Physics of Slide Wt./Spring Wt.?

[Higgins]

Can anyone with some engineering/mechanics knowledge explain the physics involved in determining the necessary weight(mass) of a pistol slide and corresponding spring wt.?

I get the basics - equal-opposite-action, conservation momentum, etc...

What I'm interested in is how to calculate necessary slide mass/spring wt. for a given cartridge, and how the slide mass/spring wt. interrelate.

Any help is appreciated. Thanks.

 

[BevrFevr]

i'm no rocket scientist but

I bet the 'equation' ain't simple. Lots of variables i would imagine like friction, method of lock up, frame material, etc. Of course if you want to set things like reality aside I image the equation gets quite a bit simpler.

Good luck with that!

 

[jungle]

In a Browning style tilting barrel locked breech pistol the design parameters generally have the objective of sufficient mass to allow the bullet to leave the barrel prior to the slide/barrel assembly recoiling to the point at which unlock begins. In a 1911 style .45ACP this is about 17 ounces for slide and barrel. They reach a peak speed of 20-25 FPS in recoil. Roughly the same weight will work for a 9mm in a 1911 because although momentum is lower, it is close enough and results in lower slide speed and therefore needs less spring weight to slow the slide at the end of recoil. Some 9mm 1911s have lightened slides to more closely approximate the momentum balance of the 1911 in 45ACP.
It is primarily the mass involved that allows the pistol to remain locked, the recoil spring acts to slow the slide at the end of recoil and then feed the next cartridge and close the slide.
Use momentum equations to balance the movement of the bullet to the mass of the slide/barrel so as to allow the bullet to exit prior to unlock and you are set. In the 1911 the barrel/slide must recoil about .125 inches to initiate unlock, while the bullet moves about five inches.
This is different for different types of actions such as blowback or gas operated. It has been estimated that a .30'06 blowback would require an 8lb breechblock to work, which is why we don't see any .30'06 blowback operated weapons. A blowback 9mm SMG will use a bolt heavier than most pistol slides, due to the higher mass required to hold it closed on firing.

 

[Doggieman]

2x + 3y = big ass slide

 

[Vern Humphrey]

Or to put it another way, calculate slide velocity using m1 * v1 = m2 * v2.

Determine how far the bullet must move to exit the bore, and how far the slide will move at the same time. Slide movement should be less than the distance needed to unlock.

slide weight = 17 oz
slide mass = 0.032996894 sl

bullet weight = 230 gr
Bullet mass 0.001020408 sl
bullet velocity 850 fps


v2 = 26.28571429

movement ratio = 0.03092437

slide movement = 0.154621849

In other words, a 17 oz slide and barrel will move back about 0.15" by the time the bullet exits the barrel. In actual practice, it will move less that that relative to the frame, since the frame will also move under recoil.

 

[NateG]

I think Bevr's got it right... the equation isn't simple at all. You could get a good starting point by calculating the initial speed of the slide under recoil (from conservation of momentum---assuming the frame stays put mass of bullet * velocity of bullet = mass of slide * velocity of slide, solve for velocity of slide), use 1/2 * m * v^2 (of the slide) to find the energy of the slide. Then figure out how far back the slide needs to move, and solve for the spring tension that gives you the same stored energy when the spring is compressed by the recoil distance as the slide had initially. (I'm not sure how the springs are rated, but the general form is E = 1/2 * k * x^2, where x is the distance the slide recoils)

So, that (or something similar) gets you the spring tension to start from. It'll probably be too high as it didn't consider the energy stored by cocking the hammer (if applicable), friction on the rails, friction from extracting the casing, etc. But it'd give you a starting point, then you could play with it from there. Unfortunately, the step with the spring tension is the problem. I just don't know what a 12# spring corresponds to. I'd try using the spring tension for "k" in the equation and see if that gives you something reasonable, but, again, I don't know what value the spring rating corresponds to)

Basically, it'll come down to trial and error, if it doesn't extract, go to a lighter spring. (Better to start too high so you don't batter the frame)

 

[JNewell]

The whole thing is enough to make you wonder why so many people insist on changing pistols from the way the design engineers at the manufacturer set the gun up...

 

[Vern Humphrey]

The whole thing is enough to make you wonder why so many people insist on changing pistols from the way the design engineers at the manufacturer set the gun up...

That's why I carry a full-size M1911, instead of a chopped and channeled micro-gun.

 

[Higgins]

Thanks everyone for your replies. Very helpful in thinking about this.

