LJA's Evidence Thread: Explanation

I've been able to clear a little off my plate today, so I'll go ahead with the explanation of all the factors at work whenever the radius of a simple firing pin stop is changed.. .and how they affect the recoil cycle of the 1911 pistol.

In the interest of a smoother flow, I ask that comments and questions be held until I've finished. I need to split this up into two or three separate segments to keep it from dragging on for too long. I've found that people who are reading an overly long text tend to skim over a lot of it... or skip it outright... the instant that they find something that they don't understand or disagree with. I won't argue these points. I've already done that too much, and I don't have to time. Instead, I'll describe the function and the physics involved, and let everytone decide for themselves whether they want to accept it or not.

First... a few things that we need to clarify and accept as simple physical and mechanical fact.

Force forward=force backward. Always. If X units of force is applied, it applies in all directions. Thus... if X units of force is brought to bear on a closed system, it's applied backward and forward in equal measure. This is known as a "vectored" force. That is... it has magnitude and direction.

Recoil is nothing more than backward acceleration in response to a force applied to a closed system in which at least one side of the system is moveable. Even if the recoil impulse doesn't produce movement, it still occurs. The reason that no movement is seen on the reaction side is because the force wasn't large enough to overcome the level of resistance offered by the reaction side. A 500-pound .223 caliber rifle may be used as a demonstration. Actgion and reaction occur, but you can't detect reaction due to the mass and the force applied to that mass.

Recoil only occurs when acceleration is applied. Once the acelerative force is removed, there is no further acceleration... and hence, no more recoil. What we perceive as recoil is mostly momentum that was conserved while the accelerating force was on the system. The actual recoil impulse is over so quickly that we scarcely have time to detect it.

When the bullet exits, the force is removed from the system, and neither bullet nor breechblock can accelerate further. All movement after the force has been removed is due to conserved momentum. For both sides, all they can do is decelerate due to outside forces that they encounter.

Any outside force that is imposed on the system while under an accelerating force will affect the rate of acceleration, and the remaining momentum that is conserved during the acceleration. Anything that has the opportunity to to affect the object's motion will do just that. Once momentum is lost, it can never be regained unless a "new" force is applied.

Force makes things move. Momentum keeps them moving. Force robs moving objects of speed and momentum.

Recoil operated pistols and straight blowback pistols are both recoil operated. Blowback is misleading. The only real difference is in the mechanical method of delaying the breech opening until the bullet has exited, and pressure drops to a safe level.

In the short recoil system, the bullet is the main delaying factor. Go back to force forward=force backward. The barrel imposes a frictional resistance to the bullet. Whatever friction the barrel offers to the bullet, the bullet offers to the barrel. The barrel moves backward while the bullet is moving forward... each one resisting the other. All other delaying forces...recoil spring... mainspring... slide mass... barrel mass... and slide to frame friction are secondary to the bullet's influence.

Small changes to the outside forces resisting the slide cause large changes to the slide's rearward travel, and the earlier in the slide's acceleration the decelerating force is applied, the larger its effect on the slide's rate of acceleration and its conserved momentum.

In an autopistol... regardless of the design... we feel very little recoil, if any, as a result of the internal ballistic event... or the "explosion" of the powder. The gun is comprised of the barrel and the breechblock... the slide. The frame is essentially a gun mount with no solid connection to the gun. None of the actual recoil impulse is transmitted directly to the frame that way it is in a fixed-breech revolver or bolt-action rifle.

Next, we'll recap the basic function of the short recoil system... just to keep it fresh.

When the pistol fires, the action/reaction begins immediately. The slide starts to move as soon as any headspace is filled by the backward-moving case when it begins bearing on the breechface. Because quick pistol powders peak rapidly, full pressure and force is brought to bear on bullet and breechface within a half-inch of bullet travel... or less... depending on the power's burn rate.

At approximately .100 inch of slide travel, the bullet exits, and the slide continues rearward on its conserved momentum. During the time that the bullet is still present and being forced through the barrel and the slide is likewise being driven in the opposite direction, its forward drag on the barrel is working to delay the slide. Remember that the slide is the recoiling part of the system... not the barrel. The barrel is forced to move backward with the slide because the locking lugs are horizontally engaged in opposition. This is the "locked" function of the locked breech.

This phase only lasts for a short time and distance... hence the name "Short Recoil" operated. Again... once the bullet clears the muzzle, all accelerative force on the system exits behind it, and the actual recoil event is over. There must be two interaction objects in an action/reaction event. When one is removed from the system, action and reaction can't exist, and any further movement of either object is due to momentum.

Momentum is a function of mass X velocity. In order to reduce the momentum, the velocity must be reduced. This reduction comes from outside forces that oppose the object's movement. Anything that can oppose the movement will oppose it.

Once momentum is lost, it can't be regained unless a new force is brought to bear on the object.

No force... No acceleration.

If momentum reduction is the goal, the best time to invoke a resistive force is at the beginning of the acceleration... before it gains a full head of steam. Once full momentum is aquired, any outside force must work harder to reduce velocity and momentum. In the short recoil system... The earliest application of outside force is applied by the bullet itself. Next in line is the hammer and mainspring. Last... and least effective... is the recoil, or action spring.

The so-called recoil spring's primary function is in returning the gun to battery. Decelerating the slide is a natural consequence of its compression by the slide's travel... but that's incidental. The two main outside forces working to slow and delay the slide are the bullet and the hammer/mainspring assembly...because they are brought to bear early in the event.

Stay tuned. The final installment is in the offing.

After a few years and dozens of pages of discussion on the reduced radius firing pin stop, we've come to understand that it delays the slide for a split fraction of a second by way of reducing the slide's mechanical advantage in compressing the mainspring... and that by doing so at the beginning of the slide's acceleration... it produces a telling effect on the slide's rearward velocity and momentum. Many people report a more user-friendly gun... either in reduced muzzle flip... smoother operation...faster split times, or whatever each shooter experiences. Sometimes, it's not really any one thing, or anything that we can put our finger on.

First, lets look at the recoil system... the separate closed system that actually produces the felt recoil in an autopistol.

Virgil Tripp's excellent slow motion and stop-action videos clearly show that the pistol moves very little until the slide hits the impact abutment in the frame... and that the bulk of muzzle flip occurs there...but the pistol does move a little as the recoil spring compresses. Force forward=force backward. Whatever force the recoil spring produces is returned to the gun, onto the slide and frame.

The reduction in slide velocity means that the spring is compressed more slowly, which spreads out the rearwaed push on the frame over a longer time... but there's another aspect of that increased time frame. Because the slide also has reduced momentum, the spring decelerates it at a faster rate... which contines to decelerate the slide, until it hits the frame. Thus, the whole would appear to be greater than the sum of its parts.

But there's another "hidden" aspect to consider... something that has a greater effect on the cycle than the immediately apparent of reduced mechanical advantage. Timing. I'm not talking about the mechanically fixed timing of the barrel. The variable timing of the bullet's exit relative to the slide's position.

We know that the bullet exits at (nominally) 1/10th inch of slide travel... but that's not chisled in stone. It can vary a little from gun to gun.

At no point in the event is the pressure..and the force that it produces... higher than at the instant of peak pressure. Pistol powders peak rapidly, and fall off rapidly. It's entirely possible for the bullet to be moving at a higher speed at some point in the barrel before exit than it's moving at the muzzle.

BUT... as long as the bullet is IN the barrel, and is being pushed by unbalanced force... the same force is driving the slide... helping it get past that little speed bump that the hammer and mainspring impose in conjunction with the FP stop radius.

The "hidden" effect of the radius is that it delays the slide at the instant that it's trying to overcome that speed bump... and the bullet is still headed for the muzzle.

There is no added speed bump... no "extra" force to impede its progress other than the friction imposed by the barrel, and that's a constant... assuming all eles is equal. The result is that the bullet exits earlier relative to the slide's position. It still requires the same amount of time to traverse the barrel... but as far as the slide knows...it got loose earlier, and thus removed the accelerating force that was helping the slide get past the speed bump... which multiplies the effect of that speed bump, and robbing the slide of momentum. So... rather than the bullet exiting at 1/10th inch of slide travel... it escapes at .090 or maybe even less slide travel. .010 inch doesn't sound like a lot... but it amounts to 10% of the short recoil function. 10% reduction in slide velocity and momentum and striking energy at the frame IS signifigant. It equates to reducing the bullet's muzzle velocity by 10%...or more than the difference between firing equal rounds in a 5-inch gun and a Commander... except there no reduction in bullet speed. The reduction is all rearward.

