jmorris
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- Joined
- Sep 30, 2005
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- 24,330
Fascinating specification...100# of crimping force. How is one supposed to measure that?
There are a number of ways one could do that but that’s not what the quoted text is saying. “100 lbs bullet pull”, is the proposed argument and could be done with a scale.
I remember a thread a few years ago where it was speculated that air pressure created upon a bullet being seated was forcing it back out of the case. I don’t remember all of the details and not sure what forum it was on but a bullet being seated didn’t create enough pressure to dislodge it from neck tension in the cases I tested.
That 130 PSI is not the “bullet pull” resistance though because that bullet doesn’t have 1 square inch of area. It’s a .452 bullet so really only has about 20.8 lbs pushing it out at that pressure.
If that bullet would not pop out of the case at 625 PSI, it would satisfy the 100 lb bullet pull number. That pressure would go up as the diameter goes down as well. So a .224 bullet with .039 area would need to resist something like 2564 psi pushing against the bottom without popping out of the case.
Olympic air rifles would be highly inaccurate if those were the operating pressures needed to achieve bullet movement accurately. You would be looking at 4000 PSI with a .177 diameter projectile for 100 lbs of force to move it.
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