Hydraulic recoil spring?

[45_auto]

Sorry, all I can go on is what you write. Can't read your mind.

If you write:

It ignores the effect that the massive bullet drag has on the barrel.

yet I show that the 1,074 pounds of drag (difference between force provided by chamber pressure and accelerating force on bullet) is accounted for in the equations, I fail to see how we can both be correct.

Either it's ignored or it's there, which is it?

 

[JRH6856]

Yes, you account for it in the equation, showing how it affects the bullet, then say it doesn't affect anything. If you don't see the inconsistency in that, I can't read your mind either.

So as I said, we'll just have to disagree on the accuracy of the description.

Its a bit like trying to do math with a Feynmann diagram or a drawing of Schrodinger's Cat. Or seeing a cat at all in equations describing the uncertainty principle.

 

[barnbwt]

"Perhaps we can agree that we agree on what is happening and agree to disagree on how best to describe it."

I hate to be a stubborn dork, but the way to best describe this (the only way to precisely describe this) is with mathematics. Either the forces/inertias add up to what we see as the net result, or they don't. Physical phenomena are not always intuitively obvious, that is why we must use math if we are to precisely describe it. Engineers are often at odds with guys that didn't take statics/dynamics classes over this stuff, but at the end of the day, there's a reason the engineers are paid to determine them; when we do our jobs right and account properly for everything, the math works every time.

Browning may have been able to mic a dimension with his fingertips (I'm calling utter BS on that myth; that's like saying Walt Disney invented all the characters himself ), but a fleet of unknown be-spectacled guys with slid-rules were determining whether the design would work right. Either that, or Browning blew up a lot of guns we don't know about before he got the timing right. As convenient as the figures line up for the 1911, my money is on them being defined arbitrarily by the man, and the machine parts designed to match. In fact, I surmise Browning did the most critical calculations, himself; just because 'design engineering' as a discipline didn't exist formally at that time, doesn't mean the same work wasn't getting done.

TCB

 

[JRH6856]

Engineers are often at odds with guys that didn't take statics/dynamics classes over this stuff, but at the end of the day, there's a reason the engineers are paid to determine them; when we do our jobs right and account properly for everything, the math works every time.

And there we are I guess. I'm not an engineer, don't seem to think like one and don't describe things the same way. I can read the language, but I seem to don't speak it well. Maybe it's my accent.

 

[barnbwt]

"Yes, you account for it in the equation, showing how it affects the bullet, then say it doesn't affect anything. If you don't see the inconsistency in that, I can't read your mind either."

Did you examine the diagram I so painstaking illustrated for all of our benefit? The friction effect on the bullet is accounted for the by the measured muzzle velocity (measured in the real world with real friction). The friction effect on the barrel/slide assembly drops out of the equations; it's still there, but an equal/opposite force opposes it. The effect of friction that is felt by the barrel/slide is that the conservation of momentum --driven by the measured-in-the-presence-of-friction muzzle velocity of the bullet-- for the whole system means the slide's motion is defined by the bullet's; recoil.

Now, if we split the barrel and slide into two separate, interfacing systems, in order to calculate the loads at their connections, then the two separate sets of equations will yield equal and opposite impacts of the friction; the barrel thrust forward, the slide back under unadulterated bolt thrust from pressure. Added together, the barrel friction terms cancel. Which is why I stated that, so long as the barrel/slide are fixed to each other, they can be treated as welded, or as a single lump of metal, like a muzzle-loader barrel

Please show a set of equations illustrating how a simple tube with a projectile recoils differently with/without friction, when the measured velocity of an equal-mass projectile exiting the muzzle is the same. If the friction itself has an effect on the system when the exterior ballistics are the same, there will be some friction variable left in the equations that will not cancel out.

Short of equations illustrating what you describe, it's your word against hundreds of years of study

TCB

 

[JRH6856]

If the friction itself has an effect on the system when the exterior ballistics are the same, there will be some friction variable left in the equations that will not cancel out.

I'm not saying friction as an effect on the system. If I'm saying anything, I'm saying it has an effect within the system.

Sometimes I feel like I see a door and I am describing the door. Everybody else is describing the floors, walls, ceiling, and windows and saying "if you do the math, you will find the door is accounted for by its absence." Which does not describe the door.

