From: bercov@bevb.bev.lbl.gov (John Bercovitz) Newsgroups: rec.guns Subject: Colt GM dynamics Keywords: physics, Colt Date: 15 Aug 90 01:04:50 GMT GM Dynamics As a result of recent discussion on the net, several people have responded privately with the request that I post an analysis of the dynamics of the government model on the net to serve as a basis for further discussion and development (a target to shoot at). I'll do the best I can but it is hoped that folks who really "know" will eventually respond and then correct and expand upon this attempt. Are folks at Aberdeen, Quantico, etc. on the net? In an effort to not exclude any potential participants, I will try to write thorough explanations. I'll assume all have had high school physics and math. These explanations may necessarily run to some length so that the article will best be broken up into several parts. Do feel free to jump in at any time. Need I have mentioned that? Part one: The first thing to get squared away is a review of impulse-momentum. 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 of government models in 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. For this first article, let's ignore the force of the recoil spring and the mass of the propellant. 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 even in the case of a bottle-necked cartridge 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 18 ounces or 510 grams. Let's say the mass of the bullet is 200 grains or 13 grams. In a government model the travel of the bullet until it clears the barrel is around 4.4 inches or 110 millimeters. 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 13/510*110 = 2.8 mm or 0.11 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 (2.3 mm) it starts to unlock. Unlocking is completed after roughly 0.2 inch (5.1mm) of travel. So the bullet has left before unlocking is complete. For this simple 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. It is amusing to extrapolate the consequences of the equation, MS = ms, further. If we have a gun in some void in intergalactic 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. * Why the force on the slide is equal to the force on the bullet even when a bottle-necked cartridge is used: Visualize the area of the bore projected axially onto the interior of the cartridge case head. The remaining annular area inside the cartridge case head is compensated by a similar annular area (more or less the shoulder of the case) around the neck end of the cartridge case. Since these two annular areas are equal, the pressure- induced forces on them oppose and cancel out. JHBercovitz@lbl.gov

From: bercov@bevb.bev.lbl.gov (John Bercovitz) Newsgroups: rec.guns Subject: Re: Colt GM dynamics Keywords: physics, Colt Government Model Date: 24 Aug 90 17:45:30 GMT A quick review: In part one it was noted that the equations of linear motion apply to recoil-operated pistols. Evidence cited for this was that stop action photos which show pistols in the initial stages of recoil do not show the pistol tipping up. Other evidence is that, unlike revolvers, borelines and sightlines of recoil-operated pistols are nearly parallel. From the equation, F = m a, the equation MS = ms was derived. This latter shows that the distance the barrel-slide assembly travel times its mass is equal to the distance the bullet travels times the bullet's mass regardless of the velocities involved. In part one the effect of the powder's mass and the effect of the recoil spring's force were ignored. So what is the effect of the powder's mass? For a specific example, let's consider 7 grains (0.45 grams) of powder and a 200 grain (13 gram) bullet. As noted before, the center of gravity of the 200 grain bullet moves approximately 4.4" (110mm) from start of motion to exit from muzzle of the 45 ACP Government Model. The center of gravity of the powder mass doesn't move nearly as far. It starts its motion in the center of the chamber volume and has moved to the center of the barrel-plus-chamber volume at the time the bullet exits. If the cartridge case has approximately the same inside diameter as the barrel, the distance the center of gravity of the powder moves is approximately half the distance the bullet moves. Proof of this is left as an exercise for the curious. (That's those who ARE curious, not those who are considered curious.) So the MS for the powder mass is 15 grain-inches or 25 gram-millimeters. This figure is less than 2% of the figure for the 200 grain bullet traveling 4.4". While the MS of the powder can't be ignored in instances of cartridges with a high ratio of powder mass to bullet mass, it seems safe to ignore it in this instance.

