Hey Tuner, xray of 1911 firing

[SDC]

Don't get me wrong, I'm not doubting your experience with the pistol, but this "bullet-pulling" force that you claim plays such an important part in the pistol is such an insignificant factor that I have to wonder how you can place any credence in it at all. Even those designs that are specifically built to take advantage of it have a hard time getting enough mechanical advantage out of "bullet pull" to do any real work.

 

[1911Tuner]

Yes, and (just like battered lugs) this can be explained wholly by the forces that are put on the lugs by the SLIDE pulling on the BARREL, not by the bullet pulling on the barrel.

God SD! Think about what you're saying...once. If all the slide had to pull backward was the barrel's mass, it wouldn't even make a dent on the lugs.
If ordnance steel was that easy to damage, frames wouldn't last a hundred rounds from the slides impacting them, and the slides' spring tunnel's would shear off in less than 25.

Even with a loose slide/lug fit, the distance that the slide would travel before hitting the barrel lugs only amounts to about .010 inch. That's a light tap compared to the slide slamming into the frame.

Back off and think about that. I'm gonna wait until peterotte and Evan provide us with some figures based on the facts instead of abstract beliefs.

 

[Owen]

The main argument here is how much drag force is on the barrel at the instant of peak pressure. All else is incidental.

That's alot harder to do. At peak pressure, you are accelerating the bullet from and unknown velocity to an unknown velocity, are probably still engraving the bullet, and have to deal with the effects of obturation.

I suspect the actual peak of that barrel pulling force is maybe several hundred pounds.

An average would be pretty easy to figure given pressure over time, barrel time and muzzle velocity data.

 

[1911Tuner]

That's alot harder to do. At peak pressure, you are accelerating the bullet from and unknown velocity to an unknown velocity, are probably still engraving the bullet, and have to deal with the effects of obturation.

Yep. That's the tough question. If it helps...try from zero to 700 fps for a half-inch of travel into the barrel. That would put the bullet base just about flush with the chamber stop shoulder. Then, plug in peterotte's findings based on the rate of acceleration that he uses. Should get us pretty close.

Glad that you're adding obturation to the soup. That makes a difference.

I suspect the actual peak of that barrel pulling force is maybe several hundred pounds.

I'm gonna go out on a limb and estimate it at a couple thousand...based on driving it to 700 fps from a standing start. I'll go even further out and guess that peterotte's gonna need about 500 pounds psi to move it a half-inch...slowly. I had a guy here Thursday...he posted earlier...who tried to push it in. I mounted the barrel in a lathe chuck, and he pushed with all he had to give...and barely marked the bullet in the leade. I tapped the barrel on the table twice and the bullet dropped. This guy ain't small. About 6"2" and over 200 pounds.

 

[Owen]

I have no earthly idea how to add obturation.

 

[peterotte]

Owen, you have just let me off the hook, so to speak, by your post. Trying to figure how to calculate the hammer resistance has me stumped! It is very non-linear without even taking the non-linear acceleration of the slide into account!

As for the bullet drag and engraving force, I will need to devise a rifled tube. I will not be able to test anything at speed. I will be looking only at engraving force and sliding friction as well static friction after engraving. These would, I am sure, represent maximum possible forward drag forces. Another factor I cannot measure is the engraving force on an already moving bullet. To do that, we would need to measure chamber pressure with bullet jump and again with bullet resting on the leade. That would require moving the leade without disturbing any other parameter etc - a bit difficult I should think.

The plan so far is to machine a tube with stepped ID's to swage the bullet in stages to get some indication of swaging forces. I have a spline cutter in our workshop and could use that to make a straight rifling equivalent but the diameter will not match the 45 ACP.

Peter

 

[Owen]

well, peak pressure is 22,500 psi or so., correct?

The diameter of the bullet is .45", the area is pi*r^2 => Area of the bullet is .159 in^2.

max force on the base of the bullet is 22,500*.159= 3579 pounds of force.

Ok, here is the messy part.

Im changing to metric, because it's easier.

230 grains= 14.9g

700fps =213m/s

assuming linear acceleration (this is an extremely risky assumption) we have an average velocity of 107m/s approx.

1/2 inch = .0127m

so the time to travel 1/2" is .0127m/107 m/s = 1.18x10^-4 seconds.

Velocity = accelration*time 700m/s = x * 1.18x10^-4

593220 m/sec^2 or 60k gravities.