Short of equations illustrating what you describe, it's your word against hundreds of years of study

To me, 45 Autos equations confirm what I am describing. You may say they don't, but if I provided equations, they would be the same equations. Like I said, I don't speak or think engineer. Even when we use the same math, we use different words.

 

[barnbwt]

"And there we are I guess."
And please don't take my line about engineers as arrogance, because God knows how many folks I've talked to take is as such. I don't purport to know how to properly run a mill or even torque a screw properly versus someone so trained/employed, and neither should a person untrained in engineering mathematics discount someone who's devoted their life/livelihood to it. The mathematical approach is sound, because it has never not been sound*, and was itself based in countless real-world measurements in the first place. The numbers really aren't pulled from 'the usual place' and we do our utmost to thoroughly account for everything such that the answer is as accurate a representation of the real world as possible.

"The bullet has to travel at a constant accelaration of 830,000 feet per second per second to reach a velocity of 830 FPS in 5 inches."
For example, this is a terrible assumption if we are examining the system before the bullet exits because in real life, we know the bullet's do not accelerate uniformly along the whole barrel length, nor does the bolt thrust remain constant. What we do know is that the end result is a measured muzzle velocity and travel distance, and that is enough to find the corresponding figures for the slide/barrel assembly (it has to do with the magic of integrals/differential equations; all the details of how the slide traveled between two points don't matter so long as we know the start and end point. It's the voyage, not the journey we're interested in, the destination being when the bullet exits or the barrel cams down, whichever is earlier)

I wasn't sure if friction affected slide timing, so I drew the diagram, summed the force vectors, and saw the terms drop out. Question answered. Whether it affects barrel lug load (it does) is a separate question, with a separate way of getting a separate answer.

TCB

*engineering methods don't fail; the engineers making assumptions about how to model systems fail. The engineering method is simply a way of solving problems analytically using principles based on countless observations of the physical world.

 

[JRH6856]

Whether it affects barrel lug load (it does) is a separate question, with a separate way of getting a separate answer.

Good. Because that is the question I have been trying to address all along. Despite the fact that everyone esle seems to think I have been saying something else. :banghead: (There, I feel better now. )

 

[barnbwt]

"If I'm saying anything, I'm saying it has an effect within the system"
And I agree. The friction applies an equal and opposite load to the lug faces of the barrel and slide. Absotively. But being equal and opposite, the term is hidden when you look at the two together, as they are when linked by the lugs.

"To me, 45 Autos equations confirm what I am describing. You may say they don't, but if I provided equations, they would be the same equations. Like I said, I don't speak or think engineer. Even when we use the same math, we use different words."
His math looks good to me, at least as far as the assumptions made can be assumed sufficiently accurate (whole nuther can of worms we needn't open ). FWIW, heated discussions on what's happening occur regularly even between engineers*. Two guys build different models to describe a system (one, or both of them often being wrong) and get different numbers, steadfastly deny their numbers are bad (both having expended considerable time and thought on the models, they take its refutation as a personal slight of their handiwork), and furiously set to proving themselves right and the other guy wrong. Eventually, the wrong guy's own proofs show he made an error, and the math changes his opinion for him.

Hilariously, both guys are so ginned up by ego-machismo at this point, that even the matching figures leave them at odds, and they "agree to agree" as asinine as that sounds.

So, JRH, Tuner, 45auto; are we all willing to agree that we agree? If so, could this be the first thread ever on THR to be locked out of high-spirited agreement?

TCB

*I liken the act of engineering to a mental analogue of manual labor, because of the amount of visualization it requires above even classical mathematics. If you picture the guys with note pads as being two guys with sledge hammers pounding rocks, the stubbornness and horse-headed-ness (and horse-assed-ness) attributed to engineers makes a lot more sense. How else would you expect two guys in disagreement over the proper way to pound rocks to act?

 

[45_auto]

Sorry, not worth the effort.

Deleted.

 

[45_auto]

Sorry, I don't see how I can agree with people who believe that the only thing keeping the breech locked on a 1911 is the vertical friction in the barrel/slide lugs, and who don't understand the M1V1 = M2V2 relationship between the bullet and slide.

In reality, I could put a miniature set of roller bearings between the engaging surfaces of the slide and barrel lugs (therefore zero friction) and it would not make a measurable difference in the performance of the gun.

