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Thread: Explanation of the operation of EGW Firing Pin Stop

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  1. #11
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    LJA,

    Do a search on You Tube for '10851Man' and I have a video posted of a rapid-fire, 4 round string from my 1927 Colt-Hartford Argentine with the EGW FPS with factory 230 grain FMJ...Robert

  2. #12
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    Great video Robert! The pistol is way cool as well. Good demo of the effects of the small radius FPS. After watching your video and some others, it does seem that the small radius FPS does magnify the cyclic differences and criticality between the original JMB 5" design and shorter slide models that came along later. Regards, George
    "Today, we need a nation of Minutemen, citizens who are not only prepared to take arms, but citizens who regard the preservation of freedom as the basic purpose of their daily life and who are willing to consciously work and sacrifice for that freedom." -- President John F. Kennedy

  3. #13
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    George,

    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...Robert





    Last edited by 10851Man; 2nd February 2010 at 09:06. Reason: Add Images:


  4. #14
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    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:

    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

  5. #15
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    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...this 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.



    Quote Originally Posted by Bob_W
    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.

    Cheers
    Last edited by 1911Tuner; 2nd February 2010 at 20:10.


  6. #16
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    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.


  7. #17
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    Tuner,

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

  8. #18
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    Quote Originally Posted by 10851Man
    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.
    Last edited by 1911Tuner; 2nd February 2010 at 20:45.


  9. #19
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    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????
    Last edited by 10851Man; 2nd February 2010 at 20:46.


  10. #20
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    Hi Tuner

    Man, you type fast

    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

    Bob

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