"Thrust Vector" and other ideas


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Jim K
August 19, 2004, 12:34 AM
Hi, Tuner and guys,

Yes, I decided a new thread might be a good idea.

First, let me say that I have a lot of respect for Jerry Kuhnhausen, although I have never met him. I have, I think, all of his books, and generally the information can be taken to the bank. In fact, some of the information I post here is from those books; I try to give credit when the information is available only in a specific book (that is, is not general knowledge), and do so with Mr. Kuhnhausen's books as well. When I have a technical problem, or a question about specifications or fitting parts, Kuhnhausen is my guide. There is none better. My only regret is that when I was doing the work, I did not have all the nice tools he has.

All that being said, his explanation of the way a 1911 type pistol operates is, to put it mildly, somewhat flawed. In his illustrations, and in the accompanying text, he states that the barrel/slide unit does not begin to move until the bullet is out of the barrel. That is simply not true; if it were, the pistol would not work. The barrel begins to move the instant the bullet begins to move. The barrel remains locked to the slide until the bullet is out the barrel, at which point it has recoiled about 1/10 inch. This has been shown repeatedly in high speed films.

His explanation of a "thrust vector" (Vol. II, Page 42, Figure 4 caption) is nonsense, as is a lot of Para 5, Page 38.

So what really happens? And how does the pistol work? First, if you don't believe in the laws of physics, go to another thread.

The answer is recoil. This is sometimes couched in terms of Newton's Third Law, "for every action there is an equal and opposite reaction." That law is now called the Law of Conservation of Momentum, but the idea is the same. When the bullet moves forward, the part of the gun in contact with the bullet, the barrel, moves back. Is pressure involved? Of course, pressure is what moves the bullet. But pressure does not directly cause the gun to function in a recoil operated pistol as it does in a blowback pistol. (No, the pistol is not kept locked by the bullet pushing forward on the barrel; that force is essentially negligible.)

The bullet's forward motion causes the barrel's rearward motion, and it is that barrel motion that shoves the slide back (through the locking lugs and the hood) and operates the pistol. Now here is where it gets interesting. If the bullet does not move, the barrel does not recoil and the gun does not cycle, period. If the barrel is blocked directly ahead of the bullet, so the bullet cannot move, the gun will not operate. It remains locked, the pressure leaks away, and nothing happens. No, the gun does not "blow up"; the cartridge case does not explode. Nothing happens. (Yes, I have done it, but under tightly controlled conditions; please do not destroy a pistol or hurt yourself trying this.)

So, if the barrel/slide begins to move when the bullet does, when does the barrel unlock from the slide? That happens after the bullet has left the barrel and pressure has dropped to a safe level. Because the bullet is much lighter (has less mass) than the slide/barrel combination, it moves 4 inches or so, while the barrel/slide has moved only 1/10 inch or so.

That movement is the distance the barrel moves before the barrel foot slips off the slide stop pin, the barrel engages the link and begins linkdown. Until that happens, the barrel and slide are locked to withstand the high pressure in the barrel. By the time the unlocking takes place, the bullet has left the barrel and pressure has dropped. Note that the barrel is (or should be) kept locked by the barrel foot riding on the slide stop pin, not by the link.

OK, how does that affect some other fond ideas? Doesn't the recoil spring keep the barrel and slide locked? Well, no. A strong recoil spring can slow down the slide after it unlocks, but no reasonable spring can keep the recoil from shoving the slide back, only mass can do that, and no spring of reasonable weight is massive enough. In fact, the pistol will function and lock for the same amount of time whether a recoil spring is present or not. (Without one, however, the slide will batter the frame, since there is nothing to slow it down. (Yes, I have done this, too. Same warning!)

Does a recoil buffer help? It helps cushion the contact between the slide and the recoil spring guide/frame, and gives some people peace of mind that the frame is not being pounded. But steel is highly elastic, and the pistol is designed to allow the slide to bounce off the frame and use some of the recovered recoil energy in the forward part of the cycle. That is why installing a recoil buffer often results in failures to feed; not enough of the slide's energy is returned to assist the forward part of the cycle. (Remember those little steel balls in the rack that bounced for a long time once started? Put a buffer between them and the bouncing will soon stop.)

Doesn't the link lock the barrel back to the slide? No. The barrel foot is made to cam up on the slide stop pin. The gun can be fired without the link; it will not unlock consistently, since gravity is not that reliable in the circumstances, but return to battery is no problem. (Same again, same warning!)

So, Tuner and guys, have at it. I know there will be tons of argument about what I have written, but your significant others will thank me for distracting you and letting them get some sleep.

Jim

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Clark
August 19, 2004, 01:03 AM
Keunhausen's book on Muasers drove me crazy when I was learning to sporterize. I am a scatter brain, and I can't read a scatter brain and get orgainized.




I did a calculation on how far cases fly.
It is interesting to me that the velocity of the bullet and gas, the mass of the barrel and slide, the travel of the slide, the spring force of the recoil spring, all do not add up to how fare the cases fly. It would if the pistol were in a heavy vice, but the hand moves back and keeps the slide from getting all the potential side velocity relative to the frame. So we need less spring than one would calculate.

link to my calculaion on far cases fly from my 1911 (http://www.google.com/groups?hl=en&lr=&ie=UTF-8&selm=be90tg%24s6i%241%40grapevine.wam.umd.edu&prev=/groups%3Fq%3DColt%2Bmass%2Bgroup:rec.guns%2Bauthor:clark%26hl%3Den%26lr%3D%26ie%3DUTF-8%26selm%3Dbe90tg%2524s6i%25241%2540grapevine.wam.umd.edu%26rnum%3D1)

1911Tuner
August 19, 2004, 07:23 AM
Hey Jim! I was wonderin' where you were hidin'.

I used to have a problem with Kuhnhausen's description of the recoil cycle too...but the more I read it and think about it, I don't think his twist is flawed so much as it is the way he describes it. Aybody who has read his manuals can attest to the fact that he's a little hard to follow. He's a great
pistolsmith, but he ain't no Hemmingway.

His description leads the reader to believe that the bullet exits BEFORE the recoil cycle starts...which is decidedly false. I'm more inclined to believe that he just didn't present it correctly. The bullet is out of the barrel before the cycle is completed, to be sure...but it's still there when the cycle BEGINS...and the momentum needed to complete the cycle is established during that brief instant that the bullet is moving from the point of origin to
the open air.

The thrust vector exists. I believe that the barrel is pulled forward by the frictional effect of the bullet, while the thrust is acting on the slide...As the
bullet moves, the rearward thrust begins to overcome the forward movement of the barrel, and the slide begins to pull the barrel backward with it. The thrust is also acting on the expanded case and the case helps
the barrel to move rearward against the slide...and it all comes together
at the right time...which helps to explain why powders with a slow burning rate won't cycle the slide reliably. The bullet accelerates more gradually
and exerts the forward pull on the barrel for a nanosecond too long...
and the whole thing comes to a halt....or nearly so.

Standin' by...

Fred Fuller
August 19, 2004, 08:08 AM
Grumble grumble.

Cross reference to other thread:
http://www.thehighroad.org/showthread.php?s=&threadid=97496

Grumble grumble.....

lpl/nc

1911Tuner
August 19, 2004, 08:24 AM
There ya go Lee! Now, kwitcherbitchin' and jump in it!:D

Higgins
August 19, 2004, 10:29 PM
I"ve told myself not to get involved in discussions like this, but here goes . . .

