"Thrust Vector" and other ideas
- Thread starter Jim K
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Jim Keenan -
And isn't there also a law that says the pressure inside a chamber (for example) must be the same everywhere?
And isn't there also a law that pressure will produce a force on a surface given by pressure times surface area?
And isn't there a law that says for every force there is an opposite and equal force?
Your description of events has the gas all moving forward for some mysterious reason and pressing only on the bullet.
Isn't it possible the that the equal and opposite force that caused the reaction is also the same type of force that caused the action?
From the law I mentioned above, if pressure is producing a force on the bullet, then it must be producing a force on the breechface also.
Isn't there also a law that says a body at rest will remain at rest unless acted on by a force?
And isn't there also a law that says the pressure inside a chamber (for example) must be the same everywhere?
And isn't there also a law that pressure will produce a force on a surface given by pressure times surface area?
And isn't there a law that says for every force there is an opposite and equal force?
Your description of events has the gas all moving forward for some mysterious reason and pressing only on the bullet.
Isn't it possible the that the equal and opposite force that caused the reaction is also the same type of force that caused the action?
From the law I mentioned above, if pressure is producing a force on the bullet, then it must be producing a force on the breechface also.
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Pressure is the same everwhere, and exerting force equally. However, remember that the breech and barrel are locked together and they are (mostly) sealed with the brass case, that's pretty much like a closed breech cannon. The ONLY thing that can move appreciably in this case is the bullet and gas, forward, sending the slide and barrel backward. By the time the slide and barrel have moved far enough, to where the case is being extracted, the only forces left working on the system, should be the rapidly exiting gasses.
Tuner,
OK, another try. In your analogy of, I'll say, "yanking" the bullet from the barrel, you seem to be convinced that the barrel will recoil.
When someone yanks a tablecloth from under some glasses (similar situation), do the glasses go flying backward? No. They move forward slightly, being dragged by the tablecloth. Even if there was a lot of friction, the glasses wouldn't go backward, they'd accelerate in the same direction as the tablecloth. This is where friction would actually pull the barrel along, that is, if we were to somehow yank a bullet out of a barrel. Now if we put a charge under the glass on the tablecloth, it would move in one direction, and the tablecloth another, but as I've been trying to convince you, in that case, the force is arising in a different manner, and applying compression bewteen the two, rather than tension on the other side (see, the push-pull makes a big difference here as well).
Maybe you can be convinced, eventually, that you don't need, indeed don't have these mysterious "unbalanced thrust vectors" from the bullet, but I'm not sure, you seem pretty stubborn
Tuner,
OK, another try. In your analogy of, I'll say, "yanking" the bullet from the barrel, you seem to be convinced that the barrel will recoil.
When someone yanks a tablecloth from under some glasses (similar situation), do the glasses go flying backward? No. They move forward slightly, being dragged by the tablecloth. Even if there was a lot of friction, the glasses wouldn't go backward, they'd accelerate in the same direction as the tablecloth. This is where friction would actually pull the barrel along, that is, if we were to somehow yank a bullet out of a barrel. Now if we put a charge under the glass on the tablecloth, it would move in one direction, and the tablecloth another, but as I've been trying to convince you, in that case, the force is arising in a different manner, and applying compression bewteen the two, rather than tension on the other side (see, the push-pull makes a big difference here as well).
Maybe you can be convinced, eventually, that you don't need, indeed don't have these mysterious "unbalanced thrust vectors" from the bullet, but I'm not sure, you seem pretty stubborn
Thegman -
But it only causes one of them to move?
Both are masses, both are free to move.
You need to back that up with some principle of physics.
What causes one mass to move when force is applied, but not another?
Here:
Take two blocks of wood. A 1 lb block on the left and a 10 lb block on the right. Put a coil spring between them. Push the two blocks towards each other compressing the spring. Now let go.
The spring must produce two equal forces, in two opposite directions. The action of one block moving left with a certain momentum, must be accompanied by the other block moving right with the same momentum (3rd law). And the force in each case was that of the spring.
The left block will move 10 times faster than the right block since it is 10 times lighter. Conservation of momentum.
What you are saying is that the spring moves to the left and presses only on the 1 lb block. Then the 10 lb block, with no spring force on it, moves to right. And you don't see anything wrong with that?
You have agreed that there is pressure on both surfaces. You have agreed that there is force in both directions.
But it only causes one of them to move?
Both are masses, both are free to move.
You need to back that up with some principle of physics.
What causes one mass to move when force is applied, but not another?
Here:
Take two blocks of wood. A 1 lb block on the left and a 10 lb block on the right. Put a coil spring between them. Push the two blocks towards each other compressing the spring. Now let go.
The spring must produce two equal forces, in two opposite directions. The action of one block moving left with a certain momentum, must be accompanied by the other block moving right with the same momentum (3rd law). And the force in each case was that of the spring.
The left block will move 10 times faster than the right block since it is 10 times lighter. Conservation of momentum.
What you are saying is that the spring moves to the left and presses only on the 1 lb block. Then the 10 lb block, with no spring force on it, moves to right. And you don't see anything wrong with that?
1911Tuner
Moderator Emeritus
Try Try Again...
Jim...You're being facetious. This HAS been fun though!
Okay gman...You've tried again, and you're right...I am pretty stubborn.
So, let's assume for a minute that you're correct in stating that a pushing force isn't going to produce the same dynamics in the breech as a pulling force, though they both accomplish the same thing....accelerating the bullet.
Let's imagine that Alliant or Du Pont has developed a magical propellant that defies the laws of physics...and directs its force in one direction
only...forward. Okay...NOW we have a bullet that's being pushed through the barrel from behind...except there is no other direction of applied force.
Forward only, and the only thing being affected by this magical force is the bullet and its frictional resistance on the barrel and its inertial resistance to
acceleration/movement. With this inidirectionally applied force...a shaped charge of sorts...would the slide still recoil? I don't think so. I believe
that the bullet would pull the barrel forward, and bring the slide with it
via the mechanical connection until the bullet exited, effectively turning the pistol into a straight-pull, bolt-action weapon.