Simplified: The stop helps to delay the slide which allows the bullet to escape earlier in the slide's cycle which removes the force from the slide earlier which leaves the slide with no force to accelerate it which gives the spring MORE time to decelerate it before it strikes the impact abutment.

The early release of pressure and force is a factor that many consider... even the people who understand the initial effect reduced radius. It's easy to grasp that. hand-cycling offers proof because it's slow enough for us to feel it, and because we're the ones providing the rearward force on the slide. When it speeds up to ballisitic level, and the cartridge is providing the force instead of hand and arm strength... it's a little harder to catch.

Think of it like this:

If you stand with your hands against a wall and shove... you accelerate away from the wall, and you continue to accelerate as long as your hands are in contact with the wall, and you continue to apply force. Once you've pushed yourself far away enough for your hands to lose contact... there is no more force... and no more acceleration. You continue to move backward on momentum.

If you stop applying force before your arms are straight... you still lose contact with the wall, but you lose it earlier... and your momentum doesn't carry you as far from the wall.

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"The stop helps to delay the slide which allows the bullet to escape earlier in the slide's cycle which removes the force from the slide earlier which leaves the slide with no force to accelerate it which gives the spring MORE time to decelerate it before it strikes the impact abutment...."

Tuner,

I have the EGW FPS on all my guns. The first one went into my 1927 Colt- Hartford Argentine and I immediately noticed the gun had a sharper initial recoil impulse, but less flip and dip of the muzzle. This was most noticeable during rapid fire strings.

I have also noticed something else about the EGW FPS; a cleaner chamber/ejection port area.

I have a Nornico 1911-A1 which is my demo/range gun that I use for training. It originally had the 5/64" radius FPS, 23lb mainpsinrg and 16.0lb recoil spring. In this conforguartion, the breech would get filthy, whereas the breech on my 1927 Colt-Hartford stayed very, very clean, even shooting the same ammo.

When I swapped the EGW FPS into my Norinco, the breech stayed very clean, but I saw more soot on the muzzle, barrel bushing, recoil spring plug and sides/top of the slide.

This, I think, is evidence of the reduction in chamber pressures when the gun unlocks with the EGW FPS.

Additionally, I used to agonize over getting the right radius on the EGW FPS. After several thousand rounds through my 1927 Colt-Hartford Argentine, I noticed the hammer made its own radius.

Now, I simply break the edge on my stone and let the hammer do the rest...Robert

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When I swapped the EGW FPS into my Norinco, the breech stayed very clean, but I saw more soot on the muzzle, barrel bushing, recoil spring plug and sides/top of the slide.

This, I think, is evidence of the reduction in chamber pressures when the gun unlocks with the EGW FPS.

I noticed the same thing on my RIA Tactical, now I have an explanation as to why this happens. Thanks.

Here's one of the Tripp Research videos that Tuner refers to. After watching it, watch it again and this time keep your eyes on the shooter's hand, and how it reacts to the slide hitting the frame at the end of its rearward travel.

YouTube

The stock Norinco FP stop radius is 7/32nds which became the standard on January 1st, 1918. Browning's original was 5/64ths. Chamber pressure must be at or near atmospheric when the breech opens. That's the whole point of delay.

The sharper inpulse is due to a more "solid" connection between slide and frame. The initial impulse is transferred faster, but channeled in a different direction. The push from the recoil spring is slower and the slide to frame impact is less intense... which results in less muzzle flip... which is why the gun gets back on target quicker.

To my hands... the small radius FPS doesn't really start to make a difference until you start firing in multiple strings.

My 7 round Tripp magazines with the flex follower will hold eight rounds. I have a photo of a target where I fired 9 rounds into it in less than 4 seconds from 30 feet and you could cover the impact on the target with a hot dog bun.

That's with 230 grain factory-loaded FMJ.

I have put off responding to the posts regarding the explanation of recoil for a while. It would only be my intention to correct some of the explanations and not to criticize the information presented. I suspect that there may be some other engineers on the forum that may also have some comments. Please don't take any comments I have in a negative way because certainly none is intended.

Just a few things on the first post: In a closed system before the bullet leaves the barrel there will be a slight recoil do to the shift in CG as explained in the link below. But after the bullet leaves the barrel the gases will cause a large recoil or a force rearward on the slide. You can not apply an acceleration to an object but you can apply a force. The acceleration produced by a force is described by the formula F=mA. It would be nice if it was that simple, but when the pressure varies over time it becomes difficult to calculate the position and velocity of the slide by hand. Maybe years ago, but now I think my slide rule is stuck and the brain cells are dying off

Towards the end of the first post, it says:

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When the bullet exits, the force is removed from the system, and neither bullet nor breechblock can accelerate further. All movement after the force has been removed is due to conserved momentum. For both sides, all they can do is decelerate due to outside forces that they encounter.

That is just not the case. I'll again refer to the link below to save some time typing.

The small radius FPS changes the magnitude of the force the mainspring exerts on the slide due to the change in effective radius from the hammer pivot point. I am sure Tuner has also mentioned that before in his previous posts. This force initially acts like a stronger recoil spring except it diminishes as the slide moves further rearward. The hammer also exerts a vertical component which affect horizontal frictional forces on the slide.

There will be a point in time when the force on the slide has dropped to effectively zero due to the drop in pressure at the muzzle. I am not sure where that occurs but it will definitely be zero at the time the case is extracted from the chamber. Then the momentum will carry the slide back through the rest of the travel as stated in the posts.

Varmit Al's

Bob

Howdy Bob.

Understood about the ejecta following the bullet, but at 5 grains of mass... typical for a .45 powder charge... at such a relatively low pressure...the added push after bullet exit won't amount to enough to effectively measure. In an overbore bottleneck rifle case, it can be signifigant. Not in a straight walled pistol case with 5 grains of mass at 20,000 psi... and that's peak pressure. Not pressure at bullet exit.

If we assume that the residual gas velocity is even as high as 2,000 fps...5 grains at that velocity would present about 1/4 the recoil impulse of a standard velocity .22 Short. (29 grains at 1,000 fps) That level of punch added to a 14-ounce slide that's already gotten a full dose of momentum just won't mean much. The compressing recoil spring would more than cancel it out.

Several years ago, my father had a Llama "Baby 1911" in .22 LR caliber. I used to buy .22 Shorts because they were cheap...t his was in the early 60s... and sneak the little gun out for some after-school mischief. They wouldn't drive the slide far enough to even eject the cases. I had to pull the slide to lock on the empty magazine... insert a fresh round into the chamber... and drop the slide in order to ready the gun to fire again. The spring was pretty badly worn, and the slide probably didn't weigh more than 4 or 5 ounces, and it wouldn't move far enough to half-cock the hammer.

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The hammer also exerts a vertical component which affect horizontal frictional forces on the slide.

Not at the outset, when the slide is suffering its highest resistance from cocking the hammer. The hammer doesn't remain in contact with the center rail until the slide is on its way. It's not a smooth cocking action. The hammer is slammed backward, and loses contact with the slide. It hits the grip safety tang and rebounds back onto the slide at a point just forward of the center of the stop...normally at the junction of FP stop and center rail...but sometimes even further forward. After it comes back to rest, friction plays a role, as does any outside force that has an opportunity to slow the slide.

Again...The slide has move (nominally) 1/10th inch at bullet exit, so the CG hasn't changed a lot. By the time the slide has reached the extraction point, the bullet has hit a 25-yard target. All necessary delay on the slide...delay for safe breech opening...has long since ended, along with any deceleration due to the small radius. Once that delay/deceleration has been accomplished...there's no difference in the operation of the gun, aside from the reduced slide velocity and momentum.

Here's a very old X-ray... probably a fluroscope photo... of a 1911 pistol in motion about a millisecond before the bullet clears the muzzle. Note the position of the link, and the rear of the slide relative to the rear of the frame. Estimate length fo slide travel at probably 090 inch... or maybe less if this picture was taken before January 1918 when the Army changed the FP stop radius to 7/32nds.

