[smithmax]

Does anyone have an idea of how fast the slide of a 9mm pistol might move? I made a stupid mistake and put my support hand in a bad spot when shooting from sandbags and lost a chunk of skin. My thumb is also starting to bruise, so I'm wondering if anyone knows how fast the typical 9mm compact slide travels back.

[cchris]

I can't give you a speed, but it seems that it would differ from gun to gun. The only slow-motion comparison I've seen was a 1911 (.45) vs. a 92 (9mm) on "Lock & Load", in which the Beretta cycled faster.

[owen]

19-20 fps is a good place to start. It varies from gun to gun of course, but full sized 9mm's i've measured have been in that ballpark.

[kanook]

depends on gun brand, spring, ammo, limp wristing, barrel, slide

[MachIVshooter]

A 1911 slide cycles in roughly 60 milliseconds, and it has to travel about 2" back and then forward again. so that's 1/3 of a foot in 60ms. Of course, that total time includes slowing down, stopping, reversing direction and accelerating again under only spring pressure. I would bet initial slide velocity to be pretty fast, maybe 100 FPS, then slowing to perhaps 10-20FPS by the time it hits it's rearmost point and impacts the stop.

Would be interesting for find out for sure, though.

[Fish Miner]

Darn fast- I did the same thing when I got my first semi. I was trying to shoot left handed and put my right hand in a bad place. It was so fast I did not know I was "hit" till I looked down and saw the blood.

First and only injury from a firearm....knocking on wood...

[Ron James]

How fast does a slide move, faster than a speeding bullet?:)

[smithmax]

It's a new Taurus 709. It looks like the slide did 1/2 of it's travel through my thumb. Thanks for the ideas of speed, I'm trying to figure out how much energy it exerted on the bone in my thumb.

[possum]

all the above are good answers, and right on the money, another big difference moght come form how hot the ammo you are shooting is as well. hot defense rounds or reloads can make that slide work alot harder and faster.

[kennjen]

An interesting question to ponder on for sure...but does it really matter ? It's kind of like asking, "Was that a 9mm or a 45acp you just shot me with ??"

I hope you are ok and no long lasting damage resulted..

[smithmax]

I was just shooting WWB, so nothing hot or special.

Kennjen, it doesn't really matter, I'm just sort of a technogeek and like to know the numbers on things like this. It's the same reason I have a Scan Gage 2 mounted on my dash in my truck, for no purpose but to see more information than I need. :)

[jungle]

Short answer, about 25 feet per second.

Long answer and how to calculate for your pistol/ammo combination:http://yarchive.net/gun/pistol/1911_dynamics.html

[owen]

jungle, he's shooting a 9mm, so he should see some lower velocities.

The spring makes almost no difference in the velocites while the slide is travelling to the rear. This was consistent among the 5 or 6 different pistols I measured. The gun I was working on was going a little slower, so we tried to manipulate the velocity with the spring...the difference was so small as to make messing with the spring to control slide velocity useless.

[kennjen]

I was just shooting WWB, so nothing hot or special.

Kennjen, it doesn't really matter, I'm just sort of a technogeek and like to know the numbers on things like this. It's the same reason I have a Scan Gage 2 mounted on my dash in my truck, for no purpose but to see more information than I need. :)
>Kennjen, it doesn't really matter,~~
>>~~for no purpose but to see more information than I need.

I know. I just had to comment on it because the sentence below :

>>"I'm trying to figure out how much energy it exerted on the bone in my thumb."

made me laugh as a fellow geek/engineer. It's some thing that may run through my head as well, even know I probably won't start a thread on it.

You are welcome to SMS me or call me, if you ever are shot. I will try to get back to you with all relative ballistic data before you pass out.

[cleetus03]

delete

[eatont9999]

Shouldn't you be telling us? :)

[blikseme300]

Slides move at the speed of ouch. The units are expressed in "@#$%"
:evil:

[jojo200517]

Well in every gun I held incorrectly the slide moves fast enough to displace any skin that gets in the way by a pushing and tearing and or cutting method. I'm not sure just how fast that is but I agree with bliksems300, its near the speed of ouch.

