Hydraulic recoil spring?
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1911Tuner
Moderator Emeritus
Then stop twanging that note, and we'll be fine.
Now then, let's try something else.
Let's do a hypothetical. I love those. You can do things that
can't be done in the real world, but they show us how things work.
Let's say that you've developed a chemical that you can use to multiply by tenfold the coefficient of friction of any material simply by smearing it on.
Smear it on a 230-grain .45 caliber bullet. The bullet's mass doesn't change. Nothing changes except the bullet's coefficient of friction, and instead of a hundred pounds of force needed to push the bullet trough the barrel, it now requires a thousand.
Load it in a case with a normal powder charge...say 6 grains of Unique...to duplicate GI hardball...and fire it in the gun.
What will happen?
Will the slide move at the normal rate to the normal speed even though the bullet is struggling to make the trip?
Think about it and read on.
Several years ago, Jim K did a demonstration by inserting a steel rod into the barrel of a 1911 pistol...threading the muzzle...and installing a screw to press the rod firmly against the nose of the bullet in a chambered round.
Then he fired the gun.
And nothing happened. The slide didn't move. The gun didn't blow up. It gave no indication of firing other than the hissing sound of the gases escaping around the breech.
Why didn't anything happen? Why did the slide not move?
Think hard.
45_auto
Member
1911tuner said:
Not sure why you're so worked up.
See Prob. 20.30
Use a velocity of zero for the projectile. Solve for the speed of recoil (velocity) of the gun (slide in your example).
What do you believe would have happened if you fired a bullet but it had too much friction to make it out of the barrel? Would the friction in the barrel keep the slide pulled forward and prevent it from cycling?
Comrade Mike
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If you can't rack the slide on modern handguns there is a world of wonderful revolvers out there.
barnbwt
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I'll have to put up a free body diagram of a slide/barrel locked together and y'all can explain what the bullet's delaying, exactly. The locked breech is done because it has the effect on the chamber of an infinetly heavy bolt body, but which obviously isn't and can decouple to eject/load. It "hides" a large portion of bolt thrust early in the cycle.
TCB
TCB
1911Tuner
Moderator Emeritus
God! Why is this so hard to grasp?
The slide in Jim K's pistol didn't move because the bullet was pushing on the rod with was pushing on the barrel which was pushing on the screw that was threaded to the barrel.
The slide didn't move backward because the barrel couldn't move backward. If he'd blocked the bullet with the rod pressed against a wall instead of involving the barrel, the slide would have moved. It wouldn't have made the full trip...but it would have moved.
In his attempt to debunk Kuhnhausen's Balanced Force Vector description, he inadvertently created one. A system in which the compelling forces and resisting forces are equal...so nothing happened.
The bullet's frictional resistance to the barrel is resistance to the slide. No matter how great or how small...it's there.
If that resistance is great enough, it can keep the system from moving at all.
The slide in Jim K's pistol didn't move because the bullet was pushing on the rod with was pushing on the barrel which was pushing on the screw that was threaded to the barrel.
The slide didn't move backward because the barrel couldn't move backward. If he'd blocked the bullet with the rod pressed against a wall instead of involving the barrel, the slide would have moved. It wouldn't have made the full trip...but it would have moved.
In his attempt to debunk Kuhnhausen's Balanced Force Vector description, he inadvertently created one. A system in which the compelling forces and resisting forces are equal...so nothing happened.
The bullet's frictional resistance to the barrel is resistance to the slide. No matter how great or how small...it's there.
If that resistance is great enough, it can keep the system from moving at all.
JRH6856
Member
barn, consider this: The 1911 locks horizontally but unlocks vertically. The barrel and slide lugs are locked by opposing horizontal forces which push the slide one direction while the barrel is pulled the opposite direction. The locking forces are greater than the shearing force exerted by the link pulling down on the barrel. Until the locking forces are released, the lug can't pull the barrel out of lock.
Without the bullet drag pulling the barrel, the breech is not locked and the link can unlock the barrel without resistance.