If the gun's CG has changed, it's not enough to be of any consequence.

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That 14.0lb recoil spring seems really light to my string musician hands...

Mainly because you've gotten used to a 16-pounder. Browning's original spring wasn't 16 pounds. It was about 14.5 give or take an ounce. 32.75 coils of .043 diameter music wire. Wolff's 14-pound spring is 32 coils of the same diameter wire. The 16 pound spring is .0445 diameter.

And...

The recoil system is also a closed system, separate and apart from the main system... the barrel and breechblock. As it compresses, it imposes its own equal and opposite action/reaction. It pushes forward on the slide and backward on the frame. The stronger the spring, the harder it pushes at any given point in its compression.

The recoil impulse feels softer (to me) with the 15.0lb and 16.0lb springs. The 14.0lb recoil spring really feels like it is slamming the slide into the frame. The gun seems to have more torque (to me) with the 14.0lb Wolff spring too.

The gun feels very smooth with the 15.0lb x 25.0lb or 16.0lb x 25.0lb combination, not the sharp 'smack' I feel with the 14.0lb x 23.0lb combination.

Am I 'feeling' things?

I don't have any info on the pressures created by different powder, but the .45's as you say are lower pressure. I have been lucky enough to not have a bullet stuck in a barrel but if the barrel stays together when a second bullet is fired, how much recoil would occur, would it cycle the slide?

There is also sufficient velocity at the muzzle when the bullet exits to allow the use of a compensator, so wouldn't that also indicate sufficient energy to create a rearward motion without the comp?

You are obviously correct about the hammer. I was intending the reference to the vertical force applying after the slide moved rearward but wasn't clear on that. Also when the hammer bounces it will not be exerting force on the slide as you mentioned. That also takes the paint off the inside of my beavertail since I didn't bob the hammer enough

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I have been lucky enough to not have a bullet stuck in a barrel but if the barrel stays together when a second bullet is fired, how much recoil would occur, would it cycle the slide?

It depends, Bob. If the bullet travels deep enough into the barrel for the slide to move to the linkdown position, the squib will very often drive it far enough to chamber another round... also very often wothout the shooter knowing it. The second round bulges the barrel if it's carbon steel... and splits it if it's stainless. I've actually watched that happen... was looking straight at the gun when the squib ejected the case, and couldn't get the shooter's attention before she fired the "killing" round. Scratch one very expensive barrel installation. The slide didn't move on the second round. The bulge in the barrel locked it up tight.

Without the bullet's continued forward drag on the barrel... it just doesn't take a lot of force to drive the slide.

But... If the bullet is completely blocked by the stuck bullet... no. The slide won't move. It can't. If the bullet bears forward on the barrel while the equal/opposite force is pushing the slide... the forces are balanced and nothing can move. As you know... things move when forces become unbalanced. Balanced force is equilibrium.

The added resistance of the small radiused firing pin stop is very small compared to the delaying force that the bullet places on it. It adds little to the total... but it adds it at just the right time, and as an engineer, you're well aware that sometimes a small thing can make a big difference... and that while nothing is everything, everything is something. Even the mass of a drop of water on the slide will have an effect.

Another delaying factor that many don't consider is that of the barrel's mass. The barrel weighs about 3 ounces, and that's added to the total reciprocating mass... and it's part of until the barrel is completely disconnected from the slide. That's about 1/5th the slide's mass. That's significant.

Bob W - you don't have to apologize for anything - your thoughts are well presented and not the least bit offensive. Our Tuner has posted a convincing rejoinder - and anything that stimulates him to show us his Yoda qualities is always appreciated. It is through the exchange of ideas in this special cyber- space that we all learn and grow.

If you haven't availed yourself of the opportunity, there is a 50+ page thread in this section of the forum in which John (The Tuner) and some bright engineers like you thrashed the daylights out of the subject. The sequel to that thread now written by Tuner is to help us mere mortals understand "how" and "why" it feels different to shoot a 1911 equipped with a small radius FPS vs. the larger radius designed by Colt long after JMB was out of the picture.

It has been established to a final conclusion by anecdotal and visual evidence that the 1911 - once returned to the system configuration designed by JMB - has noticeably different and preferable recoil characteristics. That point is beyond argument. EGW'S sales volume of the part in question would probably be sufficient evidence by itself.

I am neither an engineer nor very smart in the math and science world - but I think Tuner's final example involving a wall and pushing on the wall takes care of the subject. If you believe the gases are generating an opposing force after the bullet leaves the barrel, against what are the gases pushing to generate an equal but opposite reaction? I've shot 30-06 blanks and felt very little, if any, recoil. So, as Tuner explained, the primary application of force is the effect of the barrel resisting the movement of the bullet - which ends when the bullet is gone (I hope I got that right).

Robert - really great pictures, and really clean work on the FPS - I am impressed! My work pales in comparison.

LJA... Quick physics course. Newton's laws of Motion and Conservation of Momentum is about all you need to understand it. Simplified form.

1A) Objects at rest will remain at rest unless an outside force compels them to move.

Newton was describing inertia.

1B) Objects in motion will remain in motion until an outside force compels them to stop.

Conservation of Momentum.

Skip 2. No real need since we're talkin' straight line motion.

3) For every action, there must be an equal and opposite reaction.

Here, Newton was telling us that force forward is force backward... that whatever force is applied in one direction must be applied in the opposite direction. If you push on it, you get pushed. If you pull on it, you get pulled.

Force moves things. Momentum keeps them moving. Inertia resists a compelling force.

Things only move if the forces... the compelling and the resistive forces... become unbalanced. That is, the compelling force is greater than the resistive force. If the forces are equal... balanced... there can be no motion. Balanced force means equilibrium. Applying 100 pounds of lifting force to a 100 pound weight doesn't move the weight. Applying 101 pounds does.

The faster you try to accelerate a mass, the harder it fights. You can lift a 25 pound weight off the floor at a rate of 1 foot per minute pretty easily. Try to lift it at a rate of one foot per second, and it gets more difficult. The weight/mass hasn't changed, but the force requirement to accelerate it does.

Weight and mass aren't the same. In outer space, a baseball and a 50-foot diameter boulder weigh the same. Nothing. You can move the baseball easily. Not so for the boulder.

Newton 3 applies in all directions, with all vectored forces. Vectored meaning that it has magnitude and it has direction. If you pull a box across the floor, the box feels a frictional resistance, and the floor feels the same frictional resistance... in the opposite direction. Same goes if you push it. Hence, force forward is force backward. Always.

Conservation of Momentum means just that. If you fire a bullet in space, away from atmospheric friction and the gravitational influence of another body, it will continue to travel at the exit speed forever... or until it strikes something or until it comes under the influence of gravity.

In reference to Bob's question on whether a fired bullet into a barrel blocked by a squib... would recoil occur:

Several years ago, a savvy old pistolsmith in Maryland decided to settle an argument over Kuhnhausen's "Balanced Thrust Vector" description of the 1911's function.

He lathe-turned a steel rod, with one end form-tunred to exactly match the hardball nose profile. He threaded the muzzle of a worn barrel, and used a set screw to place the rod in solid contact with a chambered round, so that any possible bullet movement would be absolutely blocked.

He then proceeded to fire the gun repeatedly to show that if the bullet doesn't move, neither will the slide. Nothing happened when the gun was fired. No kaboom and no slide movement. Nothing.

What he didn't understand... and probably still doesn't... is that in his attempt to disprove JK's "Balanced Thrust/Force Vector" he actually created one. The bullet pushed on the rod, which was attached to the barrel via the set screw. As the force pushed the barrel forward and the slide rearward, which were locked by the lugs and thus prevented from separating. The force was equal on both barrel and slide. Neither one moved because neither one could move. They were locked in a state of equilibrium. Balanced force.

He has stated often that the bullet provides the slide with momentum... but it doesn't.

Force and the motion that results from the force provide the slide with momentum.

Momentum is a function of mass times velocity. If it doesn't move, it doesn't have any momentum. It has inertia... resistance to compelling force.