From: bercov@bevb.bev.lbl.gov (John Bercovitz) Newsgroups: rec.guns Subject: Re: Colt GM dynamics Keywords: physics, Colt Government Model, spring Date: 24 Aug 90 17:55:42 GMT One prerequisite for looking at the effect of the recoil spring is to review the mechanics of springs. There are a couple of types of wire coil compression springs. Those with large wire diameter compared to outside diameter, such as a rifle's striker spring, are a bit more complicated to describe mathematically than lighter weight springs such as recoil springs; only lighter weight springs will be considered here. Three spring properties should be reviewed: force/displacement, energy/displacement, and spring constant/physical dimensions. Springs are linear devices: the force a spring responds with is directly proportional to the amount (distance) it is compressed from its free length. The distance and force are related by the spring constant which is a property of a given spring. In the case of the recoil spring of the GM, there are three dimensions which are of prime importance other than the spring constant. They are the free length, the installed length, and the fully-compressed length. The difference between installed length and free length can be called X1 and the distance between fully compressed length and free length can be called X2. When the action is closed, the force tending to keep it closed will be the spring constant times X1. When the slide is drawn fully back, the force trying to return it will be the spring constant times X2. Work is force times distance or: integral F(s)ds. If you plot the function of force versus distance on Cartesian coordinates, the area under the curve will be this integral. Since the spring is a linear device, the area under the curve is a triangle; the area of a triangle is one half the ordinate times the abscissa. Since kx is the force of a spring, the potential energy of a spring (the work absorbed by it) is 1/2 k X^2. Therefore the work absorbed by a recoil spring when the slide moves from fully closed to fully open is = 1/2 k ((X2^2)- (X1^2)). This is equivalent to subtracting the area of the triangle associated with the installed compression of the spring from the area of the triangle associated with full compression of the spring. This article is running much too long so I'll get into numerical examples of the above in part 4 of this series. Since recoil spring manufacturers don't supply conventional spring data, it is useful to know how to calculate a spring constant. The obvious way is to put a weight on the spring and measure the spring's deflection. Without a good set up this can be difficult to do. Another approach is calculation. The formula for light weight springs is k= (G*d^4)/(64 R^3*n). G is the modulus of rigidity; its value is around 11.6*10^6 pounds per square inch for steel spring wire. It is different for other materials, such as beryllium copper or stainless steel. d is the diameter of the spring wire. R is the radius from the axis of the spring to the center of the spring wire; numerically, that would be half the quantity: (outside diameter of the spring minus the diameter of the spring wire). n is the number of active coils (don't count coils which are touching each other at the end of the spring). Try working the following example: Steel spring, outside diameter = 0.435 inch, wire diameter = 0.043 inch, number of active coils = 31.5. Ans: 2.6 lb/in. Note that the length of the spring does NOT, repeat NOT, affect the spring constant. Permanently stretching out a spring to make its free length longer only changes X1; this causes the force at the installed length to be higher, but it does not change the spring constant. From the above formulae you can see that if you cut one active coil off the return spring in the rebound slide of your S&W, you will lighten the trigger pull since X1 and X2 will be smaller but you will make the trigger force curve peakier because the spring constant will be greater. JHBercovitz@lbl.gov

From: bercov@bevb.bev.lbl.gov (John Bercovitz) Newsgroups: rec.guns Subject: Colt GM dynamics - conclusion Keywords: physics, Colt Government Model Date: 30 Aug 90 17:03:13 GMT This series is running much too long so here's the concluding article: 25 lbs in a 5 lb bag. To make up for the data compression, I'll be happy to answer wudja-mean-by's at the address at the end of this article. Part 1 of this series showed that MV = mv and that MS = ms. Part 2 showed that the effect of the mass of the powder gas is negligible during the time the bullet is in the barrel. Part 3 showed 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)). Since knowledge that kinetic energy (KE) is equal to 1/2 m V^2 is so well disseminated, I don't think it needs to be covered here. 