I've read both this post and the one started by Tuner, and you guys are making this way more complicated than necessary. (Now that I've finished writing my post, maybe I am, too). I respect the hell out of both Tuner and Mr. Keenan. They both have oodles more experience with firearms than me, and I've learned much from reading their posts in the past. However, I do believe in the laws of physics (I do, I do), and I do believe they go against Mr. Keenan's version of events and in favor of Tuner. (I've never read Kuhnhausen, so I'll not comment on that. I'll stick to Mr. Keenan's post)

It's simple. For a recoil operated pistol, such as JMB's 1911, the following is the basic sequence: A fired round in a chamber creates a lot of gas pressure very quickly. In one direction that gas pressure pushes the bullet out of the cartridge and down the barrel. Simultaneously, in the other direction that same pressure pushes back against the cartridge base, which pushes against the breech that is part of the slide. The result in that direction is the slide is accelerated rearward - and, oh yeah, the slide is locked to the barrel by the locking lugs, so the slide pulls the barrel to the rear with it, until the rear of the barrel is pulled down and out of engagement with the slide. The slide then continues on its merry way due to momentum to finish the cycling sequence. In essence: One big pressure, pushing in two equal and opposite directions simultaneously achieving two results - bullet accelerated down barrel and slide accelerated rearward. Bullet travels farther, faster than the slide/barrel because the former has much less mass than the latter and therefore is accelerated much quicker. (See Newton's 2d law: F=ma. Here the F is the same, only the m's differ, leading to different a's). Period, end of story.

The barrel does not push the slide rearward. The slide pulls the barrel rearward.

The bullet's forward motion does not cause the barrel's rearward motion. Again, the barrel's rearward motion is caused by the slide - to which the barrel is locked - pulling the barrel rearward.

And what causes the slide to be accelerated rearward? The pressure in the chamber acting against the breech face through the cartridge base.

I believe Mr. Keenan is misapplying Newton's 3d law in reasoning that the bullet traveling down the barrel somehow "creates" a force that moves the barrel rearward. Yes, the barrel - via the rifling - opposes the travel of the bullet because of friction between the rifling and the bullet. So there is a force in the barrel in the direction opposite the bullet travel. However, the rifling only offers so much resistance ("force") in oppositin to the bullet, and this resistance, as Mr. Keenan indicates, is negligible compared to the force with which the bullet is being forced down the barrel by gas pressure. The gas pressure easily overcomes the friction resistance offered by the rifling to the bullet. If the only force offered up by the barrel in the direction opposite the bullet travel is the friction of the rifling (again negligible) - a force easily overcome by the gas pressure force behind the bullet - then what force precisely exists (in terms of the bullet and barrel alone) to cause the barrel to move backwards? Things don't move in a direction without the application of some net force in that direction (See Newton's 1st law). In Mr. Keenan's bullet-barrel version, sorry to say, there is no net force - absent the slide - acting on the barrel to move it rearward. Therefore, the force causing the barrel's rearward movement must come from somewhere external to the bullet-barrel relationship - and it does. It comes from the slide pulling the barrel rearward, driven by the gas pressure in the chamber acting in the direction opposite the bullet travel.

Think of it this way: If I try to push my couch across the floor with say 5 newtons of force, the couch pushes back against me with 5 newtons of force. Same with 10 newtons, 15 newtons, etc... until I overcome the inertia and friction holding the couch in place. The couch can never push back against me with more force than I push against it, and it certainly can't move in the direction opposite I am pushing it. It can't, it's impossible. To do otherwise violates Newton's 3d law. Yet, in Mr. Keenan's version, this is precisely what he is postulating. In the "bullet moves barrel" theory, the barrel not only resists the bullet with more force than the bullet is pushing down the barrel, but this force is sufficient to move not only the barrel rearward, but also the slide (against the recoil spring and it's own mass) violently rearward. Just isn't a proper model as it's not consistent with Newton's laws.

But what about Mr. Keenan's experiment with fixing the bullet in the barrel? Fine experiment, perfectly legit result, but incorrect conclusion about what result shows. By fixing the bullet in the barrel, the bullet and barrel essentially became one unit. What force acted on the bullet acted on the barrel. Upon firing a round, gas pressure pushes the bullet to the left with force "N," and the barrel, being attached to the bullet, wants to go the same direction. But wait a minute, the barrel is also attached (locked) at the other end to the slide/breech face by the locking lugs - essentially making the slide and barrel one unit. Upon firing a round, gas pressure pushes the cartridge case base and the breech and, hence, slide to the right also with force "N" and the slide wants to take the barrel with it. So, let's see, the bullet is trying to pull the barrel to the left, while the slide is simultaneously trying to pull the barrel to the right with equal force. Equal forces acting in opposite directions cancel each other out and no net force equals no change in movement for the slide, the bullet, the barrel which are all interlocked to each other - the whole contraption stays put. What Mr. Keenan created was a closed, fixed system in which the gas pressure exerted equal force in all directions and since everything was locked up tight to everything else in a Rube Goldberg fashion, the pressure strained against it's miniature steel walled cage until it died and dispersed itself, probably in the form of heat. Think of a miniature bomb disposal unit - big round steel container police put bombs into then blow them up. Whole thing just sits there, goes nowhere despite the huge momentary bomb pressure inside. Same principle with Mr. Keenan's experiment.

As for the contention that "pressure does not directly cause the gun to function in a recoil operated pistol as it does in a blowback pistol." Poppycock. The gas pressure created by combusting nitrocellulose is the only energy source/force by which modern semi-automatic firearms (pistol or rifle) operate. It is the only energy source available to get things moving. The only difference between a blowback, recoil, delayed blowback, gas delayed blowback, or gas operated rotating bolt firearm is the manner in which they harness the gas pressure and turn it into mechanical/kinetic energy to move the slide, unlock the slide, hold the slide closed momentarily, drive the piston to move the bolt carrier which rotates the rotating bolt, etc....

Other than these minor points, everything else Mr. Keenan said appears to be spot on.

(Let me be the first to say, the above is based on my reading of Mr. Keenan's post. It's quite possible he meant to say something different and/or I misinterpreted his wording on some points. If so, I'm sure he will address my mistakes and I provide an a priori mea culpa).

R.H. Lee
August 19, 2004, 10:46 PM
Whew!

So, the reason a blank will not cycle the slide is found in Newton's 3rd law, there is no external object against which the force is exerted.

http://hyperphysics.phy-astr.gsu.edu/hbase/newt.html#nt3

RJ357
August 19, 2004, 11:19 PM
Here is the article:
Jerry Kuhnausen's article here (http://www.m1911.org/locking.htm)

Jim Keenan - In his illustrations, and in the accompanying text, he states that the barrel/slide unit does not begin to move until the bullet is out of the barrel. That is simply not true; if it were, the pistol would not work. It seems he must be wrong. The pressure would drop off drasticly as soon as the bullet left the barrel. It would be like firing a blank at that point.

1911Tuner -
Have you figured out if the forward drag by the bullet friction is greater than the rearward drag by the case friction?

Higgins -
I agree completely with your explanation for slide recoil. I do not like using the 3rd law because while it is certainly true, it does not explain the source of the force.

I think this explains it best, an analogy:

You got 40 guys standing around outside. Someone really big takes a gigantic open box and turns it upside down over these guys. They are still standing on the ground and are surrounded by the four walls of the box. They all want to get out and run to walls and start pushing. 10 guys against each wall. They all push equally hard and the walls start to bulge. The box stays in position.

