I was looking for some help concerning the proper formula to calculating center of gravity offset in a purposefully imbalance bullet. If I removed 40mg of copper from the outside jacket of a bullet on side of a bullet that weighted 11631 mg after the material was removed, and the diameter of the bullet was .284 inches, would the offset in inches just be the ratio of the material removed times the radius ? So 40/11631 * .142 resulting in a .00048835 inch offset center of gravity? I am sure I am oversimplifying the issue because of the bimetallic structure of a jacket bullet and the lead center has a greater density. And the second question is concerning the correct formula to predict the angular error from a center of gravity offset. I had read that Harold Vaughn's solution presented in chapter 9 of Rifle Accuracy Facts was incorrect as it included time of flight as a variable. I saw a formula quoted here http://riflebarrels.com/a-look-at-bullet-imbalance-and-twist/ Error in moa equals imbalance in inches divided by (.000046 * twist rate inches). Is this approach cited by Daniel Lilja the correct one to predict angular error? Any help would be appreciated with either of these questions.