F=ma F= 14.9g *593220 = 8838978gm/sec^2 or 8839 kg-m/sec^2 = 8839 Newtons.

That comes out to 1989 lbs.

there are (3579-1989) 1590 pounds of force missing.

Please keep in mind that there were a few extremely risky assumptions made.

 

[1911Tuner]

That comes out to 1989 lbs.

Hmmm...I ain't a bad guesser after all. 11 pounds short.

Owen, I'll take another wild guess and say that you can add about 3% to net for obturation...for a bullet with the core exposed at the base. Covered by the jacket would make that a little less. Just to try to cover all the angles.

Peterotte...remember that the rifling lands are about .006 inch deep...so that would make land diameter .012 inch smaller than bore diameter. I gave you the wrong figures a while back, and just thought of it. If you'd like for me to send you a 1911 barrel for the test, shoot me a PM with a place to send it and it'll be on the way Monday. That'll be easier than rifling a tube.

 

[Owen]

The 1989 pounds of force is what's accelerating the bullet. the 1590 is what we have to account for.

I'm guessing a few hundred pounds of force to engrave (I'm envisioning pushing a bullet through one of those sizing dies. It's not terribly hard to do, but I don't know what the leverage is...

So maybe 1000 pounds of force for friction?f

Anyone have a bullet sizing die to knock a .45 FMJ down a few thousandths?

Actually, knocking a .454 bullet down to .452 would probably be pretty equivalent.

 

[1911Tuner]

Anyone have a bullet sizing die to knock a .45 FMJ down a few thousandths?

Best I can do is to take a .460 diameter cast bullet down to .452 inch. Cast lead...having a lot of lubricity compared to a jacketed bullet...is kinda apples to oranges, but it's the best I've got.

The 1989 pounds of force is what's accelerating the bullet. the 1590 is what we have to account for.

Dang. And here I thought I'd hit the inner part of the 9-ring.

I'll say that actually engraving the bullet jacket with .006 inch lands will probably require at least as much force as accelerating its mass...if not more. Maybe a lot more.

Best way to know is to have peterotte make a smoothbore barrel with a .451 ID, and force a .4515 bullet through it. Then push another one through a rifled barrel and see what the difference is. We'd have to ignore the variation in diameter between his homemade barrel and the real barrel...but it'd get us close.

 

[Owen]

i think the diameter difference is importatnt, because we are concerned about % reduction. I think.

 

[Odd Job]

Surely an easier way would be:

1) Grind a few hundredths of an inch off the rearward-facing surfaces of the slide lugs.
2) Chamber a round.
3) Weld the slide to the frame.
4) Place the gun in a vice, with a long string on the trigger..
5) Have a U-shaped measuring device in contact with the barrel.
6) Fire the gun and see what force was exerted by the barrel as it moved forward against the instrument. I guess you would have to divide by two because of the welded slide.

Of course, I could also just be rambling here...

 

[1911Tuner]

1) Grind a few hundredths of an inch off the rearward-facing surfaces of the slide lugs.

Only problem with that is that...the breech would immediately open by a like amount and the case would likewise back out until the breechface stopped it.
A few hundreths of an inch would expose too much of the case at the bottom...and the case would probably blow out before peak pressure was even approached.

Valid suggestion...but it wouldn't prove much.

 

[Owen]

they have little piezo force gages that may be fast enough to catch the data.

the problem is that you have to know the order of magnitude of the forces you are dealing with, and there is a significant amount of electronical gear needed to capture the signal from the gage.

Also you don't need to divide by two.

man I wish that were so, then I'd only weigh 115 lbs!!

 

[Odd Job]

I thought you would have to divide by two because none of the gases would contribute to slide recoil...kind of like the reverse of limp-wristing?

 

[Owen]

nope

it takes a couple weeks of drawing these things out for it to sink in.

look at it this weigh (teehee)

I weigh 230 pounds. When I stand on the ground, the ground must be pushing back up with 230 pounds, because I am not moving.

If we stick a box in between me and the ground, is the box experiencing 230 pounds, or 460 pounds or force.

 

[1911Tuner]

Aweigh! Aweigh!

Odd Job...Think of it like this. Regardless of whether the slide moves or not...the forces would be the same...backward and forward. Force forward equals force backward. Just because an object doesn't move...or can't move...is no sign that a force isn't bearing on it. You just can't see the effects.

 

[Odd Job]

But surely if you prevent the slide from recoiling, then some of the energy that might have been lost rearwards, must be engaged forwards against the bullet base?

