Cutting Automatic Pistol Recoil Springs (or Not)

[MEHavey]

Dumb question du jour . . . .

1. EVERYbody says "don't do it... Get the right weight spring."

2. My understanding is the term "10-lb spring" is not a spring rate label, but that of force at full compression.

3. Whatever the spring rate is ...(lbs/in compression) that rate is a constant.

4. So if a 2-lb rate spring is fully compressed in 5 inches travel, that spring is labeled "a 10-lb spring"

QUESTION:
Why Can't that 2-lb rate spring be cut 1/2" inch to become a "9-lb spring" ?

 

[unspellable]

The point almost always overlooked, often even by aftermarket spring suppliers, is the preload, that is, how much force does the spring exert in battery? When you cut a spring you are making a bigger change in preload than in full load. This will have various effects depending on the pistol in question. In the case of a Luger it is unsafe as the Luger WILL fire from slightly short of locked battery. In the case of the Luger there is a sordid history of spring cutting and mismatched springs, a substantial role player in the Luger's street reputation for indifferent reliability. Also cutting a spring increases the rate even if it reduces the full load force in the pistol.
To quote Georg Luger, "The springs have to be right."

 

[5-SHOTS]

The problem with cutting the end of a spring is that it will have bad support on the part of the cut end. Will it work? Of course it will work, I have seen many work but it is certainly not the optimal solution. Also if you cut the original spring and something goes wrong, you will need to order a new one. If, on the other hand, you order one of those kits that contain springs of various pounds suitable for your pistol, you will have numerous options to find the best spring-ammunition combination.

 

[tark]

The point almost always overlooked, often even by aftermarket spring suppliers, is the preload, that is, how much force does the spring exert in battery?

A very good point, here. Cutting springs is never a good idea. The above mentioned condition can result in the recoiling parts moving too quickly and with too much force which will eventually lead to excessive wear and parts breakage. Find and use the proper spring!

 

[sparkyv]

I'll cut trigger and other springs, but not recoil springs because of this:

condition can result in the recoiling parts moving too quickly and with too much force which will eventually lead to excessive wear and parts breakage.

 

[chicharrones]

I've cut recoil spring length in an effort to make lower powered ammo run better in a blowback rimfire pistol. Unfortunately, the slide was a bit under sprung in battery, which meant the slide might not close the final 1/32" to 1/16" when pushing a cartridge into a fouled chamber. Light strikes were a major side effect after that modification.

 

[JDGray]

Installing a new Wolff recoil spring, will make you think you’ll need to cut it! Lol
Those things are usually quite a bit longer until they take seat. I’m with the don't cut crowd.

 

[nettlle]

I won't cut the spring. I will vary the powder charge and bullet weight to get what I want instead of cutting the spring.

 

[zaitcev]

2. My understanding is the term "10-lb spring" is not a spring rate label, but that of force at full compression.

I'm pretty sure it's a rate. So a "10-pound" spring is actually 10 pounds force per 1 inch.

BTW when I want springs to set for competition loads, I use a SprinCo or DPM set.

 

[nettlle]

You can do anything with the springs that you want. I wouldn't think the spring manufacturers or gun manufacturers are going to give a snip length. Certainly enough variables in my auto loaders as not to spend time snipping springs.

 

[Jackiesman]

What do you do if even factory ammunition won’t eject. Case in point, I bought two Walters - a P22 and a P38 for my wife. Neither will eject with factory ammo. I have cleaned it so many times that I can almost do it in my sleep. She used the P22 in qualifying for her CC, and had to manually eject each round. The range officer had me go get new CCI 40 grain ammo, didn’t work. He then disassembled it, inspected it, cleaned it, and it still had the same problem. Sometimes works well with 40 gr Winchester. For the 380 I have to load to the hot end for it to eject, and I can forget loading anything less than 90 grains. Even ordered a new spring for it. Finally gave up and got her a Glock 42, and it works beautifully. So, that said, does anyone have any words of wisdom for my Walter dilemma?

 

[MEHavey]

So a "10-pound" spring is actually 10 pounds force per 1 inch

No, "pounds per inch" is spring rate (which is admittedly how springs are normally labeled).
Pistol spring are apparently labeled for the total force they exert when fully-compressed.

