What does this show or explain about lockup?
Brian, what are my old eyes seeing at the end of the barrel? Does this 1911 have a barrel bushing or am I seeing the end of the barrel?
Do you have any idea how the X-Ray was taken? i.e. could it be a fake?
Good picture, Brian!
It shows what I've been saying for years.
The bullet is almost at the muzzle, and the slide has moved just a little... probably about .090 inch here. Look closely at the upper lugs.
See the gap at the rear? That's a big clue that many don't understand. The barrel is being pulled forward by the bullet's friction in the bore while the slide is being pushed rearward by the equal/opposite forces that are at work on the bullet base, and the slide is pulling the barrel rearward with it. The lugs are engaged... front barrel lug faces to rear slide lug faces in a shearing action... and the breech is locked under those forces.
If you go read the original patents filed by Browning, it states plainly that the breech is locked under force as the bullet is transiting the barrel. I'll see if I can dig'em up and post it here. Pretty sure I've got it somewhere.
The gap at the rear of the barrel lugs make me believe that the photograph is genuine. Detail like that is usually overlooked in fakes.
If you go to the site, there is another x-ray image of the gun and a bullet having left the barrel going through a rack of thin plates about a foot away. The slide is just starting back good, but nowhere near ejection.
One more time, Tuner. Take an old .45 barrel. Ream it out so it is larger than the bullet diameter, with just enough shoulder to support the case. Grease it good to eliminate any remaining friction. And fire the gun. Let me know what happens when there is no friction to drag the barrel forward.
Have already done that, Jim. The slide moved as usual.
Do the rope trick, jim. Do it over and over again, and watch what happens closely. It'll come to ya.
And the explanation is absolutely correct. Force forward equals force backward. Force...from the controlled explosion and expanding gasses...acts equally on the slide and the bullet. Whether locked breech or straight blowback matters not. It works exactly the same way.
Blowback:
Bullet<--{BANG}-->Slide
Recoil Operated:
Bullet<--{BANG}-->Slide&Barrel--[Barrel Linkdown Point]Slide--->
In this case there is indeed bullet friction pulling the barrel forward - but the main force is inertia. Iether way, it makes no difference.
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In this case there is indeed bullet friction pulling the barrel
forward
Indeed it must. Try driving a bullet through a barrel manually from chamber to muzzle.
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but the main force is inertia
Inertia is a resistance to a force applied. In this case, the inertial resistance of the bullet's mass is miniscule compared to the frictional resistance offered by the bore.
Let me say that Mr. Keenan has what is likely the best general gun knowledge of any gunsmith that I've ever read or spoken to. It's just the recoil question that he isn't clear on... or specifically, what causes recoil... how it manifests.
Objects don't move unless they're forced to move. Force being the operative word... and that the force must be great enough to overcome the object's resistance to acceleration. Force accelerates the bullet and the slide at the same instant... in opposite directions. As soon as the bullet's frictional resistance is broken... so is the barrel's... and the barrel is free to move backward. The only thing that has the opportunity to make the barrel reverse directions is the slide... which has been forced to move away from the bullet. The bullet and the breechblock/slide are the action/reaction pairs. The barrel is just along for the ride.
Let's pretend that we're powerful wizards, and that we can cause the slide to disappear at the very instant that the primer sparks... before the powder is lit.
What will happen? Which direction will the barrel move? Backward? No. It will move forward.
To reduce the action/reaction issue to the simplest terms:
Pick up a 30-pound medicine ball. Hold it close to your chest, and shove it away from you as hard and as suddenly as you can. The ball accelerates in one direction, and you accelerate in the other. Did the ball's movement push you backward... or did the force that you imposed on the ball with your arms cause you to move?
Later, I'll explain the rope trick that I mentioned. If I can convince Jim to do it... it'll clear things up a bit.
Wrap a length of rope around your right hand, and grip in in your left hand so that it can slip through.
The rope will represent the bullet. The gripping, left hand will represent the barrel. The wrist of that hand will represent the slide.
Pull in opposite directions, but not hard enough to cause the rope to slip. You now have an action/reaction pair of objects. The rope and your left hand. You also have a pair of forces. One on the rope and the other on your left hand. Action/reaction forces always come in pairs... even if there is only one origin for those forces. There are always two forces with any pair of action/reaction objects.
Gradually apply more force until the rope slips through your left hand, and you'll notice that as soon as the rope(bullet) slips in one direction ...your left hand(barrel) will slip in the other... in the same direction as your left wrist(slide) ...or in the direction that the force is compelling it to move.
So... As soon as the bullet begins to move through the barrel, the slide is pulling the barrel off the bullet... because they are being pushed in opposite directions from the "Ground Zero" of a small, controlled explosion that has occurred between them.
Do the rope trick over and over and over until you see it. Repeat: The instant that the rope slips through your hand, your hand will slip off the rope.
Simple... what?
Ah, yes, the old rope trick. But it is still the momentum of the bullet moving forward that causes backward recoil, not the friction in the barrel causing pressure. So if you did try the frictionless barrel, what caused the slide to move? The pressure must have been close to nothing in the barrel once the bullet was out of the case.
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But it is still the momentum of the bullet moving forward that causes
backward recoil, not the friction in the barrel causing pressure.
No, no, no! I never said that, Jim. Recoil comes from force exerted on the slide. The same force that moves the bullet also moves the slide. The bullet's forward momentum can't do anything to make the slide move backward. It's going in the wrong direction. Mass doesn't move unless it's forced to move. The only way the bullet could force the slide to move is by hitting it.
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So if you did try the frictionless barrel, what caused the slide to move?
The pressure must have been close to nothing in the barrel once the bullet was
out of the case.
Force... and the lack of friction between the bullet and the barrel actually makes it easier for the slide to move. There's nothing fighting it besides its own inertial mass and that of the barrel. It works like a straight blowback with the bore reamed out.
I think I see where you're goin'... and it's a common misunderstanding to believe that the bulk of the recoil impetus is generated after peak pressure occurs... when about 95% of the recoil force... and probbly about 90% of the muzzle velocity comes in the first half-inch or less of bullet travel. Go and unscrew the barrel from a double-action revolver and fire it. You'll notice very little difference in felt recoil. Add ballast to the gun to bring the weight back up if you wish.
A better demonstration would be to lop a 12-gauge shotgun barrel off to the same length as a fired shell and shoot it. It's illegal, but guaranteed to black your eye if you put your face on the comb of the stock.... even if you add enough lead to the stock to make it weigh the same as before the barrel amputation.
Since you won't do the rope trick... do the exercise with the medicine ball. What causes you to be pushed away from the ball? Is it the ball moving forward...or is it the fact that your arms exerted a force against you and the ball at the same time?
Mass requires force to make it move. Recoil is not a force. It's the reaction to a force applied.
Force is required to get it moving... and then it has momentum.
Look at the picture that Brian posted, Kruz. See the front barrel lug faces in contact with the rear slide lug faces?
From the looks of it, there ain't much vertical lug engagement on that old girl, either.
Looks like the barrel is ridin' the link, too.
Patent Pending-1911
Pay particular attention to lines15-25 and lines 40-60.
http://patimg1.uspto.gov/.piw?docid=...6RS=PN/0984519
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Look at the picture that Brian posted, Kruz. See the front barrel lug faces
in contact with the rear slide lug faces?
I see. The slow motion one I saw that caught the bullet leaving the barrel and a rule under the slide showing nearly 0.10 movement of the slide told me it was pressure against the breech (same pressure that moved the projectile down the barrel) that started the slide rearward.
The JMB patent told me that the slide was designed to cycle via the conservation of momentum of the bullet. (After the "body was in motion")
Therefore, I think you're both right.
A. There can be no movement without force. Mass resists movement. It must be pushed or pulled. Otherwise, it stands still.
B. Without movement, there just ain't any momentum TO conserve.
C. The slide can't conserve momentum from the bullet. They're two different objects moving in opposite directions. The bullet is moving in the wrong direction to give the slide momentum. (How can you push something by moving away from it? If you need to push your car, you don't turn your back and run from it.)
D. Once the slide is moving, THEN it has momentum... the conservation of which keeps it moving after the bullet is gone and it has no outside force driving it. That same conservation of momentum is what keeps the bullet moving after it exits the muzzle... and has no outside force driving it.
In "Fisicks Skool", one of the "classic" examples of the third law of physics... which is the conservation of energy, momentum, and angular momentum, is the cannon firing the tennis ball.
The same momentum that the tennis ball has exiting the cannon is transferred back to the cannon in the opposite direction. Since the mass of the cannon is much greater than the tennis ball, the velocity is slower going backwards.
For every action, there is an equal and opposite reaction. When a gun fires, the momentum from the moving bullet is transferred back to the gun. If the gun is being restrained and friction exists, the momentum must go somewhere... Mr. Newton said so.
Kruzr... Sorry buddy. Saying that the gun recoiled because the bullet moved is like saying that the bullet moved because the gun recoiled. Same difference.
Read it one more time...
Mass obeys Newton and tries to remain at rest until an outside force overcomes its resistance and causes it to move.
Once it's moving, it has momentum.
Bullet acceleration results from force. So does recoil. In every action/reaction event, there must be two interacting objects and two forces.
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How can you push something by moving away from it?
The slide (locked-together barrel and slide) is being pushed backward by the same pressure that is pushing the bullet forward. The bullet goes one way and the barrel-slide goes the other. If you do the math you find out that the forward momentum of the bullet is balanced by the backward momentum of the slide-barrel. The center of gravity of the slide-barrel, with the bullet in it, remains stationary. The momentum of the slide barrel is conserved; it starts out zero and remains zero. At least it tries to. As soon as the recoil spring starts to compress it becomes an unbalance external force and conservation of momentum no longer applies to the barrel-slide.
Friction does not cause the bullet to pull the barrel forward, it only slows things down. If the bullet pulled the bullet forward the gun wouldn't recoil backward; it would pull you forward.
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Friction does not cause the bullet to pull the barrel forward, it only
slows things down.
The frictional contact between the bullet and the barrel is high. Try driving a bullet through to get an idea of how high that is. If the bullet is being forced through the barrel from behind, it has to hold the barrel forward. Go back and look at the locking lugs in Brian's picture. See the gap at the rear faces of the barrel lugs? The barrel is forward as far as it can go, and the slide has been driven rearward.
Because the bullet is moving, the frictional resistance is falling... and the faster it moves, the lower the frictional force becomes. Ever notice how much easier it is to keep a heavy object sliding across a floor than it was to get it moving in the first place? This also explains why a bullet continues to gain velocity after peak pressure. It simply requires less force to accelerate it than it did to get it going in the first place. The pressure peak occurs quickly and falls off quickly because of the increasing volume of the cylinder behind the bullet.
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The slide can't conserve momentum from the bullet. They're two different
objects moving in opposite directions.
The slide (and whole gun) has to "conserve" momentum by moving in the opposite direction from the gun. Momentum is a vector quantity and is not completely described without specifiying direction. A loaded gun at rest has no momentum, and the gun+bullet system as a whole will continue to have no overall momentum. Fire the bullet north, the gun recoils south with the same momentum so as to leave zero total momentum.
Things like barrel friction and gas pressure are the means by which forces are applied to transfer that momentum. And Mr Browning et al use them to make the guns work.
PotatoJudge
July 22, 2007
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Because the bullet is moving, the frictional resistance is falling... and
the faster it moves, the lower the frictional force becomes.
Velocity makes no difference in frictional force once velocity>0. That is, once static friction is overcome the force of kinetic friction remains dependent only on the coefficient of friction between the two objects and the force one part is applying to the other.
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The slide (and whole gun) has to "conserve" momentum by moving in the
opposite direction from the gun.
Momentum is only conserved in elastic collisions, which is not the case in firearms. Better to think of things in terms of energy bearing in mind that it's conserved but can be converted to other types of energy. There are losses (conversions) all over the system (frame/barrel vibrating, deformation of parts, sound, heat, etc).
So what are all the forces that contribute to slide/barrel lockup? In a smoothbore 1911, is there lockup? I have doubts that a reamed out 1911 barrel would remove all friction considering wobble of the bullet. Is there any significance to the force the slide exerts on the barrel lugs during that first bit of travel seen in the first x-ray (is there some lockup there with a stationary barrel and moving slide, and if so how much for how long).
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Momentum is only conserved in elastic collisions
No. Momentum is always and absolutely conserved. It just shows up best in elastic collisions. Shoot a tank with a bb gun and the tank acquires as much momentum as the bb loses. It just isn't perceptable.
You're right, momentum is conserved but kinetic energy is not in inelastic collisions. In the firearm where the collisions are inelastic trying to use momentum as an argument for conserved kinetic energy can be misleading.
Some Googled reading for those who never had to endure Physics classes in college:
The Physics of Everyday Stuff:
http://www.bsharp.org/physics/stuff/recoil.html
Wikipedia:
http://en.wikipedia.org/wiki/Recoil
"...once static friction is overcome the force of kinetic friction remains dependent only on the coefficient of friction between the two objects and the force one part is applying to the other."
Only over a limited range of velocities.
Remember we are dealing with projectiles that are damaged by passage down the bore (deformed) and smear metal all the way (to one degree or another).
The pressures are high enough for obturation to occur varying the normal force that is present also.
Ah, so the deformation and loss of matter lessens the force of friction between the two parts? I guess the initial scoring of the bullet with the rifling would generate more friction than the bullet would the rest of the way down the bore.
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Only over a limited range of velocities.
What type of range?
Newton's Dictums
Brickey... I never meant that the coefficient of friction drops. I was in a rush to get to the meat of the matter.
Let's just get back to basics with Newtonian physics. I'll make a few factual statements that can be verified by any high-school physics teacher and let the folks who are following this thread draw their own conclusions. Specifically, Newton 1 and 3, since those are the two that more closely relate to the topic.
Newton 1A:
An object at rest will remain at rest until an outside force compels it to move. Here, Sir Isaac is describing inertia, and he is telling us that if that thing right there is standin' still... it ain't gonna move until you push it. Say that over and over until it sinks in. It won't move until it's forced to move.
Newton 1B:
An object in motion will remain in motion until an outside force compels it to stop. Here, he's describing the conservation of momentum, and that if that thing right there is movin'... it ain't gonna stop until somethin' makes it stop. Notice how that word "force" keeps creepin' in?
That's a major clue. Force makes it move. Momentum keeps it moving. Force stops it.
Let's skip to Newton 3... the one that seems to be so cloudy:
For every action, there is an equal and opposite reaction. That's a flat statement that doesn't elaborate much. I'll do that for Sir Isaac.
Action/reaction events require two interacting objects and two forces. It's like the Sith. There are always two. No more... No less. This is known as an "Action/Reaction Force Pair."
Two interacting objects with a force imposed on each. Remember Newton 1A. Without force, there can be no acceleration.
The bullet is an object. The slide is an object. They are separate objects. Neither one can influence the other unless there is an active vector of force between them. Force forward equals force backward whenever there is an action/reaction pair. If that force is sufficient to overcome each object's resistance... both will move... and they'll move at the same instant, with the more massive of the two accelerating more slowly.
I've heard an over-simplified description of recoil that is completely and
utterly wrong... and I've heard it from people who should know better. That
description is:
"The gun pushes the bullet and the bullet pushes the gun."
That statement is absurd. Neither the gun nor the bullet can generate force. Recoil occurs because tens of thousands of pounds per square inch of force slams into the breechblock at the same instant that it hits the base of the bullet.
Jim K
July 22, 2007
Here is something to think about. Pressure in any gun is caused by the hot gasses released by the burning powder. So pressure goes up because of the friction of the bullet in the barrel, right? Then why is there any pressure in a shotgun with a plastic sabot, where friction is very low.
Ever wonder if most of the pressure containment is due not to friction or to pushing the bullet through the rifling, but to the inertia of the bullet itself, which resists movement?
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Go back and look at the locking lugs in Brian's picture. See the gap at the
rear faces of the barrel lugs? The barrel is forward as far as it can go, and
the slide has been driven rearward.
Think of the slide as a lid on the end of the barrel because that's what it is. Think of the cartridge case as the lid's gasket because that's what it is. As the case obturates it seals up that end of the barrel and keeps the gas pressure from blowing out of the wrong end of the barrel.
The front face of the barrel lugs and rear face of the slide lugs are jammed together because that's how they keep the lid from blowing off. The same thing happens when you shake a bottle of Pepsi. The only difference is that the bottle of Pepsi uses threads, whereas the 1911 uses lugs.
This whole business about the lugs is a red herring anyway. Forget about the silly lugs. The only thing they do is keep the action from being blowback operated instead of recoil operated.
Can we at least agree on that?
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Can we at least agree on that?
But of course... as long as you understand that the frictional resistance between the bullet and bore pulls the barrel forward against both the slide lugs and the slidestop, while the slide pulls it rearward. More simply... the barrel is resisting rearward movement, but the slide just overcomes the barrel's inertial mass, and... because the bullet is slipping within the barrel... the barrel can be pulled backward along with the slide.
So... quite literally... while the bullet is being driven through the barrel, the barrel is being pulled off the bullet... by the slide.
Another example that I've used:
Let's pretend that you've invented an alloy that is so dense that a .45 caliber bullet made of it weighs in excess of a ton. Let's also pretend that this alloy offers such a low coefficient of friction that it's practically non-existent. This bullet is so heavy that a normal .45 ACP powder charge can't move it. If you were able to handload a round and get it into the gun... and fire it... would the slide move? And if we assume that it would... where would the bullet be when it was finished cycling... and why?
"where would the bullet be when it was finished cycling... and why?"
Ok, I'll bite. I'd say the slide will move and eject the casing and the bullet will remain in the chamber. The fact that the bullet is too heavy to move has nothing to do with the force exerted on the slide. So say I.
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where would the bullet be when it was finished cycling... and why?
Well, since your heavy bullet is friction free, any powder charge will move it, even a primer by itself. It won't accelerate very fast but it will move, so your initial premisses are mutually contradictory, and your hypothetical scenario cannot exist.
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Well, since your heavy bullet is friction free, any powder charge will move
it, even a primer by itself.
The bullet weighs 2,000 pounds. You ain't gonna move it with a piddlin' 6 grains of Unique... friction or no friction.
Brownie.. .KUDOS! Well... almost. The slide will cycle and drag the barrel backward over the bullet... so the bullet will be about .250 inch into the rifling.
"..so the bullet will be about .250 inch into the rifling."
Yes, of course. I thought it might move slightly, but I couldn't put my finger on the reason. Makes perfect sense.
Instead of continuing to try to demonstrate what I know here, I'll relate a short story. The clues are all there...but we must open our minds to allow it to permeate the thought process...to illustrate just how signifigant the forces acting on the barrel and breechblock really are...for those who still don't think that they are.
Sometime in 1936 or 1937, Colt modified the soft slides on the 1911 pistols by installing a hardened steel insert in the center of the breechface to forestall the peening around the firing pin hole caused by the rear of the case rim slamming into the breechface under recoil. I've seen a few of the old slides that had that damage. I've also seen stair-step deformation on barrel lugs and corresponding damage of slide lugs in older, unhardened slides... caused by the slide pulling the barrel rearward against the resistance imposed by the bullet.
