Power Factor vs Recoil spring lb

[anyducks]

I am working on some Power Factor (pf) loads and wondering if there is a valid relation ship between pf and recoil spring tension in lb.

The following are constants (I believe)
Gun weight, barrel, primer, case, and bullet weight (grains),

variables
powder charge
bullet velocity (FPS)

prediction
equillivent slide action (ejection of case)
recoil spring lb need to operate the slide

125 grain
1100 fps
PF constant 1000

grains x fps / PF constant 1000 = PF
125 1100 / 1000 = 135.5

recoil spring resistance lb = rs lb
18.5 lb

pf / rs lb = pf / rs lb cofiecent
135.5 / 18.5 = .135

at this point the slide reliably ejects the spent case and loads the next round.


Changing fps to 900 the pf drops to 112.5 (expected).
Presuming the ps / rs lb is valid then a recoil spring of 15 lb would produce .133 (ps /rs lb) and similiar reliability?

[SL1]

I'm not sure that you have all the variables, or set them up adequately.

For one thing, I think you would be better off working with the slide weight, rather than the weight of the whole gun.

Also, the buring rate of the powder probably makes some difference, even though the power factor is the same with two different powders and the same bullet.

And, without thinking about it very carefully, I am suspecting that different bullet weights will have an effect on slide velocity, as well, even though they are loaded to the same power factor.

Hopefully, somebody who has thought about it more carefully will respond and teach us both something.

SL1

[Brian Pfleuger]

I agree with SL1.

There are far more variables and it is far more complicated.

Anyways, a little story....

I was noticing a little slide peening on my G33 (357sig) and I know that it uses the same recoil spring as the 40SW, but produces considerably more energy.

So, I ordered a Wolf 21lb recoil spring and guide rod, which I expected to make a pretty significant change to recoil and the distance that the brass was ejected..... it did neither.

Then, I decided to try it with some really light loads, sub-9mm power levels. This spring is like 75%-100% over power for these loads. I figured, for sure, that the gun wouldn't even operate correctly..... wrong again. It works fine. In fact, I can't even tell the difference. The brass even lands in basically the same spot.

The ONLY thing I see different with this spring is that when I drop the slide it will sometimes jam the round against the feed ramp. Almost seems like it would be less likely to do that rather than more, doesn't it? It also destroys the round when it does that, crams the bullet right back into the case.

Recoil springs are not as simple as it would seem.

[COSteve]

HERE's a decent rifle recoil table to give you a starting place to verify your readings.

[Unclenick]

Well, the problem with recoil tables is they are in energy, which is proportional to what the shooter feels. When you are looking at the mechanics of gun operation, you are interested in momentum. The momentum imparted to the slide and barrel while they are still locked together will be equal and opposite in direction to that of the bullet plus that from the muzzle blast impulse (rocket effect) when the bullet exits.

Rocket effect can be half or more of the total recoil in some overbore rifles. If you ever wondered why muzzle brakes that simply blow outward in all directions accomplish anything, it because they relieve pressure before the bullet uncorks the muzzle, thus reducing rocket effect. They do nothing to reduce recoil from accelerating the bullet unless they are directed rearward, as the arc of a clamshell brake does. Rocket effect is a lower percentage of total recoil in pistols, but is still quite noticeable and depends on powder choice. It is greater with slower powders whose larger charges have more mass to eject, and do so at higher muzzle pressure.

The slide and barrel's combined mass being greater than that of the bullet and propellant is mainly what is responsible for determining their initial rearward velocity in reaction to bullet acceleration and rocket effect. The recoil spring then decelerates it over the distance to the frame, transferring a portion of that momentum to the frame and shooter, with the rest occurring at the point the slide impacts the frame.

So, changing springs will change the percentage of slide momentum that transfers to the shooter before the slide and frame impact. That's it. As long as a spring still has the powder to strip a fresh round from the magazine and feed it, you will have function, even if you change power factor. You will just have more frame battering.

QuickLOAD calculates rocket effect in energy, from which, if you know your slide and barrel mass and travel distance during bullet barrel time, you can work out the momentum. Then, knowing the spring's average force over the distance the slide travels back to meet the frame, you can work out how much it transfers to the frame by decelerating the slide before impact and what more force you need to keep the impact either constant or proportional. The barrel's momentum will transfer through its link, so the shape and type (pivot or slot) will determine that transfer function after the slide and barrel unlock. That part has little to do with feed function, though.

Frankly, you don't want the spring so stiff it mashes cartridges. Once you have a maximum spring that functions without mashing and that doesn't make racking the slide too awkward, if you are then trying to further reduce slide impact, a buffer system is a better way to go than further stiffening the spring. Sprinco has a good one.

