Spring length in a 1911 recoil spring

And if you bought one from Wolfe Springs it might be longer/shorter/different than that. All that matters is the amount of force required to compress it.
Just enjoy your new recoil spring and the difference in the way it makes your gun function.

23 coils sounds right.

But the old 21 coil spring could be right too, depending on how the ends are closed.

What you need to do is measure the diameter of the wire in each spring with a mic or dial caliper, and see if the old one is a different size wire stock.

It is also possible if not probable Colt has upped the spring weight a little to better handle some of todays hot SD ammo.

30 years ago, about all you could buy was 230 grain hardball, and 185 or 200 mid-range match.

rc

I had this exact question regarding my Commander-length 1911. When I compared the new one to my existing one, it was about 1/2" longer and only just shorter than my Gov't Model-length existing recoil spring. In the same shipment from Brownell's, I had ordered a lightened rebound spring kit for my S&W 342 and noticed that the lightest spring was significantly longer than my existing spring or the other two heavier springs that came in the kit. I emailed Wolff Springs asking about it. A gentleman responded with "Call me." I did.

He laughingly explained that this single question is likely the most common question asked in the history of springs and is by far the most common question that comes into Wolff. He said it's far easier to explain it by voice than writing an answer in an email.

To the point--springs are engineered to do a targeted amount of work within a general space using developed steel alloys to accomplish the task. A spring must do its work before before full compression (spring bind). The number of coils is dependent on the metallurgy before space constraints. He said a spring may settle in some after being worked a little bit.

After his reassurances that everything is OK, he said to manhandle the spring into place. The springs we were talking about were not delicate flowers and they would, indeed, do the job. After shoving the longer recoil spring into place, made a little more difficult for me because Mr. Volkmann's recoil spring plug makes spring installation from the front impossible, and running a hundred rounds through the pistol, the spring length had shortened a little bit. It was easier to reinstall than when the spring was new.

I may have bobbled the explanation some by relating it here some six months after that conversation, but I think the gist of it is accurate.

Spring weight is measured with the spring in the gun, fully compressed as far as the slide travels.

Out of the gun, fully compressed to spring stacking, would be more.


BTW: With a little ingenuity and a bathroom scale you can measure the spring weight in the gun.
Use a short length of PVC pipe for the gun muzzle to stick down into as you push it down on the scale.
Not accurate enough for NASA rocket springs, but close enough for comparison purposes on recoil springs.

rc

I would have assumed that a 14 pound spring would collapse one inch if it was supporting a 14 pound weight

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Were that the case, none of us would be strong enough to rack the slide. It would be like cocking a crossbow!

Speaking of which, it's kinda like a bow, which is measured for poundage weight at full draw.

rc

The longer looking spring is correct.
Just install it and shoot the gun.
100 or so rounds and next time you take it apart you will see that the spring as made it's "set" and is approximately the same length now as the one you took out.

Did you read the FAQ page at Wolff Gunsprings?
It might shed some light on your question.
http://www.gunsprings.com/faq#FAQ1

rc

mikemyers said:

Rcmodel - in that case, how do I find out the spring rate of a "14 pound spring"? Until you said that, I just assumed that when people talk about the "weight" of a spring, they were referring to the spring rate.

I think Monday morning I'l going to call these guys:
http://www.gunsprings.com/Semi-Auto Pistols/COLT/1911 GOV'T PISTOL/cID1/mID1/dID1#3
I guess it's no big deal..... I'll just put the new spring in and use it, but I'd like to understand "the why" behind things.

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The spring rate is the result of dividing the free length minus the fully compressed length into the spring weight. In this case, 14#. The result will be the pounds per inch the spring produces as it is compressed. This could be a useable number in the gun industry if it where implemented. But since the springs are in a controlled environment, compressed the same, in the same space, the designation between one spring and another is given in the spring's weight for comparison, of one to another.

CAW

Coil count and length provide no information about the rate (pounds per inch) of the spring.