Point 2...When the round in the chamber, the case is against the breechface...or nearly so, depending on the headspace of a given pistol.
When the powder ignites, the resulting force pushes against the base of the bullet and against the case equally. The force directed backward against the case is transferred to the slide..and no...the expanding gasses don't immediately freeze the case to the chamber walls. It occurs gradually, though very quickly because the pressure builds on a curve.
There is a vectored force on both ends. It can be no other way.
If you ignited the power charge in a free-hanging barrel, would not both the bullet and the case move? Or would the case stay in the chamber?
If by some magical mystery, such an event would allow full pressure to develop, I'd be willing to bet that the case would become the bullet, and the bullet would become the breechface...and the barrel would recoil FORWARD due to the vector of thust becoming unequal in the other direction.
This one is doable. All it would take is a barrel, a cartridge, and a fire
built under the barrel to ignite the primer, so that our experimenter could watch from a safe distance. You could even swage a bullet down to the point that it would be a very light slip-fit in the bore to make sure that it
would move....and the barrel would still recoil forward. Maybe a cast bullet
sized down to .44 caliber, and the case necked down in a .44 die...with
just a light paper patch to effect a little gas seal with minimal resistance to movement....Shall we put it to the test?
Just for the record...Pulling a barbell off the ground absolutely will exert the
asame force against the ground as lying down and pushing it up. One vector is through the feet, and the other is through the shoulders...or maybe I misread what you wrote.
Jim...You're being facetious.
This HAS been fun though!Okay gman...You've tried again, and you're right...I am pretty stubborn.
So, let's assume for a minute that you're correct in stating that a pushing force isn't going to produce the same dynamics in the breech as a pulling force, though they both accomplish the same thing....accelerating the bullet.
Let's imagine that Alliant or Du Pont has developed a magical propellant that defies the laws of physics...and directs its force in one direction
only...forward. Okay...NOW we have a bullet that's being pushed through the barrel from behind...except there is no other direction of applied force.
Forward only, and the only thing being affected by this magical force is the bullet and its frictional resistance on the barrel and its inertial resistance to
acceleration/movement. With this inidirectionally applied force...a shaped charge of sorts...would the slide still recoil? I don't think so. I believe
that the bullet would pull the barrel forward, and bring the slide with it
via the mechanical connection until the bullet exited, effectively turning the pistol into a straight-pull, bolt-action weapon.
Point 2...When the round in the chamber, the case is against the breechface...or nearly so, depending on the headspace of a given pistol.
When the powder ignites, the resulting force pushes against the base of the bullet and against the case equally. The force directed backward against the case is transferred to the slide..and no...the expanding gasses don't immediately freeze the case to the chamber walls. It occurs gradually, though very quickly because the pressure builds on a curve.
There is a vectored force on both ends. It can be no other way.
If you ignited the power charge in a free-hanging barrel, would not both the bullet and the case move? Or would the case stay in the chamber?
If by some magical mystery, such an event would allow full pressure to develop, I'd be willing to bet that the case would become the bullet, and the bullet would become the breechface...and the barrel would recoil FORWARD due to the vector of thust becoming unequal in the other direction.
This one is doable. All it would take is a barrel, a cartridge, and a fire
built under the barrel to ignite the primer, so that our experimenter could watch from a safe distance. You could even swage a bullet down to the point that it would be a very light slip-fit in the bore to make sure that it
would move....and the barrel would still recoil forward. Maybe a cast bullet
sized down to .44 caliber, and the case necked down in a .44 die...with
just a light paper patch to effect a little gas seal with minimal resistance to movement....Shall we put it to the test?
Just for the record...Pulling a barbell off the ground absolutely will exert the
asame force against the ground as lying down and pushing it up. One vector is through the feet, and the other is through the shoulders...or maybe I misread what you wrote.
Andrew Wyatt
Member
the breech is LOCKED to the barrel. it cannot move.
1911Tuner
Moderator Emeritus
Blocked the Move
Andrew said:
the breech is LOCKED to the barrel. it cannot move.
Unless they both move in the same direction...
Which they'll do as soon as the bullet starts to move...
So we're back to: Does the slide move and pull the barrel with it...or does the barrel move and push the slide? Or is it a little of both?
Aspirin! I need aspirin!
Andrew said:
the breech is LOCKED to the barrel. it cannot move.
Unless they both move in the same direction...
Which they'll do as soon as the bullet starts to move...
So we're back to: Does the slide move and pull the barrel with it...or does the barrel move and push the slide? Or is it a little of both?
Aspirin! I need aspirin!
Andrew Wyatt
Member
well, since the case is locked to the slide because the inside pressure has expanded it so its in interference fit, i'd say that the barrel pushes on the slide.
the only way the slide could recoil and pull the barrel was if the case was part of the slide and not the barrel.
the only way the slide could recoil and pull the barrel was if the case was part of the slide and not the barrel.
1911Tuner
Moderator Emeritus
Locked!
Andrew said:
well, since the case is locked to the slide because the inside pressure has expanded it so its in interference fit, i'd say that the barrel pushes on the slide.
_________________
That's the way it looks to me, except that the case doesn't instantly expand and pressure weld itself to the chamber. It blows back for a little distance...or...if the case rim is flush with the barrel hood, it's already touching the slide. So...is it a little of both? Does it start with the
case bearing directly on the slide under pressure, and then expand to the chamber walls, pulling the barrel back against the breechface like a sort of
piston? Note the very short piston stroke on the M-1 Carbine...and even
the relatively short movement of the M-14's piston. The op-rod gets
what is really nothing more than a quick bump...and the bolt cycles fully
rearward against spring pressure and even residual chamber pressure
that still has a little effect on the case. The M-14's system is timed to
tap the gas into the cylinder after chamber pressure has fallen off enough to allow extraction...but there's still a little left. (Incidentally, this is why the
gas-operated rifles don't do well with a slow powder or a pressure spike that's a little too "fat" under the curve. It taps the gas while the case is still under high pressure, and causes extraction and short-cycle problems.)
