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Post by TMan on Jul 24, 2006 21:16:35 GMT -5
Just when I thought I pretty much understood what makes a trigger heavy, creepy, and how to fix it, I watched a new video.
While at CheaperthanDirt yesterday, I noticed that they had a AGI 10/22 Trigger Job video on sale. Having been married to my wife for so long, I've gotten a little like her - can't resist a sale.
When Robert started, the rifle had a 7lb trigger. The engagement was slightly positive, and he didn't change the angle to create less positive movement, which I would have expected him to do. Instead he cut off the top of the hammer and sear, which resulted in less contact. Now I would expect that doing this would greatly reduce the creep because the sear wouldn't have to move so much to cause the hammer to drop.
He also took the hammer spring and removed some of the diameter of the spring wire on the belt sander.
When he finished he had a 3lb trigger. What gave him the 4lb drop? Was it by reducing the strength of the hammer spring or does having less metal to metal contact between the sear and the hammer result in a lighter pull? I thought that the weight of the pull was because of the weight of the hammer spring and the angle of engagement. Am I wrong?
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Post by "DoubleAction" on Jul 26, 2006 18:13:47 GMT -5
TMan; I toyed with removing the hammer tension on my 1911s and Sig Sauers and came up with significant reduction in the trigger pull weight. In removing the hammer spring ( Mainspring on the 1911) coils, I also reduced the speed and force of impact in which the hammer drops on the firing pin. Removing the engagement of the hammer to sear also removes the metal to metal drag one feels as the creep, which you already know. Altering the angle and metal to metal engagement, with the hammer and sear, will affect the trigger break but it changes nothing in the force applied between the sear and hammer . Such force between the newly formed contact areas should be that of less pressure created by the hammer spring but enough to assure a positive hammer strike. Less creep and less pull weight should not sacrafice that in which the speed and force of the hammer is compromised, nor should it affect the positive lock up engagement of the hammer and sear during cycling. The amount of force required to unlocked the sear engagement from the hammer is measured by the least amount of resistance, which is accomplished by removing the amount of friction and force between the two contact areas. The friction is that of the engagement and the force comes from that of the tension applied by the spring. These are just my own thoughts on the subject.
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Post by TMan on Jul 26, 2006 20:50:10 GMT -5
Thanks DA, but I still can't visualize...
Say that I had two pieces of metal where one was fixed and the other was movable. They are each 2" and have full contact i.e. there is 2" of metal touching each other. The movable one has a 10 lb spring. So with a certain amount of force, F1, we slide the movable one down 2", which is the creep.
Now let's replace the movable with a piece of metal 1/8", and will be centered on the 2" piece. We will reduce the creep to 15/16", but we still have the same amount of force, F1? I know that since the area has decreased, the psi exerted by the 10lb spring on the 1/8" would have increased, but what about the effects of friction? Since we would have less metal to metal contact, would the amount of force still be F1, or would we have less force: F2?
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Post by MLB on Jul 26, 2006 22:10:27 GMT -5
As I recall TMan, the force is the frictional constant times the normal force, and is independent of the contact area. Steel on steel constant is about .4 for sliding.
I am having a hard time reconciling this against the practical experience of using wider tires on a race car to provide more "grip". Despite looking through several references, wider tires shouldn't (according to theory) provide any more frictional force.
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Post by "DoubleAction" on Jul 27, 2006 20:11:28 GMT -5
Thanks DA, but I still can't visualize... Say that I had two pieces of metal where one was fixed and the other was movable. They are each 2" and have full contact i.e. there is 2" of metal touching each other. The movable one has a 10 lb spring. So with a certain amount of force, F1, we slide the movable one down 2", which is the creep. Now let's replace the movable with a piece of metal 1/8", and will be centered on the 2" piece. We will reduce the creep to 15/16", but we still have the same amount of force, F1? I know that since the area has decreased, the psi exerted by the 10lb spring on the 1/8" would have increased, but what about the effects of friction? Since we would have less metal to metal contact, would the amount of force still be F1, or would we have less force: F2? TMan; While hooking up with Jeff Quinn's website on the Taurus 1911, which you provided the link for on another thread, I took some time to look his site over. One particular area, which caught my curiosity, was called "The Poor Boy's Trigger Job". Working with the single action revolver might be somewhat different but it also involves reducing the tension on the hammer spring while achieving the goal of a lighter trigger pull. He also provides his own insight into removing the gritty "feel" ,which is felt in the trigger. www.gunblast.com/Poorboy.htm |
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Post by MLB on Jul 27, 2006 21:00:44 GMT -5
Jeff's trigger job addresses both factors in the friction equation. He smoothes the surfaces (thus reducing the coefficient of friction) and then unhooks one of the legs of the spring, thereby reducing the normal force.
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Post by TMan on Jul 27, 2006 21:19:42 GMT -5
His thing with the spring is the trigger return spring, the hammer spring, which is what exerts the pressure on the engagement surfaces is really what I'm interested in.
"Coefficient of friction", been a long time since I heard that one. Okay, so if I polish the surfaces (or "marry" them by bumping the trigger), or I apply a super slick lubricant, I will reduce the coefficient of friction. So the force to cause the two objects to slide will be reduced. My question is: does the size of the two objects matter to the force required? In other words: if I had two objects each having a square footage of 2 sq ft with a 10 lb spring. I have two other objects having a a square "footage" of 1 sq in with the 10 lb spring. Will the force to slide the 1 sq in be the same as sliding the 2 sq ft object?
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Post by MLB on Jul 28, 2006 11:24:31 GMT -5
According to all references I have, the frictional force is independent of the contact area. So in your question above, assuming the two objects have the same weight, the force should be the same.
That's what the 'ol physics book says. It is hard to rationalize though. I think I'll try it out with a rectangular box of stuff and spring gauge. I'll pull it on the flat side, and then on the end to compare.
This is starting to remind me of our recoil thread. Perhaps we should have a physics area...
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Post by "DoubleAction" on Jul 28, 2006 22:48:07 GMT -5
Maybe I would do better to surf the big hooter sites.
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Post by TMan on Jul 28, 2006 23:01:52 GMT -5
Maybe I would do better to surf the big hooter sites. Especially if you have been into the hootch. ;D |
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Post by MLB on Jul 28, 2006 23:16:11 GMT -5
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Post by "DoubleAction" on Jul 29, 2006 2:14:48 GMT -5
Howard Hughes was one who did, in fact, put forth the idea of an up lifting bra to that of his aircraft engineers. I expect those men probably brought forth their own formula, model, and end result for what Mr. Hughes had requested. Seems like everything needs a formula of some sort in order to have calculated evidence of it's legitamacy.
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