Would this help?
"Bullet Stabilization An elongated bullet flying through the air without spinning will be unstable and inaccurate. The longer the bullet is in relation to its diameter, the more spin is required to stabilize it. How much spin is required? This relationship is expressed in the Greenhill formula, a simplified verson of which is:
150 x diameter squared divided by bullet length = required spin
Example for a .45 caliber bullet .60 inches long:
150 x .45 x .45 divided by .60 = 50.6 inches
So, for the example bullet, a spin rate of 1:50.6 or faster is required
The formula can also provide us with the maximum bullet length which can be stabilized by a given barrel twist. The formula becomes:
150 x diameter squared divided by twist rate
Example for a .50 caliber barrel of 1:48 twist:
150 x .50 x .50 divided by 48 = .78 inches The barrel will stabilize a bullet .78 inches long, or shorter."
http://thehunterslife.com/forums/showthread.php?p=76345