Given that a 30-caliber rifle will cycle with a mere bump...how much of a bump would a low-intensity round like the .45 ACP require? Not a helliva lot, I'd wager.
Or...does the residual chamber pressure in the '14 actually HELP with extraction?
Andrew said:
well, since the case is locked to the slide because the inside pressure has expanded it so its in interference fit, i'd say that the barrel pushes on the slide.
_________________
That's the way it looks to me, except that the case doesn't instantly expand and pressure weld itself to the chamber. It blows back for a little distance...or...if the case rim is flush with the barrel hood, it's already touching the slide. So...is it a little of both? Does it start with the
case bearing directly on the slide under pressure, and then expand to the chamber walls, pulling the barrel back against the breechface like a sort of
piston? Note the very short piston stroke on the M-1 Carbine...and even
the relatively short movement of the M-14's piston. The op-rod gets
what is really nothing more than a quick bump...and the bolt cycles fully
rearward against spring pressure and even residual chamber pressure
that still has a little effect on the case. The M-14's system is timed to
tap the gas into the cylinder after chamber pressure has fallen off enough to allow extraction...but there's still a little left. (Incidentally, this is why the
gas-operated rifles don't do well with a slow powder or a pressure spike that's a little too "fat" under the curve. It taps the gas while the case is still under high pressure, and causes extraction and short-cycle problems.)
Given that a 30-caliber rifle will cycle with a mere bump...how much of a bump would a low-intensity round like the .45 ACP require? Not a helliva lot, I'd wager.
Or...does the residual chamber pressure in the '14 actually HELP with extraction?
Andrew Wyatt
Member
the bump is pretty small, though it is over a larger area. if you tool a 1911, and make the lockup perfect front to back with no slop whatsoever and had perfect headspace, would the gun function (barring reliability problems from the gun being too tight)? I think so.
I don't think JMB designed the gun to have that "feature".
I don't think JMB designed the gun to have that "feature".
1911Tuner
Moderator Emeritus
Bump?
the bump is pretty small, though it is over a larger area. if you tool a 1911, and make the lockup perfect front to back with no slop whatsoever and had perfect headspace, would the gun function.
Sure it would. My point is that even thought the bump would have a small influence...it would almost certainly have SOME. That's why I'm pondering on the notion that it's a combination of several functions happening in sequence and so quickly that one is hard to tell from the other...hard to say
which has the most influence. Everything means SOMEthing.
Momentum...Equal and Opposite...Gas pressure...Semi-locked/Semi delayed blowback...Balanced and then an UN-balanced vector of thrust...Is
it ALL of these things? :banghead: Aspirin! :banghead: I need aspirin!
:banghead:
One thing is sure...The force that completes the cycle has come and gone before the slide moves a quarter-inch...After that, it's up to the conservation of momentum.
the bump is pretty small, though it is over a larger area. if you tool a 1911, and make the lockup perfect front to back with no slop whatsoever and had perfect headspace, would the gun function.
Sure it would. My point is that even thought the bump would have a small influence...it would almost certainly have SOME. That's why I'm pondering on the notion that it's a combination of several functions happening in sequence and so quickly that one is hard to tell from the other...hard to say
which has the most influence. Everything means SOMEthing.
Momentum...Equal and Opposite...Gas pressure...Semi-locked/Semi delayed blowback...Balanced and then an UN-balanced vector of thrust...Is
it ALL of these things? :banghead: Aspirin! :banghead: I need aspirin!
:banghead:
One thing is sure...The force that completes the cycle has come and gone before the slide moves a quarter-inch...After that, it's up to the conservation of momentum.
Hi Tuner,
You're not so stubborn afterall!
Let's imagine that Alliant or Du Pont has developed a magical propellant... I believe
that the bullet would pull the barrel forward, and bring the slide with it via the mechanical connection until the bullet exited, effectively turning the pistol into a straight-pull, bolt-action weapon.
This is what I'm saying with the tablecloth. But don't confuse this with igniting a charge in the pistol. The internal charge essentially makes the pistol a free body diagram, momenutm is conserved, more or less, within the pistol (I guess we'd need to discount shooter's part).
A firearm with a bullet being pulled (or pushed) in only one direction, with no force on the surrounding pistol, is no longer a free body problem. So, I've convinced you that yanking a bullet from a gun won't produce any recoil?
Point 2..
If you ignited the power charge in a free-hanging barrel, would not both the bullet and the case move? Or would the case stay in the chamber?
If by some magical mystery, such an event would allow full pressure to develop, I'd be willing to bet that the case would become the bullet, and the bullet would become the breechface...and the barrel would recoil FORWARD due to the vector of thust becoming unequal in the other direction.
This is still a free body diagram. Assuming the case doesn't get stuck, its velocity will be in relation to its mass, compared to the mass of the barrel and bullet. In the real world, the case would pop out with not too much velocity, and the barrel and bullet would swing forward on the string; they'd act like the breech, as you say. But this isn't due to the vector of thrust becoming unequal! The force (thrust) is equal in both directions, the case simply has much less mass (inertia) than the barrel and bullet combination do, so it (the case) leaves with a higer velocity (becomes the projectile) due to conservation of momentum in the system.
Just for the record...Pulling a barbell off the ground absolutely will exert the
asame force against the ground as lying down and pushing it up. One vector is through the feet, and the other is through the shoulders...or maybe I misread what you wrote.
I know you said pulling the weight up, but I had the weight yanked from suspended cable to make it more similar to your tug-o-war analogy (and yes, if we followed the interactions far enough, the cable's force will transfer to the floor (or ground) somewhere).
You have agreed that there is pressure on both surfaces. You have agreed that there is force in both directions.
But it only causes one of them to move?
RJ357,
Read my paragraph again:
Pressure is the same everwhere, and exerting force equally. However, remember that the breech and barrel are locked together and they are (mostly) sealed with the brass case, that's pretty much like a closed breech cannon. The ONLY thing that can move appreciably in this case is the bullet and gas, forward, sending the slide and barrel backward. By the time the slide and barrel have moved far enough, to where the case is being extracted, the only forces left working on the system, should be the rapidly exiting gasses.