The more I look at and learn about the issue, seems to be a question of unbalanced forces acting over different times and distances -

The firing of the cartridge imparts a large impulse (F x t)(large Force, but for short period of Time) on the slide through the base of the cartridge. This impulse is enough to accelerate the slide very quickly in the first few millimeters, overcoming the countervailing spring force holding the slide forward (ignoring friction for simplicity sake). However, the impulse is gone very quickly (b/c pressure in chamber is dispersed quickly), leaving an accelerated slide that now is left to be slowed over a greater period of time/distance by the recoil spring. Does that seem right? Now I just need to work it out in mathematical terms.

This raises a question I never thought about before - Is the slide stopped at its rearward movement by the spring slowing it down and ultimately reversing the slides direction or is the slide stopped by some part of it crashing into the frame and the spring then just pushes it forward? If not the latter, then why recoil or shok-bufs to stop frame battering?

Again, thanks for any replies.

 

[jungle]

Does the spring or the slide hitting the frame stop the movement? Good question, it will vary with the pistol, load and spring-some are stopped by the spring and some by the frame. I think you will find most stopped by the frame, but by then the slide velocity has been slowed greatly by the recoil spring. Both the hammer and recoil springs slow the slide velocity in recoil as the mass inertia of the slide is overcome.
Most of you are on the right track, but the initial impulse of the cartridge firing is far greater than any normal spring could overcome and as a result the momentum of the slide/barrel mass is of primary importance in the question of maintaining lockup.
Otherwise one could build a pistol with a very low mass slide and compensate with spring weight and we know that this has not been done in a realistic or practical operating weapon of the Browning type recoil operation.
Remember that pressure is highest and acceleration lowest at the beginning of the event and pressure lowest and speed of the projectile highest at the end of the event.
Pressure does not cause slide movement, acceleration of the projectile is the basis of recoil operation. Although the pressure causes the acceleration, it is not the force that operates the weapon.
Vern brought up a good point: the recoil spring ties the slide to the frame to some extent and by extension(sorry) to the shooter thus increasing the effective mass to some degree. There is an acceleration of the projectile that is not linear and is not accounted for by using final projectile velocity in a momentum equation, but I think we have seen that it is close enough to get you into the operating range.

 

[mjolnir]

Don't forget the mass of the hammer. Small, but it's in there too.

The amazing thing is, JMB probably didn't know the difference between a pound-foot and a long walk through the country.

 

[Riphalman]

Not just the mass of the hammer, but also the force of the mainspring.

 

[.45man]

There have been some great posts here.....some well thought out points, great job guys!!

Higgins, QUOTE: "The more I look at and learn about the issue, seems to be a question of unbalanced forces acting over different times and distances - "

Yes, this is so true...lots of real world problems in life are classically defined as static-linear problems with simple equations. But life is not static, nor is it linear. Take fluid mechanics... there are equations which define fluid motions (Navier-Stokes partial differential equations, see Cool link for fellow pocket protector types). In order to solve them, you must make some bold assumptions and simplifications, otherwise it takes a Cray super-computer to even get close to solving them. That is where some testing and data analysis comes into play..... Pure brain horsepower alone cant solve these type problems, you have to rely on intuition, experience, and testing, not to mention allowing some compromises which undoubtedly must be made in the design.


mjolnir,
I know this may be getting a little off topic, but I would not be too fast to discount JMB's intellectually capabilities. I am a mechanical engineer and have a deep respect and appreciation for his ability (100 years ago) to mate concept, intuition, and practicality.

With a slide rule and some testing, this man produced some of the most profound firearms, which have changed the shape of the world to this day....

I am sure like Edison, JMB had is share of failures....but he had equally great successes.....

My hat goes off to the man!! Wish I could have spent just a day talking with him and listening to how he thinks, and visualizes things working in a fraction of a second.....

I can only imagine that there are people living today with his capability, but hamstrung by management BS, email, and political correctness

 

[mjolnir]

.45man

I think you inferred too much from my attempt at humor.

JMB was an intuitive mechanical genius of the highest order...a literal Mozart of Mechanisms. To the best of my knowledge, he had absolutely no formal engineering or scientific training whatsoever.

Also, the only rule in use at the Browning's was a carpenter's rule. I don't believe he used a slide rule at all.

Fact is, an analytical understanding of physics (or anything else for that matter) is no guarantor of any sort of creative synthetical design ability.

 

[.45man]

mjolnir,

Sorry. You are correct, it is hard to read "inflection", or intent on web-boards without using different "face icons.".......

......."Fact is, an analytical understanding of physics (or anything else for that matter) is no guarantor of any sort of creative synthetical design ability.".......

