 1 - 20 of 25 Posts

#### Glynn863

·
##### Registered
Joined
·
83 Posts
Discussion Starter · ·
Okay, in recent purchase of various 1911 parts, I have gotten two unknown 1911 recoil springs. I'm trying to determine their Force / rating. I have a NOS USGI spring for comparison. Below are the details, as measured:

USGI
Free length: 6.55"; Coil outer diameter: 0.438"; Wire diameter: 0.044"; # of coils: 30
Usually listed as a "16-lb" spring

Unknown #1
Free length: 5.88"; Coil outer diameter: 0.435"; Wire diameter: 0.046"; # of coils: 34

Unknown #2
Free length: 6.88", Coil outer diameter: 0.420"; Wire diameter: 0.038"; # of coils: 44

·
##### Registered
Joined
·
701 Posts
Are the unknown springs new? The one being about an inch shorter than the USGI could indicate used, having taken a set, or, maybe it's a Commander spring.

Assuming similar temper, you can figure the smaller diameter wire is a lighter spring, and the larger, heavier. Is the coil spacing even, or are some spaced smaller on one end? This could indicate a variable rate spring.

Since we don't know the material or temper, the only real way to tell is by testing. Such as, how much weight is needed to compress the spring(s) one inch? If not a variable spring, this would give you a value. Ther4eby, a 16-lb spring (non-variable) should show one inch compression with 16 pounds of weight.

#### megafiddle

·
##### Registered
Joined
·
610 Posts
The spring power for the unknown springs can be calculated by comparing to a spring with known power and dimensions.

Using a 16 lb Wolff govt model spring for reference, this spring has the following relevant dimensions:
free length 6.5"
active coils 31
wire diameter .045"
mean coil diameter .386"

Using a compressed length of 1.625", the rate is simply given by compressed tension / (free length - compressed length)
which equals 16.0 / (6.5 - 1.625)
which equals 3.28 lb / inch

The spring rate K, of the unknown springs is calculated by using the proportional relationship of three parameters:

K is proportional to the wire diameter (d) raised to the 4th power
K is inversely proportional to the mean coil diameter (D) raised to the 3rd power
K is inversely proportional to the number of free coils (N)

For your first spring, the wire diameter changes d in proportion to (.044 / .045) ^ 4 which equals .914
The mean coil diameter changes D in inverse proportion to (.394 / .386) ^ 3 which equals 1.06
The number of free coils changes N in inverse proportion to (30 / 31) which equals .968

So the unknown spring rate equals 3.28 x .914 / (1.06 x .968) which equals 2.92 lb / inch

The unknown spring power is then 2.92 x (6.55 - 1.625) which equals 14.4 lb
Interesting. Sounds like an ordnance power spring.

Simple, no?

I am assuming that one end of your springs are closed, and that you are counting active coils?
Confirm, and I will do the others.

-

• japsco113

#### megafiddle

·
##### Registered
Joined
·
610 Posts
Are the unknown springs new? The one being about an inch shorter than the USGI could indicate used, having taken a set, or, maybe it's a Commander spring.

Assuming similar temper, you can figure the smaller diameter wire is a lighter spring, and the larger, heavier. Is the coil spacing even, or are some spaced smaller on one end? This could indicate a variable rate spring.

Since we don't know the material or temper, the only real way to tell is by testing. Such as, how much weight is needed to compress the spring(s) one inch? If not a variable spring, this would give you a value. Ther4eby, a 16-lb spring (non-variable) should show one inch compression with 16 pounds of weight.
Recoil spring power is not measured at 1 inch of compression. That value would be spring rate.

The 16 lbs of tension occurs at about 4.9" of compression.

The shear modulus used for spring rate is based on the modulus of elasticity, which doesn't vary much for common spring wire at about 29,000,000 lbs / square inch.
It's safe to say that spring wire is spring wire, at least for our purposes.

-

·
##### Registered
Joined
·
701 Posts
True. I was unclear and implied 1 inch would be 16 pounds and that is incorrect. I had intended to say that the force applied by 1 inch compression would show differences in the springs, if any. The military print lists 2.88 lb/in for the spring rate.

