Originally Posted by prm
If I've done my math right, assuming 3,000ft/sec, they would be rotating at 270,000 RPM and 135,000 RPM respectively. I'm not sure how much that rotational force difference translates into the bullet opening, but that would seem to be a significant force. Out of curiosity, I wonder how much the force is different for a larger diameter bullet given the same rotation rate? Ex: a .243 bullet vs. a .338 spinning at the same rate. The outer edge of the .338 is further from the center of rotation. That's some serious ballistic gack to consider.



Code
Cal       KE(ftlb)  RE(ftlb)
243 Win     2100       6
25-06       2600       7
264 Win     2800      11 
270 Win     2800       9
7mm Rem     3200      14 
300 Win     3800      16
338 Win     4000      20
375 H&H     4500      19


Code
          (gr)	 mv(ft/s) twist   rpm
243 Win	   100	   3100	   10    223,200
25-06	   120	   3100	   10    223,200
264 Win	   140	   3000	    9    240,000
270 Win	   150	   2900	   10    208,800
7mm rem	   160	   3000	    9    240,000
300 Win	   180	   3100	   10    223,200
338 Win	   250	   2700	   10    194,400
375 H&H	   300	   2600	   12    156,000


KE = Kinetic Energy
RE = Rotational Energy

Rotational Energy is very small compared to Kinetic Energy

It's not that Liberals are unwilling to listen to another point of view, they are just simply amazed that another one exists.