The steel used, its temper level, the wire diameter are all needed.

mikemyers said:

Do any of you have a Colt Combat Commander, and maybe the next time you have the gun apart, you can measure the length of the spring and post it here?

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The factory weight recoil spring for a Colt Commander in .45 ACP is 18 lbs.

The factory weight recoil spring for one in 38 Super or 9mm is 16 lbs.


That's as complicated as it needs to get.


Call Wolff springs, or anyone who stocks Wolff springs, and tell them you want a factory weight spring for your Commander, and which caliber. Unless you reload, or have a comp gun, you'll be just fine with the factory standard weight.

brickeyee said:

Coil count and length provide no information about the rate (pounds per inch) of the spring.

The steel used, its temper level, the wire diameter are all needed.

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No, not needed, if you have the weight produced at full compression, the free length and the full compressed length. Those are the only factors to determine the spring rate. the weight at full compression divided by the free length minus the compressed length is the spring rate, or pounds per inch.

CAW

mikemyers said:

I plan to call Wolff tomorrow...... I've already ordered and received a new standard spring from Colt, so as far as my gun goes, I'm all set. I'd like to understand what is going on though. They'll probably have all the answers I'm looking for.

As stated up above:
"....since the springs are in a controlled environment, compressed the same, in the same space, the designation between one spring and another is given in the spring's weight for comparison..."

That makes sense. I assume that means that the spring, when it's sitting inside the gun, is pushing on each end with a force of XX pounds, where XX is the "weight" of the spring.

I'll also assume that this measurement is with the slide at one end or the other - I have no idea which.

From what I remember about spring rates from school, it's one of the most important things to know about a spring, and I'm surprised that it's so hard for me to find out that value. Maybe I'm the only person silly enough to care for a nuOr thrOmber that apparently isn't used by anyone.


(RCmodel, I did read the FAQ page, thanks for the reminder, but it didn't answer my questions, probably because nobody really needs the information I'm curious about. They're explaining how to select an appropriate spring, and I'm trying to find out what "is" a 16 pound spring - when, where, and how is that measurement taken, and what will the spring rate be..... unless that "16" actually is the spring rate?)

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The conventional recoil spring designated weight doesn't need to be in the gun to be measured. It simply needs to be compressed to the same compression length as it would be in full recoil. This is considered to be 1.625"for GM, 1.125" for Commanders, and .700" for Officers and the compressed length that Wolff uses for the rating. These are specific springs intended for specific use. Or in other words the spring designated for Government models are made specifically for them. The weight gives a clear indication of one spring being of higher or lower weight. The engineering is already done and as it turns out very limited as to what can be done in the allowable space. Springs are lineal, or increase resistance as they are compressed in a straight line. The spring rate isn't given as can be misleading if you do not also know the free length and the compressed length. So to keep these proprietary springs labeled in a manner germane to their use, the weight at full recoil compressed length is given for the purpose of ID.


Springs are interesting that is for sure, but terms and understandings can get confusing, I am included. Spring rate can not change what the spring is, spring rate is a description of the springs performance per inch. A 16# spring, meaning it can at max. produce 16# of pressure can have a range of spring rates based on it's free length. To calculate the spring rate you deduct the full compressed length from the free length, equaling the use-able or compressible length . Then divide the max. pressure by the use-able length, this is the spring rate. If the spring is fully compressed at half the number of coils it was it will still be a 16# spring, but the spring rate will be twice what it was.

So, a 16# 32-coil spring of .045" wire 6.5" long will have a compressed length of 1.44" so, the spring rate is 3.16# per inch, and if cut to 16 coils it will still produce 16# of pressure fully compressed, with a spring rate of 6.32# per inch. But never will the 16# spring be able to produce more than 16# of pressure as long as the understanding is max. compression pressure, compared to max. compression pressure. Interestingly the 8 coils cut from the 32 coils will on it's own produce 16# of pressure when fully compressed just as the 32 coil spring was, but it's spring rate will be through the roof at 12.65# per inch. But is only 1.625"long and not use able as a pistol recoil spring, too light in battery.