Suddenly, one of the walls breaks loose. the 10 guys on that wall push it outward fairly quickly. The 10 guys on the opposite wall now begin pushing whats left of box along the ground. Not quite as fast as the other guys pushed the wall, because they are pushing three times as much weight.

The 3rd law states that the box will move if the wall does. It does not say why. Looking inside the box shows why it moves.

JohnKSa
August 19, 2004, 11:41 PM
They are still standing on the groundYour analogy only works if you make this assumption. What's analogous to the ground in a gun? Nothing--the gases are fully enclosed by the cartridge and the bullet.

For it to be truly analogous, they would all have to be fully enclosed in the box like the combustion gases are. So, they're all standing on the bottom of the box and pushing against the sides. NOW when one wall gives way, what happens to the box and why?

P95Carry
August 19, 2004, 11:52 PM
I'm broadly and happily in the Higgins camp! Saves me a boatload of writing ... :)

JohnKSa
August 20, 2004, 12:02 AM
The barrel does not push the slide rearward.There is SOME rearward force transferred to the barrel by the friction between the cartridge case and the chamber wall. I agree that the majority of the rearward force is against the breech, not against the barrel.

Keenan's quote is correct: "pressure does not directly cause the gun to function". Sure, pressure causes the gun to function--it gets the bullet moving. But the recoil force is a result of the bullet (& other ejecta) motion, not directly a result of pressure. What you're saying is like saying that pumping up an airgun is the direct cause of recoil. Sure, there's no recoil without the pumping, but the pumping doesn't cause the recoil, the pellet motion and ejection of the compressed air causes the recoil motion (tiny as it is.)

RJ357
August 20, 2004, 12:21 AM
JohnKSa - For it to be truly analogous, they would all have to be fully enclosed in the box like the combustion gases are. So, they're all standing on the bottom of the box and pushing against the sides. NOW when one wall gives way, what happens to the box and why?
I think that doesn't work at all, that way. The box would never move.

To analogize a gas, I would have them running around, bumping into each other and the walls.Keenan's quote is correct: "pressure does not directly cause the gun to function". Sure, pressure causes the gun to function--it gets the bullet moving. But the recoil force is a result of the bullet (& other ejecta) motion, not directly a result of pressure. Why couldn't the slide movement be considered as the action, resulting from pressure and the bullet movement as recoil?

JohnKSa
August 20, 2004, 12:41 AM
The box would never move.It WILL! The motion of the wall as it gives way, and the motion of the guys pushing against it and now released to move will cause the box to recoil.Why couldn't the slide movement be considered as the action, resulting from pressure Well, in a locked breech gun, the slide CAN'T move based on pressure. The bullet must move and generate recoil in the slide/barrel UNIT before the slide can move.

In a straight blowback gun, I guess you could make the statement--but it's like saying that the 30 guys pushing on the three sides of the box that didn't give out are the reason that the fourth wall broke.

RJ357
August 20, 2004, 12:56 AM
It WILL! The motion of the wall as it gives way, and the motion of the guys pushing against it and now released to move will cause the box to recoil. I think this is crucial. I think this is the whole cause of all of the preceding theory arguments. I need to look at this to make sure I understand what your saying.

RJ357
August 20, 2004, 01:00 AM
Are you saying that their feet pushing back against the floor is the recoil reaction?

RJ357
August 20, 2004, 01:05 AM
Well, in a locked breech gun, the slide CAN'T move based on pressure. The bullet must move and generate recoil in the slide/barrel UNIT before the slide can move. It seems like it should work. It's easier to imagine with an extremely heavy bullet.

JohnKSa
August 20, 2004, 01:25 AM
Well, the bullet can be extremely heavy or the acceleration can be very high. Either contributes equally to a large force. In the firearm case it's the acceleration that's large.

RJ357
August 20, 2004, 01:36 AM
Isn't it simply two masses free to move in relation to each other? The only real difference is the disproportionate size. Why would one have preference as the one "acting"?

antediluvianist
August 20, 2004, 05:35 AM
You are all neglecting the little men inside the 1911 that move everything around.

1911Tuner
August 20, 2004, 06:45 AM
Jim...I don't know which offers the most resistance, but I would have to guess that the bullet does....mainly due to its having to swage down to get past the rifling.
______________________


Higgins...Great post. We're on the same page, with one exception. I don't think that it's all in the slide. The barrel eventually winds up pushing on the slide too...it just occurs a split second after the slide starts its rearward pull. Nothing is everything...Everything is something. I DO, however, believe that the slide is the primary mechanism that provides say....90%
of the action. The barrel's influence ends quickly...as soon as the link unlocks it from the slide.

I completely agree with the "Pushing against a Wall" theory. The slide and barrel being locked together, and being pushed/pulled in opposite directions with equal force can't move and will remain static until something happens to make the force unequal. That unbalancing of the force vector occurs when the bullet moves...but it occurs gradually rather than suddenly. The balanced vector isn't "broken" as per Kuhnhausen...but rather makes a transition. <------Total theory here.

I'm going to propose another experiment for Jim with a cross-post on both threads in a bit. Things are pretty chaotic around here early mornings.

Stand by...

Tuner

1911Tuner
August 20, 2004, 07:44 AM
Here 'tis. There is a duplicate post on the other thread. Mr Keenan...Ya game?
______________________

Okay....Let's recap Jim Keenan's experiment in which he threaded the muzzle end of a barrel and inserted a rod to block bullet movement. A
set screw was threaded into the muzzle, positively blocking any chance of
bullet movement.

We learned two things from this.

One is that the 1911 pistol is far stronger than many of us imagined.

Two is that if the bullet doesn't move, the slide won't move. This
SEEMS to fly in Kuhnhausen's face on his "Balanced Thrust Vector"
theory of recoil, in which he appears to state that the bullet exits and
THEN the slide recoils...which is wrong. If the bullet is gone BEFORE the slide moves...the slide won't move. There is no resistance(the bullet) to the pressure to act against the slide and cause it to move. No launching pad, if you will.

I submit that because the barrel was pushed forward...and because the
slide and barrel were mechanically connected...the slide COULDN'T
move in the opposite direction. The barrel was keeping it static via the locking lugs. it would be the same as two cars linked together by a chain
and pulling in opposite directions. Assuming equal vehicle weight, power,
gear ratios/torque multiplication, and equal traction...neither one CAN move...until something breaks to make the balanced vector UN-equal.

I propose another experiment. I'll remove the locking lugs from a USGI
Colt barrel and mail it to Jim. Hopefully. he still has the rod and screw,
so that all he needs to do is thread the barrel and go. I realize that the gun will then be blowback operated, but the point is this:

If the mechanical connection between the barrel and slide is broken,
the pressure will be free to pursue the path of least resistance, and the slide will move...probably violently. I suggest a strong recoil spring and a shock buffer. I'll include a buffer with the barrel.

Standin' by...

Delmar
August 20, 2004, 08:00 AM
That unbalancing of the force vector occurs when the bullet moves...but it occurs gradually rather than suddenly.

That I think would have to be the case, otherwise we would have to assume the powder charge would develop max pressure at the instant of firing, which is not true.

If the barrel has a part in starting the slide rearward, wouldn't there be telltale wear marks on the slide where it is impacted by the barrel?