[unspellable]

Slide velocity of a recoil operated pistol will initially be in the 10 to 20 fps range slowing considerably by the time it reaches the rear extreme of travel. This happens to be a point I've done quite a bit of research on for a pistol reliability project.

The impulse of the load is bullet mass times bullet velocity plus powder mass times powder velocity. (as a rule of thumb figure powder velocity at 1.5 times bullet velocity.

Once you have this number, divide by the slide mass and you have the slide velocity.

You will note that the recoil spring strength was not mentioned. It has so little effect on the slide velocity it's not worth including. The recoil spring's job is to slow the slide down before it hits the end of travel.

The recoil force will peak at well over 500 lbf, so a spring pre-load of 15 lbf or so is not going to have much effect.

[smithmax]

How would you figure out the poweder mass/velocity part of that?

Bullet mass is 115gr and the velocity is ~1000fps

[JohnKSa]

I did some back-of-the-envelope calculations awhile back and the results agree well with what unspellable says in his post. One correction, however.Once you have this number, divide by the slide mass and you have the slide velocity.If the pistol is a typical locked breech, recoil operated pistol you use the combined mass of the slide and barrel since the two are locked together at the moment of firing.

[wilkersk]

Short answer, about 25 feet per second.

Long answer and how to calculate for your pistol/ammo combination:http://yarchive.net/gun/pistol/1911_dynamics.html
Cool! That is just hours of fun right there!

[Guns and more]

I would think it is the ratio of the mass of the bullet to the mass of the moving parts. If the bullet weighs 1 oz. and the slide and barrel weighs 15 oz. then the slide would travel back at 1/15 of the speed of the bullet. If you know that, you should be able to calculate the velocity of the slide. Let's say the bullet goes 1000 ft/sec, then shouldn't the slide go 66 ft/sec (1/15 of 1000). I know it's fast, I don't even notice it.

[Davionmaximus]

Fass Grasshoppa..Berry Berry Fass...;)

[EOC_Jason]

Yeah, I think as a couple others have stated...

It moves fast enough for you to learn not to put your hand in the way again. ;)

[unspellable]

I did not take the time to read through the article at the link above but the guy gets into calculus which not everybody is familiar with. With out getting into calculus it works as follows.

The first step is calculating the cartridge impulse. The request above states a 115 grain bullet at 1000 fps which sounds like a lite 9 mm load, so let's say 5 grains of powder.

It's easy to measure the bullet velocity. It's not so easy to measure the average velocity of the powder gas. So as a rule of thumb for pistols we say it's 1.5 times the bullet velocity. For convenience of calculation we multiply the powder mass rather than the velocity. 115 + 1.5*5 = 122.5
We have to be consistent with units so we include a factor to convert grains to slugs. 122.5 grain * 1000 ft/s / 225218 = 0.5439 lbf*s as the impulse.

Previously I said slide mass when I should have said upper mass which includes the barrel. Let's assume the upper has a mass of 10 oz. This would be on the light end of the range for upper weights. We have to be consistent with units and convert ounces to slugs. 10 / 514.8 = 0.019426 slugs.

Next divide impulse by upper mass. 0.5439 / 0.1926 = 28 ft/s.

Given slide mass and velocity one can begin to calculate the required recoil spring strength but let's leave that for another day. It's faily simple for some types of actions, gets messy with others, the 1911 type is one of the messy ones. (Due to also shoving back the hammer.)

For reference: 7000 grain = 1 lb. 16 oz = 1 lb. 32.1740485564 lb = 1 slug (There's a valid reason fo all those decimals even though we usually round them off to about 32.174.) Note that the grain, ounce, slug, & pound "lb" are units of mass, while the pound force "lbf" is a unit of force.

[larryh1108]

Question: The upper mass does include the barrel at time of ignition but doesn't the slide move back without the barrel? The speed of the slide during recoil/rebound would be without the barrel which has remained (more or less) stationary. To say the upper mass would include the barrel would mean that the barrel's mass would also be traveling, which it's not. Yes? No?

[1911Tuner]

The recoil spring's job is to slow the slide down before it hits the end of travel.

Sorry. The recoil spring's job is to return the slide to battery. That it slows the slide as it compresses is incidental.

You will note that the recoil spring strength was not mentioned. It has so little effect on the slide velocity it's not worth including.

