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Hola amigos!! Here I am with another long technical post, I hope you will find it interesting enough to read it to the end. I would like to review the 1911 (or any short recoil pistol) dynamics from a basic physics viewpoint. Many parts are merely a rewriting of the excellent articles published by John Bercovitz at: http://yarchive.net/gun/pistol/1911_dynamics.html

The first thing we have to consider is a review of impulse-momentum, they don’t call them “recoil operated” for nothing!!
Starting with the familiar equation, F = ma:
F = ma = m dv/dt.
Rearranging: F dt = m dv. [More exactly, if F is varying: F(t) dt = m dv]
The integral of the term on the left is impulse and the integral of the term on the right is momentum. These equations are for linear action but equivalent equations exist for the angular case.
If you've ever seen stop-action photos or slo-motion films of 1911s at the process of firing, you know that during the initial stages of the slide's opening, the firearm appears to be still pointed at the target: no angular deflection is obvious though some small amount must exist. This is evidence that the first part of the cycle can be described by the linear impulse-momentum equation given above.
To make things simpler, let's ignore for a moment the force of the recoil spring and the effects of friction, hammer mass, main spring, etc., and let’s just concentrate in be balance of momentum of bullet and powder vs slide and barrel.
The force at any given time on the back of the bullet is the chamber pressure at that time multiplied by the cross sectional area of the bullet. The force at any given time on the slide is exactly equal to the force on the bullet as long as the action is locked. So integral F dt, the impulse, is the same in magnitude but opposite in sense for the bullet and the slide.
Since at any given time the magnitude of the impulse is the same for bullet and slide, the magnitudes of the momenta of the bullet and slide are also equal. Lacking subscripts, let's call M and V the mass and velocity of the slide and call m and v the mass and velocity of the bullet. It's just been shown that:
M dV = m dv
or: M dS/dt = m ds/dt
Integrating this equation with respect to time:
M S = m s for any period of time you care to name. Let's take the period of time that starts when the bullet starts to move and ends when the bullet clears the barrel:
The mass of the slide and other moving parts (when barrel and slide are locked together) is about 17 ounces. The mass of our typical bullet is 230 grains (0.526 ounces). In a government model the travel of the bullet until it clears the barrel is around 4.4”. So how far does the slide/barrel assembly travel to the rear while the bullet travels to the end of the barrel?
It would travel 0.526 / 17*4.4 = 0.136 inches. (To be more exact, one should take into account that the barrel backs off the bullet a little, decreasing the distance the bullet travels to get to the end of the barrel.) Simplistically speaking, after the slide/barrel assembly in the government model travels roughly 0.09 inch it starts to unlock. Unlocking is completed after roughly 0.2 inch of travel.

On a sidenote, of course these numbers depend on the built-in geometry of the gun: how the barrel is fitted, swing link, top lugs engagement, etc. If the barrel is not locked up at least .045" in the slide locking lugs the barrel can separate from the slide too quickly. Also in the 1911 there is a .020 (or more) flat on the barrel lower lug link ramp area (connecting arcs) that allows for a bit of rearward movement of the slide before the barrel starts to unlock. The downward movement of the barrel relative to the slide is controlled by the link, and it swings in an arc. The link is actually slightly past Top Dead Center when in battery. This link is swinging through top dead center and initially the vertical movement is essentially zero. All this means that for all practical purposes the barrel does not move much vertically (unlock) for the initial rearward motion of the slide.
In other Browning derivative pistols a curved or straight ramp cams the barrel down. The swinging link design is the stroke of a genius, but is does somewhat limits the lock time compared to a ramp (linkless) design. If you use a ramp to unlock the barrel (instead of the 1911's link), you can increase lock time and "tailor" it to your needs. On the other hand, some link less barrels have a greater unlocking angle which makes the barrel drop away quicker than the 1911 barrel; and in any case, if you have your unlock timing right then there should be no problems with either system. The principle is the same, and you can establish the exact amount the slide and barrel have to move back before unlocking.

So in a properly timed pistol the bullet would have left before unlocking is complete even if we don’t have any recoil spring or hammer to further arrest the slide backward movement. Remember our discussion here in 1911forum.com regarding “firing pin drag marks”, and how Ned Christiansen even had the guts to try this on his own gun?

In this simplified model, it has been shown that the distance the slide travels until the bullet exits depends only on the bullet's mass and is totally independent of the bullet's exit velocity. This is not true in the real world, because the force of the recoil spring, the effects of friction everywhere, hammer mass, main spring, etc.also contribute to keep the slide and barrel locked, and they do feel any extra amount of recoil energy.
It is amusing to extrapolate the consequences of the equation, MS = ms, further. If we have a gun in the outer space and this gun has its barrel's centerline going through the center of gravity of the gun, after we fire the gun, M*S will be equal to m*s for eons before anything disturbs either component. Another way of saying the same thing is that at any time after firing, the center of gravity of the two components considered as a system remains at the same point in space it occupied prior to firing.