Most commonly, we do cut springs and return them to the same task as before, and because the space of compression is the same, e.g., full recoil of a 5" is 1.625" and battery being 3.6875", the pressure produced at those dimensions will be reduced, so we say/think the spring is weaker. But it isn't because the spring is weaker, but because the spring isn't compressed as much as it had been before cutting.

Whether this helps to understand I don't know, but is relevant. If you hung 5 spring scales from each other and zeroed each, and then hung a 10# weight from the bottom one, they would all say 10# just as one would have. The difference would be that the distance the weight moved down to balance the 5 scales versus the 1 scale would be different by a factor of 5. The point of this is to see that these 5 scales which are powered by the same spring preform the same task whether there is 1 or 5, just the space it takes is changed.



For instance if you had two recoil springs that each registered 16# when compressed to the max. of 1.625". And you put them end to end and compressed the two springs together to 3.25", they would still register 16#.

So, the idea that a shorter version of an already identified spring of max. compression pressure, will somehow be more, is not in line with the facts. Just as the two 16# springs generate only 16#s together and 16#s alone, the only difference is travel distance. The spring rate of the single 16# spring will also be twice as much as the spring rate of the two springs together, pounds per compressible inch.



CAW

All that matters is the amount of force required to compress it.

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That and that the number of coils and wire diameter are such that they make the overall length when the slide is fully back short enough to not interfere. Mostly you see this with a 4.25" "Commander" spring in a 4" Kimber Pro.

F=K*dl is the "ideal" spring equation, where K is the "spring constant" (lb/in, or dyne/cm etc.) . Real springs deviate a good bit except for short length deltas.

30 year old or not, I say if it ain't broke, don't fix it!

The full compressed weight doesn't change within reason when cut, but the rate does. The rate is simply pounds per inch. The shorter spring of equal weight when fully compressed gets there in a shorter distance so the rate is higher as expressed in pounds per inch. If a spring had exactly one inch of travel and also registered 16# when compressed it's weight and rate would both be 16#. If it had two inches of travel and the weight at full compression was 16# then the rate would be 8# per inch.

CAW

The example in the link clearly supports this as when L1-L2 varies so does the rate. What doesn't vary is the weight at full compression.

Again a practical example, follow this. Two identical springs of equal weight when fully compressed. When placed end to end and again fully compressed will still register the same weight, but the traveled distance to compress them is twice as much. So the rate is also twice as much per inch, but the weight has stayed the same. The two equal springs separately will register the same weight when fully compressed as when end to end, but the rate will be half what it was when end to end as the compressible distance is half what it was.

Now place the two springs side by side and consider them one spring, the weight at full compression will be twice what it was as a single spring, and the rate will also be twice what it was.

CAW

Another example of springs weight and rate come from the sometimes confusing alteration results of automotive springs. Take an off road truck with long, or tall coils springs, and it is found that the ride is like a roller coaster, up and down, a lot. So cut a few coils off. now it still goes up and down, but not as much. Some will say the cutting of coils made the spring stronger. Others will realize the spring has the same strength, but the spring rate is higher. It gets to the same point of resistance with less travel because each coil is twisted more per inch of travel than when it was longer. Same spring and will register the same weight if fully compressed, but the cut one has a higher rate, it gets there in fewer inches.

CAW

mikemyers said:

I see what you're saying now..... explained that way, it all makes sense. The longer a spring, the less force it will take to compress it a given distance.


I think I found why we're thinking differently - this page explains it:

http://engineersedge.com/spring_comp_calc_k.htm

It isn't the actual length of the spring that counts, it's the number of active coils. Cutting a spring in half would also cut the number of active coils in half, resulting in a stiffer spring, just what you described. A longer spring, with more coils, will have a lower spring rate.

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Yes the active coils are the only ones that can add resistance, but you must have a unit of measure to relate the spring rate, and that would be the length of the active coils, not the number of coils. The number of coils times the diameter of the wire will give you the compressed length. If you wish to describe the spring in pounds per inch you must calculate using inches and pounds. You can of course use metric or any form of weight per length.

CAW