Conservation of momentum, Cocking a gun, and you

I don' think the weight of the frame factors into it, unless the difference in weight affects the ability of the shooter to rapidly press the frame forward again. I would think the friction on the rails would matter a bit, and the heavier the slide, the greater chance of this (inertial cocking) happening.

In theory, a heavier frame would matter. In practice, not so much, because frames only get so heavy. Think of extreme - 1 pound frame. Boxers throw that weight easily, and any non-disabled person should be able to too.

The rest looks good. Heavier the slide, lighter the recoil spring, less friction on rails, the easier it gets.

TenMillimaster said:

I drew that conclusion because of how the math works out.
The mass of the gun being pulled back times its velocity must equal the sum of the new momentum of the frame and the new momentum of the slide.
If anyone sees any mistakes there....

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Here's your problem. It's not a closed system. Your hand is applying a new force to the gun. The weight of the frame only matters in that it makes it harder to change the direction of frame and affects how fast you can accelerate the frame in the new direction. A heavier slide or a lighter spring would help. If you were fast/strong enough you could do this without any reward motion.

TenMillimaster said:

I checked my math again, got the sign on the velocity of the gun wrong, but the math and physics say the same; momentum of each part depends on the velocity it travels and it's mass:
if an object has momentum, but then it hits something (pool) or separates or merges with something (car crashes), the objects' new speeds must have the same total momentum


EDIT: Read your post again and whoops, noticed this:

I really hated to say "But hey I checked my math!" Now I can say this:
Yes, it would, as that's yet another system in which momentum needs to be conserved; the entire gun is moving backwards with your hand. Suddenly, you arm exerts a force on your hand which accelerates it which imparts a velocity and a new momentum to your hand. This impacts the moving frame and in the new system, the new momentum of your hand and the frame has to equal the momentum of your hand and the frame.

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As long as your hand is accelerating forward the momentum is constantly changing. This not really a conservation of momentum problem. You'd be better off looking at it as a spring problem using Hooke's law. It's still going to be a pain because the inertia of the slide complicates the problem.

Here's your problem. It's not a closed system. Your hand is applying a new force to the gun.

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Exactly. Racking the slide doesn't make the frame go forward. It essentially stays in one place due to the fact that you are holding it.

The resistance required to pull a slide back is due to springs and friction. The weight of the slide will make no difference, as the pistol slide is more or less 90 degrees away from the pull of gravity. Also at the scales we are talking about, it would take a pretty big slide for mass to be a factor in how fast you can move the slide.

If you wanted to make a slide large enough for it to continue its rearward movement on its own it would require a large force up front in order to accelerate it to that rate. There's no free lunch. Objects at rest tend to remain at rest.

The weight of the slide will make no difference, as the pistol slide is more or less 90 degrees away from the pull of gravity.

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Strongly disagree with that. We can look at it from the pov of the spring equation if you like, but -kx isn't the right one because you don't have any initial displacement. You need to look at it from potential energy.

The bigger the slide and the faster you move it, the more kinetic energy you're building up in the gun with your rearward motion. When you snap the frame forward, the slide retains its kinetic energy, compressing the spring, eventually netting something near its max KE in PE, which, in the PE = (1/2)kx^2 equation, with the only changeable variable being displacement, greater PE, greater displacement.

Strongly disagree with that. We can look at it from the pov of the spring equation if you like, but -kx isn't the right one because you don't have any initial displacement. You need to look at it from potential energy

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This has nothing to do with the spring equation. When you do work on an object, the only time the object's weight comes into the equation is if you are opposing gravity. The slide's weight doesn't affect the force needed to pull the slide back. The slide's mass does, to the extent that it affects the acceleration you can move it. Heavier slides will accelerate slower for a given force. If you want to increase the acceleration, you need to increase the force proportionately. The presence of a return spring will increase the force needed to attain that acceleration even more.

