Spring (and my mind) confusion!

[MAXM]

I need Detlef, DogmaDog and all other's experience to clear the mud in my mind...

A well-known spring manufacturer kindly sent me, for testing, some 1911 5" 45 ACP recoil springs, conventional rate.

I CAREFULLY measured them (my measures are in millimeters, 1 inch = 2.54 millimeters). In order, the first number is spring length, second is coil number, third wire diameter, fourth spring external diameter):

14 lbs: 167 35 1.08 10.54

15 lbs (spring n. 1): 167 35 1.08 10.53

15 lbs (spring n. 2): 158 32 1/2 1.14 10.68

16 lbs: 167 34 1/2 1.14 not measured

18 lbs: 169 31 1.14 10.75

for compairison:

Colt factory spring 167 32 1.09-1.10 10.95 (after 1800 rounds)

Brand XYZ, 15 lbs variable rate: 162 32 1.12 10.89

Brand XYZ, 16.5 lbs variable rate: 173 32 1.14 11.06 (after 200 rounds).

I'm confused! I can't explain. for example, the difference from the 14 lbs, 15 lbs n. 1 and 15 lbs n. 2 springs . help help help help....

[Dan Burwell]

VooDoo

Most likely material or heat treatment differences.

Dan

[Detlef]

if that much.... I wouldn't be surprised to find that they sell the exact same spring under two *spring weights*, or that they first run them all through spring gauges and then package them.

I'm not a spring expert, just a physicist. The most important measure is missing, namely the measurement of the spring constant (there's a few threads on here how to make a cheap spring gauge, I have survived 13 y of shooting w/o ever using one...)

Brian worked for a spring manufacturer at some point, I think...

Then there's always Wolff's ph no ....

--Detlef

[mcoliver]

First, probably because you got your dot wrong; 1 inch = 25.4 mm = 2.54 cm.

Beyond that, I just read what's printed on the label...sorry.

[Indiana James]

Hi MAXM,

You almost got it. I are an inginer (I'm not good with English, but I can write math equations!!!) and have studied springs.

How a spring works. First their was Newton. Newton says that

F=ma.

Force = mass * acceleration.

From this you get Hooke's Law.

F = kx

Force = Spring Constant * distance.

The Spring Constant is force divided by distance, e.g., lbs / inches.

So when we say a 15 lb spring, we mean the spring exerts 15 lbs when compressed one inch.

Now, what affects spring rate or Spring Constant:

Diameter of the wire

Wire material

Diameter of the spring = outside diameter of the spring - diameter of the wire

Number of active coils (i.e., not the "closed" coil at the end)

So with that Physics, let's talk about springs.

I'm pretty sure the manufacturer's rate their springs correctly. However, if you cut the spring the spring rate changes (aforementioned dependency - spring rate is dependent on Number of Active Coils). So, measure your springs if you need to cut them. You cut them to make them fit without binding. Brownell sells a recoil spring tester. It's great for measuring springs and seeing if they are going to bind.

Now, let's talk guns

For every force their is an equal and opposite reaction (Physics again, but I forget who proved that). So, the force that causes muzzle flip is the force from the explosion or F=ma. To simplifiy, the explisions pushes the bullet in one direction. This creates a force mass of bullet times it's acceleartion. The opposite force is the slide and you. The slide takes up some of the force in the opposite direction. The slide isn't enough to balance the forces, so you take it up and cause rotation. Your feet are the pivot point, the force pushes against your arms and your body wants to rotate back, but we end up bending the wrist or rolling the shoulders up instead. Some of us still rock backwards - oops! Burkett teaches us how not to do that.

Now, accuracy

The bullet is gone before the recoil spring compresses, therefore recoil spring rate has nothing to do with accuracy!

Now, muzzle flip

Muzzle flip is caused by the original force of the explosion causing the bullet to go forward, thus a backward force is generate. Some of that force goes into compressing the spring. The rest is resisted by you causing rotation up. So, the force generated by the slide hitting the back of the frame is very small in comparison to the forces causing muzzle flip. Remember, F=ma. When the slide hits the back of the frame it's acceleration is decreased alot by the spring. The force of the slide hitting the frame is not that much compared to the orignial force that sends the bullet down the barrel. Therefore, I would conclude that the spring rate has very little affect on muzzle flip and definitely nothing noticable to the shooter.

