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?)
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