As JohnKSA noted...it's a short recoil operated pistol. The slide moves only a short distance before the bullet exits....roughly 1/10th inch. The slide is accelerating only as long as force is applied to it. Once the bullet is gone, the force is gone, and the slide immediately starts to slow down due to outside forces provided by friction and the recoil spring. Its continued movement after the force has been removed is solely on momentum...the same as the bullet's.

Also...because it's a locked breech pistol...the bullet's drag on the barrel, and hence on the slide must also be taken into account...so it's not a simple matter of bullet mass/velocity and slide mass/velocity. The barrel doesn't push the slide, and it doesn't move on its own. The slide is driven backward and pulls the barrel with it. The bullet exerts a forward drag on the barrel...and whatever drag is imposed on the barrel is likewise imposed on the slide.

And....also as John observed...for the time and distance that the barrel is attached to the slide...while the accelerating forces are imposed on both...the barrel's mass must also be factored in. The bullet/slide/mass/velocity calculation would only apply in a straight blowback design...where the slide moves independently of the barrel...and even with that...only if the recoil spring is removed.

[unspellable]

The recoil spring's FIRST job is to slow the slide, THEN its SECOND job is to return the slide to battery. The recoil spring's required strength is primarily determined by what's required to slow the slide down. If you don't slow the slide down enough you are going to batter the pistol to death, or in an extreme case you might get the slide in your face. (This is not unknown to C96 shooters.) That's why you use a stiffer spring for hotter loads. Virtually any pistol will have a spring that's more than strong enough to return the slide to battery. The spring strength required to return to battery is independent of the load.

We are talking about a short recoil operated pistol here. The barrel and slide are locked together and recoil together until unlocking begins after the slide/barrel recoil a fraction of an inch. Recoil is determined by the momentum of the ejecta, barrel friction has absolutely nothing to do with it. It's called conservation of momentum. Before firing the momentum is zero. After firing the ejecta and the upper must have equal momentum of opposite sign in order to add up to zero.

This calculation does NOT apply to a blow back action. In a blow back it's the integral (Calculus again, darn!) of the pressure curve multiplied by the bore area divided by the slide mass. Again the spring strength has a negligible effect on slide velocity. (The area is actually the internal case head area minus he internal shoulder area if any, which is the same as the bore area.) And again bullet friction has nothing to do with it. There are "delayed" blow back actions (On the whole, not very successful although a few have been said to work.) in which cartridge case drag or some other retarding action has a noticeable effect on slide velocity.

A gas operated action is still another ball of wax, let's leave that for another day.

[LeonCarr]

I read somewhere that the slide on a 1911 .45 cycles (back and forth) in 4/100s of a second.

Just my .02,
LeonCarr

[jaydubya]

Back before I replaced the firing pin spring on my t-series Browning High Power for the first time, I would occasionally get doubles. The time between each firing was so infinitesimal that the sounds blurred into one another, and the muzzle did not even have time to climb before the second round fired. Thus my scientific analysis of how fast the slide on my BHP cycles: REAL FAST.

Cordially, Jack

Moderator, doesn't this belong in the auto section?

[1911Tuner]

The recoil spring's FIRST job is to slow the slide, THEN its SECOND job is to return the slide to battery. The recoil spring's required strength is primarily determined by what's required to slow the slide down. If you don't slow the slide down enough you are going to batter the pistol to death,

Wrong again, neighbor.

A few years back, Ned Christiansen conducted a little test in which he fired a .45 caliber 1911 with GI spec hardball repeatedly without the recoil spring in place...without damage.

Recoil spring is a misnomer. It should correctly be called the action spring...because that's its primary function...stripping a fresh round and returning the gun to battery.
It's described in the original patents if you'd like to go see.

We are talking about a short recoil operated pistol here.

I'm well aware of that. Been involved with wrenchin' on these things for nigh on 45 years now.

Recoil is determined by the momentum of the ejecta.

Not quite. The bullet and powder gasses are moving in the wrong direction to cause the breechblock to move. The bullet provides a resistance for the force to push off of...just like the breechblock. Force is what moves things. Momentum isn't a force. Momentum is the result of a force applied...assuming that the force is of sufficient magnitude to cause movement. If it doesn't move...it doesn't have momentum. Mass X Velocity. Remember?