We have showed previously that the force exerted by a spring is kX and that the energy absorbed by the spring is 1/2 k ((X2^2)- (X1^2)).
Kinetic energy of a moving mass is: KE = 1/2 m V^2.
This is the same formula that gives bullet energy: bullet energy = mass of bullet times bullet velocity squared times 2.22 divided by 10^6 where bullet mass is in grains, bullet velocity is in feet per second and bullet energy is in foot-pounds.
Now for a specific example: The moving mass (slide, barrel, etc) of a typical 1911 weighs 17 ounces (1.06 lb.) of which 3.25 ounces is barrel.
The recoil spring weighs very little, about 1/4 ounce, so I can ignore it, but technically one should add about half of its mass to that of the rest of the moving mass depending on the dynamics of the situation. One should also take into accout the polar moment of inertia of the hammer (we are not moving it straight back but rather rotating it) and the mass of the case, but this is getting complicated and there are still many more factors into play than just momentum equilibrium (springs, friction everywhere), so we’ll just continue and discuss it later.

Total capacity for absorbing energy is a linear spring is: = 1/2 k ((X2^2)- (X1^2)) ,
equal to 26.1 in-lbs. for our 16 lb. Spring.

If our 230 grain (0.526 oz) bullet leaves the barrel at 830 feet/second, then the velocity of the moving mass in the opposite direction, by MV = mv, is 25.7 fps (308 inches per second). (830fps*0.526oz/17oz = 25.7fps)

In a previous post regarding “initial 1911 slide velocity” I mentioned this very simple calculations, and added that due to the effects of friction, hammer, main spring, recoil spring, etc. probably the initial slide velocity of a 1911 (this is, the approximate velocity at the moment of unlocking, just after the bullet leaves the barrel) this velocity should be somewhat smaller, say about 20 fps. This should also be the maximum slide velocity in the entire cycle, because after the bullet and powder gasses leave the barrel there are not other forces that accelerate the slide back, only the other forcess decelerating it.

Of course everybody laughed, we all know that 20 fps is only 13.6 mph and most people can run faster than that!! Surely the slide velocity is MUCH faster!
But wait, this humble 20 fps is more than enough to cycle the gun in a very short time, after all the rearward travel is only about 1.6”, and even with a 10 fps average slide velocity the time would be 0.0133 seconds until stopping, and then say triple this time, about 0.0399 seconds for the returning stroke and we have a total cycle time of about 0.0533 seconds. I’m sure most of you have seen before a 1911 doubling or even going full auto, a very common problem if you get too carried away with trigger tuning. A small submachine gun (blowback operated, but with little bolt mass, like the Ingram) or a machine pistol shooting full auto fire at rates of aproximately 1000 rounds per second, for a cycle time of 0.06 seconds, so we are confortably in the ballpark.

Taking our guesstime of 20 fps initial slide and barrel velocity, the kinetic energy of the moving mass, by KE = 1/2 m V^2, is 6.59 ft-lb or 79.1 in-lb.

(1/2 * (1.06/32.174) * 20.0^2 = 6.59 ft-lb)

Let’s say during the time the bullet travels down the barrel the slide moves back about 0.11" (we calculated 0.136” just taking into accout the mass of slide and barrel, so we’ll guesstimate again 0.11”). By the potential energy formula for springs we see that the energy absorbed by the spring during this period is 1/2*3.23*(((2.933+0.11)^2)- (2.933^2)) or 1.06 in-lb. Comparing this with the 79.1 in-lb the slide carries, I think we can say that the recoil spring is not holding the slide locked to any large degree.

The force pushing the slide back varies quite a bit since the chamber pressure rises, peaks and falls. The peak pressure is around 14,000 psi in a typical load. The peak force, then, is the area of the bullet's base times the peak pressure: 0.16 square inches times 14000 psi = 2240 pounds. The same magnitude of force acts on bullet and slide. The chamber pressure peaks very soon, about at the moment the bullets leaves the case, and the pressure drops very rapidly, being much lower at the moment the bullet leaves the muzzle.

So how much force is the recoil spring exerting against the slide at the moment of firing?

(Average F) = k*(average x) = 3.23 lb/in * 2.988" = 9.65 pounds.

This is not much, and this is why the gun recoils very little at the moment the bullet leaves the barrel. More important, this 9.65 lbs. are not going to resist much of the backward thrust if the barrel and slide are not locked together!! Even a 10% of the peak pressure is exerting a force of 224 pounds against the breechface.