Boy isn't that the truth.....some of the most well educated folks I know, could not be creative enough to get out of a wet paper bag!!

Thanks!!

 

[1911Tuner]

Masses of Stuff

Vern stated:

In other words, a 17 oz slide and barrel will move back about 0.15" by the time the bullet exits the barrel. In actual practice, it will move less that that relative to the frame, since the frame will also move under recoil.
***********************

Actually, the bullet will exit at about .100-.110 inch of rearward slide movement. Vern's calculations are correct, assuming only the mass/acceleration of the slide vs mass/acceleration of the bullet. There are two factors missing from the equation. The weight/mass of barrel and bullet.

With the the 1911 and others that are short-recoil operated, the slide moves rearward and takes the barrel with it (via the engagement of the lugs)for a short distance...thus adding the weight and mass of the barrel to the slide...and, since the slide starts moving while the bullet is still in the bore...the weight and mass of the bullet must also be added.

Mathematicians! Do your thing!

 

[jungle]

Howdy Tuner, The 17 ounce figure was for slide and barrel and taken from our earlier discussion. I think it is pretty close. I don't know the actual weight but this was the value we used. If anyone has an accurate scale and would weigh the slide/barrel of a 1911, Glock and BHP it would be fun to use actual numbers.The slide and barrel do move backward together until unlock is complete. The bullet won't be moving backward, it's mass accelerating in the opposite direction is causing the slide/barrel to move so it goes on the opposite side of the equation.
The non linear acceleration of the bullet may account for at least part of the difference in reality and our simple momentum calculation along with friction and the recoil spring joining the mass of the frame to slide to some extent.

Here is a very good post on the spring's effect on the pistol by an engineer:

http://forums.1911forum.com/showthread.php?t=2977&pp=25

TiroFijo seems to have a very good grasp of this and does a good job of explaining it-not just the springs, but the dynamics of the recoil operation.

V/R jungle

 

[pbass]

Someone somewhere said recently that the heavier the spring, the tighter you have to hold the gun to avoid the limp-wristing effect. Now that's something I can wrap my aging brain around. Another down-to-earth post is at http://www.custom-glock.com/springtech.html

 

[1911Tuner]

Massive

Howdy Jungle...Good to see ya up and at'em.

I weighed a slide and barrel together several years ago, and the combined weight was a tick under 21 ounces (Don't forget the bushing and the recoil spring plug.)...14.5 of which was the slide. A cast slide will be a bit lighter than barstock, so that'll figure in too. Other variables would be bull barrels...bushingless/reverse plugs...firing pin stop radius...even the firing pin itself factors in. A Series 80 Colt firing pin weighs a few grams less than a
pre-Series 80 type pin, assuming similar materials.

I worked out a jackass formula a few years back for figuring recoil spring loadings.

Bullet velocity X .02. For bullets lighter or heavier than 230 grains, add or subtract the percentage difference.

Example:

850 X .02 works out to be a 17 pound spring...but GI Hardball specs called for
825 fps +/- 30. The original spring rate called for 16.5 pounds...which split the difference.

Example 2:

200 grain bullet at 900 fps=18 pound spring minus the roughly 10% difference
in bullet weight...works out to be about 16 pounds for the spring. Bump the velocity up to 1000 fps and your spring rate should be about 18 pounds.

It's not exact, but will getcha in the ballpark if you're unsure of which spring to use with a given load.

 

[mjolnir]

...the weight and mass of the bullet must also be added.

Sorry to say, Tuner, 'tisn't so...reaction to bullet movement is instant and in the opposite direction. Visualize it this way...imagine a case that's actually longer than the barrel, so that the bullet is sticking out beyond the barrel. The reaction ("recoil") effect is the same. Or, imagine a very sub-caliber bullet, so that the bullet never actually touches the barrel (ignore gas blowby). Or even no barrel at all (picture unburstable titanium case with internal rifling grooves). Ditto. The bullet is a free body when it begins to move.

Momentum discussion follows.

(MASS x VELOCITY) of everything moving in the bullets direction

=

(mass x velocity) of everything moving in the direction 180 degrees opposite (velocity would thus have a negative sign).


MASS consists of:
1. bullet
2. powder
3. negligibile amounts of primer gas & material

mass consists of:
1. slide and all its components
2. barrel
3. empty case/spent primer
4. hammer (the inertial rest mass of the hammer must also be overcome)

(Of course, just using hammer-fired, short recoil pistols as an example here). Hope that helps.