I'm not saying the pistol doesn't also move in the opposite direction, but assuming the frame is held securely, the side and barrel (and case) are the only part that can move opposite the bullet and gasses, and move together, simply becasue they are all locked together as a single unit.
I guess I'm also implying that the rest of the barrel, especially the chamber, is also 'feeling' this force from the pressure, but, hopefully, the most this part will do is stretch a little bit and won't fail. If the barrel walls do fail, we've have yet another free body diagram, this time in the form of a hand-held grenade.
You're not so stubborn afterall!
Let's imagine that Alliant or Du Pont has developed a magical propellant... I believe
that the bullet would pull the barrel forward, and bring the slide with it via the mechanical connection until the bullet exited, effectively turning the pistol into a straight-pull, bolt-action weapon.
This is what I'm saying with the tablecloth. But don't confuse this with igniting a charge in the pistol. The internal charge essentially makes the pistol a free body diagram, momenutm is conserved, more or less, within the pistol (I guess we'd need to discount shooter's part).
A firearm with a bullet being pulled (or pushed) in only one direction, with no force on the surrounding pistol, is no longer a free body problem. So, I've convinced you that yanking a bullet from a gun won't produce any recoil?
Point 2..
If you ignited the power charge in a free-hanging barrel, would not both the bullet and the case move? Or would the case stay in the chamber?
If by some magical mystery, such an event would allow full pressure to develop, I'd be willing to bet that the case would become the bullet, and the bullet would become the breechface...and the barrel would recoil FORWARD due to the vector of thust becoming unequal in the other direction.
This is still a free body diagram. Assuming the case doesn't get stuck, its velocity will be in relation to its mass, compared to the mass of the barrel and bullet. In the real world, the case would pop out with not too much velocity, and the barrel and bullet would swing forward on the string; they'd act like the breech, as you say. But this isn't due to the vector of thrust becoming unequal! The force (thrust) is equal in both directions, the case simply has much less mass (inertia) than the barrel and bullet combination do, so it (the case) leaves with a higer velocity (becomes the projectile) due to conservation of momentum in the system.
Just for the record...Pulling a barbell off the ground absolutely will exert the
asame force against the ground as lying down and pushing it up. One vector is through the feet, and the other is through the shoulders...or maybe I misread what you wrote.
I know you said pulling the weight up, but I had the weight yanked from suspended cable to make it more similar to your tug-o-war analogy (and yes, if we followed the interactions far enough, the cable's force will transfer to the floor (or ground) somewhere).
You have agreed that there is pressure on both surfaces. You have agreed that there is force in both directions.
But it only causes one of them to move?
RJ357,
Read my paragraph again:
Pressure is the same everwhere, and exerting force equally. However, remember that the breech and barrel are locked together and they are (mostly) sealed with the brass case, that's pretty much like a closed breech cannon. The ONLY thing that can move appreciably in this case is the bullet and gas, forward, sending the slide and barrel backward. By the time the slide and barrel have moved far enough, to where the case is being extracted, the only forces left working on the system, should be the rapidly exiting gasses.
I'm not saying the pistol doesn't also move in the opposite direction, but assuming the frame is held securely, the side and barrel (and case) are the only part that can move opposite the bullet and gasses, and move together, simply becasue they are all locked together as a single unit.
I guess I'm also implying that the rest of the barrel, especially the chamber, is also 'feeling' this force from the pressure, but, hopefully, the most this part will do is stretch a little bit and won't fail. If the barrel walls do fail, we've have yet another free body diagram, this time in the form of a hand-held grenade.
Tuner, in quickly reading the posts, it appears you have situation essentially clear. So, I don't know why you insist in this idea that the barrel pushing the slide is what operates the pistol. The barrel does not push the slide.
If the barrel pushed the slide, then there would be no need for locking lugs locking the slide to the barrel. What purpose would they serve if the barrel pushes the slide? Think of towing vs. pushing a car. To tow a car, you need to link it to another car (locking lugs). To push a car, no such connection needed. Just bumper to bumper and push.
If the barrel pushed the slide, the slide would move only as fast rearward as the barrel pushed it - like pushing a car. Hence, the slide would never separate from the barrel until the barrel was haulted by the link.
But there are locking lugs. Why? Because we need to lock the slide to the barrel. Why? Because if we don't, the slide separates from the barrel before gas pressure drops to a safe level. Which means? Which means that the slide is under a force accelerating it much faster than any force acting on the barrel - via cases sticking to chamber walls - and accelerating the barrel rearward.
All of this talk of cases sticking to chamber walls and imparting a rearward force to the barrel is nonsense in a short recoil weapon anyway. Unlike in a blowback or delayed blowback weapon, where reliable case movement is essential to operation of the pistol (hence HK's chamber fluting). In a short recoil weapon, case movement simply isn't required to get the slide/barrel mechanism moving.
Look at the lockup of a 1911, there is no slop between the slide lugs and barrel lugs that would allow the slide to move even a tad without moving the barrel also. Such slop which would have to be present for the case to move in the chamber at all during firing but prior to unlocking - either that of a gap between the case base and breech face, but that's not present either. So, the case in a 1911 doesn't move at all in the chamber prior to unlocking and extraction.
What's the significance of this? Sure the case walls might expand and stick to the chamber walls, but so what. The case can't move rearward at all in a 1911, so the sticking case walls can't/don't exert any pull in the rearward direction. By the time the case is able to move (unlocking/extraction) the case expansion is over and the case walls are no longer stuck to the chamber. (Think of it this way: I can put my palms inside a trash can, push the outward against the inside of the can, and then lift up the can. But if all I do is push my palms outward and never lift up, then my palm pressure never exerts an accelerating/lifting force on the trashcan in the upward direction. Same with the case expansion in the 1911. There is pressure against chamber walls, but no movement rearward possible, so no accelerating force rearward created). So, even if the case walls sticking to the chamber is capable of imparting a rearward force, in a 1911 it just doesn't happen.