1911Tuner
August 20, 2004, 08:50 AM
Howdy Delmar. You asked:

If the barrel has a part in starting the slide rearward, wouldn't there be telltale wear marks on the slide where it is impacted by the barrel?
____________________

Not unless there's a lot of fore/aft play in the barrel when it's in-battery, or
the locking lugs aren't bearing an equal load. Slap-seated lugs...or impact
equalized lugs often show damage to the barrel and slide lugs with a
stepped appearance. This can also occur whenever there's a linkdown timing problem with the barrel, but that particular damage usually produces a rolled or rounded off locking lug at the top...Flanging of the lug is also usually seen. Flanging is a burr or sharp ridge on the top of the lug that can be felt with a fingernail raking the lug from back to front.

The stepped damage on the lugs likely occurs when the bullet slams into the rifling and yanks the barrel forward against the slide lugs closer to the pressure spike...which crashes them harder due to having a longer running start. When the play is less, the barrel and slide lugs butt together earlier, before the speed and force have built up to a degree that the steel is peened on impact. Demonstrate this by whacking two hammer heads together from an inch apart...and then from a foot apart.

Stepped, impact damage to the slide lugs is worse when the total lug engagement is less than half of lug depth. Demonstrate by whacking the hammer heads together when they're offset to one another instead of
straight on with heads perfectly aligned.

When there is nearly zero play between slide and barrel lugs, they strain
against one another without the running start/momentum buildup...and
build tension between them gradually. Demonstrate by touching the
hammers together and starting pressure from that point instead of building speed and momentum before they hit.

Interestingly, it was the stepped damage in some pistols that really got me thinking about the balanced thrust vector. I'm now at the point of believing
that the slide initiates the rearward movement of the barrel...and the barrel
begins to offer some help at some point through the rearward thrust against the case through the breechface. This probably is a very small
amount of push, due to the pressure starting to fall off after the spike
and allowing the lighter barrel to be pulled rearward by the heavier slide
which has its momentum established...and because the barrel travels
for such a short distance before starting to unlock. The slide is the major player because of its greater mass and its resulting larger amount of kinetic energy.

Brian Williams
August 20, 2004, 09:33 AM
What if instead of locking the bullet to the barrel, as in Mr K's test, lock the bullet to the frame and placing the frame in a secure holding device. This would hold the bullet from moving but would not lock the barrel/slide to the bullet.

misANTHrope
August 20, 2004, 09:37 AM
I completely agree with the "Pushing against a Wall" theory. The slide and barrel being locked together, and being pushed/pulled in opposite directions with equal force can't move and will remain static until something happens to make the force unequal. That unbalancing of the force vector occurs when the bullet moves...but it occurs gradually rather than suddenly. The balanced vector isn't "broken" as per Kuhnhausen...but rather makes a transition. <------Total theory here.

Precisely the case. If we could see inside the case, and watch a slow-motion video of ignition while also having the ability to determine pressure, resulting force, etc, etc, things would look like this:

The primer ignites, and in turn the powder begins to burn. Pressure within the case builds rapidly. For a tiny moment in time, the bullet remains locked to the case via the static friction between the two, but the force on the base of the bullet resulting from the building pressure quickly overcomes this. During this moment when nothing moves, the net force on the whole system is zero. The force exerted radially on the case walls is balanced by a reaction force exerted by the chamber on the case. The force exerted on the base of the bullet is balanced by the static frictional force of the crimp. The force exerted on the case head is balanced by a reaction force exerted by the breech face.

Now the static frictional force is overcome, and the bullet starts to move. At this point, it's easier to analyze the situation using the conservation of momentum principle. Before anything moved, the total momentum was zero; momentum is equal to mass times velocity. To simplify the problem, imagine there is no recoil spring; otherwise, this adds an external force to the system and complicates things a bit. Without the recoil spring, final momentum of the system at any point- whether the bullet is still in the barrel or way downrange, we're still considering it as part of the system. Neglect air resistance, as well; it only muddys the problem unnecessarily. Condier the moment the bullet breaks contact with the barrel; at this point the bullet has momentum, since it has both mass and velocity. In order to conserve momentum, the slide/barrel (still locked at this point) will have equal momentum, but in the opposite direction. This implies that they have to move, because they have to have velocity.

If the recoil spring is replaced, it's going to exert an impulse on the slide/barrel (later just the slide after unlocking, but we're not going that far in the future), an impulse equal to force times the duration of time the force was exerted. In practice, determining this impulse would require integration, but it's enough to know the principle for now. The law of conservation of momentum actually states that the final momentum of a system is equal to the initial momentum plus the sum of applied impulses. So in reality, with the spring exerting an impulse, the slide/barrel assembly will be traveling rearward a little slower than what we'd get from the first method I described.

As far as the gradual unbalancing of the vector- put simply, the force on the base of the bullet, and therefore also on the case head, is going to vary with time, as the bullet travels down the barrel. On the one hand, powder continues to burn, increasing pressure, which would imply that force also increases, but at the same time the bullet moves, effectively expanding the area containing the gasses, so pressure is not going to increase that much. The force on the bullet at any given time is equal to the pressure at that moment multiplied by the bullet's base area, and since area isn't going to change, force is directly proportional to pressure. Not only is this force varying, but it's being applied continuously as the bullet travels down the barrel, as opposed to a single force applied at a moment in time only, like hitting a baseball into the pitcher's skull.

If I'm explaining things that are obvious to you, don't take it personally. Frankly, I just like to talk about the way things work, and how everything interacts. It's no wonder I'm an engineering major.

Side question, off-topic: Out of curiousity, what part of NC are you in?

Tamara
August 20, 2004, 09:43 AM
Well, in a locked breech gun, the slide CAN'T move based on pressure. The bullet must move and generate recoil in the slide/barrel UNIT before the slide can move.

In a straight blowback gun, I guess you could make the statement--

How does the cartridge know what kind of gun it's in? :confused:

The barrel does not recoil, unless it is being dragged rearward by the slide. The projectile does not start its forward motion as one unit with the barrel. It moves forward and engages the barrel.

Where the hell is owen when we need him anyway? He's got a slide rule. He does this stuff for a living. :scrutiny:

NMshooter
August 20, 2004, 07:13 PM
Test fixture: put pistol in vise. Affix barrel plug to vise. Slide should still be capable of motion. Load one blank cartridge. Fire. I bet the slide cycles normally, and the blank is extracted and ejected.

RJ357
August 20, 2004, 08:21 PM
A question.

During that first 1/10" of movement of the slide, does the barrel move downward at all? In other words, have the barrel lugs begun moving out of the slide notches during this 1/10" movement?

1911Tuner
August 20, 2004, 09:10 PM
Howdy RJ,

I see where you're going with your question. One thing to keep in mind is that barrel linkdown occurs AFTER the vector has done its work...not during.
if it happened during a signifigant part of the pressure curve, the case head would probably blow out.

The answer is yes. The distance from the centerline of the slidestop pin to the beginning of the convex radius on the lower lug marks the start of barrel linkdown...but the barrel doesn't fall into bed...The link stops the barrel's rearward motion and forces it to change directions. IF the linkdown/unlocking sequence is correctly timed, this occurs after the
chamber pressure has fallen off enough to allow primary extraction.

At this point in the cycle, the slide momentum needed to complete the cycle
is established. Unless the slide is oversprung, or a mechanical interference
occurs between the slide and barrel, the slide will make full travel and return to battery.