Thanks for the clear explanation and setting things straight.

#### megafiddle

·
##### Registered
Joined
·
610 Posts
Measuring the tension at 1 inch of deflection is certainly one way do it. You can then find the spring power from the free length and the compressed length of 1.625".

Note that the actual spring power also depends on the free length. A spring could have a lower spring rate, and a longer free length, and still have the same compressed tension at 1.625".

And of course you could simply measure the tension at the compressed length of 1.625" directly, if you have a good pull scale.

-

#### rickgman

·
##### Registered
Joined
·
850 Posts
My rule of thumb is that when it comes to springs is if I can’t easily identify the specific spring and it’s condition (new or used) I simply throw it away. Springs are cheap but the damage that can be done when an inappropriate spring is used can be great. Recoil springs are a consumable so throwing them in the trash isn’t exactly a big deal.

#### Glynn863

·
##### Registered
Joined
·
83 Posts
Discussion Starter · ·
OP here - Thanks for the quick replies. I don't intend on using these because I have a few extra NOS USGI springs. I was just curious; someone might want to use them.

I'll look at them closer later today. The USGI spring's ends are open coil; the other two may be closed, but I'll need to verify.

#### Jim Watson

·
##### Registered
Joined
·
22,403 Posts
Empirical test. Put the unknowns in a gun, check that they are not stacking "solid" in full recoil, and shoot.

#### Black sunshine

·
##### Registered
Joined
·
164 Posts
Here's a calculator that will get you in the ballpark. As stated above, the specific material isn't all that important...if it's some flavor steel, its elasticity will be about the same. A change from steel to aluminum or titanium, for example, would result in a significant change in elasticity....not at all likely you would have any material other than some variety of steel or sst.

#### 1saxman

·
Joined
·
16,301 Posts
Empirical test. Put the unknowns in a gun, check that they are not stacking "solid" in full recoil, and shoot.
Yes, a very important step, and one that is frequently omitted. If it stacks, its the bushing that is now stopping the recoil of the slide - that usually doesn't last too long.

#### japsco113

·
##### Registered
Joined
·
405 Posts
The spring power for the unknown springs can be calculated by comparing to a spring with known power and dimensions.

Using a 16 lb Wolff govt model spring for reference, this spring has the following relevant dimensions:
free length 6.5"
active coils 31
wire diameter .045"
mean coil diameter .386"

Using a compressed length of 1.625", the rate is simply given by compressed tension / (free length - compressed length)
which equals 16.0 / (6.5 - 1.625)
which equals 3.28 lb / inch

The spring rate K, of the unknown springs is calculated by using the proportional relationship of three parameters:

K is proportional to the wire diameter (d) raised to the 4th power
K is inversely proportional to the mean coil diameter (D) raised to the 3rd power
K is inversely proportional to the number of free coils (N)

For your first spring, the wire diameter changes d in proportion to (.044 / .045) ^ 4 which equals .914
The mean coil diameter changes D in inverse proportion to (.394 / .386) ^ 3 which equals 1.06
The number of free coils changes N in inverse proportion to (30 / 31) which equals .968

So the unknown spring rate equals 3.28 x .914 / (1.06 x .968) which equals 2.92 lb / inch

The unknown spring power is then 2.92 x (6.55 - 1.625) which equals 14.4 lb
Interesting. Sounds like an ordnance power spring.

Simple, no?

I am assuming that one end of your springs are closed, and that you are counting active coils?
Confirm, and I will do the others.

-
When I have have and arithmetic problem I going to need your help for sure. All I can say is WOW.

#### Jim Watson

·
##### Registered
Joined
·
22,403 Posts
Yes, a very important step, and one that is frequently omitted. If it stacks, its the bushing that is now stopping the recoil of the slide - that usually doesn't last too long.
I didn't do arithmetic to see if more turns of finer wire led to a longer stacked length, but it certainly must be checked. Or the springs simply discarded as somebody said he would do.