 

[1911Tuner]

Sorry

Quote:

>Sorry to say, Tuner, 'tisn't so<
*************************

Correct...I wasn't thinkin' straight. Bullet isn't part of the barrel's mass..even though it's "attached" via friction...it's moving in the opposite direction, negating its mass in the total.

I do have to argue with this one though...

>Visualize it this way...imagine a case that's actually longer than the barrel, so that the bullet is sticking out beyond the barrel. The reaction ("recoil") effect is the same. Or, imagine a .30 caliber bullet, so that the bullet never actually touches the barrel (ignore gas blowby). Or even no barrel at all.<
***********************

Three criteria for full slide cycle are:

The bullet must be there.
The bullet must move.
The bullet must be there for a long enough time.

One thing to keep in mind...The gun is recoiling as long as the bullet is in the barrel and moving. Once it exits or stops moving...recoil is over. What movement that occurs after that is purely momentum...and momentum gained in the first .100 inch of travel is what carries the slide through to full cycle.

Things that I've noticed in dealing with the chopped variant 1911s:

Light fast bullets are much more likely to produce short-cycle malfunctions than slow, heavy ones. Pondering on the "why" brought me to a conclusion
that...since the momentum of the slide is gained in such a short time, (Nominally .100 inch or so...bullet dwell time within the barrel becomes more critical as the barrel gets shorter and the bullets get lighter and faster. This, even though the 185-grain +P screamers induce higher recoil impulses on the slide than the old 230 hardball. Simply stated...Bullet exit too early in the slide's travel...say at .070 inch instead of the .100-.110 inch means that the slide hasn't gotten its full dose of momentum. I've seen too many of the little pistols puke on the 185-grain high performance ammo, but settle down and
run like Timex watches with PMC or Winchester ball. In order to tune the guns to run with the light, high velocity bullets, a lightened recoil spring, and
often increasing the radius of the firing pin stop were needed. Once done, the short-strokes, stovepipe failures to eject, and bolt-over base misfeeds ceased.

So, I have strong evidence that it's sometimes not enough to push the slide hard...you also have to push it long enough.

 

[mjolnir]

Sorry to be a boll weevil in your cotton patch, but....

From a physics standpoint, neither the barrel nor bullet are really necessary (wouldn't be much of a gun then, but that's another story).

The slide doesn't care if it's 230 gr of lead, 230 gr of Chanel No. 5, or 230 gr of concentrated ovine flatus; as long as some material object of sufficient mass is accelerated downrange, the slide will accelerate rearward.

You could, from my previous example (case held by extractor, no barrel), remove the bullet entirely and replace it with a small rocket motor (that burns only very briefly). As long as the rocket generates enough gas with enough velocity, the slide will recoil.

Conceptually, a pistol slide is identical to a small railroad handtruck, free to move back on forth on its 4 wheels on the rails. It's some distance away from a brick wall, with a [coil] spring between the truck and the brick wall.

Let's say there's a guy on board with a 16 lb. bowling ball. He throws the bowling ball downrail with all his might. The truck will "recoil" towards the brick wall. The force accelerating the bowling ball was not constant, therefore the acceleration was not constant. Once the ball leaves his hands, it has gotten all the acceleration it's going to get. It therefore continues forth at a constant velocity (Newton #1: in a gravity-free vacuum it would continue in a straight line forever). The guy+truck reacted equally and oppositely (and simultaneously). The guy/truck is much more massive, so their rearward velocity is proportionately lower. If the rear of the truck was unconstrained, they would also continue on down the rails with a constant velocity forever (neglecting the usual suspects of course).

He could just have easily let loose with a fire hose, say a brief 1/4 sec burst of 16 lbs or water. Same reaction.

The reaction force only exists as long as the two bodies are in contact.

Back to pistols. That contact is broken when the barrel links down. It's now part of the stationary frame. So, theoretically, the bullet could stop in the barrel, but that wouldn't affect the rearward impulse already imparted to the slide (practically speaking, not enough rearward velocity would have been imparted to the slide to cycle fully).

Not meaning to pick on you Tuner, just trying to clarify the physics involved.

 

[CAnnoneer]

You don't need to get the complete equation right, before doing some experimentation. The scientific approach is to start with a simple model. Then experiment to gauge agreement, adjust the model as necessary, iterate.

Assume a blowback design for simplicity. Conservation of energy and conservation of momentum give you two equations where the masses are known, the energy released is known, and the only unknowns are the two velocities (bullet and slide). Simple algebra produces answers for both. Then you can separately calculate how the kinetic energy of the slide is converted into potential energy of the spring. The desired recoil distance will determine the required spring constant.