This distance varies from gun to gun, and can occur with as little as .080 inch of barrel travel, to as much as .125 inch. In many pistols that are timed just so...the empty case will exit the chamber and eject from the port without an extractor even being in the gun.

RJ357
August 20, 2004, 09:38 PM
1911Tuner -

OK thanks

What I was wonder was, if the barrel was stuck to the slide, as from the vector force, if the slide could still move back that small amount (1/10").

Was barrel lugs the right term for the lugs on top of the barrel?

JohnKSa
August 20, 2004, 10:07 PM
How does the cartridge know what kind of gun it's in?Obviously it doesn't. The slide and barrel "know". In a locked breech gun the barrel and the slide can't unlock from each other until the whole slide/barrel unit is set in motion. In a straight blowback gun the slide is free to move without the barrel.The barrel does not recoil, unless it is being dragged rearward by the slide. Which, of course, in a locked breech gun it is. In a straight blowback gun the barrel doesn't recoil except through the force transferred to it through the frame from the slide hitting the end of travel and from the force transferred from the slide through the recoil spring to the frame before that.

1911Tuner
August 21, 2004, 04:46 AM
RJ asked:

What I was wonder was, if the barrel was stuck to the slide, as from the vector force, if the slide could still move back that small amount (1/10").

No. The barrel will only move fore/aft by the amount of clearance between the locking lugs on the barrel and the slots that they fit into in the slide. In an unworn/undamaged ordnance-spec pistol this difference is very small...about the thickness of a sheet of typing paper...up to about .010 inch (ten thousandths) if the pistol is loosely fitted. Much more than that, and the gun is barely serviceable The lugs will beat themselves to death in a fairly short time.

Tightly fitted pistols have almost none. The 1/10th inch of movement in the initial unlocking phase is made by the barrel and slide together. Hand-cycle a 1911 pistol and watch the top of the barrel hood. You'll have to cycle it very slowly and watch closely to see the barrel start to move. By about .250 inch of total slide travel, the barrel is completely unlocked and linked down...so you can see that it all happens pretty fast and in a very short space.
_______________________

Was barrel lugs the right term for the lugs on top of the barrel?

Yep...Those are the locking lugs. There are three. The two forward lugs are obvious. The third one is the "shoulder" formed at the front of the hood.

ryoushi
August 21, 2004, 11:30 AM
Ryoushi put bullets in gun, ryouchi aim gun at rotten grapefruit. Ryoushi then pull trigger and grapefruit explode in most satisfactiry manner.

Thegman
August 22, 2004, 05:14 PM
It seems this keeps going round and round, but I think it's becoming overly complicated....

Here's my (short) take (not necessarily correct, but I don't currently see were it wouldn't be):

1. Conservation of Momentum: As bullet mass moves forward, the remainder of psitol moves rearward. Barrel and slide mass can move, so they do move, rearward, proportionally to their mass WRT the mass of the bullet. If you don't agree with that, you're one of the people that apparently doesn't believe in laws of physics.

2. Friction of bullet on barrel: I think I'd more or less forget this. This can also be looked at as barrel friction on bullet. Remember, while bullet is in barrel, pistol-bullet is a single unit. Frictional forces of bullet on barrel are equal and opposite of frictional forces of barrel on bullet. Conservation of momentum, I think, is FAR more important to consider.

3. After bullet leaves barrel, slide-barrel unit have rearward momentum from #1 and at this point, the only mass left to act on recoil is the mass of the gasses from the powder. Small mass, but, after bullet leaves barrel, very high acceleration due to high pressure gradient between atmosphere and muzzle. This gas HAS to have an effect on recoil for the same reason as #1.

4. All of the other internal pulling-pushing frictional forces, ect. should essentially cancel each other out. The pistol can't pull itself "by its boot straps". Forward accellerating masses (solids and gasses), I think, are what to consider for all of the pistol's reward acceleration.

1911Tuner
August 22, 2004, 05:56 PM
Gman said:

This gas HAS to have an effect on recoil for the same reason as #1.

Absolutely. Anything that CAN have an effect WILL have an effect,
and can't be ignored...no matter how small.
__________________________

4. All of the other internal pulling-pushing frictional forces, ect. should essentially cancel each other out. The pistol can't pull itself "by its boot straps". Forward accellerating masses (solids and gasses), I think, are what to consider for all of the pistol's reward acceleration.

Except for one consideration...The gasses are pushing (accelerating) in both directions. The larger mass (barrel and slide) is moving backward
as the bullet is moving forward. Recoil begins at the exact moment of bullet movement. It just moves (accelerates) slower because of its greater mass.

The ignition of the powder is like any other explosion...even though the powder burns progressively...It produces force in all directions. Whatever
that force happens to act on is also accelerated, but at a lower rate due to
its having mass, so the faster acceleration of the unopposed gasses can
(and will) follow the path of least resistance.

Nobody can deny that less force is required to rack a slide than push a bullet through a rifled barrel, even after the initial coefficient of friction is broken. At the point that the vector begins to apply force in the direction
of least resistance and gets the slide moving, the slide's greater mass/momentum starts pulling the barrel backward with it via the mechanical connection (the lugs) even though the bullet hasn't yet exited.

The force is pushing, and the slide is straining to pull the barrel (and bullet) backward. ZIP! The bullet exits...(One side in a tug of war lets go of the rope) and the slide's conserved momentum is suddenly loosed...and that
momentum carries it through to full travel. Watch a slow-motion video of
the firing/recoil cycle. You'll see the gun move forward just a tiny bit, followed by the slide moving...slowly at first...and suddenly speed up just
as the bullet breaks free. I had to watch it 40 or 50 times before I saw
what was happening clearly.

RJ357
August 22, 2004, 06:30 PM
1911Tuner -

Thanks

You answered, "No. The barrel will only move fore/aft by the amount of clearance between the locking lugs on the barrel and the slots that they fit into in the slide. "

Sorry, I should have asked, "if the barrel was stuck to the slide, as from the vector force, if the slide and barrel together as a unit could still move back that small amount (1/10")."

1911Tuner
August 22, 2004, 07:29 PM
I should have asked, "if the barrel was stuck to the slide, as from the vector force, if the slide and barrel together as a unit could still move back that small amount.

Yes. As long as there was also an additional force acting on the slide to pull them in the same direction AND it was stronger than the force that was pulling them forward. It would be much the same as lifting a barbell
with 200 pounds on it. If you're exerting 201 foot-pounds of lifting force,
you would lift the weight one foot per minute. If Mr. Keenan had yanked on the slide at the same instant that he pulled the trigger in his blocked-bore experiment, the slide would have moved, pulling the barrel back with it until it unlocked because the vectors of force wouldn't have been equal.

RJ357
August 22, 2004, 08:00 PM
1911Tuner -

OK, thanks.

Here is what I am thinking:

There seem to be two separate meanings to the term "locked".
One is as we normally think of it in a 1911, the barrel lugs engage slots in the slide and prevent them from separating. As long as the slide is sitting within a certain distance of it's full forward position, the barrel and slide cannot separate. The pair must move back a certain distance before the lugs will disengage.

The other meaning of "locked" due to the thrust vector. Here the friction between the barrel lugs and the slide slots does not allow the barrel to link down and disengage the lugs from the slots.

