Clipping Recoil Springs?

[DougR]

Josh,

It is not the fault of any University that you fail to understand such a simple concept.

Apparently you have chosen to put your limited knowledge against the spring engineers and manufacturers that publish their information and specifications through out the internet.

I have supplied you with the facts using the spring manufacturers information.

If you choose to ignore the very people that do this kind of work for the rest of the world, then it has become your problem, not mine.

If you are truly interested, take a few minutes from your keyboard and try this simple screen door spring test.

If the test doesn't change your mind, I can't help you.

Read the web sites of the manufacturers and engineers.

I trust if you are willing to learn, rather than let this become a pissing contest, we'll all move ahead.

I'm out of here,

[josh smith]

Josh,

It is not the fault of any University that you fail to understand such a simple concept.

Apparently you have chosen to put your limited knowledge against the spring engineers and manufacturers that publish their information and specifications through out the internet.

I have supplied you with the facts using the spring manufacturers information.

If you choose to ignore the very people that do this kind of work for the rest of the world, then it has become your problem, not mine.

If you are truly interested, take a few minutes from your keyboard and try this simple screen door spring test.

If the test doesn't change your mind, I can't help you.

Read the web sites of the manufacturers and engineers.

I trust if you are willing to learn, rather than let this become a pissing contest, we'll all move ahead.

I'm out of here,

I want to see the website of spring engineers and manufactureres that you're talking about. I did search in the internet and I found informations that support what I learned from the university. I gave you examples thats proves that springs do not get stronger if you cut them. I showed you how I got my answers and what formula I used. But you didn't give me any formula and logical examples that proves you are right.

Seriously though, I want to see the website that you're talking about.

I already give the website where I found that Normal spring like recoil spring has a fixed spring rate.

If noone else going to argue with me, them I'm out of here as well, lol.

Edited April 24, 2009 by josh smith

[DougR]

OK Josh I'm back,

After I left I did a simple Google of Spring rate calculators.

It took me to a site www/efunda.com.

Engineering Fundamentals.

They offer a spring calculator that uses several parameters of spring design.

I put in the specification for a 1911 recoil spring.

Wire diameter .043"

coil diameter .430"

free length 6.5 inches

number of active coils 29

I left the other parameters alone.

I then asked for a calculation based upon these numbers.

It returned a spring constant/spring rate of 2.65 lbs/inch

I then changed the number of active coils to 28 and reduced the length .25 inches to 6.25.

Then recalculated.

The spring constant/spring rate changed to 2.76 lbs/inch. This compared to 2.65 lbs/inch with 29 coils.

The site then asked me to join if I wnted to use the calculator further.

I'm not bashful, I got out my AMEX and joined. $10 per month.

I realize the term spring constant and spring rate are indeed inter-changeable.

The formula you are using is from the simple test of hanging a weight from a spring and measuring the deflection.

Then comparing the deflection to the unweighted spring length.

The difference is the spring constant. No problem.

However it does not address changing spring length, number of active coils, etc.

The spring calculator does calculate the variables and shows the effect of changing a single parameter.

You can use this calculator once or twice before they want you to join.

Membership is cheap. I recommend you look at it.

It could even save you a trip to the hardware store.

In the interest of knowledge, I would be happy to use my membership and input any numbers you want.

I don't know if it has a print function. If it does I'd be happy to post the results.

As I said before, I'm not interested in an internet pissing contest where no one benefits.

Together, we can put away the ego's and search for the truth.

I'll be back to this site tomorrow and hopefully we can understand this together.

I feel better now, I hope you do to.

Edited April 24, 2009 by DougR

[josh smith]

Sup Doug, Im not here for pissing contest aswell, I responded to this thread because I have few knowledge about springs since I majored in Physics. I know I'm not a spring engineer so I don't expect that I am right 100%. I'm a science person so I really want to know the truth. And I will accept that I am wrong if you give me logical proof that I made a mistake. I accept defeat is I deserve it.

I put in the specification for a 1911 recoil spring.

Wire diameter .043"

coil diameter .430"

free length 6.5 inches

number of active coils 29

I left the other parameters alone.

I then asked for a calculation based upon these numbers.

It returned a spring constant/spring rate of 2.65 lbs/inch

I then changed the number of active coils to 28 and reduced the length .25 inches to 6.25.

Then recalculated.

The spring constant/spring rate changed to 2.76 lbs/inch. This compared to 2.65 lbs/inch with 29 coils.

ok, your example here is not valid. You just picked random numbers. You assumed that if you cut 1 coil you reduce the length by .25". This is wrong. If the length of 1 coil is 0.25" then for 28 coils that would be 7 inches long. With that, I don't have to say more.