I've also had a few slides to crack in the top/left corner of the ejection port adjacent to the breechface guides and behind the guides at the breechface itself. I've also seen a few crack at the front right side of the port adjacent to the rear of the first lug wall...and many have the mistaken belief that this was caused by repeated impact shock when the slide hits the frame in recoil.
Not so. The area between the first lug and the breechface come under tensile stress and shock whenever the gun is fired, stretching it like the topstrap of the revolvers. Because of the stress risers in the corners, cracking is a distinct possibility.
Often, the crack is overlooked, and the first indication that something is wrong is that the shooter notices something stinging his face when the gun fires. This comes from gas leaking and unburned particulate escaping through the crack as it's pulled open under the tensile stress imposed.
I understand that there are a few who will not be convinced, regardless of how it's presented. Most others will carefully consider it, and eventually arrive at the correct conclusion. A few others will see it immediately. The first group have their minds made up, and will not entertain the idea of considering anything else, regardless of the obvious clues. The third don't need convincing. It's the middle group that I'm trying to reach. The ones with their minds open to taking a longer, deeper look at it.
I hate to say it, but a lot of physics is semantics and non-scientific comments.
Its all about "frame or reference" and "points of origin". You would be amazed at how many times we get into discussions and we are saying the same thing, but it is because each of us has elected a different frame of reference and point of origin. It is just part of the science.
Back to the topic at hand....
I thought the X-Ray picture was fascinating in that it showed what I would have expected and two, it is impressive that they could even have made it. Think how many tries it took to get it right and how much power was used to make the image. That is a non-trivial testament to the desire to show whats on the image. I doubt that was the first try.
Now I go back to a comment I made a long time back that Tuner probably remembers. I hate to think how much Mr. Browning and his team forgot compared to what we "know" today. If we could bring him forward in time and put him on the web in this discussion I would bet he could fill in all sorts of additional points that we have not even considered. Probably not "deal breaker" issues, but more the subtlety and the finesse or art of the design.
Awwww, Alex. It's been made a lot more complicated than it really is.
Lemme try just one more hypothetical analogy.
Mount two 10-pound blocks of steel on a rail and place them close together with a stick of dynamite between the blocks. When the stick fires, the blocks will both be hurled in opposite directions an equal distance at an equal velocity. We good so far? Good.
Now... Change one of the blocks for a 50-pound block, and fire another stick.
Again... both blocks will be blown in opposite directions... except this time, one will move 5X faster than the other. We still good? A'ight!
Let's move on to the final stage.
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.
Two interacting objects. Two forces. Action and reaction.
Force moves it. (Kaboom) Momentum keeps it moving. Force stops it. (Gravity and friction)
Check the picture in this thread.
http://www.thehighroad.org/attachmen...7&d=1183767338
Bullet barely clear of barrel, slide i recoil (notice the guide rod sticking out).
Brickeyee... Yep. Probably the best photo I've seen that illustrates that the slide... and the barrel... are moving backward the whole time that the bullet is in the bore. Since the slide and barrel mass equals roughly 40 times the bullet's mass... the ratio of slide travel to bullet travel can be calculated.
To do it precisely, we'd have to weigh both and get an exact figure...but 40:1 is close enough for a rough WAG, using 4.5 inches of bullet travel from
its starting point in the chamber to the muzzle. I think you'll find that the slide will move just about .110 inch when the bullet base hits the air...which is about right for an ordnance-spec pistol.
I think I remember you writing a post once about barrel lug deformation on the front face vs slide lug/rear face. I knew by that post that you understood what was going on during the event. It takes a lot of opposing force to beat the lugs back in a straight line like that.
Brickey... Here's another one of a 1911. Look closely at the location of the base of the bullet in relation to the bore. It's about an eighth to a tenth of an inch out and the slide has moved approximately 1/8th inch.
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Pressure goes up because of bullet mass and friction. Recoil is from the
opposite reaction from forcing the bullet (mass) to move in one
direction.
Precisley... and just that simple. The force of the explosion/gas expansion drives the bullet in one direction and the breechblock in the other. No more complex than that.
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Pressure goes up because of bullet mass and friction. Recoil is from the
opposite reaction from forcing the bullet (mass) to move in one
direction.
Recoil is in the opposite direction of the bullet leaving the barrel. If you bend the barrel 90 degrees, the recoil would be opposite that.
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The force of the explosion/gas expansion drives the bullet in one direction
and the breechblock in the other.
Without the breechblock, or some other object, we would just be shooting brass the wrong way. There would not be enough resistance to move the bullet far.
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Without the breechblock, or some other object, we would just be shooting
brass the wrong way.
That's right. The brass... having the lesser mass... would become the projectile and the bullet... having the greater mass/resistance... would become the breechblock. The barrel would recoil forward.
No matter how much anyone wants to believe that the bullet drives the slide by its forward movement...
It all works the same way. Straight blowback... Locked Breech/Recoil Operated... or single-shot breakdown shotgun. The force that drives the projectile also drives the breechblock.
The LB/RO pistol operates by the same set of forces and the same mechanism as an outwardly identical straight blowback pistol. The difference... the ONLY difference... is in the mechanism used to delay the breech opening.
The blowback pistol uses slide mass and/or recoil spring strength. The LB/RO design just adds the barrel's mass to the slide for a short time and distance... and then drops it at the linkdown point when the bullet exits and pressure has fallen.
That's it. That's all there is to it. Just that simple.
On a logical plane... ask yourselves:
Assuming the same caliber... Both guns fire the same cartridge in the same manner. Primer fires... Powder burns... Gasses expand... Pressure rises... X pounds per square inch of pressure results, creating a vectored/directional force. Nothing about that event changes with a different design. Nothing.
The cartridge doesn't know what kind of pistol it's been fired in. It doesn't alter its basic operation.
The bullet does not... CANNOT... give momentum or anything else to the slide.
It's moving in the wrong direction. Momentum results from movement. Before the slide moves, it doesn't have momentum. It only has inertia.
In order to make it move, you have to MAKE it move. You have to FORCE it to move. Before the bullet has momentum... it has to move. You have to force the bullet to move, too. Those who keep quoting Newton-3 in a belief that the bullet causes slide movement are ignoring Newton-1.
"Objects at rest tend to remain at rest... until an outside FORCE causes them to move." Mass does not... will not... CAN not move... until it's pushed.
Look up the Schwartzlose blow-forward if you want to see conservation of momentum and balancing forces, etc. operating in an unobvious way.
Consider the german bent barrel attachments used in WWII, the Krummer Lauf. This was the gun and eventually a rifle attachment that bent the barrel 90 degrees for use by German Tankers.
Several Krummer Lauf devices were captured by American and British forces during the last battles of the war. Captain Philip B. Sharpe of the U.S. Army Ordnance Corps tested the 90-degree model. "Over two-thirds of the bullets were torn into two or three pieces--but at very short range it would have been deadly," Sharpe wrote after the war. "When fired from the hip as an 'around- the-corner-gun,' not more than three shots could be fired in one burst--it would spin the shooter 90 degrees in three shots--a vicious attachment."
The recoil was 90 degrees to the breech and cartridge direction. It was conserving the momentum or in other words, the equal and opposite reaction.
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The recoil was 90 degrees to the breech and cartridge direction
Because the direction of the force vector was changed. If you accelerate a car in a straight line, the reaction will be straight line. If you continue to accelerate around a curve, the equal/opposite changes, and you feel the acceleration as a side-load from the centrifugal force created by the curve.
In order to make an object move, you have to force it to move. Once it's moving, it has... and conserves... momentum.
Let's try just once more:
Remove the bolt from a Mauser rifle, and cut the length of the locking lugs down so that when the bolt is turned, camming the lugs into the recesses...there is only .001 inch of lug to recess overlap per side. The lugs are only .001 inch deep into the recesses. Everything goes to battery and locks up as normal when you turn the bolt.
What's gonna happen when you fire the rifle?
Obtain a steel pipe witha (hypothetically) perfectly sized I.D. and use it to make a double-ended muzzle loader. Obtain two lead balls that are of identical diameter and weight.
Load the two balls in such a way that they are exactly and precisely equidistant from their respective muzzles. Place a powder charge in the precise geographic center of the pipe...between the balls...and fire it. What will happen?
Clew:
Each ball is a projectile, and each one acts as a breechblock for the other.
Newton!
Finally...Newton-2 may help you to understand why your bent rifle recoils sideways. (N-2 is the hardest one for me to quote verbatim.)
http://csep10.phys.utk.edu/astr161/l...wton3laws.html
Newton
Since it's been so long ago that I could quote the textbook terminology...I looked and found it. The basic misunderstanding of recoil...and how it occurs...seems to be more widespread than I thought...so here are some copied exerpts that may clear it up.
All of Newton's laws of motion/conservation of momentum apply...not just the ones that we like. In this case, it seems to be Newton-3 while ignoring Newton-1. Here is the text. Study it hard. The answer is here. My comments are bracketed
*******************
The amount of momentum that an object has depends on two physical quantities: the mass and the velocity of the moving object in the frame of reference. In physics, the symbol for momentum is usually denoted by a small bold p (bold because it is a vector)
[ "The mass AND velocity...of a MOVING object." First it has to move. Before it moves, it is in a state of equilibrium, where its resistance to acceleration is equal to the external forces acting on it.]
The law of conservation of momentum is a fundamental law of nature, and it states that the total momentum of a closed system of objects (which has no interactions with external agents) is constant. One of the consequences of this is that the centre of mass of any system of objects will always continue with the same velocity unless acted on by a force outside the system.
In an isolated system (one where external forces are absent) the total momentum will be constant: this is implied by Newton's first law of motion.
[An object at rest will remain at rest. An object in motion will remain in motion.]
Newton's third law of motion, the law of reciprocal actions, which dictates that the forces acting between systems are equal in magnitude, but opposite in sign, is due to the conservation of momentum.
["The FORCES acting BETWEEN two systems....are equal in magnitude, but opposite in sign." (Equal and opposite.) ]
Since position in space is a vector quantity, momentum (being canonical conjugate of position) is a vector quantity as well - it has direction. Thus, when a gun is fired, although overall movement has increased compared to before the shot was fired, the momentum of the bullet in one direction is equal in magnitude, but opposite in sign, to the momentum of the gun in the other direction.
["The amount of momentum that an object has depends on two physical quantities: the mass and the velocity of the moving object in the frame of reference."]
Recoil resulting from the firing of a gun is thus explained:
["Although overall movement has increased...(from standing still to accelerating)...the momentum of the bullet in one direction is equal in magnitude but opposite in sign to momentum of the gun in the other direction." ]
**************
These two statements define it clearly. The bullet's momentum is opposite in sign/direction to the momentum of the gun. The bullet cannot give the slide momentum...andy more than the slide can give the bullet momentum. They're moving away from each other because a vectored force got between them and pushed them apart.
Kruzr, I don't understand how you think a bullet, a hunk of lead, that does nothing by itself, has the power to move the slide.
You have the gas from the explosion trying to go in 129,600 degrees, the only way it can go is in two direction, ( at least we hope so, that is why guns are made of metal and not paper) which happen to be opposit each other. One direction is pushing the bullet, the other is pushing the slide.
How can the bullet move the slide?
Tuner, Thanks for making a few things more clearer to this uneducated country boy.
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Thanks for making a few things more clearer to this uneducated country
boy.
I dunno, presspuller. Sounds like you've got a pretty good head on your shoulders.
DUnno why is this so hard to understand...
Momentum...is defined as...MASS times VELOCITY. If the object is standing still, its velocity is ZERO. Whether the object weighs 10 grams or 10 pounds or 10 tons...if it's standing still, its momentum is ZERO. If anyone can figure a way to multiply any number by zero and come up with anything besides zero, it would be a fascinating equation.
Therefore...Zero velocity equals zero momentum. It will conserve that value...zero...until an external force compels to do otherwise.
Now, here comes the law that a few want to ignore:
Newton 1A:
An object at rest will remain at rest until it's compelled to move by the application of an external FORCE. When it moves...it has velocity. When it has velocity, there is something to plug in besides zero...and it will conserve that momentum until another external force compels it to do otherwise.
Simple...
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Kruzr, I don't understand how you think a bullet, a hunk of lead, that does
nothing by itself, has the power to move the slide.
Me and all the world's physicists are just crazy I guess.
(But read this anyway: http://en.wikipedia.org/wiki/Recoil_operation)
Yeah but wikipedia is about as reliable as the national enquirer. I don't trust anything from that site.
Help me to understand this. How is the bullet that is going one direction have an effect on the slide going the other direction. Neither would do anything if it was not for the gas pushing them both.
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Help me to understand this. How is the bullet that is going one direction
have an effect on the slide going the other direction. Neither would do
anything if it was not for the gas pushing them both.
It's called Newton's Third Law of Motion. Maybe this will help: http://www.bsharp.org/physics/stuff/recoil.html (The Physics of Everyday Stuff.)
Gun Recoil
Although guns may not be everyday things for many of us, gun recoil is certainly something we're aware of (at least those of us that don't make Hollywood action films with guns whose recoil would instantly kill the person firing them!). Gun recoil is a result of momentum conservation, which is an extremely important fundamental principle. Newton was talking about momentum conservation when he wrote "every action has an equal and opposite reaction".
Momentum Conservation
Momentum characterizes an object's resistance to change in motion. If this is motion along a straight line, we call it linear momentum; if it is rotational motion we call it angular momentum. The basic idea is the same: moving things like to keep moving, and to change their motion we have to apply a force. If no force is present, then momentum doesn't change, ie. it is conserved.
Now, you might point out that a bullet coming out of a gun has a huge force on it from the exploding gunpowder. True enough, and that force is what propels the bullet forward. However, if you look at a bullet and gun together (say while the bullet is still in the barrel but already heading out at full speed), you can say there is no net force on the bullet-gun system. So the momentum of the bullet plus gun should be conserved.
If the bullet has mass mb and speed vb out of the gun, it has momentum pb given simply by
pb = mbvb
in the forward direction. To balance this momentum (and keep the net momentum of the bullet-gun system zero), the gun recoils with momentum in the opposite direction: pg = -pb, or mgvg = -mbvb
Although the bullet's mass is small, its speed is quite large, so it released with large momentum. The gun has much larger mass, so the recoil speed is much smaller, but still large enough to give a serious kick against the shooter's shoulder.
Example: Winchester .308
Let's look at an example. A Winchester .308 cartridge launches a bullet of mass 150 grains (1 grain = 64.8 mg) with a speed of 2820 ft/s (1 ft = 30.5 cm). In MKS units, then, pb = 8.4 kg m/s. This rifle has a weight of about 8 lbs, or a mass of mg = 3.8 kg. That means the recoil speed of the rifle will be
vg = - pb/mg = -2.2 m/s
This primary recoil is noticeable, but not the main recoil that one feels.
Secondary Recoil
There are actually two distinct recoils from a gun: the first, primary recoil, which I've described above, conserves momentum of the gun-bullet system. However, a larger secondary recoil comes slightly later, when the bullet leaves the muzzle: then the hot expanding gas behind the bullet shoots out of the muzzle, and the muzzle recoils further like a rocket. This is, again, conservation of momentum, but in this case is is the gas momentum out of the barrel that makes the secondary recoil. Gun manufacturers make baffles that reduce the flow of gas out of the muzzle to reduce secondary recoil. Primary recoil cannot be reduced, since it is simply associated with the forward momentum of the bullet.
Equations
linear momentum: p = mv
Summary
The total momentum of a system is conserved if there are no outside forces acting on it.
Gun recoil results from conservation of total momentum of the bullet-gun system: the backward recoil gun momentum balances the forward bullet momentum to maintain zero total momentum.
Gun recoil actually has two parts: primary recoil from the escaping bullet and secondary recoil from the escaping gas behind the bullet.
(This is like Who's on First and we keep ending up at third base. )
Mass times velocity equals momentum. If it ain't movin' it ain't GOT any momentum, and it ain't gonna have any until it moves...and it ain't gonna move until it gets pushed! See Newton 1A.
The bullet can't push the breechblock backward any more than you can push your lawn mower by runnin' away from it. You've gotta push it in the direction that you want it to go!
The image of the barrel lugs against the slide grooves is just -- wow:
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But the image of the barrel lugs against the slide grooves is just --
wow:
Yep. What you're seeing is a graphic picture of the lugs engaged under pressure... slide pulling the barrel with the barrel resisting due to the bullet's influence.
Lets see if I've got this down yet.
The force (high pressure gases), contained from going sideways by the chamber, pushes the bullet forward and the breechblock rearward. The bullet moving forward tries to drag (friction) the barrel with it. The rearward movement of the slide and the forward movement of the barrel keep the barrel lugs locked against the slides grooves (lugs?) with the front of the barrel lugs engaged with the rear of the slides grooves. When the bullet has exited the barrel there is no more forward force (drag) acting upon the barrel so the force holding the lugs locked is lessened greatly and it is now easy for the link to pull it down out of engagement.
Any comments, corrections ?
At the point shown in the radiograph, you'd see the barrel lugs forward against the slide lugs regardless, since the slide and barrel have already started their recoil cycle, and the slide is coming back. I don't think the "forward force"/friction of the bullet in the barrel is anywhere NEAR as strong as has been suggested previously, otherwise you'd feel a significant decrease in recoil if you shot the exact same load in a revolver as in a semi- auto (provided the total weights of the guns are the same). The only real factors that matter in this equation are the weight and velocity of the bullet on one side, versus the weight and velocity of the recoiling parts of the pistol on the other.
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Any comments, corrections?
Walkalong...You've got it!
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I don't think the "forward force"/friction of the bullet in the barrel is
anywhere NEAR as strong as has been suggested previously
SDC...If you'd ever seen locking lugs with stair-step deformations...barrel and slide...you'd understand how high the forces are....but you're correct in that in this picture...the forces have dropped considerably.
The front/rear forces are at their highest at the time of peak chamber pressure...which occurs at about the first half-inch of bullet movement.
This is where the majority of the recoil impulse occurs...and the majority of the bullet's final velocity, too. Beyond that point...any recoil impulse that comes is almost insignifigant by comparison.
Think of it like this: (Simplified)
A 5-inch barrel has 4.1 inches of rifling. If the muzzle velocity is 850 fps...and assuming that the 35 fps per inch...gained or lost...rule of thumb is accurate...then 4X35=140...how does the remaining 710 fps occur?
Recalculate it, using one-half inch for the peak...leaving 3.6 inches to accelerate the bullet to the final velocity. Then 3.6X35=126 fps...which means that the first half-inch of barrel will produce 724 fps.
It woild also be useful to understand the method which was once used to equalize the locking lugs...that is...get all of them bearing against their opposing lug faces in the horizontal plane. They were fitted to leave the thrid lug...the forwardmost lug...kissing about .002 inch of air, and fired with proof-level ammunition 2-3 times in order to deform the first and second lugs, and bring the third one into engagement.
Even with the softer barrel steels of the day...that takes a lot of force in just 3 rounds.
"I don't think the "forward force"/friction of the bullet in the barrel is anywhere NEAR as strong as has been suggested previously..."
Try and push a bullet through by hand.
I've pounded through a few squibs, but if that "pulling friction" WAS as strong as some seem to believe, then why isn't it apparent in a firearm that doesn't work in the same way as a semi-auto? If it actually contributed in any sort of way to the lock-up of a 1911, then you should be able to feel a definite difference in the recoil between a 1911 and a revolver of the same weight, no?
