okay lets get started.
there are a few options out there for buffering the recoil on a eaa witness, all of whicdh are pretty good to great.
starting with the basics.
one could order a wolf recoil XP (extra power pak) with a 16 18 20 22 pound springs and use these with the factory guide rod, which a ton of people do and have no problems at all, as i have HEARD of folks using a 22# spring in 9mm!! to me that seems not right, like the gun wouldn't work, but i have read some do, as i have read some use a 8-9# spring in 9mm, with this info one might say "what gives"?
i don't know, but there is a such thing as over springing a gun, which in turn creates the same problem as under springing a gun, IE--> the slide being blown back with such force, at its rear most position, it has a tremendous amount of stored kinetic energy, and when it slams back forward, you have the slide slamming just as hard in its forward movement, as the impact of a under sprung gun as the slide moves rearward after being fired.
half a dozen either way you look at it.
so some guys are perfectly content with a switch in recoil spring weight to the next size up and the gun runs fine. others like me, are never satisfied with anything, and watn the ultimate everything and anything.
i use, as stated before, a hennings cone-fit guide rod, matched spring weight for the load/caliber i shoot, along with a buffer tech recoil buffer pad thing. when i first installed this system i used the henning rod, with the factory 10mm spring, and the buffer installed. i would hand cycle the gun empty, no bullets. and i for the life of me, could not get the slide to hold open, or in other words, the slide catch notch was now to far forward of the slide catch/take down pin. it was attributed to the extra length the buffer added. so i began cutting the factory recoil spring, half a coil at a time, untill the slide would hold open. i ended up shortening the factory spring about a 1/4 inch or so, and then the slide would lock back.
so then i came here to ask if cutting coils of a spring changed it weight, i was sure it did but wanted confirmation, which i could not get, i needed to know did it maybe up the weight or decrease the weight, common sense would tell you it would decrease it not so, here read
Popular wisdom" rules. 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.
Cutting coils is limited to those types which have tangential ends. Tangential ends are those which spiral off into space. If you tried to stand the spring on end it would fall over.
more to come