The bullet is moving in the wrong direction to provide the slide with anything. The forward moving bullet can't make the slide move any more than you can push a lawn mower by turning around and running from it. What the bullet provides is a resistance for the force to push off of... and the slide provides a resistance for the force to push off of in driving the bullet. If you're suspended in mid-air and frog-kick your legs, you can't propel yourself forward. If your feet are placed against a wall... you can.

Action and reaction require two interacting objects and two forces. Remove one, and the system doesn't work. Or, another way... if the bullet is gone before the slide moves, it may as well have never been there, because one side of the equation is missing...the action side. Without action, there can be no reaction, and with balanced force, there can be no motion. If any part of the system is in motion, it's proof that the forces have become unbalanced at some point.

His assertion that: "If the bullet doesn't move, the slide won't move" is technically correct... but not for the reasons that he thinks. When the bullet moves forward... it's slipping through the barrel. When the bullet can slip forward, the barrel can slip backward. So, it's more accurate to say that if the BARREL can't move, the slide can't move. In his demonstration, the slide was mechanically blocked by the barrel and the bullet.

Incidentally, his demo also lends proof to my assertion that the bullet is the main delaying force in the design. In this, the bullet imposed the ultimate delay. It completely blocked rearward barrel movement... and the slide didn't move. Therefore, a frictional resistance placed on the barrel by the bullet would also be imposed on the slide for as long as the bullet is present.

Out of curiosity, what happened when the gun was fired in this manner? Where did the expanding gases go??

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Where did the expanding gases go?

They leaked out the breech area with a PSSSSSSSSSS.

Incidentally, after the demo, he removed the blocking mechanism, and fired the gun normally. It worked fine.

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Here's a very old X-ray...probably a fluroscope photo...

Is that the shooters hand in that ghostly image around the grip? How would they have had the shutter speed to catch the bullet in flight??? I have always been told this image depicted a static X-Ray of a barrel obstruction.?

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I have always been told this image depicted a static X-Ray of a barrel obstruction.?

No idea about the x-ray techniques used, but note the position of the barrel's radial locking lugs in the slide lug recesses.

While this could have been rigged, this position indacates the slide is in the process beginning to pull the barrel aftward.

In a static (unfired) mode, the barrel lugs would be at the back of the slide lug recesses.

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Is that the shooters hand in that ghostly image around the grip?

Yep... or rather the bones in the shooter's hand.

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How would they have had the shutter speed to catch the bullet in flight?

Dunno. It probably took several tries to catch it just right.

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I have always been told this image depicted a static X-Ray of a barrel obstruction.?

If there was an obstruction, it would show up as a solid, dark object... like the bullets in the remaining rounds in the magazine. The light-colored area behind the bullet is the gas plug that's pushing it. The "obstruction" theory was started by the people who buy into Kuhnhausen's "Balanced Thrust Vector" description in which the slide doesn't move until the bullet exits. Their minds are made up, you see... and they won't permit any factual evidence to interfere with that.

Plus... if the bullet had in fact stuck a solid obstruction, the slide and barrel would have been yanked forward, and the link wouldn't be in the position shown.

No. This pistol is firing... and this is a very old photo. Note that it's a pre-A1 model.

The photograph was probably done to settle an argument back then over whether or not the slide moves before the bullet exits. Newton says that it can't happen that way. That's all that's really needed to understand it.

To wit:

"For every action, there must be an equal and opposite reaction."

Equal being the operative word. Equal means equal... in every way. That includes setting the system in motion.

Note also the engagement of the upper lugs. The front faces of the barrel lugs are in contract with the rear faces of the slide's lugs. The slide is being driven rearward, and it's dragging the barrel with it.

The link is within a few thousandths of an inch of the start of the linkdown position.

This is how the short recoil design functions.

Disconnector is up.

Note also that... at the slide's position... it's just starting to encounter full resistance from the hammer via the FP stop. It hasn't cocked yet... but it's started to head in that direction... and the slide is still struggling to get past the speed bump.

Also of interest is the recoil system. Note that the spring will be completely encapsulated by the plug and the guide rod within another inch of slide travel... which means that the spring can't flex outward by the time the slide has moved less than half its full travel. Sorta makes the full-length guide rod "advantage" inconsequential.

From the position of the link I'd guess the slide has dragged the barrel aft about, ohhh, 0.06".

And look at all that terrible, nasty kinking going on in the recoil spring. Somebody ought to invent something like maybe a full length guide rod to prevent that, doncha think?

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Disconnector is up.

Yep. That means that... if the gun is mid-spec... the slide has only traveled about .050 inch. Disconnect is normally down at .060 inch, give or take a few thou. Evidence that the FP stop radius is 5/64ths... as per Browning's original design spec. It may be down far enough to have disconnected... but it doesn't appear to be.

Danny, mah fren... we is on the same wavelength today for sure. At .060 inch back... for all practical purposes a 16th...the disconnect has moved down far enough to prevent the gun firing out of battery. Nominally, that equates to about a 64th inch of vertical drop. I may be seeing it wrong, but it doesn't appear to be quite that far down. Of course, I'm pickin' at nits here. There's no way of knowing exactly how far it's moved, so I'll defer. The slide has moved about .060 inch. If that's accurate, the bullet will exit at roughly .075 inch of slide travel.

The point is that the barrel starts to hit the linkdown phase at 1/10th inch of rearward movement. In order for the lugs to horizontally disengage without damage... the bullet must be gone and the force relieved at some point before that.

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Evidence that the FP stop radius is 5/64ths

Wonder what my zero-radius FPS measures? It's just squared off.

You guys must have lots better eyesight than I if you can tell what the disconnector's doing. I defuzzified the original image a little bit. Didn't help me much, but others may benefit:

That 0.06" was just en eyeball estimate based on my calculations showing there's 0.048" 0.041" of barrel travel between when the link's vertical and battery. Kind of hard trying to get any good measurements for scaling those small pix when working through the monitor's CRT class - and bloodhot eyes.

I can't really make out the disconnector, but there are a lof of other clues like the grip safety being depressed. The position of the bullet and the position of the hammer explain why the breech in my 1911's stay so much cleaner with the early style FPS...Robert

And for those interested in getting some measurements, here's what that original image looks like after enlarging and with a double dose of defuzzification:

That's about all the defuzzifying that can be done before the image starts to get weird looking.

[Edited: Also note how the aft end of the bottom round is lifted up off the follower, putting itself and the rounds above in a nose-down attitude in relation to the follower top. This lack of nose support is why the top rounds in a full magazine hit farthest down on the frame ramp - and absolutely nothing can be done about it except make the cartidge column more vertical.]

I think the disconnector is up, based on its relation to the sear pin.

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Also note how the aft end of the bottom round is lifted up off the follower, putting itself and the rounds above in a nose-down attitude in relation to the follower top.

That's probably an inertial response to the gun moving in the opening stages of recoil. At this point, the gun hasn't moved much... but it has moved. Static, the rear of the round would be in contact.

Yes, while the top round already experiences friction with the slide, so it's right up against it...

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Static, the rear of the round would be in contact.

What what I thought up until a few years ago. Then I got curious about bullets nosediving down on the frame ramp (like Virgil Tripp) and found this:

The downward force on that bottom round is applied too far forward to keep it against the follower.

Think that was JMB's oversight?

Interesting. I'd think that rather than an oversight, it's something that just "is what it is" due to dimensional and geometric limits within the magazine.

rek... The top round has to be in contact with the center rail of the slide when the slide is forward. The holds the cartridge below the feed lips until it moves back and uncovers the mag well, letting the round pop up into feed position. If this didn't happen, the slide would ride over the top without contacting the rim.

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Interesting. I'd think that rather than an oversight, it's something that just "is what it is" due to dimensional and geometric limits within the magazine.

All that stuff is simply the unavoidable consequence of an angled magazine and a not-vertically-stacked cartridge column. A quick way to demonstrate it is to ask "If the bottom round in a full magazine was totally down against the follower, how could the cartridge under the top one possibly look like this (the one on the right)?"

I looked at my darlin' wife last night after reading Tuner's basic Newtonian physics course and said, "Honey, I now understand why when you shove me to get my butt off the couch you fall backwards." She didn't seem to understand the importance of that epiphany. Then, this morning I read the further elucidations of Tuner on the subject of balanced forces, and realized that I don't have to lose weight - she just needs to lose some arm strength and gain some weight to achieve the equilibrium of balanced forces. I successfully dodged the coffee mug that came flying at me (clearly the application of imbalanced forces) when I told her she needed to eat more and exercise less. At least, when she complies, she will quit trying and I can stay stretched out on the couch without interruption as I absorb the poetry and intellectual clarity of the writings to be found here.