Now, cracked frames

The frame will take more abuse. How much? If slide travel is about 3 inches, then 3 times the difference in spring rate. So, if we drop the spring 1 lb the frame will experience and extra 3 lb's of force. If the manufacurer's recommend 18 lbs and we use 10 lbs, that is only 24lbs more. I say send cracked frames back and get the manufacruers to use better materials. I think STI and SV have that covered.

Now, cycle time

Absolutley affected by the spring rate. The force from the explosion stays the same. The distant the spring travels is constant. The mass of the slide stays constant. The only variable left is time. Time must decrease. That is the time for the slide to go up and back!!! Here I'll even do the math.

F=kx (for a spring) which = ma.

Force of the spring = Spring Rate * distance = mass of slide * acceleration.

So,

kx=ma

kx = m * x/(t*t)

Spring Rate * distance = mass * distance / time squared

Now, we get time. Do some algebra

k(t*t) = m

m = mass of slide and is constant, thus k and t are inversely propotional.

Conclusion we can draw from this equation.

As k (spring rate) is increased the time is decreased exponetial or by the power of two.

So, time is dramatically decreased by lowering the spring rate.

Walla! And that is why "race gunners" use the lowest spring rate that allows the gun to cycle properly.

I may not no my geography, but I know my physics!!!

Indy

[mcoliver]

So, time is dramatically decreased by lowering the spring rate.

Indy, you lost me here. (Okay, actually you lost me after Newton says that )

From your eq. k = (m / (t*t)), doesn't that mean k is inversely proportional to t? Which makes roughly, t decreases for every increase in k?

[DogmaDog]

Ugg.

First: The "weight" of a spring as indicated by the manufacturer is NOT the spring rate. A spring rated at 15 lbs exerts 15 lbs of force on the slide when the slide is fully retracted, and some lesser force when the slide is closed.

To figure out the spring rate, you'd have to measure the force when the action is closed, and measure the distance the slide travels.

Second: Cutting coils does NOT change the spring rate. If you cut off some coils, it will reduce the length by which the spring is compressed, both when the slide is in battery, and when the slide is open, so you will reduce the amount of force exerted by the spring on the gun, but you do it by reducing x, not by changing k.

Third: "Voila!"

There's a reason I'm called:

DogmaDog

[blackdragon]

OK, Now I have a headache----Bye

Ivan

SCS Vegas

[jkmccoy]

MAXM,

Danial97 answered your questions. The apparent inconsistencies in spring rating (not rate) and wire diameter are almost certainly owing to differences in alloy/heat treatment (wire material as Indiana J. points out).

Regarding some of the other comments (this turned out to be a long diatribe, if you are really, really interested in springs you might want to read it, otherwise you could just scroll to the bottom for the "bottom line")...

DogmaD - cutting coils off of a spring does change the rate. You are correct that it also changes length (and thus preload) so the total force applied to the slide (at any position) is different, but it also changes the amount of force necessary to compress the spring a certain distance (the rate or constant). In 1911 applications the spring is long enough that cutting one or two coils will change the spring rate so little that the difference in preload will be a MUCH more noticeable effect, but the change in rate is still there. Also, although the manufacturer's rating refers to the total force to compress the spring to slide stop, since slide travel is the same (mostly, I know some guns are a little different and some people use lots of slide buffs) the rating is proportional to the spring rate.

Indiana J - Nice explanation of spring physics, but I must differ on a couple of points. (I'm just a biologist/statistician but I've had a few physics courses and I've played with a few springs - in guns, motorcycles, trucks, etc.). Despite being a biologist/statistician and being able to write/work mathematical equations I have not suffered from an inability to write English.

Fortunately we do not have to be concerned with conservation of energy in the firing of the bullet and recoil (energy is conserved, but there are many changes between kinetic and potential energy), just conservation of linear momentum. F=ma is nice, but what we really need to remember is that the velocity forward times the mass of everything moving must be equal to the velocity rearward times the mass of everything moving. Since you are far more massive than the bullet you need to move at far lower velocity to balance the linear momentum of the bullet. That is, the recoil to be absorbed by the shooter is entirely a function of the mass times the velocity of the bullet, and is not affected by the recoil spring.

The gun begins to recoil as soon as the bullet begins to move forward because momentum must be conserved. This momentum is mostly tranferred directly to the shooter and cannot be avoided. Under ideal circumstances (if you don't rock back) the momentum is absorbed by elastic stretching of muscles/connective tissue as the pistol rocks back and up. Because these tissues are not perfectly elastic some of the energy transferred will be lost (as heat) and if you had a single shot/locked breech pistol you would need to add energy (through muscular contraction) to move the pistol back (down) into firing position.