A 1911 is a simple mechanism designed to operate within limited parameters. Too little M x V means it won't unlock at all. Too much M x V means that it will unlock at high pressure, with a bullet still in the barrel.

The lugs vertically disengage at (nominally) .200 inch of rearward slide/barrel travel.

So, even if a bullet and high pressures are present when the barrel reaches the linkdown point...which can't happen, even with a heavy bullet...

With the barrel lugs engaged horizontally under this shearing force at however many thousands of pounds psi exists when the link starts to pull...I strongly suspect that the link would fail before it disengaged them and allowed the breech to open, and the he lugs would almost certainly be damaged even if the link held and succeeded in pulling the barrel down.

Uh, yeah it is. It is the friction in the locking lugs that resists the link-down and keeps the breech locked.

Pull out your 230 grain mold.

Cast an osmium bullet. It'll weigh about 460 grains.

Load it appropriately and shoot it with a muzzle velocity of 830 FPS.

Using the numbers from our earlier calculations, you'll find that the slide has recoiled approximately .16 inches. So we're still good, lugs are engaged, bullet is gone before unlocking, pressure has dropped.

Now load up another 460 grain osmium bullet so it reaches a muzzle velocity of 1,100 FPS. You'll find that the slide has traveled .21 inches before the bullet reaches the muzzle. Oops, it unlocked at .200!

You'll probably want to machine your slide and barrel from one of the super alloys to withstand the chamber pressure. Expect it to cost several million dollars (last rhenium throat we bought was about 2" x 3" with a simple hole in it and cost almost $300,000). You can believe the numbers or spend millions of dollars trying to disprove Newton.

Physics and simple geometry are all that control the unlocking of a 1911. Change the mass and velocity parameters enough and the geometry quits working.

 

[barnbwt]

But since osmium bullets and rhenium cases aren't real...a more efficient way to force the numbers to fail is to grind off two or three barrel lugs

TCB

 

[JRH6856]

A 1911 is a simple mechanism designed to operate within limited parameters.

Yes, I believe I said just that back in post #198:

JMB designed the 1911 to operated safely within a certain range of parameters. And as noted by 45 Auto as well, it was designed to account for and take advantage of the physical forces at play within these parameters. As long as these parameters are not exceeded, it works as designed. Exceed the parameters, it may or may not work safely or may not run at all. ​

 

[GLOOB]

45Auto, looks like you have left the hobbit hole to fight the demons. I know better than to stick around, but I have to point something out to you.

Using the numbers from our earlier calculations, you'll find that the slide has recoiled approximately .16 inches. So we're still good, lugs are engaged, bullet is gone before unlocking, pressure has dropped.

Now load up another 460 grain osmium bullet so it reaches a muzzle velocity of 1,100 FPS. You'll find that the slide has traveled .21 inches before the bullet reaches the muzzle. Oops, it unlocked at .200!

I have to disagree, here. When you increase the velocity of the bullet from 830 to 1100 fps, you are increasing the momentum. But you are doing so by increasing the velocity component.

So by the time the bullet reaches the muzzle (in a shorter time), the slide will have moved back the same distance. Easy oversight, and anyone can make that in the heat of an argument.

What some people here are having a hard time accepting is that conservation of momentum is real, and that it is always on. There is no timeout. There's no fudging for this or that (losses, heat, friction, pressure, etc) so that it balances out in the end. Momentum is conserved at every single point in time. The center of mass between that bullet and slide is the same at every point in time (until you add hammer and recoil spring into the equation; and to be totally clear, yes, when the locking lugs start to disengage, the slide gives momentum to the frame, and that changes things past that point).

If you believe that conservation of momentum can be delayed or distorted by friction or anything else, you might as well believe in teleportation, because that's what you would end up with. If we were living in a universe where Wiley Coyote and the Road Runner were real, then we might have to start looking and adding up all the wild and wacky factors to get an approximation. But in my universe, which I presume is the same as yours, trying to use more complex physics equations (which are derived from conservation of momentum to begin with) is not only unnecessarily cumbersome, but it invites more places to make a mistake. Save the force diagrams for when there are more than two moving parts and a single vector. Save the ballistic equations for when you are trying to figure out something that cannot be explained by COM. I mean, for chrissake, if you can't agree on the underlying fundamental principle, there is absolutely no point in arguing the derivatives of that principle on which you don't agree in the first place.