As to the X-Ray - I put Niemi's de-fuzzified image on a projector in our conference room and blew it up to 4' by 6' - and observed the following:

The bullets appear to be unjacketed lead - which may or may not be relevant to the friction (speed forward) part of the equation. I reached that conclusion based on the opaqueness to x-ray of the bullets, and the absence of any image of a jacket around them.

It also appears that the shooter is holding the gun in his or her left hand - all I can see are the ends of fingers - not the finger bones themselves.

There is something weird about the trigger and trigger position. The trigger should be the long, solid steel version of the pre-A1 design but the X-ray shows something else. If it is the long version, it pretty clearly has been pulled all the way to the rear.

The name of the shooter is Vincent.

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rek...The top round has to be in contact with the center rail of the slide when the slide is forward. The holds the cartridge below the feed lips until it moves back and uncovers the mag well, letting the round pop up into feed position. If this didn't happen, the slide would ride over the top without contacting the rim.

I realize that, I wrote what I wrote above because the top round, unlike the rest of the rounds, is not dipping.

Just corrected a dimension in Post #38 and it suddenly dawned on me that while the barrel needs to move aft 0.0413" out of battery to get the link vertical, the slide has to move back 0.0125" before it can start pulling the barrel back with it.

This means when the link's vertical, the slide has come back 0.0413 + 0.0125 = 0.0538". But I figure the link in the x-ray is about 6° back from vertical which requires another 0.029" of travel for a total slide travel of about 0.054 + 0.029 = 0.083".

That 0.083" of slide travel is getting pretty close to the point where the link goes into tension and begins pulling the barrel down out of engagement with the slide.

C'mon bullet - get legs!

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That 0.083" of slide travel is getting pretty close to the point where the link goes into tension and begins pulling the barrel down out of engagement with the slide.

"Pretty close"?

Only said that because I couldn't find my notes & calculations at the time. Anyway, located the notes and "pretty close" works out to be 0.045" because the link goes into tension and begins to pull down the barrel after the slide's moved back 0.128" from battery.

Dan... at .128 inch of rearward movement, the link is actively pulling on the barrel.

It reaches the beginning stage of linkdown... just as the link gets all the slack removed and is experiencing a little tension... at about .110 inch or so. Depending on a few things, that could go as far a .120 inch, but .110 is pretty much the norm.

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Dan... at .128 inch of rearward movement, the link is actively pulling on the barrel.

That 0.128" calculation is of the slide's movement - not the barrel. The barrel's moved about 0.013" less than that, or about 0.115".

All those calculations were for the average Gov't Model assuming the barrel's in battery at the same elevation as when the link was vertical and in compression on the way into battery. In this case, the link's loose in battery.

But if the link happens to be in tension as the gun reaches battery, the barrel need only get moved back 0.0825" before the link goes back into tension. In this case the slide needs to move back only 0.0825 + 0.0125 = 0.0950".

And for all I know, in a "rattles when you shake it" Gov't Model, it may be anywhere between these extremes with every shot.

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And for all I know, in a "rattles when you shake it" Gov't Model, it may be anywhere between these extremes with every shot.

Yes, I have one of those that will still put rounds in the 8 ring at 50 yards...

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That 0.128" calculation is of the slide's movement - not the barrel.

Ah! Okay. I see it now. We've been on slide travel for so long, I couldn't get off that.

If the link is in tension when the gun's in-battery... somethin's wrong. The barrel will try to link down before the bullet gets gone. Bad JuJu.

Incidentally, the photograph that's been the subject of much discussion illustrates the possibible problems with the sawed-off variants. With an OM- length barrel, the bullet would have escaped by that point... and nearly so even with a Commander. Force is removed earlier, which doesn't give the slide that little extra push after the peak that the longer bullet dwell time with the 5-inch barrel provides. Coupled with the stiff spring that's necessary to give the less massive slide the forward momentum it needs to return to battery... and it's easy to see how much more critical the balance of spring rates becomes in the little chopped guns.

Explanation:

Bullet acceleration is the same as with the 5-inch barrel, up to the point of escape in the shorter barrel... so the forces acting on the slide don't differ. Only the total time that the force is applied changes... and it's not for the better as far as reliable function is concerned.

Provin' once again... The devil's in the details, and it's the little things that getcha.

Tuner - now I understand why my 4" Kimber developed failures to lock after I put in a small radius FPS. You have written about using a larger radius - up to a tenth - on Commanders and shorter guns - and I can go there, but my preedickament is how to get the full effect of the smaller radius - would going to a significantly lighter spring (say - 16lbs vs. the OEM 22lbs) make enough of a difference to restore proper slide function with the small radius, or do I have to bite the bullet and start increasing the radius until the gun works with a 16-18lb spring shortened to fit the gun?

I thought for sure you would ask me how I knew that the name of the shooter in the X-ray was Vincent.

LJ... Don't be overly concerned with the recoil spring after installing the FP stop. The "recoil" spring's primary function isn't in buffering the slide. It's to return the slide to battery. If your gun does that reliably with the lighter spring... and it should... you're golden. I run my LW Commanders with approximately a 5/64ths radius and a 16 pound recoil spring.

Since the slide is less massive, and the small radius on the FP stop has cut down on its rearward velocity and momentum, it won't hit the frame any harder than it did with the stiffer spring and the old stop... and maybe not as hard. If the rearward velocity is the same as a 5-inch slide... being less massive, it won't have as much momentum, and it won't hit as hard. That's part of the beauty of the small radius. It allows a lighter recoil spring without sacrificing the decelerating and impact buffering of a stronger spring. All that results in an overall softer, more friendly felt recoil because... Force forward equals force backward. The stiffer the recoil spring, the sharper the recoil.

I'd leave the radius just as it is.

So..okay Dan. How'd ya know?

I just got back on the computer and am catching up with all the postings. Wow, there have been a lot of very interesting things discussed in one day. Not sure I can pick out all the details that have been discussed on that x-ray photo. We can take x-rays like that at work but I think they might have a problem if I walked in with a pistol One thing about the barrel and the slide: if there is pressure in the barrel then the bullet and barrel will be forced to the right and the slide and case to the left causing the upper lugs to be in the position shown. That would occur regardless of thoughts about recoil or the position of the slide on the frame. This is not saying that there isn't movement, it just explains why the lugs are touching. A blocked bullet would cause the same condition of the lugs until the pressure bleeds off.

Just so you know, I am one of those that believes that a good portion of the recoil occurs after the bullet leaves the barrel. This also means I believe there is some recoil before the bullet exits. Each will have there own beliefs and likely not to change, but we can still shoot together

I was looking to see if I could find some unbiased research papers online but couldn't find anything one way or the other regarding the recoil action in a pistol. The link below is the closest thing that I could find that talks about Newton's laws since those have been discussed a number of times. This article desrcibes rocket propulsion, but that would be the same as the condition of the pistol just after the bullet leaves the barrel. The recoil is not caused by the gases 'pushing' on the bullet. There was a discussion earlier about some "balanced thrust vector" theory but I have no idea what that was about. I think some confusion comes about because the action can be described by using either the 2nd or 3rd law as mentioned in this link.

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Just so you know, I am one of those that believes that a good portion of the recoil occurs after the bullet leaves the barrel.

Bob... buddy... What is it about action and reaction you're not understanding?

Recoil can't occur after the bullet leaves the barrel. Aside from the gasses and particulate that follow the bullet out... which doesn't amount to much in the way of force... once the bullet is gone, the action side of the system is gone. With no action, there can be no reaction. The bullet may as well have never been there.

Recoil is nothing more than acceleration backward. Recoil isn't a force. It's the result of a force applied. Acceleration that sets an object in motion means that a force has been imposed on that object, and that force is great enough to overcome the object's resistance to movement...or the outside forces that are causing it to resist being set into motion. Unbalanced forces.