The formula you are using is from the simple test of hanging a weight from a spring and measuring the deflection.

Then comparing the deflection to the unweighted spring length.

The difference is the spring constant. No problem.

The difference is NOT the spring constant.

From Hooke's Law: F=kx, F is the force, k is the spring constant and x is the distance you compress the spring. So the difference is x.

However it does not address changing spring length, number of active coils, etc.

The spring calculator does calculate the variables and shows the effect of changing a single parameter.

Results from internet calculators doesnt convince me. I studied computer programming myself, and I can make a calculator program that if you add 1 plus 1 you get 20. My point is I want the formula behind the calculator.

Since Im not a spring engineer and I only majored in Physics, so my knowledge about spring is limited.

So I researched again. and guess what? I found the HOLY GRAIL!!!, lol.

So this is my conclusion: we were both partly right. spring rate increases if you reduce the number of active coil, which you are right about. But the increase is very little. In the end the Force required to compress a certain amout on a longer spring is still greater than the force to compress a shorter spring to the same amount. So if we cut the spring shorter it then it gets LIGHTER.

Thats all I have to say. Im out of here......

heheh, jk.

I know you're not happy with that, let me give you the HOLY GRAIL.

Equation for spring constant/spring rate:

k=(Gd^4)/(8nD^3), where

k is spring constant/spring rate

G is modulus of rigidity of spring material in lbs per square inches

d is wire diameter

n is number of active coils

D is mean coil diameter.

so k is inversely proporsional to n(number of active coils), which you are right about.

But since G is very large number the effect of active coils on k is very small. And if we apply this in

actual experiment, this will only have small effect and the lenght of compression counts more.

example.

To make it more valuable I'll try to use numbers close to a recoil spring.

8 inches recoil spring. lets use 0.25" per coil(length) for easy calculation.

so theres 4 coils in 1 inches, hence 8 x 4 = 32 coils.

8" spring:32 coils

if we cut 4 coils, that makes

7" spring:28 coils

we need to find k for 8 inches and k for 7 inches.

we can find k using the internet calculator you provided or

by using the HOLY GRAIL of springs, which is k=(Gd^4)/(8nD^3).

to compare our values lets use same parameters.

d = Wire diameter .043"

D = coil diameter .430"

G = 11500000 sample from http://www.engineersedge.com/calculators/t...pring_k_pop.htm

so if we plug in n =32 for 8 inches, we get spring rate, k = 1.9316

if we plug in n =28 for 7 inches, we get spring rate, k = 2.2076

Look at that, we if reduce the number of active coils it actually increases the spring rate.

But we are not done yet.

if we plug in the recoil spring in the gun. lets say the lenght of the spring in batterry is 3" for easy calculation.

Using Hooke's Law, F=kx, where F is force, k is spring rate and x is distance if the compression.

For 8" spring, k = 1.9316 and x = 8" - 3" = 5"

So F=kx

F=(1.9316)(5) = 9.658 lbs

For 7 inches, k = 2.2076 and x = 7" - 3" = 4"

So F=kx

F=(2.2076)(4) = 8.8304 lbs

hmmm, I think that means if we cut an 8" spring to 7" long, the Force require to compress them to

batery went from 9.66 lbs to 8.83 lbs. So I was right all along that if we cut the spring shorter it gets lighter. But I give you credit for spring rate/spring constant variation.

So Doug, since you already is a member in the engineering website. try plugging in the numbers I used, lets see what you get.

I WANT TO KNOW THE TRUTH AND NOTHING BUT THE TRUTH, hehe.

later Doug.

peace

[DougR]

Hi Josh,

I've been busy with other things today, tomorrow doesn't look any better. I'll get to this asap.

I've got an idea we are going to learn some interesting things.

I should have something for you by Monday evening.

Should be fun.

[1SOW]

I shot a clipped (two coils) 13# slide spring in my 75B today. With a new the load the cases had been landing at my right foot. With the 13# spring clipped two coils it was noticeble farther away--maybe another 2-3 feet to the right. Fed and functioned well for over 200rds.

For more empirical data check Angus Hobdell's & others'replies to lowering trigger pull weight by cutting coils out of the mainspring - CZForum.com.

Edited April 26, 2009 by 1SOW

[DougR]

Hello Josh,

I’m impressed with your grasp of the math involved in these spring calculations.

I ran the numbers for the 8 inch spring, 32 active coils, .043 wire dia, .430 coil dia, through the Efunda calculator we talked about.

The calculator agreed with your numbers after I adjusted Youngs modulus of materials to a value of 150 Gpa.

Also after running the cut spring @ 7 inches and 28 active it also agreed with your numbers.