Recoil is backward acceleration in response to the bullet's forward acceleration, accomplished by an equal force applied to both. Neither slide nor bullet have momentum until they move, and neither one moves until a force is applied. Once the bullet is gone, the force is gone...and there is no more acceleration. Whatever movement is seen after the force is removed is through momentum.

Force moves it. Momentum keeps it moving after the force is gone.

Go push on a wall. Push briskly enough to cause your arms to straighten so that your hands lose contact with the wall. The accelerating force has been removed...and the action/reaction event is over. All further movement is caused by momentum that was conserved while the accelerating force was being applied.

Barrel friction has absolutely nothing to do with it.

You'd better rethink that one. Maybe this will help.

The bullet is moving...driven forward through the barrel under high frictional resistance. The barrel is being pulled backward by the driven slide at the same time. Whatever frictional resistance the barrel offers to the bullet...the bullet also offers to the barrel...in equal measure. Whatever resists the barrel's backward movement resists the slide's backward movement...because the slide pulls the barrel. The pesky thing about Newton's 3rd Law is that it works in both directions...whether pushing or pulling. Whatever frictional force is imposed on one is imposed on the other. The friction doesn't disappear just because things are moving fast...and the coefficient of friction remains the same, regardless of the velocity.

On another thread, we had a discussion about this very thing. It was demonstrated that the slide doesn't require much force to move it without the bullet's delaying influence. You can move it at its full operational speed...or faster...by hand cycling it.

Abbreviated version...10 grains of Bullseye powder was loaded into a case with a tuft of cotton to cap it off. The slide made about half its full travel...with very little pressure and a very small mass of ejecta. If the case had been bottlenecked in a .44 die, and then again in a .41 die...it likely would have made it to slidelock. If you'd care to try it for yourself, you may learn something.

The bullet has a powerful delaying effect on the slide. Moreso than all the other delaying factors combined. Believe it. That's why light powder charges often produce short recoil. There's ample force to cycle it...but with the bullet dragging on it...it doesn't work so well.

Pull on a broomhandle in opposite directions. Relax your grip with one hand, until the handle just starts to slip through the hand...but maintain enough grip to require continued force to keep the handle moving. Note that your hand also moves at the same instant that the handle slips. Hand and handle both slipping under friction...and each one's movement is resisted by the other. There's the reality of the bullet's influence.


As for the momentum issue...Momentum will be conserved...in the absence of outside force. Once the outside forces start to come into play...everything changes. The straight blowback functions without the bullet's delay...but the short recoil operated pistol is almost...ALMOST...a delayed blowback. The only thing that keeps it from being so is the fact that the slide doesn't move independently of the barrel.

[JohnKSa]

The upper mass does include the barrel at time of ignition but doesn't the slide move back without the barrel? The speed of the slide during recoil/rebound would be without the barrel which has remained (more or less) stationary. To say the upper mass would include the barrel would mean that the barrel's mass would also be traveling, which it's not.This has already been answered, but if you want to visualize what's happening, you have to think of the recoil impulse as being like a hammer hitting the slide.

Since the slide and barrel are locked together when the hammer hits, the resulting velocity of the slide AND barrel are determined by the momentum transfer to their combined weights. The fact that the barrel shortly thereafter disengages from the slide is immaterial because all of the momentum transfer has been accomplished and the velocity is already set.

If they disengaged and both kept traveling separately, they would both be traveling at the same velocity (in the absence of spring pressure or friction); the fact that the barrel stops doesn't affect the slide's velocity at all.

This is an area in which that hammer fired guns have little bit of an advantage over striker fired guns. In hammer fired guns, the slide must cock the hammer in the initial part of travel while the barrel and slide are still locked together. The force required to do that helps slow the slide velocity AND the barrel velocity.

Stopping the barrel is usually done in either design by slamming it into the frame/locking block/etc.. Striker fired guns have to stop the barrel from maximum velocity since there's nothing to significantly retard the slide/barrel early in travel. Hammer fired guns typically slow the slide & barrel a bit by making them cock the hammer against the force of the mainspring.