Once the force has been removed from the system, there can be no more acceleration. The only thing left is momentum. Can the bullet continue to accelerate after it's left the barrel? No. Is there anything acting on the bullet... any compelling force that helps it fight wind and gravity... besides its own momentum? No. The same applies to the slide.

Stand close to a wall with your hands against it at shoulder level. Push. While you're pushing ON the wall, the wall is pushing on YOU through the vectored force supplied by your arms. If you push hard enough to straighten your arms, to the point that your hands lose contact with the wall... you can't be pushed any more because the force has ended. You continue to move backward on momentum.

What you perceive as recoil in an autopistol comes from the slide's momentum as it compresses the spring and strikes the impact surface because there isn't a solid connection between the gun and the frame. By the time you actually feel it, the bullet has hit a 25-yard target.

The "gun" part of the autopistol is the barrel and slide. The frame is essentially just a gun mount, and the gun slides on rails.

Hypothetical pistol with 30-foot rails with ball bearings to minimize friction between slide and frame... no recoil spring... and fired electronically to eliminate the hammer and mainspring. You could fire the gun and feel nothing in the way of recoil.

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Recoil can't occur after the bullet leaves the barrel. Aside from the gasses and particulate that follow the bullet out... which doesn't amount to much in the way of force... once the bullet is gone, the action side of the system is gone. With no action, there can be no reaction. The bullet may as well have never been there

The slide is not "pushing" on the bullet and it does not require the bullet to be present to move rearward. You are forgetting the pressure inside the barrel is exerting pressure in every direction on everything. You used the example of pushing against the wall which excludes the pressurized gases. Maybe think of your situation where you have a hand grenade betwen you and the wall. When it goes off, does your presence have any impact on the wall?

Without getting into the physics it may be easier to use examples. If there are not any forces after the bullet leaves the barrel, how do you explain the effects of a compensator? There is a lot of pressure in the barrel at the instant the bullet leaves the barrel. Those gases certainly have the ability to produce a recoil or help minimize recoil with a compensator. Can we agree that there is a pressure existing in the barrel at the time the bullet leaves? True, once the bullet is gone it has no effect on the pistol and the pistol has no effect on the bullet. But that doesn't mean that all reactions cease to exist. You still have to account for the mass of the gas and it's velocity. Ignoring weight, would a short barrel pistol have more or less recoil? Which has the higher muzzle velocity?

There is some interaction at the time of separations due to the pressure but we will ignore those, but that is why the crown is important. That does not mean that all forces are gone, only the effects of the bullet. As the bullet is no longer there, the gases continue to escape just as in the exhaust of a rocket. In fact they could be considered exactly the same, as the gases and heat were created by a chemical reaction. So since there is no “bullet” in a rocket how does it manage to take-off? A pistol will be different from a rocket simply because it doesn't have a continuous fuel supply. The previous link had a good explantion of the physics of rockets.

Someone mentioned firing a blank round earlier. This can not be used as an example of the effects of the bullet caused recoil simply because without the bullet pressure can not develop. Just as you have very little pressure when you ignite a pile of powder on the floor but put it inside a grenade and it is a different story.

Here is a description of the action of the pistol when fired: At rest, the barrel lugs and slide are engaged on the forward edges of the slide due to the recoil spring. At the time of ignition the pressure begins to build inside the barrel. This pressure is equally applied against all surfaces inside that enclosed volume. The force on the back of bullet is defined by the pressure times the area of the bullet ( F=PxA ). The pressure develops over time and is not instant. Once the force on the back of the bullet exceeds the bore-to- bullet friction then it can start to accelerate forward while it has no clue as to what the rest of the system is doing. Toward the rear this same pressure is creating a force on the brass case according to the same formula. As the pressure builds the case starts to bulge against the chamber walls and the rearward force on the case increases until it exceeds the case-to-chamber friction at which time the case starts to move rearward. All evidence I have seen indicate the case-to-chamber friction is lower and therefore will allow the case to move first. The case will move backwards until the headspace is eliminated and then it will contact the slide. Up until now the closed system just included the barrel and contents. The pressure is pushing the slide aft via the case and the bullet will push the barrel forward until this combined motion causes the rear face of the lugs touch and lock up. Now we have created another closed system that includes the slide. In physics this would be modeled as a black box with a mass and a CG. There can be no movement unless there is an external force acting on this system. Looking at it from the outside world you would not even know what is going on. You can run around in a boat pushing on the sides and it just is not going to move! Now I should correct those statements just a little bit. There will in fact be some movement of the closed system because of the internal events that shifted the CG of the total system. That is the only way that the outside world would have even know anything occurred inside. In a closed system the total momentum must be zero as you have even stated. So if it is not that, what is moving the slide/barrel assembly? However the forces involved in the CG shift can be calculated as a shift in the system over the time involved, not an easy calculation. Imagine if it took the bullet two days to travel down the barrel instead of milliseconds the force exerted on the outside world would be different.

Now once the bullet has left it becomes an open system with pressure and mass inside the barrel. The barrel/slide assembly may now act just like a rocket or a balloon that you released and it moves reward. Since it has a limited amount of "fuel" the reward movement will soon decline until it stops. If you had a lot of slow burning powder you would still feel the force long after the bullet was gone.

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Recoil is nothing more than acceleration backward. Recoil isn't a force. It's the result of a force applied. Acceleration that sets an object in motion means that a force has been imposed on that object, and that force is great enough to overcome the object's resistance to movement... or the outside forces that are causing it to resist being set into motion. Unbalanced forces

An object accelerates as a result of a force, not the other way around. In a previous post you mention lifting a 25 pound weight with 26 pounds force. The forces as we speak of them are unbalanced and results in an acceleration upwards from the 1 pound difference. It is possible to apply a force and not have acceleration, such as pushing a lawn mower at a constant speed where you match the opposing frictional forces. Acceleration is defined as a change in velocity. And for those interested, the term ‘jerk’ (of course you may also be thinking that of me ) is defined as the change in acceleration, which actually plays an important part in felt recoil.

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Once the force has been removed from the system, there can be no more acceleration. The only thing left is momentum. Can the bullet continue to accelerate after it's left the barrel? No. Is there anything acting on the bullet... any compelling force that helps it fight wind and gravity... besides its own momentum? No. The same applies to the slide.

What force has been removed? The bullet has left and will not accelerate as you state, but you have not accounted for the remaining energy inside the barrel. Any movement in the slide prior to the exit of the bullet would be covered by the momentum part of the equation as you have stated but you still have to account for the additional effects of the remaining gases. Again those gases can provide significant forces. In a Barrett .50 those gases are said to reduce recoil by 65% through the use of a compensator. Having shot a Barrett I am glad it had one! Compensators are not 100% efficient (for one thing they still have a hole through them ) so the remaining gases are still contributing to the recoil.

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What you perceive as recoil in an autopistol comes from the slide's momentum as it compresses the spring and strikes the impact surface because there isn't a solid connection between the gun and the frame. By the time you actually feel it, the bullet has hit a 25-yard target.

The recoil in any pistol is the total of all the actions that cause the pistol to move backwards, the movement starts immediately after the trigger is pulled. If you mounted the pistol to an perfectly rigid object and then measured the reaction force with a strain gage against the mount you could see a plot of force against time. A semi-auto would show that the force is spread over a longer time and that results in a softer feel. If you had a revolver then the force would tend to spike and it would feel like a harder recoil. While looking at the force curve you would see that force starts to increase immediately regardless of the amount. As you have stated also, when there is a change in force there is acceleration and therefore movement.

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Hypothetical pistol with 30-foot rails with ball bearings to minimize friction between slide and frame... no recoil spring... and fired electronically to eliminate the hammer and mainspring. You could fire the gun and feel nothing in the way of recoil.

That is true if we define recoil as what is felt buy the shooter. Just don't stand behind the slide. But if you plugged the barrel the slide would not move much at all. What would be pushing it? Keep in mind that is completely closed.

Can we go shooting now? I think I would rather we go out so you can teach me how to shoot All this doesn't matter if I can't hit the target!!

Bob... First, the .50 caliber round has a lot more powder in it than a .45 ACP, and it's operating at a much higher pressure. I've already stated plainly that whatever gas and particulate ejecta that follows the bullet will continue to exert force on the breechblock. It's a mass It's being accelerated. It has to push in the other direction. The gasses generated in a .50 caliber have enough volume and enough mass to operate a compensator or muzzle brake. So, the .50 caliber point is apples to kumquats at best.