[zippy13]

One variable, in the dynamics of the action cycling equation, that hasn't been mentioned is the fixure of the reaction: How do we quantify the firmness and consistency (or inconsistency) of the individual shooter's grip? And, what does a stiff or loose grip have to do with actual versus the theoretical spring selection?

[Jim Watson]

So, changing springs will change the percentage of slide momentum that transfers to the shooter before the slide and frame impact. That's it. As long as a spring still has the powder to strip a fresh round from the magazine and feed it, you will have function, even if you change power factor. You will just have more frame battering.

And battering is not a severe a problem as you might think.
Many IPSC competitors use substantially reduced springs to affect the feel of recoil and their guns hold up for high round counts.

It might be better to call it the counterrecoil spring since its main function is to feed in the next round and close the slide. Mechanical lockup and inertia are the main actors during recoil, moreso than the spring.

[jmorris]

And battering is not a severe a problem as you might think.
Many IPSC competitors use substantially reduced springs to affect the feel of recoil and their guns hold up for high round counts.

Yes, I run an 8# spring in a 9mm (132 PF) and 11# in 45 (170PF).

Another addition to the information above is that on 1911 style pistols the main spring and even the firing pin stop can dramatically effect how the pistol recoils.

[Clark]

My father has dozens of gun patents. He was for 40 years chief engineer over 150 engineers and draftsmen designing guns and vehicles. He pounds of tables and screams in people's faces. I am a lazy over paid consulting engineer. I take after my mother. I called him up and asked him how to calculate this stuff.

15 pound to start and 42 pound spring at the rear [home made spring
assembly], is just right for 10.3 gr of AA#5. That is, it is just enough
powder to keep the Patriot from jamming with that spring.

The bullet leaves the gun at 1100 fps.
The bullet weighs 185gr = 185/7000 = .026 Lb.
The slide weighs .5 Lb.
The barrel weighs .1 Lb.
The spring is 15 Lb to start.
The spring is 42 lb at back.

The momentum of the bullet equals the momentum of the slide and barrel.
[1100][.026]=V[.5+.1}
V=47.7 feet per sec
Velocity of slide and barrel = velocity of slide
Energy of slide = .5mVV= one half mass velocity squared
Es=.5[mass of .5 pounds][47.7][47.7]
Mass = [wieght]/gravity= .5/32.2=.0155
Es=.5[.0155][47.7][47.7]=17.64 foot pounds of energy

The energy required to pull back the slide = [force][ distance]
Force = average force = [15+42]/2 = 28.5 pounds force
distance = 1.656"=1.656/12=.138 feet of slide stroke
Eslide = [28.5][.138]=3.93 foot pounds

BUT WAIT A MINUTE! 17.64 DOES NOT EQUAL 3.93!

Not all of the bullet momentum went into the slide and barrel.
Some of it accelerated the hand.
Only 3.933/17.64= 22% went into the slide and barrel.

How far do the cases fly:

1) For a semi-automatic case fired 5 feet form the floor, the time to
reach the floor is
t= square root ((2)distance / acceleration )= root (5'/32'/sec/sec)= .54
seconds

2) In a Colt .45 the ejector contacts the case when the slide is back
1.3". The farthest the slide can travel is 1.8" where it hits a stop. If
the spring is perfectly sized for the gun and the round, then the slide
will just run out of energy at the stop. Assume Vslide = 0 at 1.8".

3) Energy Slide at 1.3 inches = (force) (distance)= (16lb
spring)(1.8-1.3=.5")=.66 foot pounds of kinetic energy left in slide
when it hits the case

4) Energy is also = 1/2 mass velocity squared = .5 (mass of slide=
weight
of slide/ grav accel=12 oz/32 ft/sec/sec)(V squared)

5) Combining equations 3) and 4): Vslide at 1.3" = square
root(E/(.5mass)) = root(.66 ft lb/((.5)(.023 lb sec sec /ft)) = 7.6
feet/sec

6)Center of gravity [this should be moment of inertia, but that would be
work] is .25" from extractor claw and ejector hits the
case at .35" from the extractor claw.
Velocity of case = (.25"/.35") velocity of slide at 1.3" = (.25/.35)7.6
feet/sec = 5.4 feet per sec [here's your answer] = 3.6 miles per hour

7) Combining 1) and 6): Distance case travels=
(Velocity)(time)=(5.4ft/sec)(.54 sec)=2.9 feet horizontally from the
gun
And Wolff FAQ wants your empties to land from 3 to 6 feet. They must
want the slide to barely hit the stop.

[anyducks]

Great information and discussion.

Thanks