Besides, we're back to the free lunch thing. A gun that relied on inertia to complete the slide's rearward travel would either need to have a slide so heavy it would be very difficult to get going and a long travel or a very light slide and an even longer travel. Either way it's not very practical for a handheld weapon.

Just watched the video. I interpreted your question as one where you'd want want to build a gun that you cocked it by throwing the slide backward. This is why the moderators ask folks who post links to offer at least small amount of explanation.

The guy's just taking the kinetic energy of the moving slide, storing it as spring energy and releasing it. Where's the mystery? The gun is probably modded with some very light springs and I'd wager the shooter doesn't care too much about its longevity. Fun for sport, but not too practical for anything else.

This has nothing to do with the spring equation. When you do work on an object, the only time the object's weight comes into the equation is if you are opposing gravity. The slide's weight doesn't affect the force needed to pull the slide back. The slide's mass does, to the extent that it affects the acceleration you can move it. Heavier slides will accelerate slower for a given force. If you want to increase the acceleration, you need to increase the force proportionately. The presence of a return spring will increase the force needed to attain that acceleration even more.

Besides, we're back to the free lunch thing. A gun that relied on inertia to complete the slide's rearward travel would either need to have a slide so heavy it would be very difficult to get going and a long travel or a very light slide and an even longer travel. Either way it's not very practical for a handheld weapon.

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Go through it in steps.

1) Pistol in hand, motionless. KE frame = 0, KE slide = 0

2) Pistol jerked back and at end of motion prior to deceleration. KE frame = KE slide > 0

3) Pistol jerked forward following completion of rearward motion (the frame goes forward, the slide stays where it is. This is because the forces acting on the two bodies are independent. The force acting on the frame is primarily the hand + the force of the spring on the slide. The force acting on the slide is just the force of the spring). KE frame > 0, KE slide approximately = 0, PE slide (due to spring compression) > 0. Separation of slide and frame happens during this interval.

4) Hand/frame motionless, Slide chambers round. KE frame = 0, KE slide > 0 until end of motion


There's no "free lunch" issue here. The human body is doing the work.

Edit: Just read your next reply, figured there might've been miscommunication

Edit: Just read your next reply, figured there might've been miscommunication

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No worries. We were talking about two different things.

Okay, now that we settled that...how much "work" (in the physics sense) was done on the system? LOL

TenMillimaster said:

(props to THplanes for pointing out the obvious, glaring error in my 1 in the morning assumptions & getting the thought process flowing in the right direction.)

Here's your problem. It's not a closed system. Your hand is applying a new force to the gun.

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Not to be a schmuck, but I mentioned the shooter pushing against the frame in the second post.

It seems to me that the delta-V of the frame that you can generate at the end of the retraction stroke (when you reverse the direction of the frame's motion, but the slide wants to continue moving rearward) *is* affected significantly by the mass of the frame. Consider the delta-V you could generate by snapping an empty frame (no slide) back and then forward; if you can generate the same net force with your arm, then the delta-V of the frame will be larger with a lighter frame since a = f/m. And since the velocity difference between slide and frame is what will ultimately cock the slide, it would seem to be easier the heavier the slide is and the lighter the frame is, all else being equal.

Here's a video of Rob Romer showing how he does it.

http://www.youtube.com/watch?v=fKyGUYjCruY

Watch the original slow motion video. The slide is basically all the way back as he starts his hand forward. So you can use these equations to see how fast you need to move the gun back-wards to accomplish it. The real secret seems to be a near instantaneous stop, not the transition to forward movement.

For a spring P.E. = (.5)(k)(x^2)
For the slide K.E = (.5)(m)(v^2)

In this case K.E. = P.E.
You'll need to find out or measure the spring rate constant = k
I'm not sure if the 17 pound ratting on a Glock is the spring rate or something else.

This neglects the loses to friction and that the mass of the slide changes because the barrel starts as part of the slide mass. Since the barrel unlocks so early I think you can ignore it and this will get you a ball park figure on how fast the slide needs to move. Do note that if the gun is not perfectly parallel to the direction of travel the friction force may become very significant.