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 my Gold Cup weighs 17 ounces (1.06 lb.) of which 3.25 ounces is barrel. The recoil spring weighs 1/4 ounce so I 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. The recoil spring currently in my GC has the following characteristics: free length = 6.480" X1 = 2.833" X2 = 4.873" k = 2.6 lb/in So its total capacity for absorbing energy is, by 1/2 k ((X2^2)- (X1^2)), equal to 20.4 in-lbs. If my 200 grain (0.457 oz) bullet leaves the barrel at 950 feet/second, then the velocity of the moving mass in the opposite direction, by MV = mv, is 25.5 fps (306 inches per second). (950fps*0.457oz/17oz = 25.5fps) The kinetic energy of the moving mass, by KE = 1/2 m V^2, is 10.7 ft-lb or 129 in-lb. (1/2 * (1.06/32.174) * 25.5^2 = 10.7ft-lb) But during the time the bullet travels down the barrel the slide moves back 0.11" (from part 1). By the potential energy formula for springs we see that the energy absorbed by the spring during this period is 1/2*2.6*(((2.833+0.11)^2)- (2.833^2)) or 0.83 in-lb. Comparing this with the 129 in-lb the slide carried, I think we can safely ignore this effect. 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. (Same magnitude of force acts on bullet and slide, refer to part 1.) So with how much force is the recoil spring resisting this peak of 2240 pounds? (Average F) = k*(average x) = 2.6lb/in * 2.888" = 7.5 pounds. Again, I think we can safely ignore this effect. Gettin' glassy-eyed yet? Me too. 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). I haven't had the pleasure of following the paper trail on the impulses, but 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. One question is how much time this momentum transfer occupies in the firing cycle. Anyone care to pursue this? The remaining mass (17 - 3.25 = 13.75 ounces) continues traveling at 25.5 feet per second or 306 inches per second. By KE = 1/2 m v^2, this mass is carrying 104 in-lbs of kinetic energy. At the time the remaining mass impacts the frame, the kinetic energy is 104 - 20.4 = 84 in-lbs. (Remember spring absorbs 20.4 in-lbs.) Running the equation for KE in reverse, we find that the remaining velocity at impact is 275 inches per second. All this assumes that the frame doesn't move during the interim under the influence of the recoil spring. Not true and I haven't proven that it can "safely" be ignored. Anyone care to tackle this problem? I think almost all of the tilting back of the pistol occurs at the time the slide runs into the frame. This is when a recoil buffer helps. It helps dissipate some of the 84 in-lb but mostly cuts the peakiness of the impulse loading. I didn't account for the jet effect of the propellant gases leaving the barrel. Jets are easy enough to calculate if you have inputs of pressures, times, and mass flow rates. These inputs come from chemical and thermodynamic calculations and are hairy to calculate; better to just measure the momentum of the recoiling firearm and ascribe that part of the momentum which is not accounted for by the momentum of the bullet to the propellant gases. Some authorities just assign a flat 4000 fps as the velocity of the gases. In this case the impulse on the moving mass due to gas escape would be 7 grains of powder times 4000 fps (as compared to 200 grains of bullet times 950 fps for the impulse from the bullet). This is about 15% of the impulse due to the bullet. So we need to go back and add 15% to the moving mass' momentum if we accept this approach. The purpose of this series was twofold: respond to the requests for an analysis, and provoke a discussion of gun dynamics so I can learn more on the subject. If you're interested and can add to the calculations outlined above, please consider doing so. Thanks. Wudja-mean-bys to: JHBercovitz@lbl.gov

From: bercov@bevb.bev.lbl.gov (John Bercovitz) Newsgroups: rec.guns Subject: Re: Light trigger pull: the down side Keywords: spring Date: 19 Sep 90 19:33:06 GMT In article <45206@ism780c.isc.com> bercov@bevsun.bev.lbl.gov (John Bercovitz) writes: > 3) What do "16" and "20" lbs, as examples, refer to in a recoil spring? A friend lent me a W. C. Wolff Company spring catalog. I looked through it to find the answer to my question (above). Although Wolff makes wonderful springs, their catalog uses words a little loosely; notably, it seems to confuse the meanings of the words "force" and "energy". So I called up Wolff and asked them how they make the measurement to determine the rating of a particular spring. It turns out that the spring rating is merely the force the spring exerts on the slide when the slide is fully drawn to the rear. That is, the force required to compress a spring to approximately 1.6 inches in the case of a government model. As discussed before, the simplest model of a spring has two degrees of freedom so two figures are required to describe it. The present rating system of spring manufacturers does not, therefore, completely describe a spring. But one could always ask for the free length of the spring also; then you'd have it. As an aside, I found Wolff personnel extremely courteous and helpful. JHBercovitz@lbl.gov "The knowing presentation of conjecture as fact is the act of a poseur." 8-) JB