It seems that the normal locking, as in the first definition above, will guarantee that they remain locked until the bullet is gone. This is because the bullet is gone by the first 1/10" of movement. And 1/10" of movement is not enough to completely disengage the lugs.

It looks like even if Kuhnausen intended the thrust vector to work, it doesn't matter if it doesn't. The barrel and slide would be locked together anyway.

Or should I say "if Browning intended it to work".
Who originated it?

Thegman
August 22, 2004, 11:32 PM
Tuner,

I'm not sure I'm totally following what you're suggesting about the bullet-barrel getting pulled/pushed, etc. but the frictional forces between the barrel and bullet are not really doing work on the system. The friction is producing lots of heat, but probably little to no work.

Also, I'm not totally sure I know what you're saying about the gasses, but, while the bullet is still part of the system (in barrel), the gasses are adding a little to recoil, very little. If the powder charge is, say 10 grains, it's center of mass is accelerating from the cartridge case (where it is a solid) to approximately the middle of the barrel's length as it expands and fills the barrel. It's acceleration at this time cannot be any greater than the bullet in front of it. So, while the bullet is in the barrel, the gas is like an extra ~10 grain mass moving ~ 1/2 the distance of the bullet. That won't affect the system's momentum very much. Once the bullet leaves the system, the gas, with it's center of mass ~1/2 way through the barrel, will then begin to accelerate out of the system much faster than the bullet did. That momentum is now added to recoil. How much it affects the total recoil, I don't know.

As far as the slide speeding rearward due to the bullet leaving barrel; I wouldn't think that it should, not much, anyway. The final velocity of the slide/barrel, due to the bullet moving through the system cannot be more than the momentum the bullet is taking forward, and the bullet is building momentum throughout it's acceleration, so the barrel/slide are as well. I would guess that the sudden accelleration of the gas has a lot to do with what you're seeing at this time, which is where a compensator's baffles probably come into play (slowing the gasses acceleration out of the system). If one were to do some calculations on gas acceleration it would go a long way, I think, to answer that question. How about filming the sequence with various compensators?

I've never seen the video you're talking about (I'd like to), but as far as the pistol moving forward: I wouldn't say that couldn't be in part from bullet movement into the lands (I think that's what you're saying), but don't forget that the hammer just fell at that moment as well. Was the pistol in a vice during firing? Was the test also done with dryfiring to observe effects of the forward rotation of the hammer's mass? Either way, the little forward movement you're talking about isn't the recoil we're talking about, anyway, I guess.

1911Tuner
August 23, 2004, 07:21 AM
Howdy Gman,

The frictional resistance of the bullet in the barrel is pretty signifigant. If
you've ever had a bullet lodge halfway through a barrel, and tried to drive it out...You know that it takes a lost of force to get it out.

Okay...We've spoken of the hypothetical pulling of the bullet through the barrel with a cable. Take it a step further and mount this pistol in a vise.
Attach a cable to the bullet AND to the back of the slide. Start applying equal force in opposite directions, and gradually increase the force until
the bullet moves. At that point, the bullet requires LESS force to keep it moving than it did to GET it moving, while the force rearward on the slide
is unchanged. The balanced vector of thrust isn't equal any more, and the slide will begin to move, pulling the barrel with it via the lug connection, and it will continue to move at a set pace UNTIL the bullet breaks contact with the barrel. The force on the slide remains unchanged...established momentum if you will...and the slide will now accelerate quickly and move until it's forced to stop upon impact with the frame.

Imagine the theoretical tug of war in which opposite force begins, but
as one opponent starts to get tired, the balance shifts to the stronger side.
The tired tugger begins to slowly advance toward the mudhole that awaits the loser. Just before falling into the mud, he releases the rope...and the stronger man is suddenly free to move backward unimpeded.. He will move suddenly and quickly. The applied force that he has been imposing has become momentum because he was moving....no matter how slowly...
while before he moved, it was just an application of force. (Momentum
begins with movement.)

And the hammer's momentum in the video was taken into account. Holding the pistol loosely and letting the hammer fall will move the gun forward a little...but not nearly as far as the bullet's impact into the rifling would be
if the gun was held loosely and fired. Held in a normal firing grip, the hammer's effect on the gun moving would be nearly insignifigant as compared to the bullet's effect. The video's purpose was to show the
firing/recoil sequence...not the effect of the hammer's impact on the gun,
so I'd venture a guess that the gun was held in a normal fashion....gripped
fairly solidly in order to prevent a limp-grip malfunction.

Thegman
August 23, 2004, 03:23 PM
Tuner,

Respectfully, I think you're really over-thinking it (or thinking about it in the wrong way).

If you have to do all of this imagining and use analogies, but cannot find dynamic equations to describe them, the equations probably don't exist. Notice no one seems to have an equations for the 'broken force vectors'? We can all agree, I think, and calculate, slide-barrel velocity as the bullet leaves the system using m(bullet)v(bullet) = -(m(barrel-slide)v(barrel-slide)), and if we assume constant acceleration for the bullet, we can even predict the amount of slide-barrel movement. I think that's about it.

There should be a calulatable "thrust vector" once the bullet leaves the system - From the escaping gasses, but not from the bullet.

I fully agree there's significant friction as the bullet moves through the barrel (but remember that it will be different than pushing a bullet through the barrel at almost zero velocity - I suspect the frictional coefficient changes with velocity and the heat created), but your hypothetical 'pulling' the bullet analogy is seriously flawed.

--The bullet and pistol aren't being pulled from forces from outside the system, creating 'broken force vectors' like snapping cables --

--It's being pushed from within the system.--

In your tug-o-war example, the pistol is being pulled from EXTERNAL forces, that's not what's happening. The forces are internal, and that's an important difference. The barrel is getting stretched and deforming slightly, like a spring. The only rope cutting analogy when the bullet leaves would be the barrel returing to it's original length and diameter as the pressure drops.

Maybe think of your rope as a rigid pipe, closed on one end, filled with, let's say concrete. A charge is ignited at the closed end. Is there friction between the pipe and concrete? Sure. But if the friction is less than the strength of the materials, that's were the movement (the break) will occur. Assuming the pipe is much less massive than the concrete, the pipe will go one direction, and, regardless of the friction, the concrete will move a bit in the other. The pipe, won't, indeed can't, drag the concrete around from internal pressue and friction in the system. The pipe and concrete will heat up as they're -"pushed apart"- The friction will reduce the final velocity of the pipe and concrete due to this energy loss, which produced heat. But momentum will be conserved. All of the 'tugging' is producing -internal- strain on the pipe and concrete, stretching and compressing both to some extent, especially the pipe, but nothing gets dragged anywhere from the friction (other than some atoms at the interface of the pipe and concrete); the friction creates heat and robs velocity through (energy stealing) to some extent.

Rather than thinking about pulling the pistol with cables, think about a really strong spring being released between the breech and bullet - what happens in that analogy? Regardless of the frictional resitance, the gun goes in one direction and the bullet the other, right? The friction between the bullet and the barrel heats the barrel, but doesn't drag it along by its nose. The barrel stretches behind the bullet due to internal strain in the system, but the bullet can't drag the gun along by it's nose, as it's being pushed forward! Conservation of momentum is what moves the system.

Does this change you mind? If not, give me some numbers on those "broken vectors", and explain where they're coming from (no cables pulling things around ;) )

1911Tuner
August 23, 2004, 04:13 PM
Hiowdy gman,

There was a time that I could do the math and show the numbers...but it's been about 35 years or so since I've had to, so I called on the lads and gents who still have a handle on it.