Edit: After rereading the thread I discovered I forgot the energy to finish cocking the striker. You'll need the P.E. for that as well.

P.E.(recoil spring) + P.E.(finish cocking striker spring) = K.E.(slide)

RTR_RTR said:

Strongly disagree with that. We can look at it from the pov of the spring equation if you like, but -kx isn't the right one because you don't have any initial displacement. You need to look at it from potential energy.

The bigger the slide and the faster you move it, the more kinetic energy you're building up in the gun with your rearward motion. When you snap the frame forward, the slide retains its kinetic energy, compressing the spring, eventually netting something near its max KE in PE, which, in the PE = (1/2)kx^2 equation, with the only changeable variable being displacement, greater PE, greater displacement.

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There is an initial displacement on the recoil spring. Look at a 1911, for example, anyone that's ever field stripped a 1911 knows of the frustrating (and potentially dangerous) preload on the recoil spring after removing the barrel nut.

If recoil springs were not preloaded the slide would flop around like a wet noodle in the initial part of its travel. In fact, It would likely be impossible to chamber a round without an extremely heavy spring.

I checked out the video and I've never seen anything like that. Thanks.

The constant k, which is the "mass" of the spring, is determined by the length of the spring at rest vs. the length of the spring when weight is applied. Look up "Hooke's Law".

I studied some physics and a LOT of math, but I am rusty. I spent most of the time studying mathematical analysis anyway (so I'm really good with the equations). One thing you may want to consider in your dynamical system is the omission of the frame. The frame is connected to the arm, if you think about it, in which case you should factor in the arm too, no?

In conservation of energy, you will have the slide going rearward, which can be measured and displayed as kinetic energy. The other half of the equation is what is happening to the spring. As the slide spends KE, the spring stores PE or potential energy. When the slide is spent and has 0KE, the spring has maxPE and reciprocates. Basically, the values flip flop as it goes back and forth. Don't worry about friction and such unless you want to take the time to measure the friction coefficient. Your answer will be close enough without it.

Now without breaking out the pencil, I can tell you that yes, a lighter spring and heavier slide will make this easier. But do you know why a heavier spring and lighter slide makes it more difficult (to near-impossible)? Equilibrium. Think of the force being shifted along the spring. The more spring you have, the lighter the slide you need, and vice versa. There is a point along the spring where the mass of the slide and the coefficient k of the spring are the same. Then there is a point where you need no spring at all but a very heavy slide.

I lay all this out this way to avoid delving into the math. Not only would I probably mess it up right now, but I'd also make it more difficult to understand. Just understand it CAN be more involved than one would imagine. A rudimentary understanding of differential and integral calculus is a must when definitively dealing with these equations in order to understand where they come from (and wher they go!).

benEzra said:

It seems to me that the delta-V of the frame that you can generate at the end of the retraction stroke (when you reverse the direction of the frame's motion, but the slide wants to continue moving rearward) *is* affected significantly by the mass of the frame. Consider the delta-V you could generate by snapping an empty frame (no slide) back and then forward; if you can generate the same net force with your arm, then the delta-V of the frame will be larger with a lighter frame since a = f/m. And since the velocity difference between slide and frame is what will ultimately cock the slide, it would seem to be easier the heavier the slide is and the lighter the frame is, all else being equal.

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Yes and no. You are right about less force for an empty frame vs. a built frame. But you are taking the arm into consideration which makes the equation more difficult and involved (but more accurate regarding the true dynamics of the system). Assume the force is fixed for either one to make it easier. You could also include air pressure/resistance, friction, some of the energy is lost there, and you could also include the trig required to get the "true" answer since the action isn't all horizontal, and since energy is lost along the way, it may be better use partial differential equations to anyalyze this system. We could also use Einstein instead of Newton to determine all this and then evaluate it say, at the speed of light!

