always been there.

Exactly so. It's over so quickly, and moves the gun so little...that it's not noticed.

I've seen a slow-motion video of a 1911 pistol firing. As the blowby gasses exit ahead of the bullet, you can see the whole gun pitch forward very slightly and then moving backward in recoil...just before the bullet exits. It happens too fast to be attributed to the shooter pushing the gun forward...much too fast...and it shows exactly what I've been trying to explain.

When that bullet slams into the rifling under some 20,000 pounds psi...it exerts a more signifigant forward force on the barrel than one might imagine.

Keep on him tuner, I think he may come around yet.

I dunno...He's fightin' it pretty hard.

 

[1911Tuner]

if it takes 14 pounds of force to compress the recoil spring, and you're holding the pistol with 20 pounds of pressure the first time you fire it, and 7 pounds...

I knew I overlooked somethin'...

Here you have a basic misunderstanding of how the spring works. 14 pounds represents the static load...or resistance, if you prefer...of the spring at full compression as installed in the gun. What you're dealing with dynamically is the spring's rate. The two are related, but they're not the same. The rate of the spring isn't given...only the loading at full compression. The rating is the loading in pounds per inch of compression.

Now, this is where it gets tricky...

Referring to Newton-1A...Objects at rest tend to remain at rest...the harder and faster that you push on a resistive object...the harder it pushes back. Maybe a better way to say it would be...The harder you push on it, the harder it fights you. You can toss a baseball easily. If you try to throw that baseball so hard that it will break the sound barrier...it's a whole 'nother ball of wax. A clearer demonstration might be in lifting a 20-pound weight slowly as opposed to snatching it up as quickly as you can. That 20 pounds will seem more like a hundred pounds. There's a formula to figure it exactly...but it's been a long time since I wrote it out. It's sorta like the formula for momentum in reverse. Literally...the faster that you try to accelerate it, the
"heavier" it becomes.

Somebody will come along soon and do the math. It's past my bedtime, and 0400 comes early around here.

 

[JohnKSa]

Experiment: Take a 1911 and place the grips in a vise but make sure that the slide can still move freely. Put the gun into lockup.

Put a loosely fitting punch down the barrel so that the end sits flat against the breechface.

Hit the punch with a hammer as hard as you can. Make sure that the gun returns to lockup. Repeat several times.

Disassemble the gun and take a look at the locking lugs & recesses.

What will you see?

Attach a length of chain to the blocks, with about a foot of slack. Fire the dynamite. The blocks will go in opposite directions, but when the slack comes out of the chain...the lighter block will be yanked backward and start to travel in the direction of the heavier block. The heavy block will be slowed down by the addition of the lighter block's mass...but it will continue to move and pull the lighter block along with it until its momentum is expended, and it comes to a stop.

What if the chain is taut before the explosion? The explosion will apply the same amount of force to both blocks (assuming both blocks are the same size) and nothing will move.

 

[DBR]

If there was not a force resisting the bullet acting on the engagement with the rifling there would not be a torque effect during recoil. All of my handguns twist during recoil. If there is a torque effect then there is a recoil effect since it is the resistance to bullet rotation as it moves forward that creates the torque.

These are all vector forces that have both magnitude and direction.

 

[peterotte]

Much has been said of the friction between bullet and barrel. There is a high resistance to the bullet entering the barrel and there is a high torque reaction to the bullet being rotated by the rifling with a resultant forward reaction vector. But the bullet friction, once moving in the bore, is a lot less than one might first think. This is because the barrel is being expanded by the gas pressure behind the bullet and to some extent around the bullet. The major force acting on the locking lugs is still going to be the acceleration reaction of the mass of the barrel. In the x-ray one can see just how little the barrel/slide has recoiled in the time it takes the bullet to reach the muzzle. It is a 'hammer blow'. All the recoil has been imparted into the slide in that short distance! Of course the barrel is much lighter than the slide and the forward thrust against the bore of the barrel is high. But bullet friction is a constant right? So why does increasing the pressure also increase wear on the locking lugs? It's because the rearward acceleration is greater.

Peter

 

[1911Tuner]

One...Two...Three!

Hit the punch with a hammer as hard as you can. Make sure that the gun returns to lockup. Repeat several times.
Disassemble the gun and take a look at the locking lugs & recesses. What will you see?

If you've used layout fluid on the barrel lug faces....front...you'd see ink removed from the faces, indicating contact with the rear faces of the slide lugs...though in some guns without equalized horizontal lug contact, you may only see one or two lugs with signs of it.

What if the chain is taut before the explosion? The explosion will apply the same amount of force to both blocks (assuming both blocks are the same size) and nothing will move.

Because nothing CAN move. (This is the one the Jim Keenan and I have scrapped over a few times.) If the force was great enough to break the chain, they'd move...but since they were no longer locked together, they'd continue to move separately...like a straight blowback action.

The force applied must be great enough to overcome the object's resistance...where that resistance comes from inertial mass and/or friction...or a mechanical obstruction.

If there was not a force resisting the bullet acting on the engagement with the rifling there would not be a torque effect during recoil.

You're makin' progress! If frictional resistance between barrel and bullet weren't a signifigant factor, there could be no rotational torque...or at least not enough for you to detect.

But the bullet friction, once moving in the bore, is a lot less than one might first think. This is because the barrel is being expanded by the gas

The frictional resistance remains the same, regardless. Once the bullet is moving, it doesn't take as much force to keep it moving...and to accelerate it to higher speeds...but as Brickeyee noted...the coefficient of friction is constant.

If the barrel expands because of the gasses...it doesn't expand much, and it would only expand behind the bullet...not ahead of it.

 

[45auto]

There is a way to almost get the bullet out before the gun punches your hand. It will negate much of the bullet's change in impact, though it won't completely do so.