You offered:

I fully agree there's significant friction as the bullet moves through the barrel (but remember that it will be different than pushing a bullet through the barrel at almost zero velocity - I suspect the frictional coefficient changes with velocity and the heat created), but your hypothetical 'pulling' the bullet analogy is seriously flawed.

IIRC, my first hypothetical situation was to envision pulling the bullet at the
identical rate of acceleration as it would be if it were pushed by powder ignition...Which would serve to cause the slide to recoil if the bullet's movement was all that caused it. When the bullet is blown through the barrel, the vector of force is also blowing in the opposite direction.
_______________________

And:

The bullet and pistol aren't being pulled from forces from outside the system, creating 'broken force vectors' like snapping cables --

--It's being pushed from within the system

And there is no difference between the tow except for the direction.. 500
Newtons of force is 500 Newtons...whether it's applied as a push or a pull.
The analogy is valid in showing that the vector is imposing force in opposite
directions....regardless of which direction it's imposed.

If you lift a 100 pound barbell, you've applied the same force and performed the same amount of work by lifting it as you would by pushing it up from a bench-press...assuming that the distance that the weight moved is equal.

Push equally in opposite directions or pull...The amount of force is the same. Pull the bullet to 800 fps in 4.5 inches of rifled barrel, and the effect
on the barrel is the same as pushing it...even down to the stretching. If the bullet's forward momentum is the cause of the recoil, the slide should move if the bullet is pulled through...at 800 fps.

:D

Thegman
August 23, 2004, 05:47 PM
Tuner,

Push-pull is important. Let me give it another shot at convincing you.

Yes, a pull on one side is equivalent on the object being moved to a push in the other, but what is not equivalent are the affects on the surroundings. In your weight lifting analogy, PUSHING the weight up puts an opposite force on the floor (the breech of the gun), that's a correct analogy for the powder in the barrel. PULLING the weight up uses the same amount of force, in the same direction, and affects the weight the same, but it puts an opposite force on a cable attached to the celiling, THERE IS NO CEILING in the analogy for the gun system. That's the difference. There is a real floor (the breech), but no imaginary ceiling. That's the importance of push-pull, the affect on the surroundings, not the equivalent effect on the bullet.

In the firearm, where is the force originating? On an imiginary cable attached to a wall? No. It's between the bullet and the breech. The breech is real, the back end of the bullet is real, and the powder's pressure is real, and it's pushing the two apart. Final relative velocities of the system is from conservation of momentum. Does that make sense?

Your analogy of the tug-o-war implies the pistol is the rope, esentially. Something is pulling the bullet out of the barrel (the non-existent cable attached to the non-existent wall) while the opposite force is the shooter, hanging on, resisting the forward pull of this cable, keeping the gun's position esentially static. In your anaolgy for this 'broken force vector', when the bullet leaves the barrel (cable breaks), the shooter is falling back because they were tugging the other side of the rope (the pistol). That's absurd of course, but that's actually what you're saying. Give it some thought. Push-pull is very important, not so much on the bullet, but what else it affects, and how.

1911Tuner
August 23, 2004, 06:37 PM
No...In my first analogy, there wasn't anything holding the gun back...I
simply intended to paint a picture of a bullet being pulled through the barrel and expecting the slide to recoil because o fthe movement of the bullet...Oversimplification at its oversimplified best.

The second analogy...the "Tug of War" I intended to paint a picture of
one cable pulling the bullet, and another pulling on the slide...under equal force. Nothing moves under these conditions until the bullet starts to move. When the bullet starts to move, the balance is lost because of
less force required to KEEP it moving/accelerating...bit the force on the slide
remains constant. Forward "thrust" is lower than rearward...and the slide moves. The faster the bullet moves, the less force is required to keep it moving. The slide, has greater mass, and accelerates more slowly, but it keeps moving because the balance can't be restored due to the changing
conditions brought on by bullet movement and falling coefficient of friction,
so there is always greater force on the slide. Even though it's accelerating,
its greater mass...and the effect of the recoil spring...it's accelerating slower
than the bullet. The bullet breaks contact, and all that gathering momentum in the slide has to go somewhere...so it goes backward toward the path opf least resistance. Of course, it all happens in a short time frame and in a short distance.

Yes...it all makes sense, but I won't be convinced that the bullet's forward momentum is the only factor in the slide's movement. It's only a part of
the whole picture. The vector of force between the bullet and the slide/barrel unit has an effect...and I believe that effect is greater than the bullet's acceleration and momentum.

R.H. Lee
August 23, 2004, 07:00 PM
The vector of force between the bullet and the slide/barrel unit has an effect...and I believe that effect is greater than the bullet's acceleration and momentum.


Could this be tested by cutting the barrel off at the nose of the bullet, or maybe just in front of the link? Does this mean that a longer barrel will increase recoil? Or conversely, do short barrel 1911's have less recoil?

RJ357
August 23, 2004, 07:27 PM
RileyMc -Does this mean that a longer barrel will increase recoil? Or conversely, do short barrel 1911's have less recoil? This is where conservation of momentum is a good choice for analyzing things.
If the bullet has greater velocity, then the recoil will be greater. For the same weight bullet, more velocity means greater momentum. And so the slide momentum would necessarily be greater also.

Felt recoil is another matter. A longer barrel usually means a heavier gun.

R.H. Lee
August 23, 2004, 07:36 PM
If the bullet has greater velocity, then the recoil will be greater. And in order for the bullet to have greater velocity, it would have to be in the barrel with the expanding gas pushing on it longer. All the while that same expanding gas is pushing on the breechface/slide, causing recoil.

RJ357
August 23, 2004, 07:57 PM
Yes, both bullet and slide continue to accelerate as long as the bullet is in the barrel. When it finally pops out, the slide gets a small amount more from the rapidly falling pressure remaining.

1911Tuner
August 23, 2004, 08:30 PM
The amount of residual gass pressure in the barrel/breech after bullet exit is nearly insignifigant...althougn it can't be ignored. The momentum that
was imparted to the slide during the first .100-.125 inch of travel carries it
through to full cycle, assuming that the recoil spring strength isn't off the scale. Relate to light "Softball" loads with the standard spring sometimes causing short cycle malfunctions.

Jim K
August 23, 2004, 09:38 PM
Wow! And I thought I knew how the gun works.

But I am convinced. It is all magic. Browning called the 1911 a recoil operated pistol, but what the heck did he know?

Some folks just can't accept that the laws of physics really exist and really do work. I grant the effects of pressure and friction are a lot easier to grasp than the idea of motion in one direction causing motion in the other, but I can only recommend talking to a physics professor who might be able to explain better than I.

One word on blanks. No, the gas does not need to "push" on anything. That is why a rocket works in space where there is nothing to "push" against. When a blank is fired, there is a slight recoil (not nearly enough to operate an auto pistol) caused by the mass of the forward moving gas, but that recoil will be there in a vacuum. Recoil is not dependent on the presence of air or gravity.

And there is still no such thing as a magic "thrust vector" that stops recoil from taking place until the bullet it out of the barrel.

Jim

Higgins
August 23, 2004, 10:34 PM
Remind me again what the question was.:neener:

RJ357
August 23, 2004, 11:17 PM
Jim Keenan -Some folks just can't accept that the laws of physics really exist and really do work. 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.