 

[Old Fuff]

But surely if you prevent the slide from recoiling, then some of the energy that might have been lost rearwards, must be engaged forwards against the bullet base?

Not really, because by the time the slide (and attached barrel) starts to move backwards the bullet has left the muzzle. Remember in Browning's short recoil design the barrel and slide are locked as one unit until the barrel link rotates far enough to tilt the barrel down and uncouple the slide.

 

[SDC]

God SD! Think about what you're saying...once. If all the slide had to pull backward was the barrel's mass, it wouldn't even make a dent on the lugs.

That's what INERTIA does for you, and it's why things like slap-hammers can do such a good job of doing seemingly impossible things like stretching and breaking bolts. From a standing start, the slide has to accelerate to the rear, and THAT is what's causing damage such as the kind you point to as "evidence" of bullet pull.

 

[1911Tuner]

But surely if you prevent the slide from recoiling, then some of the energy that might have been lost rearwards, must be engaged forwards against the bullet base?

No more than any would be lost in a bolt-action rifle.

SD...

Lug deformation isn't as prevalent as it was in the days of soft barrel and slide steel. Modern steels have made it far less of an issue...though it still occurs. Nowadays, we see cracked and sheared barrel lugs and slides that crack at the junction of the breechface and ejection port...at the top left...and occasionally at the junction of the first lug wall and the port at the lower left.

SD...You can't bust steel that heavy by pulling on a 4-ounce barrel unless there's a helluva force HOLDING that barrel forward.

Used to be the way that locking lugs were seated and equalized by the old armorers...by fitting to leave a thousandth or two clearance between two lugs and their mating slide lugs...and firing 2 or 3 proof-level loads to set the bearing lug back and bring the others into play. I still do it, but I don't leave that much clearance. It tends to crack lugs instead of deforming them if done cold. It's known as "Pressure Seating" and you can find a description in Kuhnhausen's Volume 1.

 

[Odd Job]

Gents, allow me to ask in graphic form. May I present my finest Photoshopped barrels:



In all cases the red circle is the source of expanding gases. In barrel one, there is no breech face or slide, just a green jube-jube.
In case two, there is a solid fixed breech, like a shot gun.
In case three there are two rounds, facing away from one another.

Now here is my question:

If all three charges are ignited, am I to expect that rounds will emerge from those barrels at identical velocities?
My humble technical opinion is that they will not. Furthermore, if they did, there would be no such thing as limp-wristing (which I assume is a real phenomenon).

 

[peterotte]

Tuner

Thanks for the confidence. I am in 'Middle Earth' - New Zealand. There would be import permits and a whole lot of bother as the 'barrel' is the 'firearm'. I may be able to get hold of one from our local friendly gunshop or gunsmith. Owen has done the math and shows that the bullet drag is quite a bit higher than I first thought it might be. I really thought it would be no more than half the total accelerating force. But then I was thinking of bullet friction halfway down the bore. It is very revealing! I wonder how that friction force translates into lug face thrust, given the damping effect of the barrel inertia?

The one thing that puzzles me is that a normally loaded pistol will run for thousands of rounds with 'normal' wear on the barrel lugs but an overloaded pistol will hammer those same lugs to the point of barrel replacement. The drag does not change that much surely - or is it bullet obturation increasing the engraving force?
(A combat shottist used a 38 Super and needed the 'recoil' to qualify - so he overloaded! And went through barrels - and lived long enough to die of complications from an illness he suffered from. Poor soul).

Owen

I wonder whether you would be able to tell me just how much barrel expansion occurs behind the bullet and how far past the bullet base that expansion might extend? (I am referring to halfway down the barrel).

SDC

Great picture of the Schwarzlose 1908 pistol. I had never heard of it before. (I couldn't but help notice the bottle-neck cartridge. I would certainly have assumed that to be the source of a large portion of the 'operating' force).

Regards All
Peter

 

[Old Fuff]

If I understand the illustrations...

In #1 the bullet won't move, as the gas pressure escapes out the back of the barrel. If the jube-jube weighs more then the bullet the bullet will move, but at an unknown velocity.

In #2 the bullet will move. The velocity will depend on the gas pressure within the bore.

In #3 both bullets will move, but at a decreased velocity, because the gas is filling a larger space as both bullets move down the barrel.

Now what's this got too do with 1911 pistols?

 

[Odd Job]

It has to do with my suggested experiment with the welded slide and ground lugs.
You could say it has nothing to do with real 1911 pistols