As for the "stair-step" deformations, that sounds just as likely to be a result of improper lock-up in the first place, with ignition occuring when there's less than full lock-up of the lugs.
SDC...Keep studyin' on it. The light hasn't come on yet.
If that opposing forward/backward force will stretch the topstrap on a Model 19 Smith & Wesson...it'll deform the lugs in a straight line on a 1911 pistol.
But you're claiming that "(The first half-inch of bullet travel) is where the majority of the recoil impulse occurs...", ie. BEFORE the bullet even LEAVES the pistol. How can you even claim that this is possible? Unless the bullet leaves the pistol, you have no recoil to operate the pistol (and a squib demonstrates exactly that). I've seen a fair number of revolvers that have suffered flame cutting and topstrap erosion, but this is a different mechanism at work; if a 1911 is set up with anywhere NEAR proper tolerances, those recoil forces would also have to crush part of the barrel hood in order to crush the lugs on the barrel. If you ever have a junker frame, slide, and barrel to play with, I invite you to try an experiment I once saw with a Hi- Power; grind the locking lugs completely off the barrel, reassemble it, and see how the pistol functions as a straight blowback (in a rest, please). It's definitely hard on the pistol (which is why you want to do it with some scrap parts), but it still functions.
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BEFORE the bullet even LEAVES the pistol. How can you even claim that this
is possible? Unless the bullet leaves the pistol, you have no recoil to
operate the pistol
If you understand anything about Newtonian physics, you'll know that the recoil...essentially the reaction side of the action/reaction equation...begins at the exact same instant that bullet movement starts, and the gun is in recoil well before the bullet exits the muzzle.
Just in case you got this idea from reading and believing Kunhausen's "Balanced Thrust Vector" sheep dip...it can't happen. If the gun doesn't recoil before the bullet exits...it won't recoil.
If the bullet is gone BEFORE the gun kicks...the gun WON'T kick. That's why hollywood pretend autoloading guns require blank firing devices in order to cycle.
What little recoil effect produced by the residual pressures after burning 5 or 6 grains of pistol powder probably wouldn't knock a fly off your hand.
In order for the ballistic event to produce recoil, the bullet has to be in the barrel, and it has to move. If the bullet is missing, half of the action/reaction pair isn't present. If there's no action, there can be no reaction.
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if a 1911 is set up with anywhere NEAR proper tolerances, those recoil
forces would also have to crush part of the barrel hood in order to crush the
lugs on the barrel.
If you understood how the pistol functions, you would see that the slide is moving away from the barrel hood...so there can be no crushing.
The hood isn't a forth lug.
A "consequence" of the recoil occurring before the bullet leaves the barrel probably accounts for the different POI people have shooting the same gun/same ammo...I'd bet!
How you grip, strong/weak, one/two handed, etc.
"If you understand anything about Newtonian physics, you'll know that the recoil... essentially the reaction side of the action/reaction equation... begins at the exact same instant that bullet movement starts, and the gun is in recoil well before the bullet exits the muzzle."
No, only PART of the pistol is in recoil; the slide and barrel (locked together) on one side of the equation, with the bullet (travelling much faster, because it's so much lighter) on the other. It's not until the force of the slide and barrel is transferred to the frame (by which time the bullet is LONG down-range) that the shooter even FEELS any recoil. It's simply the mass of the slide and barrel together that balance that equation out against the mass of the bullet (the first slow, the second fast), and it's what delays the opening of the breech until after pressures have dropped. "If you understood how the pistol functions, you would see that the slide is moving away from the barrel hood...so there can be no crushing. The hood isn't a fourth lug."
Can you explain to me how "the slide is moving away from the barrel hood" when the slide and barrel (chamber, hood, lugs, and all) are LOCKED TOGETHER by the same lugs that you claim are being "crushed"? It isn't until after the link pulls the barrel down out of alignment with those lugs that the barrel has (or should, if it has proper tolerances) any motion independent of the slide.
If you watch a wave on the beach long enough you will understand what everyone is trying to say. The wave starts by rolling up the beach which is the momemtum and as the wave reaches the top of its cycle it will reach a relaxed state or conservative momentum and as the wave rolls back down the beach it will gain speed because of the resistence to the beach or being of less resistance.
When the wave hits another wave as it comes in, What happens? The wave hits more resistence and slows down, thus causing another moment of conservative momemtum which then releases that momemtum against the incoming wave. As this happens both waves push away from each other in opposite directions, that is recoil.
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A "consequence" of the recoil occurring before the bullet leaves the barrel
probably accounts for the different POI people have shooting the same gun/same
ammo...I'd bet!
Yessir. You've got it!
SDC...
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Can you explain to me how "the slide is moving away from the barrel hood"
when the slide and barrel (chamber, hood, lugs, and all) are LOCKED TOGETHER
by
I'll try...
The ballistic event is nothing more than a miniature explosion between the base of the bullet and the breechblock...transferred, of course, through the case rim.
The bullet...being an uber-tight fit in the barrel...yanks the barrel hard forward when it hits the rifling. At the same instant, the opposing force noted in Newton's 3rd...action/reaction forces always travel in pairs...is slamming the slide in the opposite direction...trying to shear the locking lugs and separate the slide and barrel. If the lugs weren't present, the barrel and slide would be separated...just like a straight blowback...but because they're locked together BY the lugs...they move as a unit. The barrel is actually part of the slide for a brief period, as it is pushed rearward. The slide pulls the barrel backward with it...as a unit...until it reaches the linkdown point...roughly .100 inch in an ordnance-spec pistol. At some fraction of an inch BEFORE the linkdown point is reached...the bullet exits. Pressure drops...and the breech can safely open. The barrel drops, and the slide continues rearward on the momentum that it aquired during that first .100 inch of travel...when it was being driven by the forces contained therein.
Got it?
Once more...
The forces acting on the barrel and slide are literally trying to rip the lugs off the barrel, and separate the two parts. Go look at a cutaway drawing...look closely at the lugs and how they relate...open your mind...and study it. It'll jump off the page atcha.
Cheers!
Lock your hands in front of your chest and pull in opposite directions.
Barrel lug deformation is always on the front face. Slide lug deformation is always on the rear face.
Study Newton-1 and 3...Look at the barrel and slide lugs...and it'll come to ya.
"If the lugs weren't present, the barrel and slide would be separated... just like a straight blowback... but because they're locked together BY the lugs... they move as a unit."
And this is what I was trying to demonstrate with my earlier "grind the lugs off of a barrel and try it" example; I've seen it work with a Hi-Power, and don't see any reason why it wouldn't work exactly the same with a 1911. If you tried it and it functions (as it does with a Hi-Power), it would convince you that this "bullet pulling the barrel forward as a part of the locking mechanism" is a figment of your imagination. The slide alone has so much more mass than a bullet that it's IMPOSSIBLE for the slide to open before the bullet has left the barrel.
As for the barrel/slide lug deformation, this is EXACTLY what you'd expect to see as soft or loose-tolerance parts wear, since it is those surfaces that grind against each other during the locking/unlocking process, correct?
If I understand all this, a bullet should actually accelerate the instant it clears the barrel due to drop in friction with the barrel. Then after some relatively short distance start to slow down due to air friction and lack of propulsion. on the side of recoil, I've heard that when the A-10 Warthogs fire their 30mm Gatlings, that the recoil slows then down a noticeable amount. Good thread; interesting and thought provoking. Thanks to all.
I don't think the bullet can speed up after leaving the barrel simply because there is nothing pushing it anymore. The pressure has dropped to almost nothing by the time the bullet gets out of the barrel.
SDC, why do you think the barrel and slide actually lock together? Simply because its a neat thing for them to do?
I have never measured it but I am sure that it takes more pressure to push the bullet out the barrel than it does to push the slide back. So I would think under the right circumstances that the slide COULD open before the bullet left the barrel if the lugs were not present.
"SDC, why do you think the barrel and slide actually lock together?"
Because that's what they were designed to do by Browning; however, DESIGNING something to work in a certain way doesn't mean that that something actually WORKS that way; as exhibit one, I give you the Blish "lock" in the 21/28 Thompson. As I've told Tuner, there's an easy way to check this out, and I've seen it done with the Browning Hi-Power; simply take the lugs off a barrel (turning the pistol into a straight blowback), and fire it that way. As long as the bullet can travel down the 5" of barrel in a shorter time than the much heavier slide and barrel can recoil over a distance of approx. 1/2", then the system will work, locked breech or not.
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Because that's what they were designed to do by Browning; however,
DESIGNING something to work in a certain way doesn't mean that that something
actually WORKS that way
When firing a centerfire round in a blowback design as you describe, what kind of flash/blow out do you have out the ejection port?
Yeah, the Browning without locking lugs and the Hi-Point will fire standard 9mm Parabellum as straight blowback. Just don't ask me to fire 9mm with an "M38" headstamp or any other +P+ or submachinegun 9mm ammo in a blowback only handgun.
The locking lugs may only be a safety feature, but safety is a nice feature to have, given the variety of ammo out there.
"When firing a centerfire round in a blowback design as you describe, what kind of flash/blow out do you have out the ejection port?"
I saw it done with C1 (Sterling) smg ammo, and it produced a fairly bright muzzle flash and some "slightly pregnant"-looking brass, but nothing really out of the ordinary at the breech end.
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If you tried it and it functions (as it does with a Hi-Power), it would
convince you that this "bullet pulling the barrel forward as a part of the
locking mechanism" is a figment of your imagination.
If you would simply take the time to look at the X-Ray photograph, you could see the figment of my imagination at work.. .and if you doubt that a bullet that is being swaged into a barrel with lands that are .006 inch smaller than its diameter doesn't hold that barrel forward... then you just haven't thought about it hard enough.
If those opposing forces will stretch the topstrap of a modern revolver, they'll surely deform those relatively small radial lugs in the 1911 pistol. I'd suggest that you try one that Brickeyee suggested. Push a bullet through a barrel from chamber to muzzle if you doubt that it happens.
You are correct on one point though...kinda.
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The slide alone has so much more mass than a bullet that it's IMPOSSIBLE
for the slide to open before the bullet has left the barrel.
The link can't draw the barrel out of engagement with the slide because of the slide's mass? No. It can't do it because...before the bullet exits...the lugs are horizontally engaged under pressure. Barrel being held forward and slide being shoved backward. That's what achieves the locked part of the locked breech. Before the gun fires, it's not actually locked. It's just held in battery by the force of the recoil spring, and...in a tightly fitted gun...a sort of wedging action. But locked up ? No.
I know that you think you understand this thing...and you seem to be pretty close...but you're still short on a few points. The part about recoil not starting until the bullet exits was the biggest clue. As we say here in the south: "That dog won't hunt."
And...If there's anything that I've learned over the course of my life...it's that, 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.
I'd suggest that you go and study on it long and hard. I'll give you my E- mail address if you'd like, so that if you have any questions, you can write to me without having to resurrect what is likely to be a dead thread.
Luck!
PcMakr...No. The bullet cant' accelerate after it exits. There is no force acting on it other than its momentum...which gravity and friction are fighting.
presspuller...He's correct. The barrel and slide are locked together by the lugs...and pressure.
Blowback designs delay the breech opening...the separation of barrel and breechblock...by slide/bolt mass and/or spring strength. The locked breech/recoil operated weapon functions under the exact same mechanics...but uses a different mechanism to delay the breech opening. Namely...by tying the barrel and slide together, the barrels mass is added to the slide's. In the case of the 1911...that equals about 40% the slide's mass. The LB/RO design is much more forgiving of spring strength.
"*sigh*
If you would simply take the time to look at the X-Ray photograph, you could see the figment of my imagination at work...and if you doubt that a bullet that is being swaged into a barrel with lands that are .006 inch smaller than its diameter doesn't hold that barrel forward...then you just haven't thought about it hard enough."
Looking at that radiograph, I don't see the BARREL pulling the SLIDE FORWARD, I see the SLIDE pulling the BARREL REARWARD (and the position of the slide on the frame supports that); if these "barrel pulling forward" forces are as strong as you say they are, what happens to them when an identical cartridge is fired in a firearm that doesn't HAVE a slide or barrel to recoil? They can't magically disappear, so they have to be so small as to be immeasurable when we're firing those guns.
"The link can't draw the barrel out of engagement with the slide because of the slide's mass?"
No, the link doesn't have anything to do with it, because it doesn't come into play until after the bullet has already exited the barrel. If you imagine a straight blowback, while the bullet is travelling down the bore, you essentially have a bullet on one side of an explosion in a tube, and the slide on the other side of that explosion. Because the slide is so much heaver (at least 30 times, in the case of a 1911 GM), it only travels a short distance before the bullet exits that tube, and the pressure drops to ambient (and this would happen regardless of any locking or delaying mechanism to affect that slide).
"The part about recoil not starting until the bullet exits was the biggest clue."
I probably should have explained what I mean by this; until the slide (with the barrel locked to it) starts to transfer its motion to the frame, from the shooter's perspective, there ISN'T any recoil. He's just holding a frame, and the little one-stroke internal combustion engine that's riding on the rails of that frame doesn't even begin to affect him until a long time (comparatively speaking) after the bullet has already left the barrel.
I understand that the link doesn't come into play until the bullet exits...unless the gun is badly out of time. We covered that. The link starts to draw the barrel down at the linkdown point...at .100 inch of rearward barrel travel...Remember?
Look...Here's your chance to prove me utterly and completely wrong.
Go get an unfired bullet. 185..200...230 grain. It doesn't matter. Take the barrel out of your pistol...drop the bullet into the chamber...and use a dowel rod to push the bulllet through the barrel and out the muzzle. Be sure and grip the barrel in your hand. No fair using a bench vise. If your theory is correct, it should be a piece of cake...even if you only use your thumb and finger to hold the barrel.
Come back and tell us what you learned.
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He's just holding a frame, and the little one-stroke internal combustion
engine that's riding on the rails of that frame doesn't even begin to affect
him until a long time (comparatively speaking) after the bullet has already
left the barrel.
I think we can disregard that statement as utterly and completely wrong...but as noted...you really don't have a clue how this thing works.
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If you imagine a straight blowback, while the bullet is travelling down the
bore, you essentially have a bullet on one side of an explosion in a tube, and
the slide on the other side of that explosion. Because the slide is so much
heaver (at least 30 times, in the case of a 1911 GM), it only travels a short
distance before the bullet exits that tube, and the pressure drops to ambient
(and this would happen
Almost. When the barrel is added to the slide, the mass is about 40 times greater.
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Look...Here's your chance to prove me utterly and completely wrong.
Go get an unfired bullet. 185..200...230 grain. It doesn't matter. Take the barrel out of your pistol...drop the bullet into the chamber...and use a dowel rod to push the bulllet through the barrel and out the muzzle. Be sure and grip the barrel in your hand. No fair using a bench vise. If your theory is correct, it should be a piece of cake...even if you only use your thumb and finger to hold the barrel.
The trouble is, this experiment doesn't tell you ANYTHING about how the force it takes to drive a bullet down a barrel is supposed to affect the cycling of a 1911; while the bullet is in the bore, that barrel may as well be a tube that is permanently WELDED CLOSED at one end. 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?
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I think we can disregard that statement as utterly and completely
wrong...but as noted...you really don't have a clue how this thing works.
How so? The force of the slide recoiling isn't transferred to the frame (and from there, to the shooter) until AFTER the slide compresses the recoil spring and or/reaches the end of its travel, all of which happens "a long time (comparatively speaking) after the bullet has already left the barrel."
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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.
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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.
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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?
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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.
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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.
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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.
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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.
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.
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It is the forward pull that holds everything locked even as the slide and
barrel recoil together.
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.
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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.
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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.
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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.
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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.
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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.
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?
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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.
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.
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.
One...Two...Three!
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
[QUOTE]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.
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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.
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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.
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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.
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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...
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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.
That (slightly restated) IS the Blish principle, something which doesn't actually exist. Blish said that he believed that this theory "worked" best when two dis-similar metals were used, but experiment after experiment has failed to show that it has any basis in reality.
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So, now you're trying to use a completely different set of dymics into the
argument in order to support your beliefs?
Not at all, I'm just using it as a similar example to point out that your argument doesn't make sense. If "bullet pull" was as crucial to the operation of a 1911 as you seem to believe, then it should produce a measurable effect in that operation. The trouble is, it doesn't. The pistol unlocks the same way and at the same point in slide travel when you fire it with live ammo as it does when you cycle it by hand.
Reading this over, I am surprised that no one has done up a full blown 3D computer model of the 1911 in operation.
That would make a really interesting PhD dissertation on exactly the forces involved in making a 1911 run and then have computer simulations that allow one to adjust the dynamics of the model.
I think this is the only way that these questions could be solved for most folks. I say most because some will always argue the simulation does not represent reality.
Blish!
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That (slightly restated) IS the Blish principle, something which doesn't
actually exist.
Not really. Let's try yet another hypothetical...
Attach a cable to a quarter-inch steel plate, 1 foot square. Lay the plate on top of a steel block that weighs 1440 pounds. Place an identical steel block on top of the plate. You now have a force of 10 pounds per square inch on the plate. Do you believe that you can pull the plate out from between the blocks by pulling on the cable...using only the force that your body can apply?
Assuming again that the barrel is indeed held forward and the slide is moving rearward...and the locking lugs are engaged horizontally in a shearing fashion...there is 20,000 pounds of force per square inch bearing on the lugs at the instant of peak pressure.
Assuming that the total surface area of the lugs is roughly a half inch square...there is an opposing force on the lugs equal to 5 tons.
Let's try another one...using a simple bolt-action rifle.
.308 caliber...52,500 CUP, which works out to about 57,000 pounds psi at peak. If you were fast enough...and could time it perfectly...do you believe that you could lift the bolt and unlock the breech at the precise instant of peak pressure?
Do you believe that poor little link could pull the barrel out of engagement with the slide...at the precise instant of peak pressure? Even at 10% of the peak pressure...when the bullet is about to clear the muzzle...there's still a thousand pounds of opposing force on the lugs.
You don't need to believe in the Blish principle to know that it would be impossible. Newton's laws will do.
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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.
In other words, wear/damage to the front faces of the barrel lugs can happen without any barrel/bullet friction as long as any significant impulse type force is directed against the breechface. Barrel/bullet friction is not necessary to account for the effect seen in the X-Ray pic.
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Because nothing CAN move.
In other words, the chained blocks provide a very poor analogy for the 1911 system. Essentially there's no analogy for the chain whether it's taut or loose at the beginning of the experiment.
BTW, the revolver analogy (stretched frames) is a poor analogy as well. The stretching is due to the actual impact of the bullet on the forcing cone-- which can generate enough force to split the cone in the extreme case. If it were due to bullet/barrel friction we would see stretched autopistol barrels and stretched rifle barrels.
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Not really. Let's try yet another hypothetical example...
Attach a cable to a quarter-inch steel plate, 1 foot square. Lay the plate on top of a steel block that weighs 1440 pounds. Place an identical steel block on top of the plate. You now have a force of 10 pounds per square inch on the plate. Do you believe that you can pull the plate out from between the blocks by pulling on the cable...using only the force that your body can apply?
No, but we're not discussing what MY BODY can do, remember? The issue is whether or not this supposed "bullet pulling" force plays any part in the operation of the pistol, and I haven't yet seen any evidence that it does. I would be very interested in seeing an experiment performed along the lines of my earlier suggestion, however (a barrel minus the lugs, so it'll work as a straight blowback).