Now here is where things change somewhat.

An 8-inch spring with a spring constant of 1.93, requires 1.93 lbs to compress it 1 inch.

To compress the 8 inch spring 3 inches requires 3X1.93=5.79 lbs.

Now we compare the above to a 7-inch spring. It began its life as an 8 inch spring. You cut 4 coils from the original 32. This reduced the length from 8 to 7 inches.

By cutting 4 coils and reducing the length to 7 inches, the spring constant went up to 2.20 lbs/inch.

With this information we know it now takes 2.2 lbs pressure to compress the 7 inch spring 1 inch.

To compress this 7 inch spring 3 inches it takes 3 X 2.2 lbs= 6.6 lbs.

6.6 lbs for the 7 inch spring compares to 5.79 lbs for the 8 inch spring when each spring is compressed 3 inches.

OK, but there is good news coming for Josh.

A 1911 recoil spring runs 6.55 inches long with 30 coils, 29 active, the rated spring constant from the ordinance print is 2.88 lbs/inch. The wire is rated higher for spring modulus. That is the wire for the 1911 spring is a bit stiffer than the wire in you example. It doesn’t change the principle of our discussion.

When the 6.55 inch stock spring is installed into the 1911 it is compressed to a length of 3.72 inches. Or to look at it another way the 6.55 inch spring has been compressed 2.83 inches. 2.83 X 2.88 = 8.15 lbs.

This same spring is compressed 2.1 inches more when the pistol is fired and the slide is fully rearward. 2.1 X 2.88 = 6.05 lbs. .................8.15 lbs plus 6.05 lbs = 14.20 lbs total at full rearward........

Now lets cut 2 coils from the stock spring. 6.55”/30 coils = .21833”per coil.

2 coils = .4366” 6.55”- .4366” = 6.1133” is the length after 2 coils cut.

The new spring constant is 3.09 from the spring calculator.

Install the cut spring into the 1911. Subtract the new compressed length from the original free length,

6.11”- 3.72”=2.3933” We have compressed the cut spring less than the original by the amount we cut,

.4366”. 2.3933” X 3.09 = 7.395 lbs initial force in battery.

So this is lighter than the stock spring in battery. 8.15lbs vs 7.39lbs.

Now let’s fire the pistol. The full rearward motion of the slide is still 2.1 inches from battery.

It will vary .050” with a shock buffer and another .050” among different 1911”s.

Measure your own 1911 to get exact numbers.

2.1 inches times 3.09 = 6.489 more lbs. 6.489 plus 7.39 = 13.879 lbs

................................Josh you are a winner. 14.20 lbs. vs 13.879 lbs.......................

Let’s take 1 coil from the 6.55 inch spring.

6.55”-.2183”= 6.33”. The new spring constant is 2.98 lbs/inch.

Install the newly cut spring into the 1911. It becomes compressed to 3.72” from 6.33”.

6.33”-3.72”= 2.61”. 2.61” X 2.98 = 7.77 lbs initial compression.

Fire the pistol, add 2.1 inches more recoil spring compression.

2.1 X 2.98 = 6.26 lbs plus 7.77 lbs= 14.028 total at full rearward motion.

Close, but the uncut stock spring is stronger by 14.20-14.028= .172 lbs. 2.75 ounces.

Josh is still the winner, by 2.75 ounces.

The great part is now we know, we really know, what happens when we cut recoil springs.

Thanks Josh

Edited April 27, 2009 by DougR

[josh smith]

Good job Doug! Now we fully understand the science of the springs because we both wanted the truth.

I agree with everything you just said. Maybe now curious people can calculate the force of a spring themselves since

we exposed the secrets of the springs. lol later Doug

JOSH

[twodownzero]

Wow you guys...

I really hated spring problems in calculus. Now my brain hurts.

[Anthony Lombardo]

What I have learned form this thread is that a recoil spring may need to be fitted so it doesn't go to solid under recoil and beat the bushing or reverse plug out of the front of the slide.

I have been doing this for years. Not all brands of 1911's are created equal. Not all guns have the same amount of tunnel space. Ismi's seem to need trimming about 25-50% of the time.

Wolff's rarely need trimming but sometimes their wire/coil diameter is to great for some plugs and they don't always compress smoothly.

There is no drop in part for a 1911, not even a spring.

If you clip a 1911 coil spring, you slightly lower the rate, but its probably not significant in the "big picture".

I found years ago if you run out of 18 lb Commander springs you can cut a 20# 5" spring and it will work OK. How's that for real world.

Please correct me if I am wrong. I am not an engineer although I did ride on a train once.

[hrt4me]

Good point