The slide pulls the barrel rearward (via the locking lugs) due to an accelerating force imparted to the slide through the case base as a result of expanding gases in the chamber. What more needs to be said? Where's the confusion?
If the barrel pushed the slide, then there would be no need for locking lugs locking the slide to the barrel. What purpose would they serve if the barrel pushes the slide? Think of towing vs. pushing a car. To tow a car, you need to link it to another car (locking lugs). To push a car, no such connection needed. Just bumper to bumper and push.
If the barrel pushed the slide, the slide would move only as fast rearward as the barrel pushed it - like pushing a car. Hence, the slide would never separate from the barrel until the barrel was haulted by the link.
But there are locking lugs. Why? Because we need to lock the slide to the barrel. Why? Because if we don't, the slide separates from the barrel before gas pressure drops to a safe level. Which means? Which means that the slide is under a force accelerating it much faster than any force acting on the barrel - via cases sticking to chamber walls - and accelerating the barrel rearward.
All of this talk of cases sticking to chamber walls and imparting a rearward force to the barrel is nonsense in a short recoil weapon anyway. Unlike in a blowback or delayed blowback weapon, where reliable case movement is essential to operation of the pistol (hence HK's chamber fluting). In a short recoil weapon, case movement simply isn't required to get the slide/barrel mechanism moving.
Look at the lockup of a 1911, there is no slop between the slide lugs and barrel lugs that would allow the slide to move even a tad without moving the barrel also. Such slop which would have to be present for the case to move in the chamber at all during firing but prior to unlocking - either that of a gap between the case base and breech face, but that's not present either. So, the case in a 1911 doesn't move at all in the chamber prior to unlocking and extraction.
What's the significance of this? Sure the case walls might expand and stick to the chamber walls, but so what. The case can't move rearward at all in a 1911, so the sticking case walls can't/don't exert any pull in the rearward direction. By the time the case is able to move (unlocking/extraction) the case expansion is over and the case walls are no longer stuck to the chamber. (Think of it this way: I can put my palms inside a trash can, push the outward against the inside of the can, and then lift up the can. But if all I do is push my palms outward and never lift up, then my palm pressure never exerts an accelerating/lifting force on the trashcan in the upward direction. Same with the case expansion in the 1911. There is pressure against chamber walls, but no movement rearward possible, so no accelerating force rearward created). So, even if the case walls sticking to the chamber is capable of imparting a rearward force, in a 1911 it just doesn't happen.
The slide pulls the barrel rearward (via the locking lugs) due to an accelerating force imparted to the slide through the case base as a result of expanding gases in the chamber. What more needs to be said? Where's the confusion?
1911Tuner
Moderator Emeritus
Straight from the Horse's Mouth
I just got back from a little drive. I stopped by one of our community
colleges to visit with some friends, and I dropped in on one of the physics professors. I asked him to...without going too deep into the laws governing the conservation of momentum and various formulae...to explain firearm recoil in the simplest terms possible. He suggested that we use the simplest gun possible..The musket. (He builds blackpowder rifles as a hobby)
From the bottom, adding one fact at a time, his explanation is this...as nearly verbatim as I can remember from an hour ago.
Acceleration of an object can't occur without a force acting on it to upset its equilibrium.
The notion that the bullet's movement alone causes the gun to recoil is absurd. The bullet and the gun must have a force between them in order for either one to accelerate, and that force must be greater than the inertial and frictional resistance to acceleration.
At that point, I mentioned my hypothetical pulling of the bullet. He liked that, and said that it was absolutely correct, and a very good analogy.
When a gun fires, the explosion of the powder provides the force to accelerate the bullet.
Recoil occurs because the bullet pushes against the gun THROUGH that force. There is no other way that it can occur. The force uses the gun to
push the bullet forward, and the bullet in turn uses the force to push the gun backward. Both occur at the same time, but because the gun has greater mass, the bullet's acceleration is greater.
He then added that a smoothbore musket will provide less recoil than an identical musket with a rifled barrel because the frictional resistance of the ball offers more resistance to acceleration and thus more force redirected backward into the rifle...even if the two rifles fire their balls at the same velocity. The harder the force has to work to overcome friction, the harder
that force pushes backward in the equal and opposite reaction.
Bullet acceleration occurs because the gun provides something for the force to use to push from.. If the breech were open and free of obstruction, the bullet would move only fractions of an inch because its mass is greater than the mass of the air behind it, and only then if it had enough freebore
to prevent frictional resistance.
Gun acceleration occurs because the bullet provides something for the
force to push from. if the bullet were absent, the gun would not recoil
because its mass is much greater than the air in the barrel.
_________________
He doesn't know a lot about 1911s, so I gave him a crash course in how
the slide and barrel were connected, and described Keenan's blocked bore
experiment. He looked puzzled and said that of course nothing moved.
Two objects were being pushed in opposite directions and that both
objects were prevented from moving by a mechanical obstruction. The
only way that movement could occur would be that if the force were great enough to destroy the obstruction.
Then, (and I swear to John Moses he said this( he said to imagine a tug of war between to identically strong men...
Here endeth the lesson...for me at least.
I just got back from a little drive. I stopped by one of our community
colleges to visit with some friends, and I dropped in on one of the physics professors. I asked him to...without going too deep into the laws governing the conservation of momentum and various formulae...to explain firearm recoil in the simplest terms possible. He suggested that we use the simplest gun possible..The musket. (He builds blackpowder rifles as a hobby)
From the bottom, adding one fact at a time, his explanation is this...as nearly verbatim as I can remember from an hour ago.
Acceleration of an object can't occur without a force acting on it to upset its equilibrium.
The notion that the bullet's movement alone causes the gun to recoil is absurd. The bullet and the gun must have a force between them in order for either one to accelerate, and that force must be greater than the inertial and frictional resistance to acceleration.
At that point, I mentioned my hypothetical pulling of the bullet. He liked that, and said that it was absolutely correct, and a very good analogy.
When a gun fires, the explosion of the powder provides the force to accelerate the bullet.