How about a practical experiment that I have actually TRIED? I don't know if you've ever seen a Richardson Guerilla Gun, but they're basically a "factory- made zip gun", sold after WW2 (Ilits Richardson was a USN lieutenant who saw Filipino guerillas using these things against the Japanese, and figured he could sell plenty of them after the war). These shotguns are simply open tubes (12 gauge size) that slide into another tube that's bolted onto a simple stock, and the rear tube has a fixed firing-pin that fires the shell when the front tube/barrel is pulled backwards into it. A typical 12-gauge shell produces MANY THOUSANDS of pounds of pressure when it fires, and that pressure is trying to blow the front and rear tubes apart, yet I have no difficulty at all holding that barrel closed during firing, simply because the pressure is gone so fast that it's not sustained for any amount of time. The same thing happens with this mystical "force" that you claim is pulling the barrel forward.
Although I think the inertial impulse force is the greater, I would suggest that the impulse force of the bullet striking the rifling, together with the torque reaction vector is quite considerable. I would also suggest that the bullet friction during it's passage through the barrel is much lower than one might think but still quite significant. All these factors (at which we are all just guessing, at least, I am) simply infuence the required spring force and slide mass to keep the pistol cycling in proper ballance. Removing the locking lugs would necessitate a much heavier slide and stronger spring, otherwise, the breach would open way too soon, and and strike the slide stop way too hard, whatever the force mechanism is.
"that pressure is trying to blow the front and rear tubes apart"
Well, the frame and outer tube would certainly be pushed backwards, but since the inner tube, the "barrel", is the same diameter as the shot column and has no rifling, there will be little or no friction between the shot column and the barrel and, therefore, little, if any, forward force exerted on it. Your experience firing it is what I would expect.
Would it were not so - I'd like a .460 Rowland but......
Would it were not so - I'd like a .460 Rowland but stretching the slide until it cracks worries me a whole lot more than pressure on the breech face or stopping the slide in recoil on the bowtie area or maybe a buffer or two.
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The M1894
[that's the year folks and the jump from this to Browning's designs shows some of the genius of JMB]
Steyr Mannlicher Blow-Forward, Semi-Automatic Pistol
This earliest Steyr Mannlicher pistol, manufactured by FAB.D'ARMES Neuhaussen, Switzerland, was designed to be self loading and to use a special rimmed cartridge in 6.5 mm caliber. The design represented an entirely new utilization of mechanical principles in automatic action called "blow-forward action". In the standard type of automatic action for low powered cartridges, the recoil (or blow-back) is utilized to drive back a movable breech face or block, Mannlicher utilized the principle of a rigid standing breech with the barrel blowing forward to extract, eject, and prepare for reloading. The blow- forward principle has been utilized in very few firearms.
A special barrel housing which carries the sight covers the entire length of the barrel (6.49 in/165 mm) when the arm is closed. A heavy recoil spring is mounted concentrically around the barrel within this housing and is compressed between a shoulder at the forward end of the casing and a shoulder at the rear of the barrel.
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United States Patent 5123329
Abstract:The present invention comprises a self-actuating blow forward firearm having a fixed breech block located to the rear of the firearm. A cartridge feeder provides a supply of cartridges to the loading area so that the firearm can be used as an automatic or semi-automatic weapon. The firearm also contains an outer receiver tube and a movable inner gun barrel which is biased toward the rear of the firearm by a spring disposed in a chamber between the receiver tube and the inner movable barrel. When the trigger mechanism is activated, the action of the bias spring causes the inner movable barrel to move backward toward the breech block striping a cartridge from the feeding device and forcing the end of the cartridge against the firing pin on the breech block. The bullet is fired down the bore of the inner movable barrel. The movement of the bullet along the bore of the inner movable barrel drags the barrel forward against the action of the bias spring. The addition of a piston device attached to the muzzle end of the inner movable barrel causes added momentum to be delivered to the moving barrel to assist the barrel in its forward movement. This added momentum can be further enhanced by including a muzzle brake adjacent the piston device. The battering of the various parts of the firearm and the cycle speed of the firearm can be reduced by use of an air pressure chamber situated between the inner movable barrel, the receiver tube, the barrel guide and a shoulder on the inner movable barrel.
I think if you re-read my post you'll see that I'm not saying that there isn't any bullet/barrel friction effect, just pointing out that it's not necessary to explain wear on the front of the barrel locking lugs.
As far as the two block example, it's an interesting physics experiment--I just don't see the application to this topic.
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I'm not saying that there isn't any bullet/barrel friction effect, just
pointing out that it's not necessary to explain wear on the front of the
barrel locking lugs.
Yes it is, John. You just haven't throught it through. How can the slide pull the barrel rearward if it doesn't have some mechanical connection to pull it WITH? What is that mechanical connection? Go look at the X-ray. The front faces of the barrel lugs are clearly being pressed against the rear faces of the slide lugs....AND...because there is a signifigant FORWARD drag on the barrel by the BULLET...it's pulling that barrel backward against resistance...against a force that is trying to push it forward...whether or not you want to believe it.
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As far as the two block example, it's an interesting physics experiment--I
just don't see the application to this topic.
Just trying to illustrate how the more massive slide can pull the less massive barrel rearward once things get moving. First, they have to be ABLE to move.
Once momentum is established, the heavier object will overcome the lighter one's resistance and cause it to change direction. If the chain is taut...and the force applied isn't large enough to snap the chain...nothing CAN move...so nothing DOES. If nothing moves, there is no momentum. None. Zip. Nada.
Because...
Momentum is the product of mass times velocity. The mass has to move before momentum is established. Mass doesn't move unless a force is imposed on it. The bullet can't push the slide backward by moving forward.
It's travelling in the wrong direction to push the slide. (It is, however, imposing a forward drag on the barrel.) The push comes from the force vector that is operating between the two objects. The bullet merely provides a resistance for the force to push off of...the same as the slide does for the bullet.
Come on Tuner, you admitted earlier on the thread that the results of my punch experiment would result in wear to the front of the locking lugs with absolutely nothing touching the barrel. The inertia of the barrel is sufficient resistance.
The barrel clearly also has inertia when the gun is fired and therefore the locking lugs will show wear from the firing process (just as from the punch experiment) even if it were possible to completely eliminate the bullet/barrel friction.
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Just trying to illustrate how the more massive slide can pull the less
massive barrel rearward once things get moving.
The slide and barrel are a single unit at the moment of firing due to the locking mechanism, therefore they move together.
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Just trying to illustrate how the more massive slide can pull the less massive barrel rearward once things get moving. ...... ..... The bullet can't push the slide backward by moving forward. ..... The bullet merely provides a resistance for the force to push off of....Nope.
The gasses acting against the rear of the case push the case back which in turn pushes the slide back which in turn pulls the barrel back which in turn is being held forward, largely by inertia. And this is Newtonian physics.
Simply put - Force = Mass X Acceleration
The slide is acceleration the barrel rearward
The barrel has mass
Therefore, a force is exerted on the locking lugs of the barrel.
If you can give me the weight of the slide, the weight of the barrel and the weight of the bullet - we know the muzzle velocity of the bullet, then I will do the math and show you how it works. I will also show you the forward drag of the bullet, the forward impulse of the bullet entering the barrel as well as the forward force vector from the torque reaction of the bullet being angularly accelerated. It will take me a few days 'though.
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The gasses acting against the rear of the case push the case back which in
turn pushes the slide back which in turn pulls the barrel back which in turn
is being held forward, largely by inertia. And this is Newtonian
physics.
Simply put - Force = Mass X Acceleration
The slide is acceleration the barrel rearward The barrel has mass
Therefore, a force is exerted on the locking lugs of the barrel.
That's what I've been saying all along...but you've forgotten one resistive force...and it's larger than the barrel's inertial resistance. I'll let ya think about it.
I'm tired of arguing the same points...so I'll just put up a few facts and let everybody figure it out for themselves. This is a little involved, and it took some time to write. All I ask is that everyone pay me the courtesy of reading it through...carefully.
Review Newton 1 and 3. They'll help. Newton 2 is a little less helpful, but go ahead and look at that one too if you like. It's related.
The straight blowback operation is simple. Force on the bullet and force on the slide pushes each in opposite directions. Any argument? No? Good.
The locked breech design works the same way. Powder burns...Gasses expand..Pressure builds...Force on the bullet pushes it through the barrel...Force on the slide pushes it rearward.
The only difference...The ONLY difference...is that in the blowback, the slide moves independently of the barrel, while with the locked breech pistol, the slide snags the barrel on its way rearward and drags it along with it. Because the barrel has mass...the slide has to overcome that. Because the two are mechanically connected, the are...for all practical purposes...one part. So, until the barrel reaches the linkdown point, the slide's mass is roughly 40% greater than it is AFTER the barrel drops. This added mass slows the slide's acceleration...giving the bullet more time to reach the muzzle. Breech opening is thus delayed. It's very nearly a delayed blowback operation...but because the slide doesn't move independently of the barrel...we have to call it locked breech.
But the barrel's mass isn't the only resistance that delays the slide. There's another, more powerful force at work...the one we've been fighting about.
The bullet is a tight fit in the barrel. Any argument? You! There...in the back. Go push a bullet through a barrel with a stick and report your findings.
Because the bullet is a tight fit...and because it stays in contact with the barrel's inner surface for the whole trip...it transfers a frictional force to the barrel in the direction that it...the bullet...is moving. Namely forward. The bullet doesn't shrink, and the barrel doesn't expand to any practical degree, so the force remains in effect as long as the bullet is in the barrel and moving forward.
Because there is a forward force on the barrel, we have to factor it in. Don't believe it's signifigant? Install a 6-inch Longslide barrel in a Commander. Lock the gun in a vise. Grab the end of the barrel and pull it forward while a friend pulls on the slide.
We have another resistive mechanism. The recoil spring. We can agree that the recoil spring doesn't provide very much resistance during 1/10th inch of compression...until we remember that, in a 5-inch pistol with a standard 32- coil/16-pound spring...there's about 4.5 pounds of preload before anything even moves. So...bearing in mind that any resistive force will become more resistive the faster and harder that we try to accelerate it...that 4 pounds of resistance will multiply.
And there is yet another one. The spring also has mass. Gotta move that, too...and the faster and harder you push on it...the harder it fights.
These last two are operating in both the blowback and the locked breech pistol. I included them to provide SD with a clue so that he can figure out why the gun moves before the bullet exits the bore...and there's one more mechanism that contributes to that. I'll leave it to him to find it. Happy huntin' m'fren.
Finally...Nothing can be discounted when trying to understand the reality. Nothing. Anything that has the opportunity to delay the slide's movement WILL delay it. Anything that CAN resist it...WILL resist it. Resistive force comes in three forms. Inertial...provided by the mass of the object. Frictional force...provided by the object's contact with anything else. Mechanical obstruction. The proverbial brick wall.
We can't see force. We can only see its effects. That we can't see it doesn't prove that it doesn't exist. The barrel can't actually be pulled forward very much...if at all...because there are mechanical obstructions blocking it. The most obvious is the slidestop pin to lower barrel lug contact. Less obvious is the locking lug engagement within the slide...which is being driven rearward. BUT...just because we can't see the barrel move is not proof that the force on it isn't there. If you push on your house, you can't move it...but the force is still there...pushing on the house. While the bullet is moving thought the barrel...it's transmitting a forward force TO the barrel...even though the barrel can't move forward. It thus becomes a resistive force for the slide...and the faster the slide moves, the harder this force works against it.
To wrap it up...I hope that by now, we're all in agreement that mass requires force to set it into motion...and that the bullet can't generate a directional force to the slide while it's moving away from it. There must be a force vector between the two...and that force must be an unbalanced force. That is...the force must be greater than the resistance offered by the object that it's trying to move. When the force is balanced...force and resistance are equal...we have equilibrium. One cancels out the other, and nothing moves. Only when the force is unbalanced do things start moving. Only when things move do we have momentum.
I'm not gonna read this whole thread, but Tuner, if that forward friction thing is SOOO huge, and the bullet's movement don't cause the recoil, then howcome my revolvers and bolt guns recoil too?
I recall something about "total mass of recoil ejecta" being the thing to consider in calculating recoil. Explains why my 125-gr bullets with 2X.X grains of WW296 kick just as hard as 146-gr bullets with 9.whatever Unique (or whatever the load was) at the same velocity. But aren't lighter bullets supposed to kick less? Yeah, but the bullet plus powder charge weighed the same.
I thought recoil forces were being applied to the launch platform long before those forces could overcome inertia and commence perceptible movement.
Tuner, nice explanation:
The "other significant force", leaving aside minor forces, would be the firing pin stop engaging the hammer to re-cock it and the subsequent compression of the main spring. This is actually a significant force re the unlocking time. This occurs almost immediately when the slide tries to move backward and its effect is probably over before the bullet leaves the barrel.
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if that forward friction thing is SOOO huge, and the bullet's movement
don't cause the recoil, then howcome my revolvers and bolt guns recoil too?
Lemme see if I can find the words to explain it...
When the bullet moves through the barrel in one direction, the barrel can also slip backward. Frictional resistance doesn't mean that it can't move. Only that its movement is resisted.
You may be better able to see it if you try this: I call it the rope trick.
Wrap a length of rope around your right hand, and grip the rope with your left hand. Pull against the rope in opposite directions with gradually increasing force...until the rope slips. Watch closely. When the rope slips through your hand in one direction...your hand slips off the rope in the opposite direction. This is action/reaction. You may need to do it several times before you realize what you're watching...but you'll eventually see it.
That's why the guns recoil SOOOO hard.
Because...
When the bullet moves forward, the barrel can move backward...so that when the force that bears against the breechblock...bolt face...recoil shield...or the closed breech of a smoothbore musket...that force shoves the gun backward. Because the barrel is attached...it goes whichever direction that the receiver goes.
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The "other significant force", leaving aside minor forces, would be the
firing pin stop engaging the hammer to re-cock it and the subsequent
compression of the main spring.
You've got it!
When the slide bears against the hammer, it exerts a linear force that transfers to the frame of the gun...rolling the muzzle upward while the bullet is still fightin' its way to open air.
There are three recoil values that concern us. Recoil energy, which is equal to the ejecta energy. Recoil momentum, which is a function of energy and gun mass. Then we have recoil velocity - which is what we feel.
I know you asked Tuner, but if I may just add, the recoil is the result of the ejecta energy regardless of the launch platform's mechanism.
Hi Peter, and thanks for using that approach...the scientific one.
The ejecta is the sum of bullet mass and that of the particulate in the gasses and smoke from the burned powder. The mass of those gasses is nearly the same as that of the unburned powder...so there is a recoil impulse after the bullet exits. It's just too small to detect during the cycle of the gun. Figuring a typical 5-grain charge at about the speed of sound...say a thousand fps here to keep it simple...and ignoring the blowby gasses that escape past the bullet at the instant of ignition...that impulse would be roughly 1/8th that of a .22 Short fired from a 2.5 pound pistol. Pretty insignifigant, but still part of the whole. With bottleneck rifle calibers, you have a little different set of dynamics at work, but that's meat for another discussion.
The nominal .45 ACP bore diameter is .451 inch. Land diameter is .005-.007 inch smaller, depending on the barrel manufacturer. Nominal bullet diameter is .451 inch. Assuming nominal dimensions, even a non-rifled bore would present a good bit of resistance to the bullet, since the plug is the same size as the hole.
Of course, bullet material is also a factor. Harder jacket alloys offer more resistance, as do harder lead cores. Lead bullets offer less for a given diameter...but alloy also is a factor. Some lead alloys are harder than others.
A basic rule of thumb is that lead bullets need to be sized to .001 inch larger than bore diameter for best performance. I size my cast bullets to .452 inch.
For your calculations, you'll need to weigh a slide, a barrel, and a bullet to determine the ratio. For all practical purposes, 40:1 will do.
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I know you asked Tuner, but if I may just add, the recoil is the result of
the ejecta energy regardless of the launch platform's mechanism.
Yep. It doesn't matter whether the gun is a semi-auto or a musket. Force forward is force backward. The semi-auto's design simply spreads the recoil impulse out over a longer time and distance. That's why a self loading shotgun's recoil is more like a hard shove, while a pump of equal mass is a sharper punch.
Oh...One more thing for Grump. You have to grip the rope a distance from the end, or you probably won't see it.
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stretching the topstrap comes from the bullet hitting the forcing cone
Think about that statement, John...and you're partially correct...but you're ignoring the fact that there still has to be an opposing force that is driving the frame away from the bullet in order to cause a tensile stress on the topstrap. Otherwise, the gun would pitch forward. There can't be one force in one direction. Force forward...Force backward. Remember?
Where you scored is...The most resistance is offered at the inception of bullet acceleration...within the first half-inch or so. Objects at rest and all. The same goes for trying to force a plug into a hole that's smaller than the plug. The harder and faster you push...the harder it resists.
Which reminds me...
Peter. One other thing. The contact resistance between bullet and barrel isn't constant. It's much greater at the very beginning, when the bullet hits the hard resistance in the leade...before it gets started into the rifling. Always harder to get an object moving than it is to keep it moving. That's why pressure peaks early and fast, and that's where 90% of the punch is delivered to the breechblock and locking lugs.
As the bullete accelerates, pressure also drops fast. There are two things at work here. The faster it's moving, the less force it takes to keep it moving...and to accelerate it. Since force forward=force backward...the force against the lugs also drops off.
And...Because the volume of the cylinder behind the bullet is increasing. So...The forces at work are sudden and high at the start...and diminish as the bullet moves.
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The front faces of the barrel lugs are clearly being pressed against the
rear faces of the slide lugs....AND...because there is a signifigant FORWARD
drag on the barrel by the BULLET...it's pulling that barrel backward against
resistance...against a force that is trying to push it forward...whether or
not you want to believe it.
You're ignoring the fact that that barrel would be forward against the lugs in any case, simply because the slide is trying to pull everything BACKWARDS; the inertia of the barrel itself is enough to make sure that the barrel will be pressed forward. And, since there's no mechanism for unlocking the pistol until well AFTER the bullet is long gone (which would be the case, locked breech or not, simply because the slide has so much more mass than the bullet), claiming that this "bullet pulling the barrel forward" hocus-pocus plays any part in the operation of the pistol is like claiming that a magical pixie sits inside the pistol, telling it what to do and when.
Not ignoring anything SD...but it sounds like you've evolved a little.
Tell me. Did you sleep through your physics class, or just skip it altogether?
Does the term "DRAG" fire up any light bulbs for ya?
Definitely; the DRAG of the slide (acting under the impulse of firing) is what makes the pistol operate, while the "DRAG" of the bullet is irrelevant, since the pistol operates identically with or without it.
As far as I can see, there is absolutely NO EVIDENCE that the friction of the bullet travelling down the barrel plays any part whatsoever in the operation of the pistol; if it DID, then the pistol would operate differently when you fire it with ammunition than when you operate it manually.
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As far as I can see, there is absolutely NO EVIDENCE that the friction of
the bullet travelling down the barrel plays any part whatsoever in the
operation of the pistol
Well...Okay then. As far as you can see is your opinion and you're welcome to it...but I know what I know.