FWIW: I am a fan of Wolff Springs -- and have had the 10-16 lb calibration set for 10-15 years.
But I went in search of a 185gr/700fps load the other day for the grandson, and ran into ejection problems even at 10-lbs.


* Gov't 1911/H&G-130/Bullseye/3.6gr for those so interested -- (Labradar'd @ 995fps ± 5)

 

[JohnKSa]

Some good answers, even one good example of what can go wrong when a spring is shortened when the actual goal is just to make it weaker.

2. My understanding is the term "10-lb spring" is not a spring rate label, but that of force at full compression.

QUESTION:
Why Can't that 2-lb rate spring be cut 1/2" inch to become a "9-lb spring" ?

Because its force at full compression is not the only factor that's important. Its force when the slide is in battery is also important to insure that the gun goes fully into battery and stays there.

In addition to the already provided answers, it's important to understand that in some striker-fired designs, the recoil spring holds the action in battery against the force of the trigger pull. If the spring force when the slide is in battery is insufficient, the trigger can actually pull the slide out of battery slightly resulting in malfunctions or potentially even unsafe conditions. That's independent of the force of the spring at full compression.

Imagine a really ridiculous case where a spring is installed that is so short it isn't even touching the slide to push it forward once the gun is in battery. Now, even if it is designed to provide identical force when fully compressed as the stock spring, it's easy to see that there are going to be problems with operation.

The same kinds of issues that would result in that thought experiment can be present (to a lesser degree, of course) with a spring that has been shortened too much. Of course, "too much" is hard to define in general.

So, that said, does anyone have any words of wisdom for my Walter dilemma?

A pretty common recommendation for the P22 is to use "high-velocity" ammunition, especially when the pistol is new.

 

[Ru4real]

QUESTION:
Why Can't that 2-lb rate spring be cut 1/2" inch to become a "9-lb spring" ?

It can be cut. And doing so will change the characteristics of operation. You are already unreliable for 185/700fps. Go for it. Might fix your reliability issue. Frame pounding isn’t a concern with 9# and 185/700.

Once you cut the spring, you will need a new spring anyway, as your grandson grows.

Limp wristing and women/kids usually run a little lighter spring anyway, at least for the people I’ve set guns up for. Incidentally, I’ve never cut a slide return spring.

 

[MEHavey]

Naah... already got the whole 10-18 lb spring set, and another ultra-light set on the way...
so no problems if the one low 10-pounder get "customized"

But you are correct in not mucking around w/ the only spring one might have....

 

[JDGray]

What do you do if even factory ammunition won’t eject. Case in point, I bought two Walters - a P22 and a P38 for my wife. Neither will eject with factory ammo. I have cleaned it so many times that I can almost do it in my sleep. She used the P22 in qualifying for her CC, and had to manually eject each round. The range officer had me go get new CCI 40 grain ammo, didn’t work. He then disassembled it, inspected it, cleaned it, and it still had the same problem. Sometimes works well with 40 gr Winchester. For the 380 I have to load to the hot end for it to eject, and I can forget loading anything less than 90 grains. Even ordered a new spring for it. Finally gave up and got her a Glock 42, and it works beautifully. So, that said, does anyone have any words of wisdom for my Walter dilemma?

A buddy bought a Sig Mosquito that just will not function. It came with 2 recoil springs, neither of them work with any ammo, might get a few in a row to cycle. The CPL instructor had to manually cycle the pistol while his wife took the class.

 

[lysanderxiii]

Dumb question du jour . . . .

1. EVERYbody says "don't do it... Get the right weight spring."

2. My understanding is the term "10-lb spring" is not a spring rate label, but that of force at full compression.

3. Whatever the spring rate is ...(lbs/in compression) that rate is a constant.

4. So if a 2-lb rate spring is fully compressed in 5 inches travel, that spring is labeled "a 10-lb spring"

QUESTION:
Why Can't that 2-lb rate spring be cut 1/2" inch to become a "9-lb spring" ?

Because that's not how it works.