Without the bullet drag pulling the barrel, the breech is not locked and the link can unlock the barrel without resistance.
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1911Tuner
Moderator Emeritus
Let's go back to the chemical hypothetical for a minute.
Instead of increasing the coefficient of friction 10 times...let's say that it doubles it with each successive application. one application doubles it. Two applications makes it four times the original, and so forth. Now, we can at least keep the bullet moving.
Let's start with two applications. Now, the bullet's coefficient of friction is four times that of an untreated bullet.
The first thing you'll notice is reduced bullet velocity. We can all see that and accept it.
Now for the big question.
What will the effect be on the slide's rate of acceleration and velocity? Will it slow down, or will it remain the same as with an untreated bullet?
Before you let your beliefs cloud your thoughts...remember that force forward is force backward...whether that force is compelling or resisting acceleration, it will compel or resist in both directions in equal measure.
Now, let's keep applying the chemical until we reach a point that the bullet won't move at all, or until we achieve the same conditions that Jim K had with his rod and screw.
Somewhere in the middle, the bullet keeps moving slower and slower...and so does the slide.
Because whatever you do to one end of the system, you do to the other end.
Instead of increasing the coefficient of friction 10 times...let's say that it doubles it with each successive application. one application doubles it. Two applications makes it four times the original, and so forth. Now, we can at least keep the bullet moving.
Let's start with two applications. Now, the bullet's coefficient of friction is four times that of an untreated bullet.
The first thing you'll notice is reduced bullet velocity. We can all see that and accept it.
Now for the big question.
What will the effect be on the slide's rate of acceleration and velocity? Will it slow down, or will it remain the same as with an untreated bullet?
Before you let your beliefs cloud your thoughts...remember that force forward is force backward...whether that force is compelling or resisting acceleration, it will compel or resist in both directions in equal measure.
Now, let's keep applying the chemical until we reach a point that the bullet won't move at all, or until we achieve the same conditions that Jim K had with his rod and screw.
Somewhere in the middle, the bullet keeps moving slower and slower...and so does the slide.
Because whatever you do to one end of the system, you do to the other end.
I'm gonna try again. It's too bad I had to sleep and your hypothetical questions have already been answered. They are more or less true and obvious. But they still don't answer this assertion:
Let's presume, for the sake of argument, that a bullet which is twice as heavy will have twice the friction, and that twice the friction means half the acceleration and resultant muzzle velocity that it would otherwise have. Let's also presume, for the sake of argument, that by doubling the mass, we also halve the resultant acceleration and resultant muzzle speed. So for the sake of argument, let's try to imagine that a bullet that is twice as heavy will result in 1/4 of the muzzle speed.
Now, let's examine what happens to the slide/barrel in this hypothetical. The slide will move back with 1/4 the speed, too, right? Force in equals force out?
Well, think about it. That's false. If this were true, momentum would not be conserved. The half of the speed drop caused by friction is completely balanced out by a reduction in slide velocity. The slide will go back half as fast. But the change in speed that is due to the doubling in mass DOES NOT reduce the slide velocity. Half the speed with double the mass is the SAME MOMENTUM it started with.
What happens is the slide moves back half as fast. Bullet moves 1/4 as fast. A bullet that has twice the mass only makes it half as far from the starting point by the time the barrel hits the locking lug. This is the exact same answer, no matter what the powder charge or friction. Well, so long as you are not factoring in the recoil spring, which you have already conceded that we can ignore, for sake of argument. As you can see, conservation of momentum has not been cheated.
Does this strike a chord? Are we playing in harmony, here?
Let's presume, for the sake of argument, that a bullet which is twice as heavy will have twice the friction, and that twice the friction means half the acceleration and resultant muzzle velocity that it would otherwise have. Let's also presume, for the sake of argument, that by doubling the mass, we also halve the resultant acceleration and resultant muzzle speed. So for the sake of argument, let's try to imagine that a bullet that is twice as heavy will result in 1/4 of the muzzle speed.