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As far as I can see, there is absolutely NO EVIDENCE that the friction of
the bullet travelling down the barrel plays any part whatsoever in the
operation of the pistol; if it DID, then the pistol would operate differently
when you fire it with ammunition than when you operate it manually.
Well when you can cycle it manually in the same time frame it does under shooting conditions then I guess it will not matter.
Since it's pretty clear that you haven't taken a look at the original patent specifications, I've copied a few lines...verbatim...of what is contained therein. Here, the designer...the inventor...The man himself is telling you how the pistol works. He didn't go into great detail, but common sense, logic, and a basic understanding of physics will provide the rest of the story.
Of course...Browning was another moron who believed in pixie dust...so how can you take him seriously?
The text from the original patents:
"The presentation and introduction of a loaded cartridge into the chamber of the barrel-and CLOSING AND LOCKING of the breech-are automatically effected through or by the energy of the recoil of the breech bolt-or that part which AT THE TIME OF THE FIRING OF THE SHOT closes the breech of the barrel."
Skip over to:
"For the additional reason that-at the commencing of the recoil-that the light barrel may readily yield to, and move rearward with the breech bolt."
Can anyone explain how the Schwarzlose blow-forward action worked with a standing breech and movable barrel, if the bullet does not drag the barrel forward as Tuner claims?
Or is that another can of worms?
Tuner, you know as well as I do that Browning's patent says nothing about using "bullet pull" to keep the breech locked; that's what the LUGS are there for, remember?
OTOH, the Schwarzlose clearly DOES use some element of friction between the bullet and barrel to operate, but it was designed to operate as a "blow- forward" design (the energy of the cartridge firing is directed against both the bullet and the barrel, and the bullet being substantially lighter, the bullet ends up being propelled faster). If the 1911 operated in a similar manner, Tuner might have a point, but since the pistol is locked and unlocked by a mechanism that is totally UNRELATED to "bullet pull", I remain unconvinced of his hypothesis.
Hmm, or how a blank would work without drag on the barrel. I haven't read the entire post so pardon me for stepping in late. But there's a lot of typing gone by.
Peterotte,
"If you can give me the weight of the slide, the weight of the
barrel...then I will do the math and show you how it works"
I weighed the slide & barrel from my PC1911 using a trigger pull gauge and got 14oz for the slide and 3.3oz for the barrel. These may not be exact and they vary from gun to gun anyway but they'll give you something to work with.
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but since the pistol is locked and unlocked by a mechanism that is totally
UNRELATED to "bullet pull", I remain unconvinced of his hypothesis.
But the only time it is locked is during firing. When the bullet is pulling the barrel against the slide lugs.
Any other time it is not locked.
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But the only time it is locked is during firing. When the bullet is pulling
the barrel against the slide lugs.
Any other time it is not locked.
No, the pistol is locked at any point between full battery (when the slide is forward) and the moment of unlocking (the point where the link pulls the rear of the barrel down and the locking lugs out of their recess in the slide). And, as JohnKSA's experiment shows, the pistol doesn't require any sort of "bullet pulling force" to operate the way it does.
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No, the pistol is locked at any point between full battery (when the slide
is forward) and the moment of unlocking (the point where the link pulls the
rear of the barrel down and the locking lugs out of their recess in the slide)
uh...if it was locked then you could not open it by hand. Just as you would not be able to open it during that fraction of a second that the bullet is going down the barrel.
THEN it is locked.
I'm not so hot on the workings of a pistol but I have a question. When the cartridge is discharged, the case expands tightly against the chamber, not so? How much 'grip' does the rearward-moving case have on the chamber, and is this a significant rearward force applied via the case to the barrel?
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the case expands tightly against the chamber, not so? How much 'grip' does
the rearward-moving case have on the chamber
With a straight-walled pistol case...enough to seal the chamber and prevent most of the gas blow-by, but not enough to keep the case nailed in position. Again...Different set of dynamics at work with a bottle-necked rifle case operating at nearly 60,000 pounds psi. Good question, though.
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I think that what you and I were doing is thinking of bullet drag down the
barrel? It's bullet resistance as it gets swaged into the leade AND the
initial static friction that Tuner is refering to (Am I correct there Tuner?).
That's where 90% or more of the force comes in, but the bullet is causing a forward drag on the barrel as long as it's in the barrel and moving.
If anything CAN have an effect, it WILL have an effect. Nothing is everything, but everything is something.
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that's what the LUGS are there for, remember?
No no no! The lugs are there to keep the slide and barrel from separating under pressure. The lugs don't lock the gun when it's static/in-battery. The just mesh together...like gear teeth.
Re-read this part...carefully this time.
"The presentation and introduction of a loaded cartridge into the chamber of the barrel-and CLOSING AND LOCKING of the breech-are automatically effected through or by the energy of the recoil of the breech bolt-or that part which AT THE TIME OF THE FIRING OF THE SHOT closes the breech of the barrel."
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but it was designed to operate as a "blow-forward" design (the energy of
the cartridge firing is directed against both the bullet and the barrel
No. That ain't what it said in the description posted...but I guess that now the people who engineered it don't know how it works any better than Browning understood the 1911.
Here's a cut and paste exerpt of the blowforward description. You may want to go back and take another look.
"The bullet is fired down the bore of the inner movable barrel. The movement of the bullet along the bore of the inner movable barrel drags the barrel forward against the action of the bias spring."
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uh...if it was locked then you could not open it by hand. Just as you
would not be able to open it during that fraction of a second that the bullet
is going down the barrel.
THEN it is locked.
It's physically locked at any point during that travel, unless you believe that two or more solid pieces of metal can pass through one another, and it has nothing to do with whether or not a bullet is passing down the bore; the fact that the slide is locked to the slide during part of the firing cycle is a nice feature and all, but the fact is, it would still operate as a straight blowback because there is no way that the slide can open as fast as the bullet can clear the barrel.
Apply that thinking to what you're arguing for just a second, Tuner. You're claiming that this impressive "bullet pulling" force is what's ACTUALLY "locking" the 1911 during operation, yet the various blow-forward designs (Schwarzlose, Hino-Kinomura, a few others) show that this force is next to nothing; if it WASN'T, you wouldn't be able to COCK them by pulling the barrel forward.
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I think that what you and I were doing is thinking of bullet drag down the
barrel?
That's what Tuner is trying to convince me of, but failing If something like this mystical "bullet pulling" force has enough energy to do what he CLAIMS it does, we should be able to see some evidence of it in operation, for example the open-bolt 20mm Oerlikon, or when using different bullet/projectile weights.
"...it would still operate as a straight blowback..."
If I may jump in, I don't think that there is any question about whether it could function as a straight blow-back action: it probably could, at least for a little while. The point of locking the barrel & slide together is to reduce the speed of the recoiling slide enough to keep the gun from beating itself to pieces and from venting gas through the ejection port. At least, that is my understanding.
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it would still operate as a straight blowback because there is no way that
the slide can open as fast as the bullet can clear the barrel.
This just gets better and better...
It's not necessary for the slide to open very far to blow a case head...which is why delaying breech opening is critical. All it has to do is back away from the barrel far enough to let the case back out about .015-.020 inch while the pressure is still on.
Read that again, and understand that the function of the locking lugs...the ONLY function...is in preventing the slide and barrel from separating before the bullet exits. That occurs horizontally. There is no positive lockup before the gun fires. None.
And it DOES function exactly like a straight blowback. The only difference is that the slide doesn't move independently of the barrel until the barrel links down...after the first 1/10th inch of travel.
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that's what the LUGS are there for, remember?
No no no! The lugs are there to keep the slide and barrel from separating under pressure. The lugs don't lock the gun when it's static/in-battery. The just mesh together...like gear teeth.
The slide most certainly IS locked to the barrel when the pistol is in battery, unless you believe that the metal of the barrel can move through the metal of the slide.
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Re-read this part...carefully this time.
"The presentation and introduction of a loaded cartridge into the chamber of the barrel-and CLOSING AND LOCKING of the breech-are automatically effected through or by the energy of the recoil of the breech bolt-or that part which AT THE TIME OF THE FIRING OF THE SHOT closes the breech of the barrel."
And it's clear to me that you're reading something into this description; this simply says that the energy of firing a shot is used to operate the mechanism of the pistol; that's all, and it makes no mention of "bullet pull" to do so.
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but it was designed to operate as a "blow-forward" design (the energy of
the cartridge firing is directed against both the bullet and the barrel
No. That ain't what it said in the description posted...but I guess that now the people who engineered it don't know how it works any better than Browning understood the 1911.
Here's a cut and paste exerpt of the blowforward description. You may want to go back and take another look.
"The bullet is fired down the bore of the inner movable barrel. The movement of the bullet along the bore of the inner movable barrel drags the barrel forward against the action of the bias spring."
In the blow-forward design, you have three important parts: a standing breech (which cannot move), a barrel (which can move) and a bullet (which can also move). Since the firing of a cartridge CAN'T move the standing breech, that means it has to move the other two components in the equation, the bullet and the barrel.
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And it's clear to me that you're reading something into this description;
Lemme go look at it again...
"The bullet is fired down the bore of the inner movable barrel. The movement of the bullet along the bore of the inner movable barrel drags the barrel forward against the action of the bias spring."
(sic) "The movement of the bullet along the bore...drags the barrel forward."
Not a lotta room for interpretation there...
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It's not necessary for the slide to open very far to blow a case
head...which is why delaying breech opening is critical. All it has to do is
back away from the barrel far enough to let the case back out about .015 inch
while the pressure is still on.
If this was going to happen, why don't we SEE it happening with designs that are meant as blowbacks from the get-go? We definitely see it happen when someone gets over-active when they're "throating" a barrel , but that's equivalent to having a hole in the chamber when you fire a round.
"why don't we SEE it happening with designs that are meant as blowbacks"
They use lower pressure cartridges?
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Lemme go look at it again...
"The bullet is fired down the bore of the inner movable barrel. The movement of the bullet along the bore of the inner movable barrel drags the barrel forward against the action of the bias spring."
That's from a description of the blow-forward action, not of the Browning locked-breech action; as I said earlier, if this "bullet-pulling" force was the immense factor it's claimed to be, we wouldn't be able to COCK a blow- forward pistol with our bare hands. Somehow, I don't think a pocket pistol that you'd have lay down and cock with your legs (ala the PIAT) would have made it into production.
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They use lower pressure cartridges?
Nope; the 20mm Oerlikon (which uses a bottlenecked case to boot) operates as a straight blowback, and I don't think you can consider it a "low pressure cartridge" by any stretch of the imagination.
It's not pressure, but recoil impulse, energy, and momentum that precludes using a straight blowback design for higher-powered rounds. Since the blowback utilizes slide mass and/or recoil spring resistance to delay the breech opening, anything stouter than .380 would make for a top-heavy, clumsy pistol. The High-Point pistol is a good example. The slide is massive.
But...For a mounted or crew-served weapon, the sky is pretty much the limit. It doesn't need to be portable, so the breechblock can be as big as it needs to be...and one could keep adding weight to it until it wouldn't move at all. Then again, you could do the same thing with a locked breech/recoil-operated weapon too.
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If this was going to happen, why don't we SEE it happening with designs
that are meant as blowbacks from the get-go?
Well...We do...if we have enough experience to actually understand how they work. All it takes for a blowback to do that is for the owner to neglect changing the recoil spring for a little too long in guns that opt for high spring tension to perform that function in order to keep the slide mass low. These guns have horrendously strong recoil springs for their size and caliber.
It's not pressure, but recoil impulse, energy, and momentum that precludes using a straight blowback design for higher-powered rounds. Since the blowback utilizes slide mass and/or recoil spring resistance to delay the breech opening, anything stouter than .380 would make for a top-heavy, clumsy pistol. The High-Point pistol is a good example. The slide is massive.
Which returns me to the example I posted earlier, of the Hi-Power which has had it's locking-lugs ground off so the pistol works as a blowback; it certainly works, but if the short-recoil system works without battering the slide or frame, why not use it? At the least, it shows that this "bullet- pulling" force isn't required for the pistol to operate.
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At the least, it shows that this "bullet-pulling" force isn't required for
the pistol to operate.
I never said that it's necessary for it to operate. I said that it's necessary for it to operate properly...as a locked-breech design.
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No no no! The lugs are there to keep the slide and barrel from separating
under pressure. The lugs don't lock the gun when it's static/in-battery. The
just mesh together...like gear teeth.
Gee whiz, Wally! Last time I had any tilt-barrel-locked slide/barrel combo off of a frame, the two were solidly locked together (in that all-important bore- axis "longitudinal" direction) whenever the barrel was up enough for the locking lugs to engage.
Sounds like some people are getting confused because pulling back on the slide when the thing is put together makes the slide act like the "unlocking" lever, so to speak. Take the slide stop out of your 1911 and tell me how well you can unlock the barrel/slide by pulling back on the slide, eh?
A little math please? a .452 bullet has a bit more than .0160453 square inch area. That means it takes about 6.23 of such circles to equal a square inch. IF it takes about 200 lbs of force to push the bullet down the barrel (educated guess from my limited "slugging" experience), then the "PSI equivalent" is about 1,246 PSI on the bullet base needed to push it through. Assume, with all dangers, a powder-puff load of 12,500 PSI, which *is* enough to obturate any lead bullet, and there's another 12,250 fps at peak pressure being used to overcome the bullet's intertia.
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Take the slide stop out of your 1911 and tell me how well you can unlock
the barrel/slide by pulling back on the slide, eh?
Of course it won't "unlock." The slidestop crosspin is what the link uses to draw the barrel vertically out of engagement with the slide. That's why the crosspin is inserted through the open hole in the link. Now...Let's see if you can tell us why the slide will only travel about a quarter-inch with the slidestop removed. Hmmm?
Don't help him! Let him answer.
Time's up, Grump. You may be excused. I suggest that you take some time studying the gun. eh?
Okay...This has been a hoot, but I'm gettin' bored. Time to drop the rest of the story. I'd hoped that, with all the clues, it would have been clear to the ones who cannot see...so I'm gonna go ahead and spell it out. Those that see it need no further explanation, and there's no sense in continuing to say the same things over and over.
The locked breech/recoil operated 1911 is almost...ALMOST...a delayed blowback pistol. The only minor detail that keeps it out of that class is the fact that the slide doesn't move independently of the barrel.
There are six factors that work to delay...resist...the slide's rearward trip to the stop surface.
They are:
Slide mass. The inertial resistance of the slide itself, including all the attached hardware.
The recoil spring. A variable resistance with a dual function.
The barrel's mass. Since the slide has to accelerate it backward, it must overcome the barrel's inertial mass.
The hammer's mass...which includes the attached hardware.
The mainspring's resistance...including the mainspring cap and the inertial mass of the spring.
The forward drag that the bullet exerts on the barrel by virtue of its tight fit and forced passage through the barrel. This force can't be underestimated because it's doubtless the most powerful and resistive of the six. Pushing the bullet through the barrel requires more force than is required to overcome the resistance offered by all the others combined.
This last one is very easy to demonstrate...if the non-believers will simply take the time to experiment with it...and think about it.
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The forward drag that the bullet exerts on the barrel by virtue of its
tight fit and forced passage through the barrel. This force can't be
underestimated because it's doubtless the most powerful and resistive of the
six. Pushing the bullet through the barrel requires more force than is
required to overcome the resistance offered by all the others combined.
And do you actually have any EVIDENCE to support this? Going by what we see in blow-forward designs, that "bullet-pulling" force is only enough to compress a recoil spring, and certainly doesn't deliver the thousands of pounds of force that you'd have us believe.
Holeshot. The slide in either one would start moving immediately...spring or no spring. The difference is that the locked breech pistol wouldn't allow the slide and barrel to separate until well after the bullet exits. The blowback...as long as the spring is good...is supposed to delay the slide long enough to let the bullet leave...but install a 3/4 power spring and watch the fireworks.
Ned Christiansen fired a 1911 pistol repeatedly without a recoil spring about 5 yeas ago to show that the spring didn't have any effect on unlock timing and barrel drop.
SDC...I'd suggest that you take a course in applied physics at a local community college, but at this point, I don't think you'd believe that either...and if you'd go do the cork and pipe experiment, you'd have all the evidence that you need.
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SDC...I'd suggest that you take a course in applied physics at a local
community college, but at this point, I don't think you'd believe that
either...and if you'd go do the cork and pipe experiment, you'd have all the
evidence that you need.
I could do that (to go with my other degrees), but if this is such a self- evident concept, it should be simple to prove, right? Which cork and pipe experiment in particular are you thinking of? I've seen at least half a dozen "cork and pipe" experiments, but don't recall any of them having any relation to what you're claiming. The biggest thing you could do to buttress your claim is explain how it is that this "bullet pulling force" is simultaneously so weak that it can only pull a couple of ounces of barrel against a spring (as in the various blow-forward designs), yet still be so powerful as to exert thousands of pounds of "locking pressure" against the lugs in a short-recoil design. You said yourself "They know that if they're fired in a blow-forward pistol that they have to drag the barrel forward to make it work, see? But if they're fired in anything else, they know not to do it. Simple!"
I'm gonna leave ya with one final clue...or actually remind you of something that you seem to be in denial about.
If it CAN affect it...it WILL affect it. You keep trying to ignore an involved force, and a formidable one...but do carry on. Hear?
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Take the slide stop out of your 1911 and tell me how well you can unlock
the barrel/slide by pulling back on the slide, eh?
Of course it won't "unlock." The slidestop crosspin is what the link uses to draw the barrel vertically out of engagement with the slide. That's why the crosspin is inserted through the open hole in the link. Now...Let's see if you can tell us why the slide will only travel about a quarter-inch with the slidestop removed. Hmmm?
Perhaps we're talking past each other, but the slide don't go back without a link because it's still locked to the barrel. Physical interference. Maybe you're talking about some "sticky" effect of the pressure of contact, but I still submit that of the 19,000 or whatever PSI of a .45 ACP round's peak chamber pressure, bullet drag on the barrel is truly there, can be measured, and ain't likely to be more than 2,000 PSI of it. The rest is gas pressure pushing equally in all directions including against the bullet and the breech. Bullet is moving, so the slide/barrel assembly moves backwards under recoil.
Betcha that high-speed photography of a 1911 firing a nice high-pressure blank will still show the slide pulling backwards and engaging the barrel's locking lugs. No bullet pull there. The design lets the barrel/slide assembly be locked together just through simple geometry.
"Almost" being a blowback is as nonsensical as (but in the opposite direction from" being a "little" pregnant. The pistol is a locked breech design, and the speed of the cycle is such that there is some residual chamber pressure left when the case clears the chamber on extraction. You can test the extraction "boost" of that pressure by firing the pistol without an extractor. Do .38 Supers work better this way than .45 ACPs?
I'm not the mathematician to calculate the resistive effect of the slide mass on its rearward movement. I'd be far more convinced if there were some real numbers attached to this argument. Why is it that I, a stupid attorney with not one college-level physics class, am the only one (it seems) to apply some rough numbers to this? Use all those degrees you have, guys!
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This force can't be underestimated because it's doubtless the most
powerful and resistive of the six. Pushing the bullet through the barrel
requires more force than is required to overcome the resistance offered by all
the others combined.