I'll cheat since I've read it before.

Use a 23 Lb mainspring and a FP stop with little radius. That will "delay" the barrel/slide combo a "tad"...greater resistance!

Heavier recoil springs have little to no effect in delaying the slide if I recall.

 

[SDC]

It is the forward pull that holds everything locked even as the slide and barrel recoil together.

This statement is plainly ludicrous; the slide and barrel remain locked together because there's no way for them to UNLOCK until the link pulls the barrel down out of engagement with the lugs.

 

[SDC]

Watch somebody that has not shot a big bore revolver before, espescially one handed. Stand behind them and watch that revolver twist in their hand. That the force of the bullet pulling on the barrel.
Same thing with a big bore rifle shot from a rest. Watch it and see if it does not rotate. Again the bullet pulling on the barrel.

But those aren't the "pulling" forces at discussion here, those are "twisting" forces caused by the rotational inertia of the bullet as it's forced to turn by the rifling.

 

[1911Tuner]

This statement is plainly ludicrous; the slide and barrel remain locked together because there's no way for them to UNLOCK until the link pulls the barrel down out of engagement with the lugs.

Mornin' SD! Sorry. I overslept.

Well...I had hopes, but it looks like we're back to the initial observation that you simply don't understand how the gun functions. I'll make a stab at explaining it.

Let's assume for the sake of argument that Brickeyee and I are dead on. Okay?

The barrel is being held hard forward. Under the same forces that are driving the bullet, the slide is being slammed rearward. The lugs engage in a shearing direction...under some 20,000 pounds psi at peak. AT this point...the breech is locked. Before the pressure hits, the breech is NOT truly locked.

You can do a very simple demonstration to see how this pressure lock occurs.

Stand in front of a door that opens away from you...and close it. Turn the knob, and you can turn it easily. Place your hand two feet above the knob and push hard so that the "lug" is forced against the striker plate's edge. Try to turn the knob.

And be careful. I added "statement" to your comment above to prevent having it appear that you attacked brickeyee instead of his statement.
A minor detail...but I just wanted to keep it completely clean.

 

[SDC]

If you've used layout fluid on the barrel lug faces....front...you'd see ink removed from the faces, indicating contact with the rear faces of the slide lugs...though in some guns without equalized horizontal lug contact, you may only see one or two lugs with signs of it.

And this is what I'm trying to get through to you; John's experiment produces exactly the same effect (mashed surfaces on the front of the barrel lugs) WITHOUT any "bullet pull" that you claim is a RESULT of "bullet pull" when live ammo is fired. Since you get the same effect from a force applied to the breechface only, this tells me that THAT is the real mechanism at work here.

 

[1911Tuner]

Since you get the same effect from a force applied to the breechface only, this tells me that THAT is the real mechanism at work here.

*sigh*

SD... Are you awake yet?

In an action/reaction force pair, you can't have force in one direction without force in the other. Can't happen...unless you've figured a way to rewrite the natural laws of physics.

There is a forward force on the bullet that is equal to the force on the slide. There is a high resistance between the bullet and the barrel. Think about it. Go try to push a bullet through a barrel from chamber to muzzle with a stick. Which direction will the barrel move?

John's hypothetical worked because it used the inertial mass of the barrel to take off the ink and by moving the slide quickly. The harder and faster that you try to move an object...the harder it fights you.

 

[SDC]

Mornin' SD! Sorry. I overslept.

Well...I had hopes, but it looks like we're back to the initial observation that you simply don't understand how the gun functions. I'll make a stab at explaining it.

Let's assume for the sake of argument that Brickeyee and I are dead on. Okay?

The barrel is being held hard forward. Under the same forces that are driving the bullet, the slide is being slammed rearward. The lugs engage in a shearing effect...under some 20,000 pounds psi at peak. AT this point...the breech is locked. Before the pressure hits, the breech is NOT truly locked.

Good morning (This has to be one of the most interesting threads I've been on in years; you could even drop the "D" if you'd like, since these are just my initials )

In effect, you're arguing that the "Blish principle" actually EXISTS, that metals under pressure "stick together"; just as it was proven in the TSMG that this supposed "principle" doesn't work (by grinding the lugs of the H-shaped "lock" off, and allowing the gun to work by straight blowback), you can do exactly the same thing by grinding the lugs off of a 1911 barrel and allowing IT to work as a straight blowback. The only way you can claim that a 1911 in battery ISN'T "locked" is if you believe that two solid pieces of metal can pass through one another; you can shear them off or crush them through overpressure, but they ARE locked until that point in the recoil cycle where the link yanks the rear of the barrel down out of engagement with the slide.

 

[SDC]

There is a forward force on the bullet that is equal to the force on the slide. There is a high resistance between the bullet and the barrel. Think about it. Go try to push a bullet through a barrel from chamber to muzzle with a stick. Which direction will the barrel move?

You've convinced me that there's friction between a barrel and a bullet passing through that barrel (something i don't think anyone seriously disputes); what I remain unconvinced of is the claim that this friction does anything useful to the design of the pistol, any more than the vertical friction between a piston and a cylinder in a car engine has anything to do with moving that car down the road horizontally.

 

[1911Tuner]

In effect, you're arguing that the "Blish principle" actually EXISTS, that metals under pressure "stick together";

Nope. Not at all. What I am suggesting is that you can press two objects together with enough force that it's extremely difficult to pull them apart at 90 degrees to the direction of that force...which is how the barrel disengages from the slide...vertically.

what I remain unconvinced of is the claim that this friction does anything useful to the design of the pistol, any more than the vertical friction between a piston and a cylinder in a car engine has anything to do with moving that car down the road horizontally.

So, now you're trying to use a completely different set of dymics into the argument in order to support your beliefs? Interesting...