Now, let's examine what happens to the slide/barrel in this hypothetical. The slide will move back with 1/4 the speed, too, right? Force in equals force out?
Well, think about it. That's false. If this were true, momentum would not be conserved. The half of the speed drop caused by friction is completely balanced out by a reduction in slide velocity. The slide will go back half as fast. But the change in speed that is due to the doubling in mass DOES NOT reduce the slide velocity. Half the speed with double the mass is the SAME MOMENTUM it started with.
What happens is the slide moves back half as fast. Bullet moves 1/4 as fast. A bullet that has twice the mass only makes it half as far from the starting point by the time the barrel hits the locking lug. This is the exact same answer, no matter what the powder charge or friction. Well, so long as you are not factoring in the recoil spring, which you have already conceded that we can ignore, for sake of argument. As you can see, conservation of momentum has not been cheated.
Does this strike a chord? Are we playing in harmony, here?
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JRH6856
Member
Since the barrel and slide start out with the locking lugs in contact, it appears you are no longer talking about a 1911 so I'm not sure what you are talking about.
1911Tuner
Moderator Emeritus
I'm a little confused by this.
The 1911 has three locking lugs...none of which have anything to do with barrel linkdown/unlock timing.
And all three barrel lugs are in contact with the lugs in the slide the instant that the bullet enters the barrel throat...and if the barrel hood is closely fitted to the slide...they're in firm contact before the gun fires.
And you're still in denial about the bullet's influence on the slide's acceleration.
And my hypotheticals really haven't been answered. Danced around a bit maybe...but not answered.
Sorry, I meant by the time the locking lug/link hits the locking block. Or w/e the terminology is when the barrel starts to unlock.
Really? Can you guide me to a specific question? Gimme the post # and the a few lines to go off of.
If you wonder how come I get so snarky, it's because I picture this applying to you. Let's try to figure out why two smart adults can disagree on something that appears to be so obvious and self-evident. I really thought I did this with my last post, but I can do this all day. So bring on your hypothetical.
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JRH6856
Member
Two things I am seeing in this discussion.
1. The COM arguments give to counter the "broomstick theory" all appear to describe a different set of condtions than those that exist in a 1911. In GLOOBs sitcom, to create a close to accurate analogy of the 1911, Gary needs to be standing on the ice and Willis needs to be standing on the bank with each holding one end of the broomstick.
2. The arguments claiming that friction/drag resistance doesn't apply to the barrel/slide when they are locked overlooks the fact that they are locked by the friction/drag resistance.
1. The COM arguments give to counter the "broomstick theory" all appear to describe a different set of condtions than those that exist in a 1911. In GLOOBs sitcom, to create a close to accurate analogy of the 1911, Gary needs to be standing on the ice and Willis needs to be standing on the bank with each holding one end of the broomstick.
2. The arguments claiming that friction/drag resistance doesn't apply to the barrel/slide when they are locked overlooks the fact that they are locked by the friction/drag resistance.
1. Ok good. I like that you're thinking this through. But to add friction to the analogy, YOU still have to be on the ice as well. If you remove the recoil spring from the 1911, the slide just rides on the frame with practically no resistance. YOU are the that slide. You are fixed to one end of the broomstick, which represents the barrel. Gary holds onto the broomstick sort of lightly, so there's friction.
When you and Gary push off from each other, you both will stop moving at some point, due to friction. No matter where you stop, he will be twice as far from the starting point as you are. If he pulls on the broomstick until the two of you are sitting face-to-face, again, you will be back to the same spot on the ice where you started. No matter where you look, your collective center of mass will remain the same.
This is why Tuner is saying that as you increase friction, both the bullet and the slide lose velocity. He is describing conservation of momentum. (Gary and you are both moving slower, but you are still maintaining the same 2:1 ratio in velocity.) When you get all the way down to zero bullet velocity, the slide doesn't move at all. This is not a surprise, at all. The whole point of a locked breech is so that momentum IS conserved because there ARE no outside factors. Bullet mass*velocity=0. Then slide mass*velocity = 0. Conservation of momentum is pure and unmuddied by those factors which figure in to a High Point. Every truth he states actually supports the claim I have made, in so far as I can tell.