Come on, that's like saying "everybody KNOWS". It's a logical fallacy and unsupported by anything more than weak argument. Wish I could remember all that calculus as it applies to instantaneous acceleration problems. Your 22-lb recoil spring force dwarfs in contrast with the power of inertia when the time of force applied are measured in milliseconds!
I beg to differ. Remove the link from your barrel...assemble it completely...and manually draw the slide rearward. The barrel will fall.
Psssst! Grump! Keep lookin'. You've been wrong since you jumped into this one...and you get wronger with every post.
PS. You can't calculate it. The coefficient of friction is an empirical measurement...it has to be measured experimentally, and can't be found through calculations. If you'd paid closer attention in your high school physics class, you'd have known that.
It goes up...or down...depending on several factors. Materials of the contacting surfaces are only two. The load that is placed on either or both is the biggie. i.e. You can drag a board across a floor easily. If you place a 2,000-pound weight on top of it...it ain't so easy.
Try it upside-down, Tuner. Gravity substituting for a link doesn't prove the barrel and slide are unlocked when forward and in battery.
A "Thought Experiment" for this tiresome thread
Okay, brainflash (or fart, depending on your prejudices).
Two Yugos, parked and stationary. One is on clean dry pavement on earth, and one is on the moon on a perfectly flat sheet of water ice.
Propel your Crown Vic crash test dummy car into the rear of each of them, impact speed of 45 mph.
I betcha the difference in damage caused by "road drag" resisting the acceleration of the Yugo will be real, measureable, and tiny compared to the effects of inertia vs. inertia.
Where's the physics guy when we need one?
The time over which the force is applied is key here.
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Try it upside-down, Tuner. Gravity substituting for a link doesn't prove
the barrel and slide are unlocked when forward and in battery.
Oh, I know. I just wondered if you did.
Your moon test is about as closely related to what we're discussing as a linebacker slamming into a quarterback. Impact momentum are the wild cards.
Grump...I know you're trying', and I'm tryin' to help you see this. I think that the reason you can't is because...with everything else moving backward, and happening so fast, you can't grasp how anything moving forward can have such a profound effect. I had to ponder on it long and hard before I understood it.
So I'll try again...just for the sake of understanding.
We've already established that the barrel is pulled backward by the slide. Right?
Read this next part carefully. Read it more than once if you need to. It usually takes three times for an unfamiliar idea to take root.
As the barrel is drawn rearward, the bullet is moving forward in the barrel. Both surfaces...bullet OD and bore ID...are slipping over one another at the same time. Bullet driven forward through the bore...bore surface pulled backward off the bullet. That it only lasts for a split fraction of a fraction of a second doesn't mean a thing.
While the bullet is in the bore and moving forward...and the barrel is being dragged backward...the cohesive friction between bore and bullet surface are resisting both parts and in both directions. Are ya with me here?
When you pull the cork through the pipe, the pipe's surface is resisting the cork's forward movement, and the cork's surface is resisting the pipe's rearward movement...EQUALLY.
So...Whatever force that is resisting one is resisting the other...in equal measure. If it takes X pounds of force to push the bullet in one direction, it requires the exact same amount of force to pull the barrel backward across the bullet's surface...because...the two events are occurring simultaneously.
Go and push a bullet through a barrel with a rod and understand just how much force is required, and then consider how much would be required to not only push it through...but to push it through in the time frame involved. No fair starting it with a hammer. Just drop it into the chamber...place the end of the rod against it...and push. Lock it into a vise if you want. Unless you have the upper body strength of a Sumo wrestler, you won't get the full diameter of the bullet into the leade. Go get help. Have two or three grown men try it.
Correct me if I am wrong, Gentlemen. I am not a physics major.
As an example, lets accelerate a 3.5 oz barrel to 32 ft/sec in 1/1000 sec.
The force to do this will make the barrel seem to weigh 3,500 oz, or about 219 pounds. Double this to account for bullet drag. let's say 440 lbs of force on 1/4 square inch of lug. 16 times 440 will give 7040 pounds on the lugs.
Feel free to correct my math, laugh, hurl insults or whatever.
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16 times 440 will give 7040 pounds on the lugs.
Sounds about right. Of course, it will be higher at the beginning, when resistance is highest...and because the force required to accelerate the bullet will fall with an increase in bullet velocity, it will be lower just as the bullet is about to exit...but that sounds reasonable for an average number
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Go and push a bullet through a barrel with a rod and understand just how
much force is required, and then consider how much would be required to not
only push it through...but to push it through in the time frame involved. No
fair starting it with a hammer. Just drop it into the chamber...place the end
of the rod against it...and push. Lock it into a vise if you want. Unless you
have the upper body strength of a Sumo wrestler, you won't get the full
diameter of the bullet into the leade. Go get help. Have two or three grown
men try it.
I've tried this and barely scored the jacket of a standard diameter bullet with an average barrel. Couldn't even drive it into the rifling.
Now, do it again, and explain to me how it is that this "huge force" is so weak that all it can do is carry the weight of a barrel forward against a recoil spring in any of the "blow-forward designs". If this "bullet pulling" effect was as strong as it's claimed to be, you either wouldn't be able to COCK one of those designs, or it would be a single-shot, because firing a shot would tear the barrel out of the receiver and launch it downrange.
It's a simple question, so it should be simple to answer; do these laws of physics NOT APPLY in the blow-forward design, or is this "bullet-pulling force" simply being given WAY too much credit? The blow-forward principle has even been used in a .308-class rifle, the 7.5x53mm Swiss AK53; if a 1911 depends on "thousands of pounds of force" generated by the "bullet pull" of the .45 Auto cartridge, the "bullet pulling forces" generated in the AK53 must be on the order of MILLIONS of pounds, right?
I can by no means explain things as simply as Tuner but I would think the amount of force required to push a bullet thru a barrel is about the same regardless of size.
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I would think the amount of force required to push a bullet thru a barrel
is about the same regardless of size.
Theoretically true at least...but there are variables that would affect the final figure.
Bore diameter to bullet diameter. Land diameter to bullet diameter. Jacket material and thickness. Core material and hardness. All these things increase or decrease the bullet's resistance to being forced into the bore.
A basic law of physics states that we can't have force in one direction without an equal measure of force in the other. This is the essence of Newton's 3rd Law. Force forward equals force backward. Whatever force is imposed in one direction will be imposed in the other.
NEWS FLASH!!
This law applies to resistive forces as well as compulsive forces.
So, it follows that if X units of force are resisting in one direction, then X units of force will resist in the opposite direction. Force forward equals force backward.
I.E. If X units of force are required to pull the cork through the pipe, then X units are required to pull the pipe off the cork, because whatever resistance that the pipe offers TO the cork, an equal measure of resistance will be offered to the pipe BY the cork.
This is easy to understand if we perform the actions one at a time...slowly. If we anchor the pipe and pull the cork...or anchor the cork and pull the pipe...it requires the same force irregardless of which is being pulled. It becomes harder to understand when both events occur at the same time. It becomes harder still if both events are occurring at the same time and in a very short time frame...and it becomes even harder if they occur within a very short distance. It still happens...even though we can't SEE it happen.
There's no way to explain these basic truths away, except in your own mind. As to that...I'm firmly convinced that cats are aliens from another galaxy, sent here to enslave us and make us do their bidding...but until I can offer solid evidence of that...it remains no more than a theory.
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If this "bullet pulling" effect was as strong as it's claimed to be, you
either wouldn't be able to COCK one of those designs, or it would be a single-
shot, because firing a shot would tear the barrel out of the receiver and
launch it downrange.
No more than a blowback would be impossible to cock or that it would destroy the slide and shove it through your head. I don't know much about the blowforward device, but it would require some sort of positive impact surface to limit the barrel's forward travel...like the slide has in a blowback pistol...it would seem.
But, you have evolved. At least now you're admitting that the bullet does exert a forward drag on the barrel during its passage. That's something, at least. All we have to do at this point is to get you to understand the magnitude of the forces involved.
Go push a bullet through a barrel. We'll wait...
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I've tried this and barely scored the jacket of a standard diameter bullet
with an average barrel. Couldn't even drive it into the rifling.
Now, just imagine what sort of force is required to drive that bullet to 800+fps in a little over 4 inches of barrel, and it'll give you an idea of what the locking lugs are enduring...shot after shot.
You don't seem willing to admit that the actual FORCE transmitted to the barrel in a blow-forward design (and, by necessity, in any OTHER design) can't possibly be much more than the force that's required to cock that firearm in the first place. Does it take hundreds, or thousands of pounds of force to cock a blow-forward design? No, of course not, otherwise it would never be adopted. That means that the force that's transferred to that barrel during firing must LIKEWISE be much less than what you're claiming.
And neither does it take thousands of pounds of force to cock a blowback pistol...but if the slide didn't have a mechanical limit, the county coroner would have to pull it outta your head before callin' anybody down to ID your body...
I haven't done the math yet. Had to load up some rounds for a shooting session (rifle).
There is one facter that reduces the magnitude of the forward drag force acting on the locking lugs and that is the inertia of the barrel. (Very short time duration). I am planning on setting up a rig to actually measure the force required to drive the bullet through the barrel. I will use a hydraulic ram and simply record the pressure. This will take a while, so don't anybody hold your breath waiting!
If I may just add a comment here. It is difficult for anyone to picture in our minds the forces at play during the firing of a pistol. Things happen so fast and move so quickly. Even the way one holds the gun can effect it's operation. Some guns, if not held firmly, won't cycle properly.
Picture this, if you have a one pound weight resting on the floor, it will exert a force of one pound on the floor. If you now attach a string to it and pull upwards with a force of almost one pound, the weight will exert a force of almost nothing on the floor. Now if the string were to snap at this point, the weight will exert a momentary force of two pounds on the floor. (Suddenly applied load).
Well to be honest when I first opened this thread there was a LOT about the internal working during firing of a recoil opperated pistol that I did not completely understand, but you (Tuner) made it so simple to understand that the light came on completely.
While I can not put it into words as good as you or could never get the point across as good as you, I can say I understand it complety thanks to you.
Peterotte..When you measure the forces involved...remember the mention that you made of suddenly applied force...and that during the live operation of the gun, the bullet is not only being forced through, but it's being accelerated to over 800 fps in 4.1 inches of barrel.
Please, please...go on to explain the differences in kinetic friction and static friction...and that it requires more force to get something moving...or slipping...than it does to keep it moving.
One final point, if I may...before I turn it over to you.
The forces involved...some 20,000 pounds per square inch...are only operating for a very brief time. The resistance to initial acceleration is a big part of what causes pressures to peak so early and rapidly. At the instant of peak...whatever force is involved is what is borne by the lugs...engaged in a shearing action. As the bullet accelerates ever faster, the force required to keep it moving drops...and since force forward equals force backward...so does the force exerted on the breechblock...and consequently on the lugs.
So...Peak force on the lugs isn't maintained for any longer than peak pressure...and because pressure drops as the volume of the cylinder, or area behind the bullet increases...net force also drops.
So..If you can vary the speed of the bullet as you press it through the barrel...it would be helpful to report the difference between pressing it through slowly and pressing it through quickly.
Before the bullet leaves the barrel we have two cylinders acting in two directions under gas pressure: the cartridge case pressing back against the slide, and the barrel (temporarily plugged by the moving bullet) pushing its lugs forward against the locking grooves in the slide. Or at least that is what I see in the X-ray.
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And neither does it take thousands of pounds of force to cock a blowback
pistol...but if the slide didn't have a mechanical limit, the county coroner
would have to pull it outta your head before callin' anybody down to ID your
body...
You're not getting the point here; in the Schwarzlose (the only blow-forward design I have personally handled), it takes maybe 3 pounds of forward pressure on the slide/barrel to load and cock the pistol. In use, that operation is performed by the "drag" of the bullet passing through the barrel. So, what happened to all these "thousands of pounds" of pressure that you insist is produced by the act of firing a bullet down a barrel? That "extra" force can't simply disappear, it either has to go somewhere, OR there wasn't enough there to worry about in any event, and this unseen, immeasurable force is the equivalent of a snipe hunt.
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That "extra" force can't simply disappear
Huh....ever heard of a thing call recoil?
Try to keep up; "recoil" is force that's directed BACKWARDS. The force that Tuner is trying to convince me is so vital to the operation of a 1911 is caused by the bullet DRAGGING THE BARREL FORWARD (you know, the OPPOSITE direction).
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all these "thousands of pounds" of pressure that you insist is produced by
the act of firing a bullet down a barrel?
That pressure is trying to go in all direction but thanks to modern metallurgy it doesn't. It goes in the only direction it can.
While I have never fired one of these that you talk about I can just about bet the farm that there is rearward recoil when fired. That would account for the effects of some of that pressure that is forcing the bullet down the barrel.
SD...Have ya ever heard of a little thing called an impact surface? It's what stops the recoiling slide in an autopistol. It's made up of the rear face of the slide's spring tunnel and the front face of the frame rails. The recoil guide rod flange is sandwiched between the two surfaces. The slide strikes it and comes to a stop. Thus, the unnecessary energy and momentum of the slide is absorbed and causes about half of the muzzle rise that we experience as recoil.
The oddball design that you're referring to would require one for the barrel as well. Otherwise, it would launch downrange for a pretty fair distance. I had that happen once when firing an old USGI Colt. The barrel had flame cut so badly that it separated just forward of the chamber. I found the forward section of the barrel buried up in the 100-yard berm...with the bullet still stuck inside it.
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While I have never fired one of these that you talk about I can just about
bet the farm that there is rearward recoil when fired. That would account for
the effects of some of that pressure that is forcing the bullet down the
barrel.
There's no DOUBT that there's recoil generated by firing a shot; this is a necessary reaction to the bullet being propelled in one direction (and recoil, therefore, goes in the opposite direction). But, Tuner is claiming that this other force is working in the same direction that the bullet is travelling, and that this force is the result of the bullet "dragging" the barrel along with it. I agree fully up until this point. But, he further claims that this is an immensely powerful force, so powerful that it can beat and pound the lugs of a 1911 to the point where they don't lock properly anymore. Well, if it's that powerful, how come when we look at a design that actually USES that force, the best that that force can do is move an ounce or two of barrel against spring pressure, no more than we can easily do by hand? Firing a bullet down a bore certainly produces SOME "barrel drag", otherwise blow- forward designs wouldn't work, but it's ludicrous to claim that this force is anywhere NEAR as powerful as Tuner makes it out to be.
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SD...Have ya ever heard of a little thing called an impact surface? It's
what stops the recoiling slide in an autopistol. It's made up of the rear face
of the slide's spring tunnel and the front face of the frame rails. The recoil
guide rod flange is sandwiched between the two surfaces. The slide strikes it
and comes to a stop. Thus, the unnecessary energy and momentum of the slide is
absorbed and causes about half of the muzzle rise that we experience as
recoil.
The oddball design that you're referring to would require one for the barrel as well. Otherwise, it would launch downrange for a pretty fair distance.
EXACTLY. Yet, all that this pistol requires to stop the forward movement of the barrel is a lightweight spring, an ounce or two of metal in that barrel, and a lightweight takedown piece that would quickly be battered beyond usefulness if the forces you claim are being directed against it were ACTUALLY being directed against it. If these "bullet-pulling" forces were as strong as you claim they have to be, either the recoil spring would have to be so strong that we wouldn't be able to cock the pistol, the barrel would have to be so heavy that the pistol wouldn't operate as a semi-auto, or the barrel would be launched off the end of the receiver with the very first shot fired. Since none of these things happen, the only conclusion I can come to is that you're imagining this force to be much greater than it actually is.
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Firing a bullet down a bore certainly produces SOME "barrel drag...
SD...Have you tried to manually push a bullet through a barrel? Actually tried it?
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But, he further claims that this is an immensely powerful force, so
powerful that it can beat and pound the lugs of a 1911 to the point where they
don't lock properly anymore.
Where did I say that? Do you have the first clue how the pistol works?
Don't spin my statements to suit your argument.
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Yet, all that this pistol requires to stop the forward movement of the
barrel is a lightweight spring, an ounce or two of metal in that barrel
That's about all that it takes to stop the slide on a 1911...or haven't you noticed? Cut the spring tunnel off a junk slide and weigh it.
Keep in mind that the 1911's slide is massive.
Study this picture carefully. Enlarge it if you need to. Notice that the lugs are flanged at the front corners, and just below the flanges, you'll see a distinct stair-step shape. That's straight-line damage, and not the result of a barrel linkdown timing problem...which causes the lug corners to become rounded.
The damage here was caused in part by excessive endshake....fore and aft play between barrel and slide lugs...which gave the slide lugs about .010 inch of running start before engaging the barrel lugs. The other issue here is that the barrel lugs only provided about 70% vertical lug engagement. The two issues combined resulted in peened barrel lugs within about 500 rounds.
The final result of this deformation is that the working headspace increased to well beyond the maximum limit. The pistol wnet to battery easily on the standard NO-GO gauge...and allowed a .020 feeler gauge to slip between the breechface and the gauge. The stair step on the worst lug was .035 inch deep.
Okay...back. I decided to split this mini-tutorial into two pieces. Partly because I had to go deal with the dogs, and partly because I didn't want my man SD to go into overload.
The first thing that you need to understand is that the primary function of the recoil spring....which should correctly be named the "Action Spring" is NOT stopping the slide. Its primary function is returning the slide to battery. The same would apply to the blowforward thing. Read that again.
Stopping the movement of the reciprocating part is the function of the impact surface....not the spring.
Moreover...The spring's effect in delaying the slide is only a part of the whole...and a pretty small part, because it only compresses a measly 1/10th inch past its preload before the bullet exits.
Now then...If you want a blowback 1911, it can be done. Remove the lugs, and it'll function as a straight blowback...but if you want to fire it without blowing the case due to early breech opening...you'll need about a 40-pound recoil spring...so good luck in manually racking the slide.
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SD...Have you tried to manually push a bullet through a barrel? Actually
tried it?
Yes, I have tried it, and it's impossible to do it without some sort of tool. This has NOTHING to do with the issue you're trying to convince me of, since the levels of force involved are vastly different between firing a bullet down a bore and manually pushing that bullet down that bore. It's a red herring that you are clinging to because you want to avoid answering the question I've put to you; "If this force is as powerful as you claim it to be, then how come those designs that actually USE this force to operate, 1) work as blowforwards, and 2) only require a little bit of weight and spring force to do so?"
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Where did I say that? Do you have the first clue how the pistol works?
You say it not only several times previously in this very thread (beginning at your post #49 "It takes a lot of opposing force to beat the lugs back in a straight line like that") and even in the very REPLY above. Yes, I know how the pistol works, and I know how it's described to work; I've never read, seen, or heard ANYONE (other than yourself) make any claim that the pistol makes any use of this "bullet pulling force" to operate the way it does.
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That's about all that it takes to stop the slide on a 1911...or haven't you
noticed?
Yes, and that's part of what I've been arguing; I'm confident that the 1911 would not only operate as a blowback (if, for example, you were to grind the locking lugs off of a barrel), but that this "bullet-pulling force" plays no part whatsoever in the operation of the pistol. However, you claim that this "bullet-pulling force" is one of the things that makes the 1911 work the way it does, and there is no evidence at all for this idea. As proof, I point to a design which actually USES this "bullet-pulling force" to operate (the Schwarzlose 1908), and it turns out that this "bullet-pulling force" is nowhere NEAR as powerful as you're trying to make it out to be. If it was, the Schwarzlose 1908 wouldn't be able to operate the way it does while still being constructed the way it is.