From: bercov@bevb.bev.lbl.gov (John Bercovitz) Subject: Re: GM trigger problems Organization: Lawrence Berkeley Laboratory, Berkeley In article <33094@mimsy.umd.edu> megatek!avf@UCSD.EDU (Andy Funk) writes: #What difference is there between dropping the slide on an empty chamber #(purported to damage custom trigger jobs), and dropping the slide when #chambering a round (as when firing the pistol) with regard to trigger #damage/stress? Of course this is my question and that of several others, too. Dropping the slide like this is harder on other parts of the pistol, so I don't do it, but how it's harder on the trigger, I don't know. Just for reference, though, some of the recoil spring's stored energy is used to strip the top round from the magazine thereby leaving less energy to accelerate the slide. The result, then, is that the slide doesn't slam home as hard as it would if it _didn't_ have to strip and feed a round. The impact of the slide slamming home is taken up by the shaft of the slide stop. So I imagine that the slide stop's frame hole will be elongated in time by the slamming of the slide. Another area which takes up the impact is the barrel- hood/breech-face contact. I think this area would also be beaten up by habitual slide-dropping. I also don't like to do this on "mechanical aesthetic" grounds: that sort of clanking is against my religion. 8-) John Bercovitz

From: bercov@bevb.bev.lbl.gov (John Bercovitz) Subject: Re: GM trigger problems Organization: Lawrence Berkeley Laboratory, Berkeley In article <33070@mimsy.umd.edu> charlesg@tybalt.caltech.edu (Charles Grosjean) writes: #............. it seems that if you drop the slide without holding the #trigger down, then the disconnector is in the way of the slide, that is, it #is pushing against the rear of the slide harder than it is after you've pulled #the trigger. .......... I'm not sure if I'm exactly on point here, but if your slide has more trouble going forward when the trigger is not pulled than when it is, you've got a small fit-up problem in the disconnector area which can lead to a little more slop in the trigger pull. When you aren't pulling the trigger, the disconnector sticks up out of the frame higher than when you are pulling the trigger. So if there isn't enough ramp on the back side of the disconnector, or if the disconnector sticks up farther than it should, or if that teeny-tiny ramp at the bottom of the breech face is missing, then the disconnector gets banged hard before it pops down into its hole. If this keeps up, over a long period of time the disconnector hole will be beaten into an oval shape - it will be stretched forward. Since the upper end of the disconnector moves forward when you pull the trigger, there will be a slightly longer distance to pull before the upper end of the disconnector runs into the front of the now-elongated hole. For a fix, I would caution you against removing material from the upper very top of the disconnector unless you're really sure of yourself; this is a really good way to go full-auto in the most unreliable way. And remember: the third one goes up yer nose. At least it will if you're not used to occasional doubling. Oh yeah - if you want to check for this problem most easily, turn your muzzle bushing, let the recoil spring plug out, and cycle your slide. It's a lot easier to feel the problem when you're not simultaneously trying to overcome the force of the recoil spring. John Bercovitz

From: bercov@bevsun.bev.lbl.gov (John Bercovitz) Subject: gm recoil springs This weekend, faced with an enormous amount of reloading, I opted to see if I could characterize differences between various recoil springs used in Colt Government models. The springs I looked at were: the stock spring used in the 45 ACP Gov't Model, the 18.5# "HD" spring supplied with the Wilson- Rogers one-piece full-length recoil spring guide, and the dual-spring set up that comes with the Colt Delta Elite. I had a little success, but not a lot: I can infer some of the effects of the various springs, but I don't know whether these are the essential effects or merely incidental. I'd appreciate others' speculations. Gov't model background: As I've shown before in various articles, the slide under recoil initially carries 70 to 150 in-lb of kinetic energy according to how hot the load you're firing is (slow 45 ACP target or smoking 10 mm). Of this, 10 to 30 in-lb is given up to the recoil spring according to its type, some is lost in