2. I do not disagree with this locking lug friction thing. But if the locking lugs are still engaged strongly (i.e. bullet still in barrel) when the barrel starts to cam, and if, as Tuner has suggested, the cam link might even shear off before the locking lugs disengage... well I don't think the gun is working as intended, anymore. And the next shot will be quite different. If it's working properly, the bullet is already gone and pressures have started to drop, so this friction in the locking lugs is not very significant in the big picture, is it? Why is this pertinent to the question at hand? Are you saying that a heavier bullet will still not unlock the gun too soon, but instead it will break it? Then ok, fine, you win. I will agree that you are correct.
When you and Gary push off from each other, you both will stop moving at some point, due to friction. No matter where you stop, he will be twice as far from the starting point as you are. If he pulls on the broomstick until the two of you are sitting face-to-face, again, you will be back to the same spot on the ice where you started. No matter where you look, your collective center of mass will remain the same.
This is why Tuner is saying that as you increase friction, both the bullet and the slide lose velocity. He is describing conservation of momentum. (Gary and you are both moving slower, but you are still maintaining the same 2:1 ratio in velocity.) When you get all the way down to zero bullet velocity, the slide doesn't move at all. This is not a surprise, at all. The whole point of a locked breech is so that momentum IS conserved because there ARE no outside factors. Bullet mass*velocity=0. Then slide mass*velocity = 0. Conservation of momentum is pure and unmuddied by those factors which figure in to a High Point. Every truth he states actually supports the claim I have made, in so far as I can tell.
2. I do not disagree with this locking lug friction thing. But if the locking lugs are still engaged strongly (i.e. bullet still in barrel) when the barrel starts to cam, and if, as Tuner has suggested, the cam link might even shear off before the locking lugs disengage... well I don't think the gun is working as intended, anymore. And the next shot will be quite different. If it's working properly, the bullet is already gone and pressures have started to drop, so this friction in the locking lugs is not very significant in the big picture, is it? Why is this pertinent to the question at hand? Are you saying that a heavier bullet will still not unlock the gun too soon, but instead it will break it? Then ok, fine, you win. I will agree that you are correct.
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1911Tuner
Moderator Emeritus
Just go back and read everything that you didn't read earlier. Pay particular attention to Jim K's demonstration.
No such thing, and I apologize if I gave that impression. I was just making an observation based on so many others who would like to ignore that influence. i.e. Whatever resistance (through friction) that the barrel offers to the bullet's forward movement...the bullet offers to the barrel's rearward movement...in equal measure. It requires 100 pounds of force to start a jacketed bullet into the rifling, and a steady force to keep it moving to the muzzle. It only follows that it would require the same force to pull the barrel backward off the bullet. They're moving in opposite directions at the same time. It has to work that way.
And few seem to get this...and I can't figure out why. It's Newton 3 in reverse and no more complicated than that.
The friction between the lugs isn't really what's in question here, other than an aside. The bullet's forward drag on the barrel due to the frictional resistance that exists between them and the influence it has on the slide's acceleration backward is what it's about.
This is what you seem to be trying to ignore, and this is why your heavy bullet/early unlock theory doesn't work. The equal resistance in both directions involves time, too. The longer the bullet is in the barrel, the longer it resists the slide. The heavier bullet is a longer bullet. A longer bullet offers more surface area to the barrel. More surface area creates more friction. More friction creates more resistance. More resistance means the slide moves slower. The heave bullet will still clear the muzzle before the barrel drops.
The reason everyone ignores it, is because in a locked breech you CAN ignore it. That's the whole point. You said it yourself. Forward friction, rearward friction. In equal measure. Add that up and you get zero.
It's in a blowback gun where you can't ignore friction when calculating slide momentum.
That would be nothing. I have read everything you have posted. I can see what it is you are missing. And I have pointed it out and explained it. I also recapped the Jim K thing, specifically, and there are no surprises, there. If there are posts of mine you have not read, I would urge you to go back and do so. If you have already read and understood them, then well, there it is. We're at what you might call an impasse.