1. The spring rate will change when you shorten a spring.
2. The pre-load will be reduced.

 

[MEHavey]

Spring rates are spring rates -- regardless of length.
That's how springs are defined: 'X' pounds per inch of compression.
It's called the Spring Constant.

If I have a 10-inch spring rated as 10-lbs/inch -- and compress it two inches: --> 20-lbs pre-load
If I cut that 10-inch spring to 8 inches but compress it that same two inches again: --> same 20-lbs pre-load

Now... some quick analysis: (New Spring)
CONSIDER:
2-lb (rate spring) uncompressed 7" long (Gov't 1911) --> 3" remain outside of gun uncompressed
Compress that 3" to fit/lock into rod guide (i.e., ...assembled): 3" x 2lbs/in --> 6lbs pre-load
Compress another 2" upon firing: 2" x 2lbs/in = 4lbs more
6+4= 10lbs total at end of slide movement.
(We have a classically-defined "10lb" spring")

NOW:
Cut ½" off -- start now with 6½" spring uncompressed.
Compress remaining 2½" outside to fit into gun --> 2½ x 2lbs/in = 5lbs preload
Compress same additional 2" on firing --> 2" x 2lbs/in = same 4lbs
5lbs + 4lbs = 9lbs
(Effectively becoming a " 9-lb spring ") upon firing

Someone let me know where my logic/math has gone off the rails

 

[JohnKSa]

The math looks right. Just make sure that 5lbs of preload is enough to make everything work right.

Cutting the spring reduced the full compression force by 10%, but it also reduced the preload force--and by a larger percentage--17%.

Think about what happens if, instead of shortening the existing spring, a spring that is 8" uncompressed and has a rate of 1.5lbs/in is installed.

The new spring, when fully compressed will have a force of 9lbs, exactly the result obtained by shortening the 2lbs/in spring, but the preload force with the new spring will be 6lbs which is exactly the same as the preload of 6lbs for the original unmodified spring.

Maybe that's important, maybe it's not, but it does highlight the fact that there can be a difference between shortening a spring and using one with a different spec.

There are definitely some cases where I wouldn't want to reduce the preload even if I did want a lighter full-compression number. Fortunately there's a way to get that--but not by just cutting the existing spring.

 

[unspellable]

Spring rate is NOT independent of length. If you take a given spring and cut it, the rate is increased. Exagerated example: Say a given spring has 20 coils and is rated at 10 lbf/in. Apply ten pounds force and you compress it one inch. Now cut it to 10 coils. Compress it one inch. You have half as mnay coils so each coil is flexed twice as much. You now have a 20 lbf/in spring. Put half a spring in any reasonable pistol and the preload is gone or nearly so.

 

[MEHavey]

Within the minor length difference of 7% shown in the example,
spring rate is effectively still a constant. I wouldn't be adverse to
minor "spring-tuning" now having done the math.
-- BUT --
As also noted above, I've got Wolff's ultra-low calibration pac arriving
Tuesday.

 

[lysanderxiii]

Spring rates are spring rates -- regardless of length.
That's how springs are defined: 'X' pounds per inch of compression.
It's called the Spring Constant.

For that spring.

If I have a 10-inch spring rated as 10-lbs/inch -- and compress it two inches: --> 20-lbs pre-load
If I cut that 10-inch spring to 8 inches but compress it that same two inches again: --> same 20-lbs pre-load

Now... some quick analysis: (New Spring)
CONSIDER:
2-lb (rate spring) uncompressed 7" long (Gov't 1911) --> 3" remain outside of gun uncompressed
Compress that 3" to fit/lock into rod guide (i.e., ...assembled): 3" x 2lbs/in --> 6lbs pre-load
Compress another 2" upon firing: 2" x 2lbs/in = 4lbs more
6+4= 10lbs total at end of slide movement.
(We have a classically-defined "10lb" spring")

NOW:
Cut ½" off -- start now with 6½" spring uncompressed.
Compress remaining 2½" outside to fit into gun --> 2½ x 2lbs/in = 5lbs preload
Compress same additional 2" on firing --> 2" x 2lbs/in = same 4lbs
5lbs + 4lbs = 9lbs
(Effectively becoming a " 9-lb spring ") upon firing

Someone let me know where my logic/math has gone off the rails

Your understanding of springs is a bit lacking.