After the slide travels a short distance, the barrel drops out of it, impacting the frame. Though the coefficient of restitution of steel is high, eventually this energy is burned up in multiple collisions with frame and slide (equivalent of totally inelastic collision). By the principle of conservation of momentum, the total momentum of the system (frame, slide, and barrel) must be the same before and after the barrel's multiple collisions with frame and slide. Put another way, the momentum the barrel carries as a component will be found in the system after all is said and done, and this momentum of the barrel is transferred to the hands via the frame.

The remaining mass (17 - 3.25 = 13.75 ounces) is the slide alone, and continues traveling at a little less than the original 20 fps. By KE = 1/2 m v^2, this mass is carrying 64 in-lbs of kinetic energy. At the time the remaining mass impacts the frame, the kinetic energy is 64 – 26.1 = 37.9 in-lbs. (Remember the recoil spring absorbs 26.1 in-lbs.) Running the equation for KE in reverse, we find that the remaining velocity at impact is 15 fps or 184 inches per second.

All this assumes that the frame doesn't move during the interim under the influence of the recoil spring, which is not true, but is pretty close. Almost all of the tilting back of the pistol occurs at the time the slide runs into the frame.

Once again, all these calculations are obviously not exact because we are disregarding the effects of friction, hammer and main spring, how rigid is the pistol mount (or grip), etc. But if we could measure the slide velocity in the real world and then compare the results with these crude calculations we could validate them if the correlation is good. Knowing that many of you won’t trust the results of this mathematical model, this is just what I did, but let’s leave that for another post.
 

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Interesting...hopefully tirofijo is still around to help me out on the impulse problem.

I'm guessing the impulse the slide imparts on the frame causes majority of the flip we see in recoil? Meaning keeping the momentum the same, a system with a heavier slide will push on the frame more at a longer time than a lighter slide?

So, if I were to lighten the weight of the slide, will this equate to a lesser impulse imparted on the frame (even with the increased speed)? Or is it the same impulse but imparted at a shorter time? :scratch:
 

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Dynamics....wow

YOU GOT WAY TOO MUCH TIME ON YOUR HANDS...PERHAPS YOU SHOULD GET A HOBBIE! :rofl: :rofl:
 

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Discussion Starter #5
Hi, the recoil impulse is always the same.
Sorry, I had a longer response but the new forum software is giving me a lot of problems.
 

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So your saying that in outer space the 1911 can cause cosmic strings and if we sat on top of the slide as it cycles we would essentially have a time machine? Huh?
 

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TiroFijo said:
Hi, the recoil impulse is always the same.
Sorry, I had a longer response but the new forum software is giving me a lot of problems.
Best technical post I have seen on this forum. However, I will disagree with you on one very tiny-minor detail. You mention that you should have accounted for the mass of the spring, but since it was small you could ignor it. I agree with ignoring the mass of the spring, but for a different reason. The mass of the spring is accounted for in the spring rate term (k). If you added it back in, you would be double counting it. See I told you it was a minor detail :rock:
 

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cra955,

Muzzle flip is actually a result of torque (a simple explanation). Torque is measured by a perpendicular force (vector) imparted a certain distance away from a pivot point.
- Translated into the 1911, the pivot point will be your hand/grip/wrist).

The closer the force is to the pivot point, the less torque you will get.
- For the 1911, the force is being imparted on the plane of the barrel/bullet. Note that the higher you can put the pivot point, the less torque will occur meaning the higher the grip on the pistol, the less muzzle flip you will feel.
 

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So does all this mean that a 16 pound recoil spring combined with a 23 pound mainspring and a firing pin stop with a small radius on the bottom edge is what is required to properly time a 5 inch barreled Government model firing a 230 grain bullet at a nominal 850 feet per second?

J.M. Browning was a real firearms genius and he did not even have a pocket calculator! :)
 

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TiroFijo,

I showed your post to the guy behind the counter at a gun store. After purusing it for 30.28 seconds (P), he screamed, ran a distance of 50.97 feet(D), and dropped dead from a cerebral hemorrhage. Assuming a negative correlation between the time spent reading and the distance traversed before onset of the cerebral hemorrhage, and excluding factors related to the subject's general health, physical condition and the possible effects of screaming, then the fraction D/P would represent the distance the subject could run after reading your post before onset of cerebral hemorrhage, not taking into account the subject's momentum. Accordingly, by my computations, each additional second of reading time would reduce the distance traversed by 1.683 feet. This would further lead to the conclusion that after 60.56 seconds, the subject would be unable to run due to the onset of cerebral hemorrhage before forward motion could be initiated.