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JRH6856
Member
Uh, yeah it is. It is the friction in the locking lugs that resists the link-down and keeps the breech locked. But more importantly, it is the friction/drag of the bullet on the barrel that creates the force opposing the rearward momentum of the slide that locks the breech. Without the friction, the breech unlocks as soon as the link initiates link down. With the friction in the equation, the link down is delayed until the breech is unlocked which occurs when the bullet leaves the barrel and the opposing forces are no longer in play.
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Oh... ok.
I likey.
So you are suggesting that the barrel and slide move back and contact the locking block/cam-thingy while the bullet might still be in the barrel. And then the friction keeps the lugs from moving, because pressure is still too high.
What you are missing here is that if the locking lugs do not slide apart, the barrel and slide will stop moving. Dead in it's tracks. That momentum is then transferred to the frame and to the shooter.
Once the bullet leaves the barrel and the pressure subsides enough for the lugs to disengage, well, where does the slide get the momentum to continue cycling?
I have some work to do, but I'm basically here all day. Fire away.
I likey.
So you are suggesting that the barrel and slide move back and contact the locking block/cam-thingy while the bullet might still be in the barrel. And then the friction keeps the lugs from moving, because pressure is still too high.
What you are missing here is that if the locking lugs do not slide apart, the barrel and slide will stop moving. Dead in it's tracks. That momentum is then transferred to the frame and to the shooter.
Once the bullet leaves the barrel and the pressure subsides enough for the lugs to disengage, well, where does the slide get the momentum to continue cycling?
I have some work to do, but I'm basically here all day. Fire away.
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JRH6856
Member
GLOOB said:
But they do slide apart. The horizontal lock delays the vertical unlock but does not stop it. Just like the friction between bullet and barrel does not stop the movement of either, the friction between the jugs do not stop the vertical movement, it just slows it down. The mass of the slide retains sufficient momentum to complete the cycle of the action. The link is not really necessary for unlocking. Take it out and the gun will function, but it will beat itself to death. When the pressure drops, the barrel will drop out of lock on its own, but not until then and not quickly enough. Without the link to pull the barrel out of battery, the lugs will be damaged because they will still be in partial contact when the lower barrel lug hits the impact abuttment in the frame.
Hand cycle without the link and the barrel will drop out of battery as soon as the lower lugs are pulled off of the crosspin of the slide stop.
My Llama IIIa .380 is reduced model of the 1911, but it is straight blowback. It has no lugs and no link. It needs no link because it has no slide lugs. Earlier models had both lugs and link, but Llama realized they were not needed in a .380 as the slide mass and springs were sufficient to keep the breech closed.
JRH6856
Member
No.
Link-down does not start until the slide has moved 0.1 inches. At which time the bullet is 0.1 inches out of the barrel.
And now we have come full circle.
How long it takes the bullet and the slide to reach these respective positions is determined by the actions of all of the forces we have been discussing. But time is not the determining factor. Until both slide and bullet reach these positions, the breech is locked, and the barrel axis does not change.
So you've said nothing. That's the operating principle we were already working under all the while.
When you double the mass of the bullet, the slide will move back nearly 0.2" an inch by the time the bullet exits.
I thought you were suggesting that at 0.1" the friction of the lugs will temporarily retard the slide until such time as the bullet has exited.
It was already obvious and self-evident to the rest of us that the barrel will not start to drop until the link forces it to, at 0.1". Even if gravity is enough to drop the barrel before 0.1" when you hand cycle, you do not need friction to keep it from dropping by gravity. This happens so fast, the effect of gravity will be practically nil. (I agree there's friction there, holding it in place, on top of that!)
So really, your point was already taken. I have used a discrete distance to describe where the barrel starts to unlock in every one of my examples and hypotheticals. I have physically measured this point and posted this information for two of my own handguns. I have never suggested the barrel starts to unlock immediately. And you are frustrated with me?