That means that in a hammer fired gun you can either go with a lighter recoil spring (making racking the slide easier) or you can just stay with the heavier recoil spring and get a little less slide velocity and a little less barrel velocity which will help with durability.Not quite. The momentum of the bullet and powder gasses are moving in the wrong direction to cause the breechblock to move.Ok, it's more explicitly correct to say that "the momentum of the recoiling mass is equal to the momentum of the ejecta.", but it is not incorrect to say that recoil is determined by the momentum of the ejecta. You calculate recoil velocity (which is the topic of this discussion) by calculating momentum of the ejecta and dividing by the mass of the recoiling mass. So it is correct to say that recoil (which in the context of this discussion means recoil velocity) is determined by momentum of the ejecta.The momentum of the bullet and powder gasses are moving in the wrong direction to cause the breechblock to move. Force is what moves things. Momentum isn't a force.The two momentums are equal and in opposite directions as you say. The fact that they are in opposite directions is a function of the way conservation of momentum works, it is not evidence that the momentum of the ejecta is not related to recoil nor does it imply that momentum is a force.

The momentum of the ejecta and the momentum of the recoiling mass are equal per conservation of momentum. That's a given and not open to debate (for at least the last few centuries). :D

NOW, getting into the forces that create the momentum is a bit more complicated but fortunately it's not required to answer the question about the slide velocity.

The comment about barrel friction not entering into it is also correct. One need not consider barrel friction to accurately and precisely calculate recoil velocity. Conservation of momentum makes it unnecessary to get into that if all you want to know is how fast the recoiling mass is moving.

The barrel friction is all wrapped into how fast the ejecta exits the muzzle and that's all you need to know to calculate the velocity of the recoiling mass.

[1911Tuner]

The two momentums are equal and in opposite directions as you say.

Well...not quite, John. Momentum is conserved in the absence of outside force. The only way that the bullet and slide momentum could be equal is if there were no outside forces acting on them...or if the outside forces were equal on both...which they're not.

The only significant outside force working against the bullet is barrel friction.
The slide has bullet /barrel friction...the recoil spring...the mainspring...the hammer's inertial mass...the barrel's inertial mass...and the slide to frame friction. The momentums can't be equal under those circumstances.

One need not consider barrel friction to accurately and precisely calculate recoil velocity. Conservation of momentum makes it unnecessary to get into that if all you want to know is how fast the recoiling mass is moving.

Sure you do. As we demonstrated in the other thread...it doesn't take a lot of force to move the slide when the bullet is absent.

Look at it like this:

If the same level of force were applied to the slide without the bullet in the barrel to resist it...the slide and frame would self-destruct in short order. The rearward velocity would be off the scale.

The bullet's frictional drag on the barrel is what stretches revolver frames and cracks autopistol slides at the junction of the breechface and the ejection port...which I've seen several times through the years. That's a pretty signifigant force.

I realize that it's hard for many to understand the bullet's role in the slide's delay...but it's there.
Believe it.

The bullet is moving forward through the barrel. The barrel is moving backward over the bullet. Both are occurring at the same time...like the broomstick demo. The barrel is resisting the bullet's passage...and the bullet is resisting the barrel's movement.

The only way that it can be negated is if the bullet doesn't move until the slide cycles...and that ain't gonna happen.

Or, the bullet starts...stops and waits for the slide to cycle...and takes off again...and that ain't gonna happen.

Or...The slide doesn't move until the bullet is gone...as per Kunhausen's Balcanced Trust Vector description...and that can't happen.

[AKElroy]

How fast does the slide move?

It's all relative. Are we assuming it's not moving when not firing? The math could get really ugly; are we calibrating relative to the rotation of the Earth on it's axis (which varies in ground speed relative to ones distance to the equator), our travel through our orbit of the sun (which also varies relative to which stage of the arch in which we happen to be traveling through our seasonal eliptical arch), our solar systems' travel around the black hole at the center of our galaxy, or our galaxies' travel with regard to the known universe? When one adds the additional complexity of gravitational time dialation, with time being crucial to establish velocity, then one must also calculate the mass of the object on which the measured object rests, and at what altitude. (The two cecium fountain clocks that exist; one in boulder, CO and the other in England, are in fact nearly a second off from each other, and both are exactly correct given that Boulder is nearly 1 mile above sea level higher than its british counterpart.