#### Evil_Ed

·
Joined
·
131 Posts
Not sure if it's helpful but per Brent (formerly at Colt), Colt does not rate by weight, but by wire diameter and coil count (direct quote)..so comparing measurements to a known factory and known good Colt supplied recoil spring could at least get you a ballpark?

• Black sunshine

#### sevenL4

·
Joined
·
3,063 Posts
I made a recoil spring gauge out of a piece of 1/2 inch PVC pipe and a few pieces of hardware from my rust collection. The most expensive part was the fish scale that cost about \$15.00. There's plenty of examples online.

#### log man

·
Joined
·
14,943 Posts
For some reason spring discussions can become testy, lol. The way gun springs are currently rated as a #16, for example is the force it produces at the full recoil position of the slide. So a 5"GM #16 will register #16 at 1.6125", a Commander spring will register at 1.125", A Kimber Pro .940" an OACP at .700" a Detonics at .500" .

Glock flat wire springs are rated on the full recoil space of either the 19, or 17 as most Gocks no matter the model fall under one or the other of available recoil compression spaces, which are 1.105" for the 17, and .920" for the 19. Note, no one to date is making flat wire recoil springs specifically for the 1911, however they can be used to great advantage as they have more coils.

Something to consider also is the number of coils relates to the compressible length, not to the compression strength, as each coil of equal helix, or the distance of each coils spacing being equal, produce the same resistance at full compression. When a full size 1911 #16 recoil spring is compressed to 1.625" it is very near stacking and producing as much force as it is capable, #16.

So, realize also that if you cut, say a #18 so it registers #16 at the full compressed recoil space of 1.625" it will differ from the uncut #16 when in battery, and be less. So always use a springs whose force at full recoil allows full function, and has the most number of coils so when going to battery it also has sufficient force to smoothly chamber and close.

To clarify and rate recoil springs I highly recommend building a spring tester, you will be able to compare one to another in both the full recoil space as well as the in battery space. The in battery space for the GM it is 3.6875", Commander 3.130", OACP 2.5625", Kimber Pro 2.815", and the Detonics 2.1875". This is without a buffer which will increase the in battery force, but not the full recoil force.

LOG

A full size 5" GM #16 recoil spring compressed to the full recoil space of a 1911.

• japsco113 and Heavy Industries

#### Glynn863

·
##### Registered
Joined
·
83 Posts
Discussion Starter · ·
OP here - the USGI spring has open ends. Unknown #1 has one open end and one closed end (not ground). Unknown #2 has closed ends (not ground).

Here's a calculator that will get you in the ballpark. As stated above, the specific material isn't all that important...if it's some flavor steel, its elasticity will be about the same. A change from steel to aluminum or titanium, for example, would result in a significant change in elasticity....not at all likely you would have any material other than some variety of steel or sst.

I went to this website and played with some numbers. Since the USGI spring is rated for 16 lbs, I plugged in its data. The website said it has 10.74 lb max load. That was the only result in lbs. The USGI spring had a rate (k) of 3.036.

If I do a little math interpolation (since I know the USGI spec), I figure Unknown #1 is an 18 lb spring, and Unknown #2 is an 11 lb spring.

• Black sunshine

#### megafiddle

·
##### Registered
Joined
·
610 Posts
OP here - the USGI spring has open ends. Unknown #1 has one open end and one closed end (not ground). Unknown #2 has closed ends (not ground).

I went to this website and played with some numbers. Since the USGI spring is rated for 16 lbs, I plugged in its data. The website said it has 10.74 lb max load. That was the only result in lbs. The USGI spring had a rate (k) of 3.036.

If I do a little math interpolation (since I know the USGI spec), I figure Unknown #1 is an 18 lb spring, and Unknown #2 is an 11 lb spring.
The ordnance recoil spring is spec'd at 13.55 lbs at 1.81" compressed length. Wolff, and many here, use a compressed length of 1.625" for govt model recoil springs. At a compressed length of 1.625", the ordnance spring is about 14.2 lbs.

Using the same method of comparison (as I used earlier) to calculate your USGI spring from ordnance spec, I get about 14.3 lbs. I can say your USGI spring is not 16 lb, but closer to the original ordnance weight of 14.2 lb.

