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.