What was the question again?

[JohnKSa]

The bullet's frictional drag on the barrel is what stretches revolver frames and cracks autopistol slides at the junction of the breechface and the ejection port...which I've seen several times through the years. That's a pretty signifigant force.Without getting into all that you can still calculate recoil velocity accurately and precisely.

REGARDLESS of what happens INSIDE the system before the ejecta exits, after the ejecta exits then all you need to calculate the velocity of the recoiling mass is the principle of conservation of momentum.The only significant outside force working against the bullet is barrel friction.That force is a factor while the bullet is still in the barrel, but once the bullet leaves the barrel all you need to know to calculate the velocity of the recoiling mass is conservation of momentum.

The barrel friction was part of the system that determined the muzzle velocity of the bullet--therefore any contribution due to barrel friction is already taken into account in the ejecta momentum figure.The momentums can't be equal under those circumstances.You're complicating the issue by trying to figure up all the forces before the bullet exits. That is a very complicated problem.

You don't need to do that if all you want to know is the recoil velocity. Conservation of momentum gives that to you on a silver platter.

All those "outside forces" you keep talking about are not "outside forces" at all. They are VERY much "INSIDE forces". An outside force would be something that acted on the bullet once it left the barrel therefore affecting its momentum at that point. An outside force would be something that acted on the firearm from the OUTSIDE thereby changing its recoil velocity or effective mass. The ejecta (including the bullet) is not an "outside force" acting on the firearm in any practical sense as far as calculating recoil velocity is concerned.

Assuming nothing is acting on the ejecta other than the act of firing (which includes everything that the firearm does to it) and assuming that nothing is acting on the firearm other than the act of firing (which includes everything that the ejecta does to it) then conservation of momentum applies once the ejecta is clear of the barrel. That is what is meant by "in the absence of outside forces".

Trying to apply conservation of momentum and figuring all the forces on the bullet and firearm while the bullet is still in the gun is a nightmare. It can be done but there's absolutely no point in doing so if all you want to know is what the velocity of the recoiling mass will be.

[larryh1108]

Ohhhhhhhhhh, my head hurts! :banghead: :what: :confused: :scrutiny: :barf:

However, all else being equal, the intelligence on these boards is mind numbing. Thanks to everyone who likes to contribute to these boards. Sharing your knowledge and experience is priceless!

[smithmax]

I'm amazed by how well you guys can discuss. It's mostly over my head, but I'm trying to soak it up and understand it. Thanks!

[WingRider]

:what:+1 on the info. I'm 35yrs past my last calculus class. But great info just the same. I love THR:D You can learn MORE than you ever wanted to know:neener:

[1911Tuner]

I may be misunderstanding what you've written John...but calculating one side's velocity by using conservation of momentum would require that you know at least one velocity...the bullet's...and to assume equal momentums for both sides.

But both sides can't be equal in this instance...whether the bullet is inside the barrel or not.

If momentum is a function of Mass X Velocity...and it is...and the slide's velocity is reduced as soon as the force is removed...and it is...and the slide's velocity loss is determined by the amount of resistive force imposed on it...then all you have to do in order to change its velocity at any given point in its travel is to change the recoil spring's load. i.e A 20 pound spring decelerates the slide more rapidly than a 10 pound spring, and reduces its momentum, while the bullet's momentum is constant, and only affected by air friction and gravity once it exits the barrel...assuming equal velocities and bullet mass for both shots and equal/constant wind speed and direction.

And...Calculating the slide's velocity and its momentum by using conservation of momentum formulae is tricky at best...due to the fact that the resistive force on the slide changes with each fraction of an inch that it moves. Doable, I'm sure, because there are some real math whiz kids in the audience...but it would likely require 3-4 pages of equations to determine something that doesn't really answer the question: "How fast does the slide move" because the question is sort of open-ended.

Speed at which point? Speed at the terminus of its rearward travel...or its speed just before that point? Speed at the halfway point...or at the beginning? Speed as it returns to battery...or just as it gets to battery?