When the slide begins to move rearward some of the rearward momentum is transferred to the slide instead of directly to the shooter. Although the mass of the slide is much less than the mass of the shooter the velocity is so much higher that a lot of momentum can be transferred. Because the spring is resisting the rearward movement of the slide it transfers momentum to the shooter (the force required to compress the recoil spring places force on the shooter through the frame of the pistol). If you had a recoil spring with an infinitely large spring rate all the force would be transferred to the shooter (i.e. a single-shot pistol). A spring with an infinitely small spring rage would transfer no force to the shooter.

Neither a recoil spring with an infinitely large nor infinitely small spring rate will work in an autoloading pistol. The spring stores kinetic energy as it is compressed and must return sufficient energy to move the slide forward at the end of the cycle and push the next round into the chamber. The spring is nearly perfectly elastic (all the energy stored in compressing the spring as the slide moves rearward will be put into moving the slide back forward). The amount of energy stored is proportional to the spring rate times the distance compressed. Since the distance the slide moves (from locked breech to full recoil) is about the same the only thing we can change is the spring rate. A recoil spring with a higher rate (heavier spring) will store more energy (assuming that it is compressed the same distance) and thus will have more energy to move the slide forward and load the next round.

My disagreement with Indiana J is mainly in that he has discounted the losses owing to inelastic collisions... Imagine a 1911-type pistol and two very different recoil springs (one very heavy and one very light). Both springs allow the slide to recoil completely and contact the frame. The rearward momentum imparted to the slide is the same for both springs (it's a function of the load being fired, not the spring). So at maximum recoil the two springs have absorbed/stored very different amounts of kinetic energy (much more energy is stored by the heavy spring). So what happened to the energy/momentum of the slide that was not stored by the light spring? It is transferred to the frame of the pistol because the slide hits the frame at much higher velocity. Some of that energy/momentum is transferred to the shooter, but much is absorbed by the inelastic battering of the slide/frame. Very light springs certainly could result in frame damage. If you use a shock buff there is some elastic compression of the shock buff (much like compression of a spring) that will go into pushing the slide back forward, but the buff also absorbes much of the energy in an inelastic collision.

Now the slide is at full recoil. The energy to move the slide back forward is strictly a function of the rate/weight of the recoil spring. The only energy available is that energy stored in the compressed spring. You must have enough energy to move the slide back into battery, to strip the next round from the top of the magazine, and to overcome all of the friction associated with those actions. Imagine that our light spring stores just barely enough energy to perform those functions. The slide will move forward, load the next round, lock into battery, and there will be no energy/momentum left. The muzzle of the pistol will remain pointing up. The elastic rebound of tissues in the shooter's hands/arms/shoulders will push the pistol back towards the point of aim but because there are lots of inelastic losses in those tissues the shooter will need to add energy (through muscle contraction) to get back to the same spot. That means the pistol will end up pointing above the original point of aim. Our very heavy spring has plenty of energy stored to return the slide to battery and load the next round, but when the slide locks forward there is lots of energy left over which is transferred to the frame (through another inelastic collision) and then to the shooter. This tends to force the pistol down when the slide goes forward and if not opposed by muscular contraction results in the pistol pointing below the original point of aim.

Our goal is to have the sights return exactly to point of aim at the end of the recoil cycle so we must balance the spring rate between being so heavy that the sights are forced below the point of aim and having the spring rate so light that the the sights don't come all the way down to the point of aim.

Bottom line...There is no single answer to "What is the right recoil spring weight?" There are too many variations in firearms and shooters for one combination to be correct for everyone. There is a huge amount of experience available on this forum. Read everything you can. Understand everything you can. Then try enough springs to find what works for you. The experience available on this forum can provide a good starting point and a lot of useful considerations, but it can't tell you what spring rate will work best in your pistol for you. Just because your buddy gets great results with a 10# spring doesn't mean that it is right for you. You might find a 12# spring more comfortable. Find what works for you.

I'm not even getting into the discussion of slide speed/cycle time. Figure it out for yourself.

Cheers,

Kelly McCoy

A42081

[froglegs]

Just my .000000002 cents Springs are cheap and fun to play with try a bunch of different ones and see what you like