Recoil occurs because the bullet pushes against the gun THROUGH that force. There is no other way that it can occur. The force uses the gun to
push the bullet forward, and the bullet in turn uses the force to push the gun backward. Both occur at the same time, but because the gun has greater mass, the bullet's acceleration is greater.
He then added that a smoothbore musket will provide less recoil than an identical musket with a rifled barrel because the frictional resistance of the ball offers more resistance to acceleration and thus more force redirected backward into the rifle...even if the two rifles fire their balls at the same velocity. The harder the force has to work to overcome friction, the harder
that force pushes backward in the equal and opposite reaction.
Bullet acceleration occurs because the gun provides something for the force to use to push from.. If the breech were open and free of obstruction, the bullet would move only fractions of an inch because its mass is greater than the mass of the air behind it, and only then if it had enough freebore
to prevent frictional resistance.
Gun acceleration occurs because the bullet provides something for the
force to push from. if the bullet were absent, the gun would not recoil
because its mass is much greater than the air in the barrel.
_________________
He doesn't know a lot about 1911s, so I gave him a crash course in how
the slide and barrel were connected, and described Keenan's blocked bore
experiment. He looked puzzled and said that of course nothing moved.
Two objects were being pushed in opposite directions and that both
objects were prevented from moving by a mechanical obstruction. The
only way that movement could occur would be that if the force were great enough to destroy the obstruction.
Then, (and I swear to John Moses he said this( he said to imagine a tug of war between to identically strong men...
Here endeth the lesson...for me at least.
Hand_Rifle_Guy
Member
Finished? Not on your tintype...
Well, I'm glad you've got it solved to your satisfaction. I also thought it'd been covered thoroughly enough to get all that across, but sometimes you just gotta keep pluggin' until you latch on to the right analogy, and then it's like: Eureka!
Now, get thee hence to yonder thread and forthwith dispense thy profuse wisdom.
Cryptic remarks ain't a'gonna fly. Chamber fluting's got it's own set of action-defining bugaboos. It's not nice to leave your fellow HighRiders hangin'.
Well, I'm glad you've got it solved to your satisfaction. I also thought it'd been covered thoroughly enough to get all that across, but sometimes you just gotta keep pluggin' until you latch on to the right analogy, and then it's like: Eureka!
Now, get thee hence to yonder thread and forthwith dispense thy profuse wisdom.
Cryptic remarks ain't a'gonna fly. Chamber fluting's got it's own set of action-defining bugaboos. It's not nice to leave your fellow HighRiders hangin'.
1911Tuner
Moderator Emeritus
Eureka!
Howdy Hand_Rifle_Guy,
I had it down all along, I just wasn't able to simplify it to that degree. gman and I were on the same page...we just started from different points in the book on the wayto it.
Okay...So everybody's satisfied that the burning gasses do provide the
force for the recoil stroke. The only thing left to determine is:
Does the case transfer the force directly to the slide, pulling the barrel with it after it overcomes the forward thrust of the bullet and rifling friction?
Does the case expand, seal the chamber and pull the barrel backward, dragging the slide with it...after the bullet's momentum is overcome by the slide's?
OR...is it a little of both?
Aspirin? Anybody?
Howdy Hand_Rifle_Guy,
I had it down all along, I just wasn't able to simplify it to that degree. gman and I were on the same page...we just started from different points in the book on the wayto it.
Okay...So everybody's satisfied that the burning gasses do provide the
force for the recoil stroke. The only thing left to determine is:
Does the case transfer the force directly to the slide, pulling the barrel with it after it overcomes the forward thrust of the bullet and rifling friction?
Does the case expand, seal the chamber and pull the barrel backward, dragging the slide with it...after the bullet's momentum is overcome by the slide's?
OR...is it a little of both?
Aspirin? Anybody?
Hand_Rifle_Guy
Member
Why do they always have to be so DIFFICULT?!!
RJ357, .45 ACP operating pressure runs in a range of about 17,000 PSI to 19,000 PSI depending on the load. "+P" ammo runs as high as 21,000 PSI.
Tuner, Your Definitive Answer, no qualifier this time. Dis be Da Troot, So Hep Me Ahmighty:
The case sets back before expansion provides the gas seal, so I will establish that th FIRST BIT of applied rearward force is initiated against the slide, which moves back, taking up what little slack that exists between the slide and the barrel, and proceeds to exert the initial rearward force to the barrel, thus dragging it along.
There. I said it. The slide drags the barrel.
For about a millisecond, whereupon the slide, case, and barrel become "glued" together into what amounts to a solid unit by expansion pressure, and stay that way until the change in angular momentum of the barrel brought about by the link geometry drags the barrel free of the slide's influence, and it bangs to a halt. The slide continues on it's merry way by virtue of momentum, dragging the now unpressurized case from the chamber and into the ejector, which pivots the case off of the extractor claw's grip and out of the gun. The slide next crashes into the frame or a Shok-buff and stops, whereupon the recoil spring starts to shove it forward again to repeat the whole tedious proccess over again. (I called it tedious? It takes a whopping, what, 5,6 milliseconds to cycle? I gotta shut the heck up.)
It is worth noting that the initiating conditions are subject to a different order of operations when subjected to a significantly differently timed burn/expansion curve by virtue of a different powder in a given load. This is only a difference in 1/2-millisecond increments to the components involved which initiates no change to the overall outcome.
If THAT doesn't put paid to this thread, Tuner, so help me I'll...I'll...I'll put vaseline on the toilet seats and short-sheet the beds! Really I will! See if I don't! Nyaah!
Don't you make me come down there... Edited for typos.
RJ357, .45 ACP operating pressure runs in a range of about 17,000 PSI to 19,000 PSI depending on the load. "+P" ammo runs as high as 21,000 PSI.
Tuner, Your Definitive Answer, no qualifier this time. Dis be Da Troot, So Hep Me Ahmighty:
The case sets back before expansion provides the gas seal, so I will establish that th FIRST BIT of applied rearward force is initiated against the slide, which moves back, taking up what little slack that exists between the slide and the barrel, and proceeds to exert the initial rearward force to the barrel, thus dragging it along.
