Then in ball-wear formula (25), T = 6.9/K Log10 Da/Db; but from (29), K = Rt/Wt. Then T = 6.9Wt/Rt Log10 Da/Db T is 1 day, Wt is the original weight of the ball charge, and Rt is the ball wear for one day. Then Log10 Da/Db = Rt/6.9Wt are all known, and it is only necessary to solve for Db, the diameter of the balls to be added.