In a large volume bottleneck rifle case, that push can be signifigant. In a straight walled pistol case with a 5-grain powder charge, operating at 20,000 psi... it's not. The added push against the reaction side of the system would have about as much effect as slapping a Grand national car as it streaks past you on the backstretch at Talledaga. It's been affected by an outside force...but it doesn't change the outcome of the race.

Again... 5 grains at a very liberal estimated 2000 fps exit velocity would produce about a fourth the impulse of a standard velocity 22 short. (29 grains at 1,000 fps)

Since the pressure is low, and the bullet is subsonic... I would suspect that the exit velocity of the gas would be closer to the speed of sound... but I'll allude to 2000 fps just to give it the benefit of the doubt.

I also understand that the slide can be pushed without the bullet... and it will move... but it won't move much. The whole point is that action and reaction require two interacting objects with a compelling force between them. Once that force has been removed, there can be no further acceleration on either end. The remaining force inside the barrel at bullet exit will provide a little more push on the breechblock after bullet exit... but not much. If the slide doesn't move until after the bullet is gone... the slide won't move... at least not enough to notice, and it surely won't make the full trip rearward. It can't. There isn't enough force remaining.

Let's take it from the top.

The bullet exits at 1/10th inch of slide travel... or less. That's where the term "Short Recoil Operated" comes from. Actual recoil that results from the acceleration of bullet and slide only lasts for a short time and distance. The gun doesn't depend on anything "extra" that comes from the gaseous ejecta leaving behind the bullet because it's insignifigant. The force generated is too small to affect it to any detectable degree. For all practical purposes... and for the purposes of the gun's function... recoil ends when the bullet clears the muzzle. Whatever the magnitude of force that's imposed on the system after that is too small to make any difference.

Bullet and slide move at the same time, assuming zero headspace. If ther's headspace, bullet and case move at the same time... and when the few thousandths of headspace is filled... the slide starts its journey. So, "Most of the recoil" can't happen after the bullet exits. By the time you are actually able to detect recoil in the gun, the bullet is long gone... because the "gun" moves on a rail, connected to the frame by a spring. The gun moves very little until the slide hits the impact abutment in the frame, and that impact is what everybody thinks is the recoil... but it's not. It's the result of a moving mass hitting a solid object. BY the time the slide hits the frame, the bullet has struck a 25-yard target. I'd estimate that all remaining force within the barrel is well past gone by that point. No?

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Tuner - "gaseous ejecta" - that must be high-tech redneck talk for doing something impolite in mixed company. Gaseous ejecta through a variably constricting canine nozzle, aside from being noxious, probably does not make your pups move any faster, but it might make you recoil!

Said many times that if there's ever a market for dog hair and dog sh... uh... spoor... I'll be a tycoon. heh

I ain't no JMB. I've studied his pistol for goin' on 50 years, truin' to get into his head and crack his code...and it seems like somethin' new jumps out at me every day. One of those: "So THAT'S why he did that!" moments.

That said, I do understand simple Newtonian physics well enough to apply it to the short recoil system and know that 5 grains of gas and particulate ejecta that follows the bullet out the pipe just ain't gonna cycle the slide. It'd be like tryin' to drive a railroad spike with a tack hammer.

What's funny is that whenever I get involved in discussions on recoil and how it functions in the 1911... I can predict what the counter arguments will be almost verbatim. First, rifle calibers compensators are tossed in... then compensators... then rocket propulsion will eventually find its way into the debate.

This is partly why I've begun to distance myself from the forums lately. Weather and other circumstances provided me with a little time this past week.... but I almost wish I'd just played Scrabble instead.

JMB designed a combat weapon. One that works in any environment, works without being cleaned, works when dropped in the mud and most importantly, works when the targets shoot back.

I have one of Castillo's ball-bearing triggers in my Nornico range mule and it's a nice piece. It doesn't require fitting, which is nice, and it's well made and sturdy. Is it an improvement over JMB's design???? Nope, just easier to install.

I also have an Aftec Extractor that I bought because I wanted to experiment with one on my Nornico. Is it any better than JMB's original design??? Not really. It is more complex, requires more machining to produce, is more expensive and has tiny little springs to lose, but it really isn't any better.

In fact, it didn't have any influence over the extraction & ejection pattern in my Norinco whatsoever. It works just as good as the cheap Brownell's 'house brand' extractor that I put in it did, but I have noticed no changes with the Aftec, other than having to 'round-off' the right side of a perfectly good EGW FPS to install it. The Aftec does seem to be harder on the cases as I see a lot more brass in the gun now, but Aftec said that's the nature of the beast.

I have been on the range with another officer when he started having a FRTB with his Springfield Operator. A flick of the wrist would send the slide home, but the problem progressed. We cleaned it, lubed it with CLP, and it worked fine for another 150 to 200 rounds or so.

He called Springfield and spoke to a tech about it. He was told that the FRTB was due to debris affecting the close tolerances of the gun. The tech advised thay they don't recommend firing the Operator's more than 50-75 rounds without cleaning. Better than JMB's design??? Not at all. Not even close. In fact, the Springfield Operator doesn't group any better on a combat course than my 1927 Colt Hartford Argentine.

When I let him shoot my old Colt, he asked me, "What kind of ammo you shooting? That's some really low recoil stuff!" When I told him to put one of his Springfield magazines in my Colt with his own loads and try it, he just looked at me and said, "That's got less kick than my gun!"

Better than JMB...Not even close.

I have fired my 1927 Colt-Hartford Argentine over 1,000 consecutive rounds in a single session, without cleaning or re-oiling, to the point it was too hot to hold onto. I now have over 3,750 rounds, both factory and handloads, without a failure of any kind through a gun that is "too soft" and "wasn't designed to last more than a few thousand rounds."

A lot of people are trying to improve on what JMB did and it just isn't going to happen. As far as a combat weapon is concerned, one that you may have to bet your life or the life of someone else on, you cannot beat the original ordinance spec 1911...Robert

5th February 2010 Quote:
That said, I do understand simple Newtonian physics well enough to apply it to the short recoil system and know that 5 grains of gas and particulate ejecta that follows the bullet out the pipe just ain't gonna cycle the slide.

Well, I haven't done anything for 50 years, but I have argued physics with Nobel Prize winners, and they settle things like that with insults to each other's personal hygiene and broken beer bottles.

Wait, that's the guys at the biker bar. The physicists run the numbers.

5 grains of gas weighs (roughly) 1/40th as much as the bullet. For it to have a comparable effect, momentum-wise, as the bullet, it would need to be moving 40 times faster, or roughly 32,000 fps (assuming it all moves that fast, which of course wouldn't be true). For it to comprise 10% of the recoil force, it would need to be moving at 3,200 fps. Now, before the bullet leaves, it can't be moving faster than the bullet, so it would need to accelerate after the bullet exits. I'm not going to do the calculus to work out what the pressures would need to be, but color me skeptical.

Turning to rifles, I'm no expert on the 50 BMG, but my .270 loads involve about 50 grains of powder under a 130 grain bullet, which means that the gas alone could provide about 1/3 of the recoil even if it only moves exactly as fast as the bullet. It's a different beast altogether.

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5 grains of gas weighs (roughly) 1/40th as much as the bullet.

Recalculate that. For a 230 grain bullet... 1/47th. Don't forget to add the mass of the powder charge before the bullet exits. It's part of the total.

For centerfire rifles in the .223-.308 class, the gas velocity has been calculated at just under 4,000 fps for the brief instant after the bullet has cleared the muzzle and before it gets to one-half caliber forward of it. I have my doubts that it's that high, but I'll allude to it for the sake of argument. I'd also guess that the higher the bullet escape velocity, the higher the gas velocity by way of at least keeping up with the bullet. Let's split the difference and say that the gas velocity for that split nanosecond is halfway between the bullet velocity and your figures necessary for 10% of the impulse. That would place it at my first estimate of 2,000 fps assuming 800 fps bullet exit velocity... which I still have doubts about, given the low pressures... but it'll work.