So when someone thinks that "oh there's not enough higher math or diagrams for GLOOB's explanation to be right," it must be because they do not believe in what is probably the most fundamental principle in Newtonian physics. I didn't take any higher physics, but I did ace basic Newtonian physics in high school and college, at the top of my class. I don't remember 99% of it, but I do know you always, always start with COM in a case like this. This is the yardstick against which you compare all other calculations in order to catch all the mistakes you are going to inevitably make. If you do not believe in COM, you do not believe in physics. Is it possible there are exceptions? Yes, it is possible. Maybe on some quantum level or alternate model of reality. But Newtonian physicists agrees that momentum is ALWAYS conserved at ALL times for sake of now being able to compare apples to apples. It's how the rest of the house of cards is defined. And so far as we have the ability to detect and observe and measure, this principle is always correct, on a macro level.

If you don't think COM explains enough... if you think there's a greater construct that more accurately describes the universe on a macro level... some universal theory which explains both Newtonian physics and quantum physics... which it quite possibly exists but we can't comprehend it, yet... Then that's fine, too. But you are no longer playing in the same sandbox. And you're in relatively small company. And with what you have started with, your model is already noticeably less accurate... But maybe the "locking lug" school of physics will be able to explain something that ordinary physics cannot, one day. Who knows?

BTW, barnbwt. I am honored to be left off your invite list. If I were to join the team, you and Tuner would both have to play right field. And have designated hitters.

 

[barnbwt]

"BTW, barnbwt. I am honored to be left off your invite list. If I were to join the team, you and Tuner would both have to play right field. And have designated hitters."
Sorry man, I tried to remember everyone (I'm a visual learner, and the no-profile-pictures format here precludes that) and listed all the posters that showed up on page...8 is it now? (sheesh )

GLOOB:

"What some people here are having a hard time accepting is that conservation of momentum is real, and that it is always on. There is no timeout. There's no fudging for this or that (losses, heat, friction, pressure, etc) so that it balances out in the end."

"If you believe that conservation of momentum can be delayed or distorted by friction or anything else, you might as well believe in teleportation"

Beware of absolutist statements when in the presence of bored nerds. There is also a little thing called Conservation of Energy which is a far more accurate (complex) representation of our world. When you slow to a stop on a sheet of ice, you have not sped up the world by a corresponding equal amount; you also evolved a chunk of that momentum transfer in the form of heat. Conservation of momentum only applies in a loss-less transfer of kinetic energy. Throw heat, electrical, potential, or even nuclear in the mix and bets are off. A bouncing ball will bounce forever without impact/aerodynamic losses. All that said, this paragraph is just me being a cute smartass, because our particular problem doesn't happen to require a specific consideration for friction, because...

-we've already accounted for it in the momentum equations by using a real-world measured muzzle velocity for the bullet
-the gun's operation is effectively frictionless (slide motion, etc.)
-at the time of barrel unlock, the friction is no longer present since we've shown the bullet has already exited (absent the most wild of circumstances which would probably result in unsafe operation)
-and I suspect we could show that even fully-loaded barrel lugs wouldn't develop enough friction to slow the link's authoritative camming force downward appreciably

Thus, the question we're asking about cam-timing does not require a friction term so long as we base our numbers on measured muzzle velocities; done.

"So when someone thinks that "oh there's not enough higher math or diagrams for GLOOB's explanation to be right,""
I'm inclined to think you are not accounting for all necessary factors. If the factors weren't necessary, they wouldn't result in the numbers coming out different (they don't in this case, but if we used a muzzle velocity estimated for a frictionless barrel, they would be 10% or so off; that is not quite 'negligible' by my standards)

"If you don't think COM explains enough... if you think there's a greater construct that more accurately describes the universe on a macro level... "
Equations are just tools, GLOOB, you have to determine which to use for the job. COM is the simplest, most basic description of motion between bodies, so it's good to start with (all jobs start with a big effin' hammer ), but you also need to construct your model to match reality as closely as possible, until adding additional factors does not change the result (your chisels, then scrapers). If we were firing an Americium bullet with a half-life of .0004 seconds or whatever, we had better incorporate the mass change of the bullet as well as the hellacious amount of heat imparted to the driving gas if wanted to know what the gun would actually do (you know, besides melt and/or be vaporized in a nuclear explosion ). If we're interested in intra-black hole ballistics for some stupid reason, we'd need to account for Reletavistic effects. Personally, I see all 'what if' scenarios not grounded in practical reality as 'navel gazing,' but they can be illustrative of principles