The point is that we can simplify the system down to the spring and the slide and then CLEARLY see the relationship. All to often clarity is lost in science and we are left with more arguements than we started with.

Basic physics. Newton figured it out about 500 years ago. You have to accelerate the gun fast enough to overcome the recoil spring force.

F=MA (Force = Mass x Acceleration)

Press the slide of your pistol on a scale and cock it, you'll see how much force is required to move it to the rear of the frame against the recoil spring. For example, my 1911 with a 16 pound recoil spring requires 16 pounds of force to move the slide to the rear (imagine that). This is the "F" in the equation. This is how much force you must apply to overcome the recoil spring and cock the pistol.

The other side of the equation tells us that the mass of your slide times the acceleration you can apply to it has to at least equal the recoil spring force to cock your pistol if you want to cock it like the videos.

Weigh your slide then divide the weight in pounds by 32.2 to convert to mass. The english system unit of mass is the slug. Weight is a force, it equals Mass x Acceleration. Acceleration due to gravity is 32.2 ft/sec^2 (feet per second squared), so you have to eliminate it to get the mass in slugs.

For example, if your slide weighs 1.61 pounds, it has a mass of 32.2/1.61 = .05 slug.

This is the mass in the equation.

Since Force = Mass x Acceleration, it'll now be easy to figure out how much acceleration you have to apply to the gun to get the slide to move all the way back.

1G of acceleration = 32.2 ft/sec^2, like we discussed above.

If you accelerate your pistol at 1G, the slide will apply a force of 32.2 ft/sec^2 x .05 slug = 1.61 pounds. That's why the slide weighs 1.61 pounds on a scale, because gravity is accelerating it at 1G toward the center of the earth. Obviously, you'll have to apply more than 1G to a 1.61 pound slide to deliver 16 pounds of force. About 10 times more, as a matter of fact.

Since you know the Force required and the Mass of the slide, simply put them in the equation and solve for the Acceleration.

F = MA

16 = .05 x A

A = 16 / .05 = 320 ft/sec^2

Since 1G = 32.2 ft/sec^2, 320 ft/sec^2 is about 10G.

So you have to apply 10G's of acceleration to the gun to get it to cock.

If your arm "stroke" is 2 feet, how quick do you have to extend the pistol over a distance of 2 feet to achieve 10G?

Well, we all know from our basic engineering Statics class that Position = Initial Position + Initial Velocity + Acceleration x (time^2)/2

Since Initial Position and Initial Velocity are both zero,

Position = A(t^2)/2

We know the final position is 2 feet and the acceleration is 320 ft/sec^2, so

2 = 320(t^2)/2, solving for t gives

squareroot of 2(2)/320 = t, so squareroot of 4/320 = .11 second

In other words, if you can extend your gun with a 1.61 pound slide 2 feet in .11 seconds, it will cock the gun.

A heavier slide means more mass, so you need less acceleration to deliver the same force to the recoil spring. For example, a slide that weighed twice as much, say 3.22 pounds, would only require half the acceleration to deliver the same force. Moving a gun with a slide that weighs 3.22 pounds 2 feet in .22 seconds would cock the gun against a 16 pound recoil spring.

Bottom line, divide your recoil spring force by the weight of your slide and you'll know how much acceleration (in G's) you have to apply to get it to cock. As a side note, if you ever go to a planet with that much gravity, your slide will go fully to the rear if you point your gun straight up.

All you guys and your theroy make my head hurt. Why don't you just pick up the gun and practice until you become profecient...........chris3

All you guys and your theroy make my head hurt.

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Sorry about that. However, it's not "theroy", it's the same physical science that put men on the moon.

Why don't you just pick up the gun and practice until you become profecient

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Because of the very same reason you'll see very few people practicing till they become "profecient" at throwing a 100 MPH fast ball.

If you don't genetically have the inherent strength, reflexes and fast twitch muscle response, no amount of practicing will get you there.

Still nice to know what it takes to accomplish it, though.