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Yes, I have tried it, and it's impossible to do it without some sort of
tool. This has NOTHING to do with the issue you're trying to convince me of,
since the levels of force involved are vastly different between firing a
bullet down a bore and manually pushing that bullet down that bore.
Vastly different indeed. It requires much more force to drive it through when it has to be accelerated to 800 fps than it does to push it through slowly with a hydraulic press. The harder/faster that you try to accelerate an object, the more force you have to use. Any little league baseball pitcher understands...that if he wants to speed up his fastball...he has to throw it harder.
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However, you claim that this "bullet-pulling force" is one of the things
that makes the 1911 work the way it does, and there is no evidence at all for
this idea.
There's ample evidence. You simply refuse to believe what you're seeing.
Swartzlose
I can see the impact/stop mechanism clearly. Can't you?
Well...You're still evolving. You've at least admitted that you can't push a bullet through a barrel without a powerful tool...so we're making progress.
Now...Let's try this from a different angle. Shall we?
If you have to impose a thousand pounds per square inch of force to push the bullet in....would it also make sense to believe that the barrel itself is under a thousand pounds per square inch of force in the same direction? That, if the muzzle was in turn pressing against another object...that THAT object would also be under a thousand pounds per square inch of directional force?
Think about it! You're close!
No argument? Outstanding!
Now then...Let's go a step further, and add locking lugs.
If you have to place a thousand pounds psi to push the bullet into the bore...and the barrel also is under the same measure of forward force...and the barrel lugs are bearing against the slide's lugs in the same direction...would it be reasonable to assume that the same measure of directional force is transmitted to the slide through the lugs? Now, this is a direct connection, with no buffering between the lugs...and it's a 1:1 ratio, so there's no mechanical reduction.
Therefore...would it be reasonable to assume that whatever force is imposed forward on the barrel...and transmitted to the slide...that it would require a like force in the opposite direction to pull the barrel backward AGAINST the force that is holding the barrel forward?
Would it also be reasonable to assume that the contact between the barrel and slide lugs is likewise under that same measure of force that is driving them in opposite directions while literally trying to shear the lugs?
Have you ever seen a barrel lug crack in the bottom corner... or even pull loose from the barrel, bringing a hunk of the barrel with it? I have...Many times.
The reason the slide doesn't move much relative to the bullet is because of the relative velocities. The bullet is honking out of the barrel at 800 fps. The barrel slide combo is crawling with a peak velocity of 20 fps or so.
In the time it takes the bullet to travel 5" (assuming linear acceleration, so the average velocity is 400 fps, and 10 fps) the slide has traveled .010 inches
Bullet 400fps = 4800 ips. 5"/400ips = .001 seconds
Slide 10fps average* .001 seconds = .01 inches of travel.
At everypoint the bullet and the slide have exactly the same momentum.
That's why the slide barely moves before the bullet leaves.
The lugs are necessary to prevent the barrel from shooting down range with the slide. Ever notice that all blow back pistols have berrels fixed to the frame?
Also, on a tilt barrel like the 1911 or glock, the lugs are necessary to pull the barrel to the rear so the barrel can tilt to allow the slide to unlock.
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The barrel slide combo is crawling with a peak velocity of 20 fps or so.
Good post, Owen, and pretty accurate. There's more delaying the slide than inertial mass alone, though. There are six different things at work. Seven if you count the frictional resistance offered by the frame and slide rail interaction...but that one is so miniscule that it can practically be ignored.
Using the momentum equations allows you to ignore all the other forces. It turns it into a blackbox system that can be analyzed. Considering that most of the other factors affect the internal ballistics of the gun itself (ie, a high engraving force can actually increase velocity because of the jump in pressure) that I'm not sure its a valuable exercise. As an engineer, if I were to spend the weeks necessary, im pretty sure my boss would be pissed. In a previous life I actually spent a few days drawing diagrams of the forces on a bunch of different parts. Trying to work all those forces together into a coherent mathematical system became a mess.
It was an interesting exercise, and I think the visiting VIP's were impressed with my whiteboard and the reams of math on my desk, but in the end, I'm not sure I learned much beyond the fact that most components of guns are radically overbuilt and that running an all-inclusive mathematical model of a recoil operated pistol would probably choke a supercomputer.
We can take an actual pistol, with a specific lot of ammunition, and take some measurements. We can get a pretty good pressure curve for the ammo, but if we do what Jim wants to do by boring out the rifling, that pressure curve is useless.
Add in the complexity from the fact that the range of ammunition a gun is expected to run well with is like expecting a V-8 to run on every flammable fluid from parafin to acetelyne, just by changing the muffler. ( there are something like 900 commercially available loadings for 9x19)
What do we know? we know that a 230 grain bullet at 800 to 850 fps, or any ammo with a similar muzzle momentum will run a stock 1911.
BTW, I think he's right. I'm pretty sure if you give the breach face a good whack, while never touching the barrel the gun will cycle just fine. All of the factors you are talking about go into calculating exactly what the necessary whack on an existing gun is, but its the hard way around to get there.
The force Tuner is talking about does exist. By worrying about the friction in the barrel you are approaching the dynamic system from the work and energy direction, instead of the momentum direction.
Using the momentum analysis is like using the stock market to measure the economy. All the factors are being accounted for, but its a black box. No one is sure what factor accounts for how much.
Trying to measure all the forces and frictions is like trying to measure the economy by interviewing every single person in the country. You know exactly how much each factor accounts for, but you spend so much time doing it that you never ever wanna try again. Economy? Screw that, i'm gonna watch the game!
As an aside, man I bet the recoil on that blow-forward feels funky.
I'm thinking about how to calculate the actual frictional force of the bullet down the barrel. Anyone have a mathematical model of the .45 ACP hardball pressure curve.
If we take the work done by the pressure (Pressure x Area x Distance) and subract the muzzle energy (1/2 MV^2) the difference should be the work done by friction to slow the bullet down. The average friction force should be the difference noted above, divided by the length of the barrel.
So actually, we are looking for is the integral of the pressure curve from 0 to t, where t is the time of the bullet exiting the barrel, times the area of the bullet, times the length of the barrel.
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BTW, I think he's right. I'm pretty sure if you give the breach face a
good whack, while never touching the barrel the gun will cycle just fine
Sure it will. By setting the slide in motion, it pulls the barrel backward with it...just like pulling on the slide from the rear. It doesn't realize what force set it into motion, or where it came from.
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Using the momentum equations allows you to ignore all the other forces.
There's where we differ. I'm of the mind that nothing is everything, but everything is something...and nothing can be ignored or discounted.
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We can take an actual pistol, with a specific lot of ammunition, and take
some measurements. We can get a pretty good pressure curve for the ammo, but
if we do what Jim wants to do by boring out the rifling, that pressure curve
is useless.
I actually did that...and the gun cycled just fine. It was a demonstration to show that most of the recoil force is generated at the outset of bullet acceleration rather than depending on acceleration after the pressure peak.
It worked largely because the lack of bullet drag on the barrel didn't work against the slide. The pistol essentially functioned as a straight blowback with a varying slide mass...losing the barrel after it linked down.
The pressure curve generated with pistol powder with a quick burn rate is more like a spike, with very little area under the curve. Even slow numbers like Olin 296 and Hodgon's H-110 are pretty peaky, according to some of the graphs that I've seen. They peak early and fast, and drop off quickly as the bullet moves forward and increases the volume of the cylinder. Slow rifle powders are another matter. They peak pretty early too...but the area under the curve is much larger.
Owen...One last thing. Peterotte is devising a test to determine how much force is required to drive a bullet from chamber into the barrel. If you...or he...can use that and plug in his rate of acceleration to determine how much would be required to accelerate it to...say...700 fps in a half inch of travel, it would be of great help.
The main argument here is how much drag force is on the barrel at the instant of peak pressure. All else is incidental.
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There's ample evidence. You simply refuse to believe what you're
seeing.
Such as? The things that you've pointed to as "evidence" for these extreme "bullet-pulling forces" (such as deformation on the front face of the barrel lugs) are far more explainable by other mechanisms such as barrel inertia; not only are they NOT "proof" of this force you're claiming is a vital component of operation, they can be duplicated by means that have no "bullet pull" whatsoever (like John KSA's rod and hammer experiment, or even converting a 1911 to work via air pressure).
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I can see the impact/stop mechanism clearly. Can't you?
There is a physical forward limit for the slide travel in the Schwarzlose, but the pistol never reaches that limit in operation (unlike the 1911). So, this leaves me wondering again where all these "thousands of pounds of barrel- pulling pressure" magically disappear to when you fire a cartridge in the Schwarlose. They don't go into the rearward recoil (wrong direction). They don't go into the frame (otherwise they would cancel out the REARWARD recoil, and you'd have a magical recoilless pistol that everyone and their dog would be carrying today). Most importantly, they don't go into the barrel, otherwise that barrel would be torn off the frame and launched downrange, and who wants a single-shot semi-auto pocket pistol? Any "bullet-pulling" force that this pistol generates is soaked up entirely by a lightweight barrel and a pitiful recoil spring, and if that's all it takes to counteract those "bullet-pulling" forces, that tells me they weren't much to begin with.
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If you have to impose a thousand pounds per square inch of force to push
the bullet in....would it also make sense to believe that the barrel itself is
under a thousand pounds per square inch of force in the same direction?
ONLY if that barrel wasn't locked to the breechblock/slide; the force you're talking about now isn't JUST pulling the barrel forward via barrel/bullet friction, it's also pulling the barrel BACKWARD via the locked slide/barrel lugs. This cancels out the "bullet-pulling force" you're arguing in favour of, and leaves the pistol operating exactly the way Browning describes it in his patent.
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Have you ever seen a barrel lug crack in the bottom corner... or even pull
loose from the barrel, bringing a hunk of the barrel with it? I have...Many
times.
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.
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.
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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.
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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.
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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.
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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, 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.
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.
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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.
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.
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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.
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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.
I think the diameter difference is importatnt, because we are concerned about % reduction. I think.
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...
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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.
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!!
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?
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.
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.
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?
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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.
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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.
Quote: 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.
Gents, allow me to ask in graphic form. May I present my finest Photoshopped barrels:
PICTURE 404
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).
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).
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?
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
I have been looking on the 'net for a value of the rate of pressure rise in the 45ACP chamber. Couldn't find it. Can someone help?
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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.
Exactly so. The pistol was designed to operate at 17,500 CUP...which translates roughly into 20,000 pounds psi...and within certain parameters for recoil energy and bullet and slide momentum. While it's over-engineered to a certain point...it'll only stand so much foolishness before problems start to show up. The design is amply strong to handle pressures well in excess of standard USGI hardball... but pressure is only a part of the picture...and while modern metallurgy has rendered the gun stronger and more durable...it's still the same basic design as it was nearly a hundred years ago. The locking lugs aren't massive, and their maximum vertical engagement rarely exceeds .052 inch....and if all three aren't bearing an equal share of the load...it doesn't take long to do irreversible damage to the barrel with heavy loads. The heavier the bullet for a given pressure...the faster the damage occurs. Yes. Bullet obturation...known as "Bump Up" to we who cast our own...increases bullet drag, but probably not enough to warrant much concern unless the ammunition is already overpressured...and the higher the pressure...the more it bumps...or the more solidly it plugs the bore. Obturation of the bullet helps to seal the bore against gas cutting along the sides of the bullet. The result...if the bullet is sized correctly...is zero or near zero lead fouling in the bore, so it would have to have some effect in driving up the forces involved.
Odd Job...You can make a 1911 function as a straight blowback without adding slide mass...but the recoil/action spring would probably have to be in the 40+ pound range. If you fired the gun without making some change to delay the slide...you'd blow the case before peak pressure was ever a reality.
If you ground the lugs off and welded the slide, you'd convert it into a closed breech weapon...much like a muzzle-loader. Same deal.
Force forward is force backward. No loss in velocity due to escaping gasses, and all the rearward force against the breech would result in...recoil. Remember. All that happens is that the bullet and the breechblock are simply being forced apart at the same time.
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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).
It's part of the advertising that Schwarzlose used in its advertising, but it's an "artist's conception" and has at least two things wrong with it that I can see; one, it's chambered in .32 Auto, so the round isn't bottlenecked the way the artist has portrayed it here, and two, the standing breechface isn't shown tapered from bottom to top the way it is in real life. As this pistol cycles, the barrel moves forward and ejects the empty out the left side, then it picks up a fresh round as the recoil spring forces the barrel back against the breechface.
Re: Odd Job's drawings, I'd also have to say that these bullets would not emerge at the same velocities. In barrel one, both the bullet and ju-jube would move, but the ju-jube (being lighter) would travel considerably faster. In barrel two, the bullet would be propelled at the expected velocity for any other firearm of that calibre with that barrel length. In barrel three, both bullet A and bullet B would be propelled out the opposite ends of the barrel, at identical velocities to EACH OTHER, but at different velocities from the previous two examples.
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Odd Job...You can make a 1911 function as a straight blowback without
adding slide mass...but the recoil/action spring would probably have to be in
the 40+ pound range. If you fired the gun without making some change to delay
the slide...you'd blow the case before peak pressure was ever a reality.
I might have said the same thing about the Hi-Power, before I saw it being done, stock slide, stock spring, and all.
Ummm..... Tuner? Are you sitting down?
I just did some tests I thought up. I do not know what to make of the results. Maybe I should just go back and keep redoing them untill the results come out 'right'.
I made a bullet jacket from a spent 22lr case (In a die I had previously made). I loaded up a hornet round, filled to the base of the neck, inserted the jacket, loaded it into my rifle and fired it into my firing tube. Phuut! (Does that sound about right)? SDC, I hope you are reading this! OK, back one step. I put a piece of folded up nylon sun screen on top of the medium. This had a nice, melted hole in it from the still burning powder. So far so good. I loaded up a air rifle pellet the same way with a paper sabot and same thing. Lets go back some more. I also put a piece of 16mm fibre board under the nylon. Both air rifle pellet and empty bullet jacket made holes bigger than a 22lr hv. Zero pressure signs at this point. No way of gauging that level of pressure anyway.
Next, I filled some AAA shot onto a jacket, weighed it - 3.2 grams - loaded it and fired it. THUDD! Good pressure signs on the primer - almost full pressure.
Still not good enough, so I took a piece of turned brass that I had used to guage the bore of my Remmie 22lr and cut it to length. (You can see where this is going)? Weighed it - 2.3 grams - a bit light but what the heck. Made a cardboad wad, placed that over the powder charge, seated the brass 'bullet' using a paper sabot and fired it into the firing tube. Now the sound effects come again. THUDD! Near full pressure signs on the primer! And no melted nylon.
I don't know what the Heck to make of that! That's all wrong? Isn't it? But it's not over yet. The lady person still has to sing!
More test still to do.
This is fun! I just had to tell you guys.
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Go and push a bullet through a barrel with a rod and understand just how
much force is required, and then consider how much would be required to not
only push it through...but to push it through in the time frame involved.
Came back in after a long break. Not sure I'll bother coming back.
What's missing from this thread:
*Pressure/time/velocity curve for any .45 ACP load, or anything else chambered in the 1911. All I can remember is that we're dealing with a primer ignition to bullet exit time of much more than a millisecond.
*Consistent acknowledgment of the role of inertia on the bullet AND the slide AND the barrel. Both sides in the disagreement seem to have done this.
*Hard quantification of the masses involved. My scale capacities have a big gap between bullet weights and slide weights and aren't that precise for the heavier weights anyway. A nice graph of Slide-XXX grams/oz, Barrel-YYY grams/oz, and bullet-ZZZ grams/oz would be nice.
Tuner: What frosts me in this discussion is your one-sided cherry-picking of ideas. Your quote above illustrates my point and is the reason for my "thought experiment" with the Yugos. To "push it through in the time frame involved" unavoidably includes inertia. As the rough-and-ready quick calculations posted recently show, there's a certain component of all that gas pressure that is NOT used to overcome barrel friction. It's used to accelerate the bullet against its own inertia. That's constant, whether through a tight barrel or from an oversize one through which the bullet literally rattles--the bored-out barrel proposed by one of us THR people. Then you explain the low slide velocity with its own inertia resisting the acceleration, and talked for days about barrel pull as if that were the only large/significant/whatever adjective force resisting the bullet's forward motion.
I apologize for not citing names--it's late and I'm tired and I want to get this over with. The bored-out barrel idea is great. So are the no-locking-lug experiments, but I see Tuner and others ignoring the report(s?) of others who have seen at least one John Moses-by-God Browning-designed tilting barrel locked breech design cycle with no locking lugs and no case blowout--at higher pressure than .45 ACP, by the way.
NOW, KIDS!!! This inertia effect applies to EVERYTHING--the bullet, the barrel, the slide, and eventually to the frame some time after bullet departure. I say the barrel and slide are locked together by geometry, physical interference. A slide/barrel can fire just fine and stay locked just fine without a frame attached. The barrel pull by bullet friction contributes a part of the pressure on the lugs, but not ALL of it. Part of that gas pressure is pushing against a fairly static breechface--agreed, are we, that the entire gas pressure is pushing back full strength? Whether the slide has begun to accelerate or not, that force is there.
The "forward" component of the pressure is being relieved by bullet movement. Part of that is used to swage the bullet to the lands and grooves, part to overcoming friction, and part of it is devoted to accelerating the bullet. Inertia resists the slide's movement back, resists the barrel's movement backwards as pulled by the lugs, and similarly resists the bullet's movement forward.
The gases do all the work, just like my Crown Vic hitting the Yugo. My point was what about the bullet's inertia? For the acceleration rates involved, inertia has to be a larger factor than friction--to my educated but uncalibrated guesses.
To say nothing is everything and everything is nothing or however it turns out, then to ignore others' arguments about the forces involved has continued to annoy me and entice me to jump back in...
Anyway, the SAME inertia arguments about why the slide doesn't go so fast backwards also apply to the Swartslose pistol's barrel unit. The force imparted to it is also applied over a very short time period, and its mass is far greater than the bullet, so whatever thousands of pounds of barrel pull (whatever the number) exist will be just as unavoidably LESS than the 3 lbs needed to cock the thing, as the 22 lbs or whatever it takes to cycle the slide on a 1911 pistol.
Problem for Tuner is, there's 100% of the gas pressure pushing backwards against the slide, but less than 100% of the gas pressure being used to swage the bullet and overcome friction.
Problem for CBC or whoever is, the less-than-full chamber pressure force which ends up being applied to pull the barrel on a Swartslose pistol is being applied over a very short time period as well.
HOWEVER, I believe that comparing the mass ratios of the various parts involved will be truly instructive. IF the Swartslose pistol barrel unit weighs the same as a 1911 slide, one would expect it to reach the same velocity as a 1911 slide during the firing cycle, AND the 'operating' spring would need to be the same draw weight. But what if that part is lighter? Betcha it is. How about a firm report of the spring weight? IF that is also lighter, AND the reciprocating part velocities are similar, then there's a huge problem with Tuner's theory--bullet pull would have to be something less than what you could call "casehead press".