various forms of friction, and the rest is still there when the slide bangs into the stop (which may or may not have a shock buffer for frame protection). Spring background: Fundamentally, a spring has two characteristic numbers: its free length and its spring rate. Its force is proportional to its deflection (its deflection is the difference between its free length and its compressed length); the proportionality constant is the spring rate. Here are the data; symbols are defined below. Stock Colt Wilson-Rogers Stock Delta 45 ACP Gov't 18 1/2# "HD" Elite dual Model with buffer springs X0 6.53 5.92 NA X1 2.75 2.33 NA S 2.18 2.05 2.12 X2 4.93 4.38 NA K 2.65 4.30 7.15 F1 7.29 10.0 7.65 F2 13.1 18.8 22.8 TSE 22.2 29.6 32.3 KEC 7.60 8.72 11.7 VC 83.1 89.1 103 NA is "not applicable" because the Delta has 2 springs; it actually has one value for each spring X0 is the free, unrestrained, length of the spring in inches X1 is the initial compression of the spring (free length less installed length) X2 is the final compression (free length less fully- compressed length) S is the stroke of the slide (X2-X1) K is the spring constant in pounds per inch F1 is the force in pounds on the spring when it is compressed X1 F2 is the force on the spring when it is compressed X2 TSE is the total energy in inch-pounds required to compress the spring to compress it from X1 to X2 KEC is the kinetic energy the slide has on its return at the point in time that the bullet contacts the feed ramp VC is the velocity the slide has in inches per second when the bullet contacts the feed ramp Observations: The Delta's dual-spring set up is the equivalent of a very stiff spring with little initial compression. This results in a "peaky" force curve: it starts relatively low and rises relatively high. The other two spring types are effectively longer with consequent flatter force curves. The Wilson-Rogers spring absorbs approximately the same amount of energy as does the Delta dual spring. What effects does using the Delta's peaky spring assembly have compared to using the Wilson-Rogers spring? Well for one thing, the slide using the Delta spring will have a little higher velocity at the time the case runs into the ejector. This will result in smarter ejection. The slide has higher velocity because less energy has been removed from it because spring forces on it are lower during the early part of its travel. Another effect is that on its return, the slide will have more velocity at the time it is pushing the cartridge's bullet up the feed ramp. This is because the slide will have been operating in the high-force range of the spring's force curve: the peaky area. What about cycle time? Qualitatively speaking, as the slide goes rearward in recoil, the Delta's peaky spring set allows it to move at higher velocity during the early part of the travel; it doesn't "pull the velocity out of the slide" until later. So the slide's average velocity to the rear will be higher. On the slide's return forward, the Delta's peaky spring set gets the slide up to a given speed sooner resulting in higher average slide velocity forward. Because of these two similar effects, total cycle time should be lower. Are these minor effects the whole point of Colt's putting the dual spring set in the Delta Elite? Have I missed a subtler (to me) effect? I could use some help here. Thanks Notes: The reason for the different values of S is that the recoil spring guides (some with buffers) are different thicknesses. The Wilson-Rogers, with the thickest buffer & guide, has somewhat lower VC than it would have if the slide had more of a running start at the job. The Wilson-Rogers spring started out with 6.2 inch free length. It now seems to be stabilizing at 5.92. So the spring was overloaded and is now operating at yield stress. This may or may not be a good situation; I don't know. The Delta Elite has a recoil buffer built into its recoil spring guide. The buffer's effect is to eat up a lot of the slide's energy when it is rammed rather than allowing the slide to bounce off of the frame stop and thereby regain some of its velocity (but with the velocity's sense reversed). The Delta Elite's spring set is composed of a stock gov't model spring plus a small-diameter stiff inner spring which is compressed very little in the initial, installed position. The W.C. Wolff company sells variable rate springs. These should have an effect similar to the effect of the Delta's spring set only more so. This is because the force curve is no longer straight-line; it is now "swamped" in the middle so forces will stay even lower, relatively speaking, at the beginning of the slide's travel in recoil, and will be relatively even higher at the end of the slide's travel. JHBercovitz@lbl.gov (John Bercovitz)

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