That is a simple and easy calculation which will be a good first approximation. Then stick a gun in a vice at a range and fine-tune the spring to avoid FTE's etc.

 

[1911Tuner]

Boll Weevils

Da Quote:

>From a physics standpoint, neither the barrel nor bullet are really necessary (wouldn't be much of a gun then, but that's another story).<
***********************

Ahhhh...We may have to carry this to another thread, since it's veering slightly high/left of the original topic...but I beg to differ. The barrel IS necessary. Not for recoil impulse, but for proper function. In your rail car analogy, you're assuming straight blowback. The locked breech/recoil operated pistol doesn't function like that.

See Jerry Kuhnhausen's well-versed, but completely incorrect "Balanced Thrust Vector" theory of operation and stydy it closely to see that it can't work that way.

Understand this: If the bullet doesn't move, the slide won't move. If the bullet exits BEFORE the slide moves...the slide won't move. If the bullet CAN'T move...the slide won't move. Jim Keenan proved that much by threading the muzzle of a 1911 pistol, and inserting a steel rod fully against the bullet...and locking it all together with a set screw. He fired the gun repeatedly, and the slide didn't budge. No damage was done to the gun, and when reassembled with a standard barrel, it functioned perfectly. Of course, with any action, you'll get an equal and oppostite reaction. In the gun with no barrel, though...it won't be enough of a reaction to drive a 14-ounce slide backward under the tension of the recoil spring. Won't happen, Kemosabe.

The 1911, et al...does not lock until it fires...and then it locks under pressure as the barrel is held forward by the friction of the bullet in the bore with the gasses thrusting against it...and the same gasses slam the slide rearward. The barrel and slide being driven in opposite directions with the locking lugs engaged in a ballistic tug of war...Front barrel lug walls
against rear slide lug walls. When the bullet exits, the pressure drops...and it unlocks. The link does NOT unlock the barrel from the slide. It only draws the barrel down and out of the way of the slide.

 

[mjolnir]

..If the bullet doesn't move, the slide won't move. If the bullet exits BEFORE the slide moves...the slide won't move. If the bullet CAN'T move...the slide won't move. Jim Keenan proved that much by threading the muzzle of a 1911 pistol, and inserting a steel rod fully against the bullet...and locking it all together with a set screw. He fired the gun repeatedly, and the slide didn't budge. No damage was done to the gun, and when reassembled with a standard barrel, it functioned perfectly.

I'm surprised Mr. Keenan wen't to the trouble, as that result would be expected from the physics. If the bullet is not free to move, it undergoes no acceleration, and there is no movement in any direction. The exploding gases within the case push outward in all directions equally. The reactions from the case (and barrel) and bullet base are equal and in the opposite direction. There is not net unbalanced force, thus no accelerations.

Back to the sled...you fill the hollowed-out bowling ball with powder, seal it, and assuming the bowling ball is strong enough to contain the pressure, you ignite it. The sled/guy move in no direction at all, due to no unbalanced forces.

As far as blowback/short-recoil, the physics are essentially the same for both. Imagine a 14 lb. separate weight sitting on the front of the sled. The other end of the weight is connected with a chain to the railway. There 's enough slack in the chain that after the sled travels about a foot, the chain tightens and pulls the weight off the sled...that's short recoil.

The main purpose of the barrel travelling with the slide is to "delay" the separation of breech and cartridge...that is, the bullet has exited the barrel before the slide has moved very far...the bullet has high velocity, low mass; the slide has high mass, low velocity, thus hasn't travelled very far before pressure is released into the open atmosphere...it's a timing operation. (sidebar: it also lowers the necessary mass of the slide...JMB accomplishing multiple missions yet again).

Note the slide begins to move the very instant the bullet does...the reaction is simultaneous...the bullet's displacement is many inches for the slide's fractions of inches due to the reasons stated above. It's not quite correct to say the slide is delayed, rather it's acceleration is much lower than the bullet's (F=ma, a=F/m, F is equal to the bullet's F but in the opposite direction, m is much higher than the bullet's).

Of course, with any action, you'll get an equal and oppostite reaction. In the gun with no barrel, though...it won't be enough of a reaction to drive a 14-ounce slide backward under the tension of the recoil spring. Won't happen, Kemosabe.

Well, nothing's impossible! You're right, with standard gun stuff, it wouldn't move, but you could do it. You could lathe out a chrome-moly case with extra thick walls to contain the higher pressures and weld the base to the slide face. To compensate for the missing bullet and barrel mass, you'd probably have to use a low-grade explosive to produce the necessary velocity in compensation.

With that, I leave the cotton fields, while the gettin' is good.