Is there any reason this would not be safe?

[M27]

I recently changed the recoil spring on my 1911



top spring is the new one, bottom is the old

they are identical replacement parts

here is my real question, and why I put this in the reloading section

Would it be safe if I cut down the old spring so I could make supper low power loads that still cycle

I was thinking 200gn SWC on Trail boss as 600fps

could it do any damage to the pistol

are there any problems I'm not seeing

[G-forceJunkie]

It might work, but it would be better to just get a lightweight spring. Cutting a spring does not make it softer, it actually increases it "spring rate." It may feel softer initially because it has less preload on it, but the rate has been increased.

[gschoelles]

I used to do that back in the day for some specific shooting needs. Takes some work and always run the chance of beating up the frame too much.
You won't be the first, so research it well.

[shooterbill]

If your load is way less than a starting load, you can get weird pressures. The more the spring gets used, the coils lose their tension.

[sofbak]

It might work, but it would be better to just get a lightweight spring. Cutting a spring does not make it softer, it actually increases it "spring rate." It may feel softer initially because it has less preload on it, but the rate has been increased.

Could you explain this please? How does free length of a spring affect its constant or spring rate?

[DRAB_81]

Competition shooters cut coils off of reduced power recoil springs all the time, so I would think you'd be fine. They usually cut coils off of new springs, so I can't say for sure. Maybe do a search over on Enos Forums, or post the question there.

Have you tried shooting that load with the old spring the way it is?

[sargenv]

I'd reccomend against cutting coils off a spring.. I'd say get yourself a pack of several weight springs and try different ones to get to your liking.. If the spring is cut, the weight of the spring changes and you can batter your gun if you aren't careful.

[G-forceJunkie]

Could you explain this please? How does free length of a spring affect its constant or spring rate?

Its technical, but here is one explanition from Google I'll cut and paste:

Cutting coils does increase the spring rate. Let me explain why.

The strength of a spring, leaf or coil is a function of the cube of the steel used. Keeping with the subject of your question, coil springs, the diameter of the wire and the length of the wire will give us the amount of steel used.

For this whole discussion we will be talking about springs with the same wire diameter and the same inside diameter. The only thing that will change will be the length of the wire used to wind the spring.

The longer the wire is the lower the spring rate. As the wire get shorter, such as when cutting the coil, the spring rate increases.

So everyone has a clear understanding lets describe what "rate" is. Rate is the amount of weight it takes to deflect a spring one Inch.

A very common mistake is to think that spring rate is how much a spring supports. How much weight a spring is designed to support is called "Load" or "Designed Load" or"Load Rate". This is cover in Spring Tech 101.

Rate and Load Rate are two totally different animals.

The calculation to find the rate of a coil spring is:

11,250,000 times the wire diameter to the 4th power divided by 8 times the active number of turns times the mean diameter cubed.

Active turns are the number of turns of the spring that do not touch anything. Any part of the coil which makes contact with anything becomes inactive, that is it no longer functions as part of the spring.

The mean diameter is the inside coil diameter plus one wire thickness. Or the outside coil diameter less one wire thickness.

Let's say for example a 1967 Mustang GT front spring is made from .610 wire and has an inside diameter of 3.875" and has a free height of16.145" (not installed) and is deflected down to 10.5" (load height) when loaded to 1,519 Lbs. (load rate) This spring has a spring rate of 269 Lbs.

This spring has 9.33 total coils but 1.33 coils touch the spring seat so they are inactive leaving 8 active turns. (I know this from the Ford blue print).

The mean diameter is 3.875 + .610 (The inside is the important diameter because it is the inside of the spring which is used to locate the spring on the corresponding suspension parts. The outside diameter is not considered because it will change with a change of wire diameter)

Do the math-

11,250,000 x (.610 x .610 x .610 x .610) / 8 x 8 active turns x (4.485 x 4.485 x 4.485) = 269 Lbs.

Double check the math - 16.145 - 10.5 = 5.645 deflection. 1,519/5.645 = 269

Now if we cut say 1/2 turn off this spring the active turns become 7.5.

So 11,250,000 x (.610 x .610 x .610 x .610) / 8 x 7.5 x (4.485 x 4.485 x 4.485) = 287 Lbs.

While the rate is increased the load is unchanged. Rate is the amount of weight required to deflect the spring one Inch while load is the amount of weight the spring will support at a given height.

[noylj]

The top competition shooters do not consider banging up the frame as a problem. The gun, for them, is no more than a tool to win and it is replaceable. By year's end, they have tens of thousands of rounds through it and are ready to move the whatever the next "best" thing is.
With the cost of proper length springs of reduced weight so cheap, I wouldn't consider chopping up a spring and possibly beating up the frame or otherwise damaging my guns. Mine are NOT replaceable.