TCB

 

[GLOOB]

Beware of absolutist statements when in the presence of bored nerds. There is also a little thing called Conservation of Energy which is a far more accurate (complex) representation of our world. When you slow to a stop on a sheet of ice, you have not sped up the world by a corresponding equal amount; you also evolved a chunk of that momentum transfer in the form of heat. Conservation of momentum only applies in a loss-less transfer of kinetic energy. Throw heat, electrical, potential, or even nuclear in the mix and bets are off. A bouncing ball will bounce forever without impact/aerodynamic losses. All that said, this paragraph is just me being a cute smartass, because our particular problem doesn't happen to require a specific consideration for friction, because...

It's true that when you slow down you do not speed up the world by a corresponding amount. You do transfer your exact momentum to the world. Whether you hit a brick wall or just gradually come to a stop, it doesn't matter; that momentum you had is transferred. Where did you get your momentum to begin with? You pushed off against the ground!

If you built a giant railway that went all the way around the earth's equator, and you put a massive 10 million ton train on that rail, then you made it go east for 100 revolutions around the earth, what happens? The train will have temporarily robbed the earth of some of its rotational momentum, so long as it remains in motion. You will temporarily slow the earth's rotation. Of course, when that train finally stops, that momentum is restored, and the earth will return to its original rotational speed. But you would have forever changed the time that the sun rises and sets by just a fraction of a second. To get the clock back to where it was, you will have to move that same train for 100 revolutions to the west.

Conservation of momentum only applies in a loss-less transfer of kinetic energy.

FALSE.

Kinetic energy can be transformed into say heat in a collision or interaction between objects, but it is conserved, overall. Momentum is conserved. It doesn't matter if the objects bounce apart (perfectly elastic collision would mean no kinetic energy loss), or stick and crush, or smush into different shapes, or heat up a lot or a little (kinetic energy lost in an inelastic collision). Momentum is always conserved.

 

[barnbwt]

"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."
Put that gun in a Ransom rest and bolt the rod to a wall/table instead of the barrel (allowing it and the slide to recoil) and you'd have the 'extreme case' of an infinitely massive bullet. Bullet won't move one iota. Barrel will

...and then you'd be looking for a new gun (if that's what happens when 1911's unlock under pressure; I have no idea how sturdy the design is against breakage in that worst of scenarios). I suspect this this why Jim K agreed to do the experiment the way he did

TCB

 

[barnbwt]

"Energy can be transformed into say heat in a collision or interaction between objects, but it is conserved. Momentum is also conserved"
Momentum is an expression of energy stored in a moving mass; that's why kinetic energy and momentum contain the same terms. They're two sides of the same coin, completely interchangeable, convertible, how ever you want to say it. Same idea, different units. Momentum is directly comparable to linear potential energy, like from a spring or elevation change. If KE is not conserved in a system because friction spirits it away, neither can be momentum. I won't even go into entropy which most assuredly does not conserve energy (even that transfer of heat out of the system incurs a penalty, which is the next-most precise equation to describe a system).

For the subject matter at hand, though, this truly is arguing angels on pin heads. For the purpose of the 1911, the only friction force worth nothing is the component contributing to barrel thrust (which swaging/taper also contribute to), and we've already accounted for it by using a measured muzzle velocity. This is why I like guns; there is actually very little obnoxious math to contend with, compared to nearly any other discipline. Everything is so efficient and so rapid, the losses are inconsequential and it very nearly boils down to simple machine mechanics

TCB

 

[GLOOB]

If KE is not conserved in a system because friction spirits it away, neither can be momentum.

Completely wrong.

For one thing, kinetic energy doesn't have a vector. Momentum does. This is because velocity has a vector. And when you multiply a vector by itself, the sign cancels. It's always positive. This makes kinetic energy a scalar.

So here's an example:

When two objects of the same mass are moving towards each other at the same speed, the momentum of the system is zero.