I'll take a guess that it's very close to the 1/4th to 1/5th the impulse prvided by a standard velocity .22 Short. 29 grains at 1,000 fps. So, let's have a show of hands. Who thinks that a .22 CB Cap will drive 18 ounces... don't forget the barrel mass that's there for a 10th of an inch of travel... will drive the slide to full cycle against a 16 pound recoil spring and while compressing a 23-pound mainspring... even ignoring the hammer's mass. Anybody?

Remember the argument. It was stated that there was a belief that "most" of the recoil occurs after the bullet exits... even though Newton says it can't happ'n Cap'n, and in spite of the photograph showing evidence to the contrary.

And now. There isn't a muzzle obstruction in the picture. What would be the point of setting up a complicated experiment/demo like that... with an obstruction in the bore? To capture the instant the gun blows up? Naw. That dog won't hunt.

I actually do have a substantive question for you - when the Army requested the radius change in the FPS in 1918 or 1919, was there any more to the story than alleged difficulty in racking the slide with the hammer down? Seems like there would have to be more - since my wife has no trouble racking the slides on her 1911'S that I have put small radius FPS parts in. I can't imagine it was that tough for our doughboys to operate. Is it possible that the Army wanted higher slide velocity and stronger recoil springs to assist in ejecting and chambering rounds and to minimize the effects of spring wear? Gotta be something more to it - or the Army was just plain dumb.

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when the Army requested the radius change in the FPS in 1918 or 1919, was there any more to the story than alleged difficulty in racking the slide with the hammer down?

Nah. It was due to the complaints of hard manual operation. Soldiers, being preoccupied with surviving, were skittish if anything that slowed getting the weapon ready to go... even for a split second. They learned that split seconds could mean the difference between living and dying... so the change was made.

And never forget that Military Intelligence is an oxymoron. "There ain't any reason for it. It's just policy."

Whenever I get into a brawl over short recoil operated pistols such as the 1911, I've found that many physicists and mathematicians nit pick the insignifigant things... such as the resudial pressure within the barrel issue. I know about that, and have acknowledged it many times in the past during these debates. The whole point that I try to make is that it doesn't mean anything in the practical or mechanical sense in the function of the pistol. The force generated by such is so small that it can quite literally be ignored...because it doesn't make or break the function. It's like takin' a leak in Lake Erie tryin' to change the water's salinity.

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...when the Army requested the radius change in the FPS in 1918 or 1919, was there any more to the story than alleged difficulty in racking the slide with the hammer down?

Maybe the fact that some of that slide racking was being done on horseback in less than favorable conditions?

Even More Evidence re: 4" 1911's

Haooy to report that the 4" Kimber's mysterious refusal to lock-back with all slide locks save and except for the OEM Kimber part has been resolved - Thanks Tuner - you were right on the money again!

The stock action spring weight is 22lbs - the Wolff XP spring made for the 4" Kimbers is also rated nominally at approximately that weight but is a few coils longer than the stock spring. which, for reasons unkown to most of us, is a Colt Officer's Spring and wears out rather quickly.

I replaced the Wolff XP spring with an 18lb Commander spring with 2.5 coils clipped off. No more mysteries or problems. So, I now have all the benefits of the FPS AND a lighter spring with perfect function. What was vexing me before was that the gun would lock by manually cycling the slide, but not after the last round was fired. So, after about thirty rounds at the range today with four different mags and four different slide locks, I can report that if the slide locks by manually cycling it, it will lock after the last round is fired. One slide lock (Ed Brown) would not lock the slide back manually or by firing the last round with one magazine (Checkmate 7rd), but the same mag locked the gun with the other three slide locks.

I think the gun would work fine with a 16lb spring, but the ejection pattern was fine with the 18.

The relevance of this post is to let other owners of 4" Kimbers and similar guns know that the small radius FPS is great and works to re-shape the recoil of the gun, but a lower spring rate is required. In this instance, the radius on the FPS is quite small to maximize hammer spring delay. George

If you wanna feel the recoil really get tame, try the 16-pound spring. Force forward equals force backward.

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You mean like 18# of force springing forward pushes back at the rate of 18#?

Yep. To simplify Newton's 3rd law:

In a closed system wherein there are two interracting objects and a common force...whatever the magnitude of the force is in one direction, the magnitude of that force will be equal in the opposite direction. Actually, there are always two objects and two forces, even though the forces originate from a single source.

Using the "pound" rating for a spring is a little misleading, because it only gives one piece of information during one point in its function... and that only for a brief instant... most often at its minimum compressed length as installed. Another way...

The spring provides a maximum of 16 pounds of resistance when compressed as far as it can be within this machine. Not at the point of the coils touching... but at its minimum compressed length as installed in the gun.

The other, more pertinent information isn't given... that concerning the rate of the spring. The rate determines how fast the spring builds resistance or unleashes its stored energy... but people understand and readily accept "pounds." If the label on the spring were to state:

"This spring provides X pounds/ounces of resistance per Y units of compression for a maximum loading of Z at length abc... the phones wouldn't stop ringing long enough for the manufacturer to make springs. Pounds, they understand.

Here's a hypothetical for you to study on: Understand first that perfection only exists in the mind of God, and that there are no perfect dimensions.

Imagine a double-end cannon, with a muzzle at each end, and the firing chamber located precisely in the geographic center of that barrel.

The bore is perfectly round, concentric, and precisely equal diameter from one end to the other. The surface finish is also perfectly smooth and consistent over its length.

Two cannon balls... also perfectly round, precisely the same diameter, and precisely of equal surface finish, and of precisely equal mass/weight.

The balls are loaded in the barrel precisely equidistant from their respective muzzles, and the gun fired.

The balls will accelerate at equal rates, and they will exit their respective muzzles at precisely the same instant... at perfectly equal but OPPOSITE velocities.

They'll exit at the same speed. "Velocity" indicates direction... like "Vectored" indicates that the force has magnitude and direction.

"Outside Force" is any force that exists outside of the closed system, and it suggests a resistance to motion on that system. Many things can comprise an outside force, but the two constants are friction and gravity. As long as the system operates on planet Earth, those are the two that you can't get around. They're always there...always lurking...and always working to bring any moving object to rest and back to a state of equilibrium.

Now, to simplify Newton's 1st Law that states:

An object in a constant state of motion will remain in that state UNLESS AND UNTIL an outside force compels it to change. An object at rest...in a state of equilibrium... will not move unless it's forced to move. An object in motion will not stop until it's forced to stop. Inertia and momentum.

The bullet from a gun fired in the vacuum of outer space will continue to travel in a straight line at its exit velocity forever... unless it encounters another object or falls under the effects of gravity. Conservation of Momentum.

Moreover... Motion is the result of unbalanced forces. That is... the compelling force must be great enough to overcome the object's inertia, and the outside force of friction before it can set the object in motion. If the forces are balanced... compelling and resistive forces are equal... the object is in a state of equilibrium, and will not move.

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If you wanna feel the recoil really get tame, try the 16-pound spring. Force forward equals force backward.

Are you saying less recoil with heavier springs? And, less 'dip' going back into battery with lighter springs??? And 14.0lb x 23.0lb is optimum overall for a GM??

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Are you saying less recoil with heavier springs?

yes less dip going back into battery with lighter

recoil spring(slide doesn't slam home as hard causing muzzle to dip).

Yes...14#x23# is optimum WITH small radius FPS.

Thanks Tuner - and yes that helps a lot. I dig it when your mind catches on fire and you write your Physics 101 explanations and examples - and in a way where I understand the concept.

It's funny - I read a lot of older posts that you put up (particularly in the 2006 53 page thread on the small radius FPS) - and you mentioned "force forward = force backwards" more than once.

It was not until my awareness got stretched enough this go-round (and with your first three posts in the "More Evidence" thread) that I actually comprehended that a heavier spring produced greater recoil. I had always thought the opposite.

And now, with your clarification of spring rate over time, I understand that most of the backwards force (equal and opposite reaction) occurs during the early part of the "un-coiling" of the spring, followed by the slide slamming home, which transmits another force - to which there must be an equal and opposite reaction (muzzle dip) as their is when the slide hits the frame abutment in recoil (muzzle flip). I think both of these forces must be vectored in some way to produce upward and downward motion - is that right?

This is fun Tuner - I dig it.