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Clark, I too have refined some calulations. Please explain the action lenght in your calulations. I assume you used 2.5" for a pistol action. With the H&R 700 wouldn't it be bolt + hammer + distance traveled to cock the hammer? Or 3.235 + 1.55 + 1.83 = 6.615". This is a long action. Also, if I were to do this, the shoulder on the 5.7 round would have to be set back .065", this would equate to a seven percent reduction in case cappacity and velocity. So using a velocity of 2300 fps. Does this make sense?
5.7 Short calulations:
assume: Peak chamber pressure = 50kp/i/i
assume: average chamber pressure = 25kp/i/i
assume: Peak bullet velocity = 2300 f/s
assume: Barrel length = 22i = 1.83f
assume: brass case inside diameter = .297 i
calculate force from chamber = PA = [25kp/i][.297i/4][.297i/4][3.14]= 430 pounds
calculate time of chamber force = 2 1.83f/[2300f/s]=.0016 s
assume: action 6.615i long = .551 feet
assume: spring force = 18 pounds
calculate spring energy =fd=18 .551 = 9.918 footpounds
calculate distance chamber pushes bolt = E/F = 9.918 fp/430 p = .023 feet
This means the bolt will be accelerating back for .001 seconds until it has gone .023 feet back and then it will be slowed down by the recoil spring for 6.592 inches where it just runs out of speed as it reaches the back of the action.
calculate bolt peak velocity = 2D/t = 2 .023 f/.001s = 46 f/s
calculate mass of bolt = 2 E/VV = 2 9.9 fp/ 46 f/s / 46 f/s = .009 pss/f
calculate weight of bolt = GM= [32.2 f/s/s] [.009 pss/f] = 0.2898 pounds, or 4.637 onces.
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