The kinetic energy is 1/2 mass x velocity squared for each object. The total would the sum.

If they collide and just stop, all that kinetic energy turns into heat.

If they bounce off of each other in a perfectly elastic collision, they will rebound in exactly the same speed in the opposite direction, hence kinetic energy is not lost. There is no heat produced in this kind of collision.

In either case, momentum is zero at all times.

 

[GLOOB]

Lest you think this is just a sneaky way to cancel momentum with vectors, here's another example.

You take an object mass X (M1) moving towards a stationary object of mass
X, as well, (M2).

The velocity of M1 is V. The velocity of M2 is 0.

The momentum of the system is X*V. The kinetic energy of the system is (1/2)X*V*V.

When the objects collide, the momentum will be the same.

The kinetic energy will be dependent on the elasticity of the collision. In a perfectly elastic collision, M1 will completely stop. and M2 will move in the original direction at a speed of V. Kinetic energy is the same. Momentum is the same.

In a perfectly inelastic collision, both M1 and M2 will carry on as one, with a speed of V/2. The kinetic energy will be X(V/2)(V/2) = (1/4)X*V*V. The kinetic energy will be one half what it started with. This is lost to heat. Yet, the momentum is still the same.

You guys really think I'm a crazy coot that makes this stuff up as I go, don't you?

 

[barnbwt]

Having been made aware of video evidence that shows the slide slowing abruptly during link-down, I've modified/extended my theory regarding the 1911's function a bit

Here's my new theory, and wait for it --it still doesn't need friction to work! What angle does the 1911's link sit at relative to the barrel when in battery, if you draw a line through the pivot points?

Okay, it's basically top-dead center; it actually pulls down with zero pounds when it first starts camming; probably what makes them such soft shooters*, in part (the Hi Power cam ramp slams into a cross pin, for comparison)

And by the time the lugs are disengaged, the link angle is pointing quite a bit backwards. This is where the new theory comes in. The link can only apply load along the direction defined between its pivot points. As the barrel starts camming down, the link's load vector points more and more forward. It's actually pointing forward at ~30deg or so when the lugs are still engaged, as best I can tell, which means a chunk of the load pulling the barrel down is also pulling it forward while the lugs are engaged! That explains everything! That even explains why over-loaded guns damage their links in addition to their extractors!

Now, at this point, the bullet's gone and there's no risk of kaboom, so the only design impact of this last-minute slow down of the barrel/slide is to reduce the energy that must be absorbed by the recoil spring.

 

[barnbwt]

"Lest you think this is just a sneaky way to cancel momentum with vectors, here's another example."
Well, it isn't sneaky, but it's looking at two things instead of one thing. If you define the system as closed and encompassing both items, the sum is zero. If you examine only one object in an open system so that force vectors and other factors may be applied, the sum is no longer zero (which is why the individual object moves, while the sum of the two may not).

I think we're arguing the same point on different premises; agree to agree?

"In a perfectly inelastic collision, both M1 and M2 will carry on as one, with a speed of V/2. The kinetic energy will be X(V/2)(V/2). The kinetic energy will be one half what it started with."
At the risk of being wrong since I haven't done this particular kinematic experiment in quite some time; what if these collisions involved heat or potential energy storage (like loading up a recoil spring that's caught on an empty mag, for example)? There'd be an extra energy term that has to be accounted for by the velocities of the two components, which means the portion remaining to calculate system momentum would not be quite the same (right? )

Not to push you off, or anything, GLOOB, but this is kinda drifting wild from the 1911-specific question we were trying to solve (I think), so we could continue in PM if you'd like. This stuff's important for proper gun design, so I/we obviously have a vested interest in making sure both of us are correct at the end of the day

TCB

 

[barnbwt]

Anyone pining for straight blow-back weapons after all this crap? I feel a need to go shoot my Skorpion, now

TCB

 

[oneounceload]

Perhaps the Boberg bears some consideration for the OP's wife, or maybe reading www.thecorneredcat.com on how to easily work a stiff slide

 

[barnbwt]

Or an R51 (hey, mine can be racked with a single hand, and mah momma had no trouble with it, unlike a PX4 she could maybe rack once a day, after several 'warm up' attempts --no she won't take my advice on technique/Cornered Cat, so whatever )

TCB