Thegman
August 24, 2004, 04:26 AM
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 ;)

RJ357
August 24, 2004, 08:08 AM
Thegman -The ONLY thing that can move appreciably in this case is the bullet and gas, forward, sending the slide and barrel backward. 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
August 24, 2004, 08:13 AM
Jim...You're being facetious.:D This HAS been fun though!

Okay gman...You've tried again, and you're right...I am pretty stubborn.:p

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
August 24, 2004, 01:50 PM
But it only causes one of them to move? Both are masses, both are free to move.

the breech is LOCKED to the barrel. it cannot move.

1911Tuner
August 24, 2004, 01:56 PM
Andrew said:

the breech is LOCKED to the barrel. it cannot move.

Unless they both move in the same direction...:scrutiny: :D

Which they'll do as soon as the bullet starts to move...:scrutiny:

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! :D

Andrew Wyatt
August 24, 2004, 02:43 PM
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.

1911Tuner
August 24, 2004, 03:12 PM
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? :D

Andrew Wyatt
August 24, 2004, 03:23 PM
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".

1911Tuner
August 24, 2004, 03:44 PM
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.

:D

Thegman
August 24, 2004, 04:52 PM
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.

Higgins
August 24, 2004, 05:14 PM
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?

1911Tuner
August 24, 2004, 06:15 PM
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...:D

Here endeth the lesson...for me at least.

RJ357
August 24, 2004, 06:45 PM
Thegman -

OK, thanks for clarifying.

1911Tuner -

You don't know how glad I am to hear that.

Hand_Rifle_Guy
August 24, 2004, 07:43 PM
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 (http://www.thehighroad.org/showthread.php?s=&threadid=98340) 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
August 24, 2004, 08:03 PM
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.:D

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?:p

RJ357
August 24, 2004, 08:14 PM
One way you might tell is to look at the case head and see if there are any breechmarks on it. That would indicate that it had been pressed hard against it.

Also, just how much pressure is in a .45 standard load (230 gr)?

Hand_Rifle_Guy
August 24, 2004, 08:39 PM
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! :neener:

Don't you make me come down there...;) Edited for typos.

RJ357
August 24, 2004, 09:04 PM
Hand_Rifle_Guy -

OK thanks

If I calculated it right, that's over 2500 lbs of force rearward on the case. The force on the case walls would also be quite large, but the rearward force will win out?

P95Carry
August 24, 2004, 11:06 PM
What a momentous ''can-o-worms'' thread!:)

HRG ..... I'll put vaseline on the toilet seats Oh - I'd thought it was super-glue!!:D :neener:

Clark
August 25, 2004, 02:20 AM
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.

Thegman
August 25, 2004, 04:19 AM
...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.

1911Tuner
August 25, 2004, 04:30 AM
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?:D

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... :p

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
August 25, 2004, 04:57 AM
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.:cool:

sm
August 25, 2004, 05:21 AM
I have enjoyed this thread folks and the other one similar. Okay- I have enjoyed all the mind boggling threads Tuner tosses out. :)

Hey - I just shoot the darn things. :p I shot a 1911 the other day - brass goes yonder, I had to bend over and pick it up. I shot a Model 19 and didn't have to. Since the ground was wet....I kept shooting the K frame.

Simple, lazy, something to do... Oh ...speaking of keeping it simple. The Target - pc of cardboard with a hole in it. If I am hitting no holes in target. If I miss , easy to see and where / what I did wrong. Gee - I wouldn't want to have to actually staple up another pc of carboard [ drag out cardboard, drag out stapler] ...that's work....incentive to not miss. :D

Carry on....

Thegman
August 25, 2004, 06:12 AM
Hi Tuner,

I'm not saying there's not more recoil, I fully believe it, I just disagree as to how friction is causing the increase :)

If you could, print up what I wrote and have him look at it when you see him, I think he'll agree.

One other thing. The rifled barrel is also imparting angular momentum to the bullet. I don't know how much that affects felt recoil, but I suppose it does rotate (torque) the gun. But a black powder rifle generally has a much slower twist, like 1/66, so not as much angular momentum there.

Anyway, I'd better get some sleep.

Night

niemand
August 31, 2004, 12:05 PM
Very good discussion!

Have "Long History" M1911/Recoil-Operated Arms. In response to queries, have ever described as; "Friction-Delay Tilting Locks".

Two experiments; 1. (On Old Slide) Create Fine/Sharp Horizontal Grooves upon Breech-Face, You will discover "Friction" of Cartridge-Case Head upon Breech-Face in "Pressure Hesitation-Lock".

2. Look to Mainspring, as well "Firing-Pin Stop". Note Radius at bottom of said "Stop". Now change then this angle. Consider now Role of this part in "Locking". (Did these for Years in "Magnumising" M1911 Arms, only Drama is need larger diametre Hammer-Pin, Better material of said Part, as well, "Bushing" should Frame of inferiour Heat-Treating).

Have found (Contingent upon how One "Times" the issue) "Locking-Lugs" utterly superfluous to process (M1911). Think "Over Top Dead-Centre" however minutely in this case.

If Posting repeats any previous, apologies!

1911Tuner
August 31, 2004, 03:14 PM
Howdy niemand! Welcome to THR and this little quagmire that I got started...even though I knew where it would go...and where the hello is
Nirgendwo??? :scrutiny:

Funny you should toss the firing pin stop radius and mainspring rating into the middle of this. I've been screamin' about the possible negative effects of light mainsprings for over 35 years. The small radius firing pin stop was actually part of the original design, and was changed on request from the
Army in order to make the slide easier to hand-cycle with the hammer down. Seems that the lads were havin' a hard time of it.

EGW makes oversized firing pin stops with a square bottom to allow the
smith to not only fit the stop to the slide, but also delay the slide's rearward movement and barrel linkdown timing a little. This, in turn,
allows a lighter recoil spring with no increased battering of the recoil impact surfaces. Neat. 15 bucks from Brownells and top-quality steel.

Grump
August 31, 2004, 03:56 PM
much that affects felt recoil, but I suppose it does rotate (torque) the gun.
Yup, Colt-barreled pistols torque into the palm of a right-handed shooter. I shot bowling pins right-to-left because my Smith torqued the other way.

Nirgendwo???

Methinks that's a variation of "nowhere" in German, auf Deutsch. Look at our new friend's name for a clue on that... We shooters iz clever, ain't we?:D

niemand
August 31, 2004, 04:21 PM
Richtig! "Niemand von Nirgendwo". "Wishing to Protect Identity of Guilty". (Ich bin deutsch)

With seriousness, Am Maker of Rifled Barrels, and more, Designer of Defence Munitions. (Got "Joe Zambone" up/running with "Magsafe")

As to "Firing-Pin Stop"/Radius Has been for long My "Easy Demonstration of How Arm Locks".

For Thirty-Years have argued against "Light Mainsprings for Trigger-Pull improvement" had once fixture created "Testing Slide Velocity" wiith divers Mainsprings/Firing-Pin Stop Radii. (Had over Years many "Loud Discussions" Armand Swenson Favourite of Lot)

I recall (Dimly) something of this "Radius-Change For Troopies", yet question; No-One perhaps thought the Hammer to Cock? Or perhaps that solution "Too Simple"?

Note to those RE: "Barrel Torque Effects", some Arms (Such as "Obregon" or French "MAB") utilised in part this effect as Barrel would in Firing rotate to Unlock.


Many Thanks for Kind Welcome!

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