Let's look at the USGI recoil spring:
Wire dia - .043
OD - .430
Free length - 6.55
Total coils - 30
Music Wire

Doing the calculation for that spring, you get: (If you don't know how to calculate the spring parameters, click here)
A spring rate of 2.825 lbs/in
Max load - 10.278 lbs
And, with the spring initially compressed to 3.72 inch, (this is the distance under the slide when closed) the force holding the slide shut is 7.995 lbs.

Chop off 1/2 inches (or 3.21 coils), now your spring looks like this:
Wire dia - .043
OD - .430
Free length - 6.05
Total coils - 26.78
Music Wire

Doing the math for that spring:
A spring rate of 3.164 lbs/in
Max load - 10.278 lbs
and it takes 34.9906 inches of wire.
And, with the spring initially compressed to 3.72 inch, the force holding the slide shut is 7.372 lbs.

Your max load hasn't changed, the rate has increased, but the initial force (preload) had decreased.

 

[lysanderxiii]

If you really want a "9 pound spring", you need to reduce the wire diameter.

I see where you are confused. A "10 pound spring" is not a spring that when compressed to solid produces a force of 10 pounds.

A USGI recoil spring, when compressed to solid produces more like 15 pounds.

solid length = 0.043 x 30 = 1.29

So, the total compression is:

Solid length - free length, or

6.55 - 1.29 = 5.26

Total compression x spring rate = force, or

5.26 x 2.825 = 14.86

 

[MEHavey]

Your max load hasn't changed....

So you are saying preload for your 6.55" original spring is 8 lbs
and 2" more compression upon firing adds 2.825 x 2" = 5.7lbs...
TOTAL: 13.7 lbs at end of slide travel

And the 1/2" cut spring results in a preload of 7.4 lbs ?
and 2" more compression adds 3.164 x 2" = 6.3lbs....
TOTAL: (drumroll please) 13.7 lbs at end of slide travel.

Interesting. ...

No matter where you go... there you are.*
Thank you for the update to (however lacking) my presumption of constant rate.




* due credit to to B.Banzai

 

[lysanderxiii]

So you are saying preload for your 6.55" original spring is 8 lbs
and 2" more compression upon firing adds 2.825 x 2" = 5.7lbs...
TOTAL: 13.7 lbs at end of slide travel

Uhhhhh, not quite.

If your spring has been compressed to a length of 3.72 inches, and you compress it another 2 inches it has now been compressed 5.72 inches. The spring started out 6.55 inches, so it should now only be .83 inches long, but that cannot be, as the solid length is 1.29 inches.

That is my error, the pre-load length in and M1911 is 3.83 inches with a pre-load of 7.684 pounds

6.55 (free length) - 3.83 (space under the barrel) = 2.72 (initial compression)

2.72 x 2.825 = 7.684 pounds (holding force)

The final compression length when the pistol is fired will be 1.70 inches, as that is the space available in the slide shroud.

6.55 (free length) - 1.70 (space in shroud) = 4.85 (compression)

4.85 (compression) x 2.825 (rate) = 13.701 pounds

And the 1/2" cut spring results in a preload of 7.4 lbs ?
and 2" more compression adds 3.164 x 2" = 6.3lbs....
TOTAL: (drumroll please) 13.7 lbs at end of slide travel.

Using the information from Post #22, and assuming the same M1911.

With a free length of 6.05 and a pre-load length of 3.82 inches, the pre-load will be 7.02 pounds

6.05 (free length) - 3.82 (space under the barrel) = 2.22 (initial compression)

2.22 (compression) x 3.164 (rate) = 7.02 pounds (holding force)

The final compression length when the pistol is fired will be the same as shown above as the shroud is what stops the slide, and that will be 1.70 inches.

6.05 (free length) - 1.70 (space in shroud) = 4.35

4.35 (compression) x 3.164 (rate) = 13.763 pounds

When you clip coils, if you keep the initial and final compression positions the same, the initial load drops and the final load increases.