Regards,

Dr. Irwin Cory :D
 

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bud41 said:
HJK!!!!

:biglaugh:

also no flame to first poster, I just never liked trig and phyics!

:barf:
Thanks HJK. That's the best laugh I've had in a long, long time. And I must admit, by the time I finished Tiro's post, I forgot the earlier parts of it. But then, if my algebra teacher in high school hadn't been my fishing parther, I'd have flunked the course!

Bob
 

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I've said it before and I'll say it again: Dang Paraguayan rocket scientist!

I actually did make it through the post and it all seems sound to me. Tiro, this should be easy for you: What then is the velocity of the barrel, along its unlocking arc (dictated by the linK), when it impacts the frame? I want to know the down velocity, the rearward velocity, and the actual radial velocity, given the nominal 20 FPS slide velocity, which probably at the moment of unlocking has not yet been attained, right?

Great post! I was just down you way and shot a match in Venezuela, wish I'da thought of inviting you
 

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Discussion Starter #18
Yes Ned, cra955 resurrected an OLD thread (10/30/2001) :)
80% of the people were (and still are?) :rolleyes: ??, laughing :rofl: , :barf:, sleeping or :hrm:
and the other 20% were fighting over this trivia.

If the slide/barrel are travelling back at 20 fps while unlocking, the velocity of the barrel at the moment it impacts the frame is: 20 fps minus some small velocity loss due to friction everywhere, the compression of the recoil and main springs, and the flex of the grip.
The unlocking is pretty abrupt, occurs in a very small linear displacement of the slide. Unlocking starts after about 0.09"-0.10" of rearward displacement, is completed after about 0.20", and the lower lugs hit the frame after about 0.27"-0.28" (...these are just my crude, "ballpark" measurements).
Since bullet exit is at about 0.13" of rear travel and our maximum slide/barrel velocity is about 20 fps a little after this due to the contribution of the muzzle blast, it should be pretty much the same after when the lower lugs hit the frame since there's only 0.15" more of linear displacement.
Since there are no other forces acting on the system (the link pulls down the barrel as a reaction to its movement), the barrel lower lugs will be moving at 20 fps (radial velocity, tangent to the arc). I don't remember the link angle at the lowermost position, but I'm sure your can measure it. The down velocity would be: 20fps*sin angle; and the rear velocity: 20fps*cos angle. Remember these are just the velocity components of the barrel feet/chamber area, most of the mass of the barrel does not follow the same arc.
 

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Discussion Starter #19
And since we're having so much fun :hrm: let's calculate the torque due to the angular acceleration of the bullet by the barrel rifling:

Net torque = moment of inertia * angular acceleration

The polar moment of inertia for a cylinder (pretty close for a handgun bullet) is:
Ix = 0.5 * bullet mass (lbs) * bullet radius^2 (feet)
Ix = 0.5*230/(7000*32.2)*(0.451/24)^2 = 1.8^-7 foot-pounds

The actual number would be a tiny bit smaller, since it is a RN bullet and the jacket is of a lighter material.

This bullet accelerates to 830 fps in about 1.0 miliseconds.
If the barrel twist is 1-16", then rotational velocity at the muzzle is 622.5 revolutions per second (rps) = 37350 rpm = 3911 rad/second.

Assuming a constant acceleration the average angular acceleration would be:
3911 / 0.001 = 3,911,000 rad * s^(-2)

And average torque would be:
1.8^-7 foot-pounds * 3,911,000 rad * s^(-2) = 0.704 foot-pounds
This is 8.45 inch-pounds

The peak torque would be much larger because acceleration is much larger in the first few inches of bullet travel.
 

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shimbii said:
cra955,

Muzzle flip is actually a result of torque (a simple explanation). Torque is measured by a perpendicular force (vector) imparted a certain distance away from a pivot point.
- Translated into the 1911, the pivot point will be your hand/grip/wrist).

The closer the force is to the pivot point, the less torque you will get.
- For the 1911, the force is being imparted on the plane of the barrel/bullet. Note that the higher you can put the pivot point, the less torque will occur meaning the higher the grip on the pistol, the less muzzle flip you will feel.
Shimbii, yes this part I can understand.

However, I'm interested in the amount of force the slide weight plays when it acts on the frame causing the torquing on the hands. Since Tirofijo responded that the impulse is the same, I'm guessing with the lightened slide, it will be acting at a a shorter time...?

If that is the case, it's actually good for most shooters because they'll need lesser force to time the recoil and bring the sights back on target. Hopefully in 2 weeks time I'l have an answer to my question once I get my slide back from my smith.
 
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