So now that you've made your full-circle non-point, what was your point?
When you double the mass of the bullet, the slide will move back nearly 0.2" an inch by the time the bullet exits.
I thought you were suggesting that at 0.1" the friction of the lugs will temporarily retard the slide until such time as the bullet has exited.
It was already obvious and self-evident to the rest of us that the barrel will not start to drop until the link forces it to, at 0.1". Even if gravity is enough to drop the barrel before 0.1" when you hand cycle, you do not need friction to keep it from dropping by gravity. This happens so fast, the effect of gravity will be practically nil. (I agree there's friction there, holding it in place, on top of that!)
So really, your point was already taken. I have used a discrete distance to describe where the barrel starts to unlock in every one of my examples and hypotheticals. I have physically measured this point and posted this information for two of my own handguns. I have never suggested the barrel starts to unlock immediately. And you are frustrated with me?
So now that you've made your full-circle non-point, what was your point?
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JRH6856
Member
No, it will not. Oh yes, it will in my Llama, because that is the action you have been describing all along: straight blowback. But you have not been talking about the 1911 and your constant reference to a "unlock/cam thingy" or "cam/locking block" make me question your basic understanding of how it works.
The link is not a "locking block" The link does not lock (or unlock) anything. The locking blocks are the lugs on the frame and slide and the barrel does not have to move to make contact. The lugs are in contact before any movement is initiated.
No need to retard the slide at that point because the bullet has already exited.
Ok on a Glock, (also browning tilt) the slide locks to the barrel via the barrel hood. So that's the locking lug equivalent.
There's a barrel lug on the bottom of the chamber. This hits a locking block that is pinned into the frame.
That locking block is what I'm talking about. At 0.09" on a Glock 27, the barrel lug hits that locking block and the block deflects the barrel down, and a fraction of an inch later it stops the barrel altogether. That locking block is what I'm referring to in my 80's sitcom. I misspoke and called it a locking lug once, I apologize.
I know the 1911 has a moving link or whatnot. Cam, link, I do not know the names for the parts. However it is called or works, it does so in an equivalent way to a Glock.
It's so hard to not be snarky, but when you boil it down, you seem to be possessed by a blind faith that doesn't allow room for reason. You take my explanations as attacks on your faith. You belittle my understanding of a 1911, and you make wild statements out of left field when I am trying so hard to follow your mysterious threads of logic to where they lead. But whenever we get there, you yank the rug back and start over by restating your core beliefs, which you have yet to prove or apply in a meaningful fashion.
There's a barrel lug on the bottom of the chamber. This hits a locking block that is pinned into the frame.
That locking block is what I'm talking about. At 0.09" on a Glock 27, the barrel lug hits that locking block and the block deflects the barrel down, and a fraction of an inch later it stops the barrel altogether. That locking block is what I'm referring to in my 80's sitcom. I misspoke and called it a locking lug once, I apologize.
I know the 1911 has a moving link or whatnot. Cam, link, I do not know the names for the parts. However it is called or works, it does so in an equivalent way to a Glock.
Yes, this.
????? Where did this come from? Are you just trying to take my argument and turn it around backwards? Well, this is a square peg, and you're trying to put it into the color green. This doesn't even begin to make sense unless you can show me why you think this? (It makes no sense to me how you wanted to twist my ice analogy by standing on the bank, either. That was kind of scary, wondering how you got there!)
So you contend. You have made no attempt to demonstrate your reasoning to anyone else. Even where I thought you were trying to support this stance, you went back and retracted your statement to make it into something that is obvious and self-evident and that which we ALL already know and take for granted. And which does nothing to support your belief.
It's so hard to not be snarky, but when you boil it down, you seem to be possessed by a blind faith that doesn't allow room for reason. You take my explanations as attacks on your faith. You belittle my understanding of a 1911, and you make wild statements out of left field when I am trying so hard to follow your mysterious threads of logic to where they lead. But whenever we get there, you yank the rug back and start over by restating your core beliefs, which you have yet to prove or apply in a meaningful fashion.
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