There. I said it. The slide drags the barrel.
For about a millisecond, whereupon the slide, case, and barrel become "glued" together into what amounts to a solid unit by expansion pressure, and stay that way until the change in angular momentum of the barrel brought about by the link geometry drags the barrel free of the slide's influence, and it bangs to a halt. The slide continues on it's merry way by virtue of momentum, dragging the now unpressurized case from the chamber and into the ejector, which pivots the case off of the extractor claw's grip and out of the gun. The slide next crashes into the frame or a Shok-buff and stops, whereupon the recoil spring starts to shove it forward again to repeat the whole tedious proccess over again. (I called it tedious? It takes a whopping, what, 5,6 milliseconds to cycle? I gotta shut the heck up.)
It is worth noting that the initiating conditions are subject to a different order of operations when subjected to a significantly differently timed burn/expansion curve by virtue of a different powder in a given load. This is only a difference in 1/2-millisecond increments to the components involved which initiates no change to the overall outcome.
If THAT doesn't put paid to this thread, Tuner, so help me I'll...I'll...I'll put vaseline on the toilet seats and short-sheet the beds! Really I will! See if I don't! Nyaah!
Don't you make me come down there...
Edited for typos.P95Carry
Moderator Emeritus
What a momentous ''can-o-worms'' thread!
HRG .....
HRG .....
Oh - I'd thought it was super-glue!!
Clark
Member
There is a thrust vector in my pistol.
I powered my car to work this morning with a piston.
I held my pants up today with a belt sinch jerk.
I learned to speak English with lips mouth tingle.
The color of the Sky is dim spectral clear.
There is a thrust vector in my make up phase random say go blah blah.
I powered my car to work this morning with a piston.
I held my pants up today with a belt sinch jerk.
I learned to speak English with lips mouth tingle.
The color of the Sky is dim spectral clear.
There is a thrust vector in my make up phase random say go blah blah.
...He then added that a smoothbore musket will provide less recoil than an identical musket with a rifled barrel because the frictional resistance of the ball offers more resistance to acceleration and thus more force redirected backward into the rifle...even if the two rifles fire their balls at the same velocity. The harder the force has to work to overcome friction, the harder that force pushes backward in the equal and opposite reaction.
Tuner,
We're almost on the same page now...
...but, I don't think the physics professor is totally correct, WRT friction. I don't think the friction itself is creating more recoil simply due to more force being required to push the bullet down the bore.
Remember the professor's puzzled look about Keenan's experiment? No bullet movement, no slide movement, becasue everything was mechanically locked? If friction is slowing the bullet, it's actually becoming a form of a mechanical lock in the bore. Now, before you get all flustered with me, give this some thought:
If we raised the friction in the barrel enough (for instance a tapering internal bore diameter) to allow the bullet to move, for instance, 2 inches down the bore at a low velocity, with little accelleration, and then stop, and this movemnet required, say, 100,000 psi, we'd essentially have Keenan's experiment (albeit even more pressure). The gun's breech would be pushing really, really hard against that slowy moving bullet, and the gun might explode, but it wouldn't recoil appreciably.
As far as friction from rifling causing more recoil becasue the gun is pushing harder on the bullet:
For equal velocities, the smooth bore probably will have less recoil, but that's due, I think, to the requirement of considerably less powder and less pressure to reach that velocity. I think the greater mass of higher pressure gasses are what's causing the increase in recoil in the rifled bore. Of course, friction is what is causing the requirement for higher pressures, so in that way, I guess it is responisble. But not because the gun has to push harder on the bullet, but because their is a requirement for more gas, at a greater pressure to attain a given velocity. The gun's harder push on the bullet due to friction creates greater static forces in the gun, especially the barrel, but not more recoil.
Taking only the bullet and the gun into account (not the gasses), the gun's recoil velocity, I think, will be calculated due to the bullet's mass and it's attained velocity; the amount of force required to get the bullet to that velocity will not be part of the equation. But in the real world, if we need more force to push the bullet due to friction, that will require greater amounts of gasses at ever higer pressures.
If you give this note to the physics professor, after a couple seconds of thought, I think he'll agree.
Tuner,
We're almost on the same page now...
...but, I don't think the physics professor is totally correct, WRT friction. I don't think the friction itself is creating more recoil simply due to more force being required to push the bullet down the bore.
Remember the professor's puzzled look about Keenan's experiment? No bullet movement, no slide movement, becasue everything was mechanically locked? If friction is slowing the bullet, it's actually becoming a form of a mechanical lock in the bore. Now, before you get all flustered with me, give this some thought:
If we raised the friction in the barrel enough (for instance a tapering internal bore diameter) to allow the bullet to move, for instance, 2 inches down the bore at a low velocity, with little accelleration, and then stop, and this movemnet required, say, 100,000 psi, we'd essentially have Keenan's experiment (albeit even more pressure). The gun's breech would be pushing really, really hard against that slowy moving bullet, and the gun might explode, but it wouldn't recoil appreciably.
As far as friction from rifling causing more recoil becasue the gun is pushing harder on the bullet:
For equal velocities, the smooth bore probably will have less recoil, but that's due, I think, to the requirement of considerably less powder and less pressure to reach that velocity. I think the greater mass of higher pressure gasses are what's causing the increase in recoil in the rifled bore. Of course, friction is what is causing the requirement for higher pressures, so in that way, I guess it is responisble. But not because the gun has to push harder on the bullet, but because their is a requirement for more gas, at a greater pressure to attain a given velocity. The gun's harder push on the bullet due to friction creates greater static forces in the gun, especially the barrel, but not more recoil.
Taking only the bullet and the gun into account (not the gasses), the gun's recoil velocity, I think, will be calculated due to the bullet's mass and it's attained velocity; the amount of force required to get the bullet to that velocity will not be part of the equation. But in the real world, if we need more force to push the bullet due to friction, that will require greater amounts of gasses at ever higer pressures.
If you give this note to the physics professor, after a couple seconds of thought, I think he'll agree.
1911Tuner
Moderator Emeritus
Da Troot!
ROFL!
