Base theory is right, but we need to clean up the math. We have to get a SWAG of the angular velocity of the shotgun. That, not linear velocity of the target, is the needed number. We will, however, get it from the target's linear velocity. Note, these are approximations, not exacting calculations. We need only have a good approximation to understand this issue. OK, lets take a reeeeealy fast bird, say 100 feet per second (over 60 mph) in close, say ten yards (30 feet). The angular velocity of the bird would be approximated by 360 degrees divided by the time the bird would require to fly one lap. The circumference of a 10 yard circle is 2 X pi X radius = 2 X 3.14 X 30 feet = 185 feet. 185feet/ 100 feet per second = 1.85 seconds. 360 degrees/1.85 seconds = 195 degrees per second. The shotgun barrel must match this angular velocity or a little faster. A 1 inch shot charge will transit the muzzle in 1"/1200 fps = 1"/1200 fps x 12 in/ft = 1"/14,400 ips = 0.00007 seconds. The angular shift in the direction the barrel is pointing between the exit of the first pellet and the last pellet is 195 degrees per second X 0.00007 seconds = 0.01 degrees = 0.01 degrees X 60 MOA/degree = 0.4 MOA. Riflemen will tell you that 1 MOA is approximated at 1" at 100 yards. So, 0.40 MOA = 0.4" at 100 yards or about 0.2" at 50 yards. Ergo, the shot pattern is bent by the swing of the barrel, but only a trivial amount.
Math check from all interested would be appreciated.
Last edited by Rocketman; 07/24/10 01:45 PM.