Much simpler to time the full cycle and figure the average speed for the distance...sort of like the old VASCAR/TDS units that Officer Friendly used to cause your insurance bill to go up. TDS. Time...Distance...Speed. And even that only calculated the average speed of the vehicle as it traversed the timing points.

So...Maybe it would be better to see how quickly the slide moves from dead stop to dead stop, and go from there. There's a rocket scientist from Paraguay on another forum who calculated it at about 22.5 fps average speed on the recoil cycle firing a 230-grain bullet at 830 fps and using a 16-pound recoil spring. Mainspring wasn't mentioned, nor the firing pin stop's radius...but I'm assuming standard spring rate and 7/32nds radius on the stop. Change either of those three things...and the slide's average speed changes...and as its velocity changes...so does its momentum.

[James T Thomas]

I have an approximate calculation made by another, more simple method.

-For all of us who have developed eye strain.

In order for a phenomenon not to register on our brains, it must occur in less than 1/25th of a second. That is the "flicker" speed of the shutter on the old fashioned movie projectors that would enable the viewer to percieve continuous motion from projected frames of film.

With me so far?

My presumption: the slides must be moving at 1/25th of a second or less; not to be seen. I have seen some move; just barely, and so my conclusion that they cycle at approximately this quick of time.

Being that velocity is simply defined as the speed divided by time, then:

1/3 foot divided by say 1/30 second = 10 feet per second.

I'm sure many may travel faster than this, but we are discussing what would be common case.

Now for those with eyes like Jelly Bean Bryce or who see the bullet trail through the raindrops, they may see that old slide slipping along.
Not me.

[larryh1108]

That makes sense but where does the 1/3 of a foot come in?
That's 4" and by using 4" you imply that the slide moves 2" in
each direction. That seems a bit short. Did you figure it one way instead of round trip? if so, then it comes in at 20'/sec which is close to some of the other posts using advanced calculus and a slide rule!
However if the OP wanted to know the speed of the slide in it's rearward motion only then the forward motion would be a moot point and all the algorithms here would be invalid, no? I've lost track if the answers were for a complete cycle or a 1/2 cycle. Sigh.
;)

[James T Thomas]

One third of a foot must be considered because of my premise that the slide motion; total was so quick it was not seen.

Not quick in velocity, for an arrow at much higher velocity can be seen, but the time length, the duration of event.

Therefor, the motion fore and aft, to and fro is used.

Four inches of travel.

[JohnKSa]

If momentum is a function of Mass X Velocity...and it is...and the slide's velocity is reduced as soon as the force is removed...and it is...and the slide's velocity loss is determined by the amount of resistive force imposed on it...then all you have to do in order to change its velocity at any given point in its travel is to change the recoil spring's load. i.e A 20 pound spring decelerates the slide more rapidly than a 10 pound spring, and reduces its momentum, while the bullet's momentum is constant, and only affected by air friction and gravity once it exits the barrel...assuming equal velocities and bullet mass for both shots and equal/constant wind speed and direction.At the instant that the ejecta leaves the barrel the momentums are equal. As soon as anything (friction, spring pressure, etc.) affects the velocity of the recoiling mass or as soon as drag affects the velocity of the ejecta then they are no longer equal.

What the conservation of momentum gives you is the velocity of the recoiling mass at the instant that the ejecta exits. If you want an continuous plot of the slide velocity from the beginning of travel to the end of travel as it works against friction and spring force then things will get complicated unless you make some assumptions to simplify the problem.Speed at which point? Speed at the terminus of its rearward travel...or its speed just before that point? Speed at the halfway point...or at the beginning? Speed as it returns to battery...or just as it gets to battery?This is a job for high-speed (slow motion) video work. Trying to beat it to death analytically for exact figures will be painful.

HOWEVER, if you want a pretty accurate idea of slide velocity you can rapidly bound the problem with some simple calculations. The first one would be to calculate maximum slide velocity using conservation of momentum. From there it's possible to make some assumptions and figure the slide velocity at the point that it impacts the slide by neglecting friction, taking the maximum velocity and determining how the spring force(s) will affect it.

It's trickier to calculate forward velocity since you get a "bounce" effect from the impact of the slide on the frame--assuming the slide is still traveling at significant velocity when it makes contact with the frame.