Okay...Hand_Rifle_Guy! I know ya said it, bro...and I've agreed
with that theory for a while...off and on...until it came back to square one. Ain't that the way it usually happens? Go off in search of happiness and wind up back where we started from?
HOW-EVAH<-----(Reb-Bonics) Once it starts pushin', I believe that the
case gets pushed back into the chamber before the pressure gets too high...rim flush with the hood...and the hood also bears on the slide.
Of coruse, it still depends on the expanded case pullin' on the barrel,
which would launch out the back if it weren't for the slide holdin' it in there,
so everything not only means something, everything depends on everything else.
_________________
RJ! The pressure that expands the case against the chamber walls doesn't peak instantly...It increases gradually (although very rapidly on a real-world time frame) so the case would be set back as far as the slide would let it before it pulled the barrel back against the now locked-to-the-barrel
slide...and get pushed back in until the rim is flush with the hood. Note the
small gap between the hood and slide on most factory 1911 pistols.
P95 mah fren! Can-o-Worms threads are mah specialty...
It breaks the monotony and gets the synapses firin'. This has been a lively
discussion and it drew some excellent responses from all corners. It also
revealed the fact that we really do have some sharp folks on this board.
It's been a long time since I studied physics...longer than it has since some
of the members were born, and I've re-learned a lot from'em. Part of that
is that recoil is simple physics. Equal and opposite reaction/Conservation
of Momentum...in a simple gun such as the muzzle loader...but when a tilting barrel and a reciprocating slide are added to the equation, it gets a bit more complex.
Excellent thread...even if it did prompt some headaches.
Cheers all!
ROFL!
Okay...Hand_Rifle_Guy! I know ya said it, bro...and I've agreed
with that theory for a while...off and on...until it came back to square one. Ain't that the way it usually happens? Go off in search of happiness and wind up back where we started from?
HOW-EVAH<-----(Reb-Bonics) Once it starts pushin', I believe that the
case gets pushed back into the chamber before the pressure gets too high...rim flush with the hood...and the hood also bears on the slide.
Of coruse, it still depends on the expanded case pullin' on the barrel,
which would launch out the back if it weren't for the slide holdin' it in there,
so everything not only means something, everything depends on everything else.
_________________
RJ! The pressure that expands the case against the chamber walls doesn't peak instantly...It increases gradually (although very rapidly on a real-world time frame) so the case would be set back as far as the slide would let it before it pulled the barrel back against the now locked-to-the-barrel
slide...and get pushed back in until the rim is flush with the hood. Note the
small gap between the hood and slide on most factory 1911 pistols.
P95 mah fren! Can-o-Worms threads are mah specialty...
It breaks the monotony and gets the synapses firin'. This has been a lively
discussion and it drew some excellent responses from all corners. It also
revealed the fact that we really do have some sharp folks on this board.
It's been a long time since I studied physics...longer than it has since some
of the members were born, and I've re-learned a lot from'em. Part of that
is that recoil is simple physics. Equal and opposite reaction/Conservation
of Momentum...in a simple gun such as the muzzle loader...but when a tilting barrel and a reciprocating slide are added to the equation, it gets a bit more complex.
Excellent thread...even if it did prompt some headaches.
Cheers all!
1911Tuner
Moderator Emeritus
The Perfessor
Mornin' Gman,
Quote:
If friction is slowing the bullet, it's actually becoming a form of a mechanical lock in the bore.
___________________
Except that the bullet is moving, whereas Jim's wasn't. That's the difference. Movement in one direction results in movement in the other.
The rifling restricted the ball's acceleration, giving the force more time to act in the opposite direction...and a more solid platform to push from.
For an equal velocity/equal bullet mass, the recoil in foot-pounds would be equal...but the dynamics and effect of those foot pounds is different...Sharper. So I suppose that what he really meant was felt recoil....and just simplified it due to time constraints. (He had an evening class due to start in a few minutes)
Remember that the old guy builds muzzle-loading rifles...flintlock and caplock...as a hobby. He made that statement after my physics lesson
was over and the talk turned to his lifelong passion. I'm supposed to
go and have a look at some of his work one day this week. He brought that smoothbore vs rifled bore part up during that part of the visit because
that was his observation after having a barrel maker supply him with a
69 caliber rifled barrel so that he could have a heavy thumper that he
could use for boar hunting...and he wanted it to be more accurate than
his smoothbore Brown Bess clone. He's used a patched round ball in both
weapons...and he says that the difference is dramatic. The Bess' recoil is more like a push, while the rifle is punishing, and not for the recoil-sensitive. I can only take his word for it until I get a chance to try'em both.
When a man tells me that his dog will bite, I tend to believe him.
Mornin' Gman,
Quote:
If friction is slowing the bullet, it's actually becoming a form of a mechanical lock in the bore.
___________________
Except that the bullet is moving, whereas Jim's wasn't. That's the difference. Movement in one direction results in movement in the other.
The rifling restricted the ball's acceleration, giving the force more time to act in the opposite direction...and a more solid platform to push from.
For an equal velocity/equal bullet mass, the recoil in foot-pounds would be equal...but the dynamics and effect of those foot pounds is different...Sharper. So I suppose that what he really meant was felt recoil....and just simplified it due to time constraints. (He had an evening class due to start in a few minutes)
Remember that the old guy builds muzzle-loading rifles...flintlock and caplock...as a hobby. He made that statement after my physics lesson
was over and the talk turned to his lifelong passion. I'm supposed to
go and have a look at some of his work one day this week. He brought that smoothbore vs rifled bore part up during that part of the visit because
that was his observation after having a barrel maker supply him with a
69 caliber rifled barrel so that he could have a heavy thumper that he
could use for boar hunting...and he wanted it to be more accurate than
his smoothbore Brown Bess clone. He's used a patched round ball in both
weapons...and he says that the difference is dramatic. The Bess' recoil is more like a push, while the rifle is punishing, and not for the recoil-sensitive. I can only take his word for it until I get a chance to try'em both.
When a man tells me that his dog will bite, I tend to believe him.
- Status
