Originally Posted by
popper
Gear, I looked at Larry's pressure curves, the peak is (I assume) when the Cb is getting 'sized' and accelerated, then drops off rapidly. You and I differ on which end of the bbl 'uses' the most energy, I think it is the chamber end. My reasoning is: 1) assume the CB has constant acceleration (linear increase in V). 2) the CB has momentum. 3) F=M*A. M doesn't change, F decreases so A must also. I think the result is a logarithmic velocity curve and the same for the land force. I did a simple calc for the actual land thrust force, given the .004 land height and ~1" bearing surface and came up with ~200k #. That can't happen, Pb can't take that force. I'm trying to re-write the external ballistics program to try to figure out the bbl dynamics - I'll see how far I get.