I don't believe that bullet pull is "needed" for "proper" functioning of a 1911 or any locked-breech pistol. It's there, but not needed. I'd even venture to say that somewhere on this Board, we might have a person who fired (or witnessed firing of) a 9mm in a .40 S&W gun, in which the gun at least partially cycled. Betcha a "zero bullet pull" 1911 firing a .452 bullet down a .460 bore would definitely cycle. The slide is going to be pushed backwards by the gases against the inertial resistance of the bullet, it will get accelerated fast enough to pull the barrel backwards against the link, swivel the barrel down, and continue the slide's rearward travel as the barrel stops going backwards and down.
I don't have the facilities to check this. Since Tuner has so many boxes of .45 parts laying around, I will throw down the gauntlet and ask that he test it for us. Now, it might be necessary to use a heavier bullet to get the proper chamber pressure to get the proper bullet speed to produce the proper RECOIL for the RECOIL-operated pistol, but once those parameters are met (someone loan the man a chronograph if he doesn't have one), the pistol will cycle just fine. Remember, without the bullet speed, the gas pressure will not produce the recoil force on the slide which is "proper" for "proper functioning".
Based on the numbers posted above, it appears that I was wrong in my position that barrel pull is only a tiny (tiny being "small fraction" or whatever--I was thinking less than 10%) part of the forces applied by the expanding gases. It's apparently bigger than I thought but still not important to functioning of a 1911 pistol. If anyone can check or refine those calculations, please do. Since force forward has to equal force backwards, Tuner, only that force which exists from gas pressure but which is "missing" from the bullet acceleration is available to be barrel pull. Just don't ignore the role of inertia on all sides of the equation, because even the tiniest movement of the slide before bullet exit is going to add to the lug engagement pressure because of the barrel's own resistance to being pulled backwards by the slide.
I'll stand back and take another peek in a couple of days.
Grump , don't go away yet. I suspect you are dead on! SDC too.
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...barrel pull is only a tiny (tiny being "small fraction" or whatever--I
was thinking less than 10%) ...
That was my original thought. Then I thought it must be more like 60%.
I have done some more tests.
Test #1 Drill out the core of a hollow point match bullet and load it as usual. Jacket mass - 0.7 gram Charge mass - 0.68 gram AR2205 Minimal flattening of primer Jacket recovered from sand medium. (Pretty ripped up). Muzzle blast minimum Test #2 Drive lead bullet down bore with brass drift untill it can be pushed by rod. Charged case chambered with cardboard wad over powder. Bullet seated into case from muzzle end. Bullet mass - 3.63 gram Charge mass - 0.68 gram AR2205 Normal flattening of primer Muzzle blast normal Bullet not recovered. Only lump of dry sand found. (In damp sand). Test#3 Lead bullet seated onto cardboard wad over powder as normal. Bullet mass - 3.63 gram Charge mass - 0.68 gram AR2205 Normal flattening of primer Muzzle blast normal. (Could be a shade louder than in test#2)? Bullet not recovered. Only lump of dry sand found. (In damp sand).Folks, this was from a hornet rifle, not a 1911, I know. I do not know how to interpret my observations but it sure looks a lot like the pressure generated by bullet drag & swaging forces is closer to 10% rather than the 60% or so I was thinking. But make no mistake, driving that bullet into the bore required a mallet and brass drift. Also, it does appear that the primer flattening is slightly more with the normal bullet. (Primer flattening is not a accurate pressure indicator).
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Betcha a "zero bullet pull" 1911 firing a .452 bullet down a .460 bore
would definitely cycle
Grump, someone definately needs to try that one.
Tuner, what do you make of all this? Would you be able to do some tests with a 1911?
SDC, you should be enjoying this . It's sure been fun doing the tests so far.
More tests to be done. (It's not over ......).
Peterotte...10% may be pretty close...but 10% of a peak operating pressure is still about 2,000 pounds of force required to drive the bullet...which would exert a healthy forward drag on the barrel...and it seems to correlate with Owen's first calculation.
I'd have figured it a bit higher, but so far, so good.
SD and Grump...The description of the locking/lug engagement mechanism is described in Browning's original patents. All you need to do is read it through a few times and let it sink in. I hate that I've apparently alienated a couple of folks here. That truly wasn't my intent...and all ribbing was done in good nature. hard to see it via the printed word. Face-to-face would make it more plain...but here we seem to be anyway.
Keep testing, and we'll call Owen back into it later today...hopefully. The only explanation that I can find for your findings is that the resistance to the projectiles wasn't high enough.
Grump...Of course a bullet with zero pull would cycle the action. Force is still required to overcome the bullet's inertial mass. Owen showed us that with his calculations...that it would require nearly 2,000 psi to accelerate the bullet's mass to 700 fps without any other resisting factor. So...We appear to be in agreement at least on that point.
Thanks for the testing, peterotte. My only contribution to this discussion is that I don't believe that "bullet drag" plays any sort of practical role in the operation of a 1911 pistol. Does the fact that aerodynamic engineers have "proven" that a bumblebee can't fly stop that bumblebee from flying? No, so they're obviously missing something somewhere in their calculations. Ditto for "bullet pull"; since actual working firearms that actually operate on this principle EXIST, and they don't require anything near what the math claims they should in order to operate, that tells me something is missing in those calculations.
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Problem for CBC or whoever is, the less-than-full chamber pressure force
which ends up being applied to pull the barrel on a Swartslose pistol is being
applied over a very short time period as well.
Exactly; that's what makes this talk of "thousands of pounds of bullet-pulling pressure" null and void, because if there actually WAS thousands of pounds of bullet-pulling pressure being exerted in this design, the Schwarzlose would not only not work, it would be a single-shot because the barrel would go downrange with the bullet.
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someone definately needs to try that one.
ALready did it...and as expected, it will cycle. There's ample recoil force to do that, even under very low pressures. That's why squibs will cycle the action...provided that the slide moves far enough for the barrel to reach the linkdown point.
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applied over a very short time period as well
No argument there, either. As I've tried to point out...peak pressure occurs early and quickly, and drops quickly...and pressure falls off as the bullet accelerates and increases the volume of the cylinder behind the bullet. It naturally follows that whatever forces are at work at peak pressure will also drop...so the high forces against the barrel...and the lugs...only lasts for as long as the peak. It's variable...falling rapidly following the pressure peak.
And...so goes the horizontal lockup and slide delay. And...since about 90% of the total muzzle velocity occurs within the first half-inch of bullet travel...the same applies to the force.
It's a little like the small-radius firing pin stop's effect in delaying the slide. It only lasts until the hammer is cocked...but it occurs at a time when that's most effective...before the slide has had a chance to gain most of its momentum.
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....but 10% of a peak operating pressure is still about 2,000 pounds of
force required to drive the bullet...which would exert a healthy forward drag
on the barrel...and it seems to correlate with Owen's first calculation.
Quite so. It's hard to 'picture' the magnitude of the pressures and forces involved. (And the short time durations). The force I used to drive a lead bullet into the muzzle of a hornet is large relative to my strength but quite small relative to the forces in the chamber. A copper jacket must require a heap more force to engrave. Certainly, the copper jacket produced a lot more pressure than did the spent 22lr jacket, going by the primer and the degree of still burning powder exiting the muzzle. With the copper jacket, there was no melting of the nylon.
By the way, that jacket was moving pretty fast! If my 22 air rifle pellet loads are anything to go by, muzzle velocities of very light projectiles are very supersonic even at miniscule pressures. Funny that the powder actually burns up in the barrel! - With the air rifle pellet I mean.
Clark...Most slide failures occur at the top, left corner of the port..adjacent to the breechface guide blocks. A reinforced port wouldn't do a lot to help with the Rowland. If somebody would machine a slide and leave a small fillet in those corners, it would be a major shot in the arm for the souped-up conversions.
www.guns.connect.fi/gow/QA17.html
Check out the slide on that thing. Was the rotary bolt locked to the barrel during firing?
"When the breech-bolt was blown back after unlocking.."
It looks like the bottom left detail is the actuator track that rotates the bolt, but you really need to see the entire patent (and maybe have it translated).
Wildey uses a similar curved track machined on the inside of the slide to rotate the bolt for locking.
I have read and tried to follow this thread from the beginning. I am getting more confused by the page.
The way I see it the barrel and slide are a "closed system" until the barrel links down. The bullet drag in the barrel is resisted by the barrel lugs against the slide lugs as is the gas pressure accelerating the bullet. The rest of the recoil action/reaction is simply momentum, friction/force (of the guns parts) and inertia. The bullet drag might have a minute effect on the recoil motion of the slide until the barrel lugs and slide lugs lock up. After that time the barrel and slide are for all intents one piece where the pressure of the case head pressing on the breach face equals the total of forces acting on the bullet/barrel combo in effect trying to force the barrel forward from the breach. Because this is a closed system, the only recoil effect is action/reaction of the bullet mass/gas/slide.
Other than acting as a resisting force in keeping the slide and barrel tightly locked together until the bullet leaves the barrel - which the mechanics of the swinging link does anyway I don't see how bullet drag is relevant to the 1911s fundamental operation.
The recoil cycle will not be directly influenced by what is happening in the "closed system". As Tuner said, it does act sort of like a blow back system but the blow back is not what is delayed. The mass of the barrel is added to the slide mass and the release of breach pressure is what is delayed - not slide recoil motion. Mostly by the delay of the vertical motion of the barrel vs the slide before the barrel tilts down and unlocks but also by adding the inertia of the barrel to the slide in the initial rearward motion.
Where am I wrong?
DBR, I agree with you.
There are three ways to analyze motion. You can use momentum methods, energy methods, or Newtonian motion equations. Generally what happens is is you switch from one method to the next, to get pieces of information you need but don't have.
While trying to determine the velocity of the slide, the momentum method is simple and gets you what you are looking for right away. What it doesn't tell you is the magnitude of the forces involved. For instance, if I wanted to know what the load on the lugs was, I couldn't get that from just using conservation of momentum.
Analysis of the force on the lugs requires using Newtonian Equations (f=ma, x=vt+1/2at^2), and then the whole problem becomes very complex. Some big nasty integrals and all that.
To get that information, I would have to start drawing freebody diagrams of each component.
Tuner's position is that the friction in the barrel is a major force, and the implications of that force carry though to the rest of the gun. Some of the effects are going to be, a much slower slide velocity than if there was no friction, much higher friction forces on the locking lugs.
I agree that the friction in the barrel is a major factor in determining slide velocity of short recoil operation system, but I also don't believe there is a whole lot that can be done to manipulate that factor.
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I agree that the friction in the barrel is a major factor in determining
slide velocity of short recoil operation system, but I also don't believe
there is a whole lot that can be done to manipulate that factor.
Exactly. It is...and will remain...largely an unknown.
Owen...et al...A brief, slow-motion description of the best explanation I can come up with, based on several observations and a few discussions with engineers who are also familiar with firearms and ballistics.
Bang! Powder lights...gasses expand, and the driving forces start to overcome the bullet's inertial mass and frictional resistance imposed by the case.. Bullet goes from static friction to kinetic friction, which...working in conjunction with the bullet's fledgling momentum...reduces the magnitude of the force required to keep it moving.
At the same instant, the slide also starts to move...but because of its greater mass...accelerates sluggishly by comparison. The slide is gaining momentum. At this point, there is no delay imposed by the bullet's resistance, because...as CA noted...the momentum end of the equation essentially cancels it out.
Now...The resistance of pushing a .451 diameter plug into a .451 diameter hole requires quite a bit of force alone, and ...since the bullet's inertial mass is still fighting...the two become pretty formidable. The fly in the ointment is the land diamater...which is .010-.012 inch smaller than the bore, or groove diameter...and all of a sudden...we have a NEW force to contend with, and with it comes a new equation.
For those who don't understand what a "New Force" is...An object that is set in motion by a force can't accelerate any further after that force is removed. Only by application of a new force can it be compelled to move faster, and only by applying a new force can it be brought to a stop. Thus...A bullet that exits the muzzle can only slow down unless a new force is applied at its base...and in the absence of a restraining force...the bullet will theoretically continue at its present velocity forever. Fire a gun in the weightless void of outer space. It'll never stop, and it'll never slow down unless it's compelled to do so by the application of an outside force.
Because the bullet's initial resistance was overcome by the rising pressures and the force generated by that factor...the increasing pressure hits a short, but still rising plateau. Because the bullet has encountered this new resistant force...and because nothing happens instantly, and everything takes time...the bullet stops for a tick. This may only last a half a millisecond. Maybe more...Maybe less. We have no way of calculating it...but we can see evidence of it on a pressure graph. There's a slight, steep rise...a short, shallow rise...then a sudden, violent climb to peak.
During this plateau phase, the bullet has stopped. When the bullet stops...so does the slide. When everything stops...all momentum is lost, and it all has to start over again. But because the bullet has moved forward, and can't move backward...it has a slight head start toward its destination. By the time the bullet begins to overcome the new resistance, it's moved maybe a quarter-inch. That's 5% of its total distance.
I've caught this slide hesitation after watching extremely slow-motion/stop- action videos. You have to watch closely and you have to watch it several times to be sure that's what you're actually seeing...but it's there. The blow-by gasses escape ahead of the bullet...before obturation and bore seal. The slide moves for a very short distance...maybe .a 32nd inch...hesitates...then suddenly accelerates rearward as the full force of recoil bears on it. Much easier to see is the gun pitching slightly forward just before the slide starts to move backward in the violent recoil phase. Gasses...slide movement...gun forward movement...slide stops...slide recoils...bullet exits. Just like that.
It's right there...at the point that the bullet encounters the land diameter and stops...that the slide is delayed. The bullet overcomes the resistance, and screams toward the muzzle under the full force of the pressure peak...while the slide sluggishly fights to regain its lost momentum. And it's right...THERE...that Kunhausen's balanced thrust vector exists.
Balanced. Suspended briefly in a state of equilibrium. So...By the time that the bullet has fully entered the rifling, and the slide has regained most of its lost velocity and momentum...10% of the bullet's trip is complete.
All this is very brief...taking maybe a millisecond from start to finish for the cycle of static friction to kinetic friction...back to static, and finally...back to kinetic. By the time the bullet has reached roughly the mid- point of the barrel...the momentum cancels out any resistance offered to the slide by the bullet's passage...even though opposing forces still bear on the lugs...and the bullet finally exits. (Now, your momentum/resistance equations will balance out.)
Adding up the effects of the delaying forces, we also have to understand that they don't all bring their full effects to bear at the same instant. More like handing off the baton in a relay race. First, of course, is the inertial mass of the slide and barrel. Then the bullet slamming into the rifling...the big one...Then comes the hammer and mainspring. Finally...after the momentum has reached the point that its effects counteract the resistance...the recoil spring comes into its own and begins to bear hard on the slide after the bullet is long gone.
Everything that has the opportunity to slow the slide WILL slow the slide. Nothing can be discounted, no matter how small or insignifigant. Not even a fly that happens to get smacked by the back of the slide as it moves. If it CAN affect it. It will.
A final clue here is this...and it requires thought to fully understand what it's saying:
A balanced force system is a necessary condition for equilibrium, but it is not a sufficient condition to ENSURE equilibrium.
That's a riddle of sorts...
Here's a thought, for believers and non-believers alike, with regards to barrel pull's effect on the gun unlocking:
Consider some IPSC open guns.
These guns are, operationally, 1911's, shooting 38 super +P (really, more like +P+). Way more pressure than .45 ACP.
They have radically lightened slides.
They run with recoil springs as light as 8 or 9 pounds.
Given just those factors, these guns should be blowing out brass and hammering themselves to pieces. Yet they do neither.
Why is that?
(Hint: It begins with "C" and ends with "ompensator.")
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They have radically lightened slides.
They run with recoil springs as light as 8 or 9 pounds.
Good points...but remember that if the comp is attached to the barrel, it adds mass to the reciprocating assembly...so it gets a little closer to square one. The other point is that the recoil spring doesn't have a lot of effect in delaying the slide at the very beginning of the cycle...and the linkdown and breech opening won't happen until the barrel is drawn back far enough to reach the linkdown timing point. "Time" and "Timing" are related, but they're not the same. Time is a function of distance. Timing is fixed, and it will occur at the same point, regardless of how fast or slow the slide moves. It only has to be delayed long enough for the bullet to exit.
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if the comp is attached to the barrel, it adds mass to the reciprocating
assembly
Absolutely. The mass of the compensator does come into play. As does the gas hitting the compensator baffles. Like you say, if it CAN affect it, it WILL.
Slide mass, barrel mass, recoil spring rate, and barrel pull from bullet drag or a compensator all play a part in the complex balancing act that is a functional pistol. Change any one factor, and you have to account for it elsewhere, or there are consequences. (failure to function, destroying brass, abusing the gun, etc)
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The other point is that the recoil spring doesn't have a lot of effect in
delaying the slide at the very beginning of the cycle
Still right there with you. I'm just trying to illustrate a point about bullet drag / barrel pull, which I'll get to in a second.
By the time the action has unlocked, the bullet and the gases propelling it are out the barrel and gone. The remainder of the slide's travel is powered only by inertia. It's during this part of the slide's travel that the recoil spring comes into play. The spring slows the slide down as it gets compressed. Too low a spring rate, and the pistol gets battered and wears prematurely. Too high a spring rate, and the gun may not cycle.
Now, given the same round firing in the gun, there is a fixed amount of energy being produced. The more energy required to get things to linkdown & unlock, the less energy left over afterward for the slide to keep moving. Increasing the barrel weight, bullet drag, or forward thrust from a compensator require more energy to move the barrel back to linkdown, so the slide will have less energy once the gun is unlocked. You then need to lower the recoil spring rate to balance things out.
Of course, it is theoretically possible to add so much weight and pull on the barrel that the recoil spring required to balance things out is too weak strip rounds from the magazine. It all goes back to that complex balancing act.
But I digress.
Consider that the standard recoil spring for a 38super is 14#. Conventional wisdom says to run a heavier spring with hot ammo, so a gun shooting +P+ should probably have a 16-18# spring, if not more. Yet by adding a compensator to the equation, we can cut the recoil spring rate in half.
To make a long story short (too late!), here is my point:
We know that bullet drag is significant enough to operate a blow-forward design. Given what the compensator example shows, how could bullet drag not have been accounted for in the design?
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By the time the action has unlocked, the bullet and the gases propelling it
are out the barrel and gone. The remainder of the slide's travel is powered
only by inertia.
Just nitpickin' a bit...and I know what you meant...but it's momentum that keeps the slide moving rearward. Inertia resists.
You're right. I knew I was going to screw up a physics term in there somewhere.
nah, inertia and momentum are the same thing.
Inertia is an idea, not a quantifiable number. I guess you could say it is the same as mass.
Inertia is the idea that an object at rest requires some sort of force to go into motion. Once the object is in motion, some sort of force is required to change the velocity of the object.
I'll concede to that. I've always thought of inertia as "Mass at Rest" instead of in motion.
If you are talking about that Schwarzlose being a blow foward design, go to post 295, click on the www thing and read the description of that pistol and its' operation. Nothing mentioned about blow foward.
If there is such a hand gun, could someone post a link, please?
Iron Bottom, your post shows the Schwarzlose 1898; the blow-forward design is the Schwarzlose 1908 (see post #236 for a picture of the 1908).
Inertia is the property of a body that resists acceleration (or deceleration) and is only mass related.
Momentum is a product of mass and velocity. The two are quite different.