I also tried the spring calculator. It reported a spring rate of 2.935 lb / inch. This is very close to the 2.92 lb / inch that I calculated earlier.

You cannot use the maximum load reported there. It's not measured (calculated) at the compressed length you are interested in, which is 1.625". The compressed length they are using is far short of compressing down to 1.625". Their compressed length can also vary with different springs, so you can't simply scale their maximum loads to a known spring. (I believe scaling is what you meant by "interpolation").

You can use the spring rate though, which does appear to be pretty accurate. Use the calculator to the find the spring rate for each spring. Then simply multiply (free length - 1.625) by the spring rate. You should get a pretty accurate spring power value that way.

You may already know, but be careful counting active coils. It's easy to be off by one.

Also, fully compress the spring several times before measuring free length. A new spring will shorten a bit and "set" after being compressed the first few times.

-

• japsco113

#### megafiddle

·
##### Registered
Joined
·
610 Posts
I used the spring calculator at thespringstore.com to verify the Wolff 16 lb spring specs:

16 lb Wolff govt model spring:

free length 6.5"
active coils 31
wire diameter .045"
coil diameter .431"
mean coil diameter .386"

spring rate from rated power and free length 3.28 lb / in
spring rate from the spring calculator 3.304 lb / in

spring power from the Wolff spec 16.0 lb
spring power from spring calculator rate 16.1 lb

That's pretty good agreement.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

The calculator at thespringstore.com has been unaccesible at times. I only have comparisons for your USGI spring so far:

USGI:

free length 6.55"
active coils 30
wire diameter .044"
coil diameter .438"
mean coil diameter .394" (coil OD - wire diameter)

d = (.044 / .045) ^ 4 = .914
D = (.394 / .386) ^ 3 = 1.06
N = (30 / 31) = .968

as calculated in proportion to the Wolff 16 lb:

spring rate = 3.28 x .914 / (1.06 x .968) = 2.92 lb / in
spring power = 2.92 x (6.55 - 1.625) = 14.4 lb

as calculated from the spring calculator rate:

spring rate = 2.935 lb / in
spring power = 2.935 x (6.55 - 1.625) = 14.5 lb

Also pretty good agreement.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

For the other two springs, these are only in proportion to the Wolff 16 lb.
If I can access the spring calculator again, I expect it will also be in agreement:

Unknown #1:

free length 5.88"
active coils 34
wire diameter .046"
coil diameter .435"
mean coil diameter .389"

d = (.046 / .045) ^ 4 = 1.09
D = (.389 / .386) ^ 3 = 1.02
N = (34 / 31) = 1.10

spring rate = 3.28 x 1.09 / (1.02 x 1.10) = 3.19 lb / in
spring power = 3.19 x (5.88 - 1.625) = 13.6 lb

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Unknown #2:

free length 6.88"
active coils 44
wire diameter .038"
coil diameter .420"
mean coil diameter .382"

d = (.038 / .045) ^ 4 = .508
D = (.382 / .386) ^ 3 = .969
N = (44 / 31) = 1.42

spring rate = 3.28 x .508 / (.969 x 1.42) = 1.21 lb / in
spring power = 1.21 x (6.88 - 1.625) = 6.36 lb

-

#### Glynn863

·
##### Registered
Joined
·
83 Posts
Discussion Starter · ·
OP here - Thanks for that.

So a USGI spring, by US Ordnance values / specs, is actually closer to a 14-lb spring, rather than 16-lb?

I have never changed a spring in any of my 1911's. My 1983-vintage Series 80 Government Model is still stock. I bought a NOS Colt Government spring from Midway for a Remington R1 I'm pseudo-custom building. Unknown #1 came from a Remington Rand M1911A1 as I received it, along with a 2-piece long guide rod. I bought a trio of NOS USGI springs (contract date 1984) off eBay, and have used one in the RR, so I have two spares.

I already knew that the Unknown #2 was a lighter spring just by a "hands-on" examination; I didn't know of a numerical value to associate it with.

1 - 20 of 25 Posts