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Here's a brain strainer:

I recently posted about some hammer spring surgery on my Beretta Bobcats. I would prefer a shorter version of the spring, to keep both flat ground ends, but that's not available.

Us gun nuts are always cutting some spring or other, and I've wondered about the physics of it. I can understand where cutting a compression spring would apparently weaken it, but does it really? I think the spring would still have the same strength, if its "working distance" was also reduced the same amount.

So in cutting a Bobcat hammer spring (or any hammer spring), I haven't actually weakened it, I've just reduced the amount it compresses in use, which weakens its effect. It would still take the original amount of pressure to compress the spring completely?

A progressive spring (uneven coils) or an expansion (stretching) spring would be another set of rules, I'm sure.

Technically you weakened it. Spring have a rating of load/in.

http://en.wikipedia.org/wiki/Spring_%28device%29

Thanks, blue. Probably more than anyone ever needs to know about springs.....

Post number 18 on this page of the 1911 Forum broke it down more into layman's terms:

http://forums.1911forum.com/showthread.php?t=104764

I had a 16 coil spring, so that's 14 active coils. I removed about 2.5 coils, so I weakened it about 18%. That's as far as I'm comfortable with, so I'll hope it's good. If not, I have new springs on the way from Brownell's.

I may give up and use my local indoor, $10 an hour, Billy-Bob shooting range to try them tomorrow. They'll likely have a cow if I empty a mag fast, but that's part of a function test. I can go with 14 full mags ready, to save time.

I can't stand the place, but it's not worth a 70 mile round trip to my usual Forest Service range if it's raining.