I’m here again to pick your knowledgeable brain! I have always wondered at what distance does the same shot size fired at 1150 F.P.S and 1400 F.P.S Equalize? In other words I know that at 20 yards a 1400 FPS pattern requires less lead than a pattern fired at 1150 BUT at what distance do they catch up and remain the same?? Example: Would my lead be the same on a target at 60 yards no matter what the muzzle velocity??
The slow load never catches up with the fast load. By that, I mean that the fast load will always take less time to reach the target than the slow load. Therefore the fast load will always require less lead. Until, that is, until they hit the ground. And then the faster load will have gone further.
Here are some time-to-distance numbers for 7-1/2″ shot taken from Lowry’s “Ballistics For Windows” program. I think that 35 mph would be a good middle ground target speed to start with.
7-1/2 shot with an 3′ MV of 1150 fps and the lead on a 35 mph 90 degree crosser:
time to 20 yards: .0602 sec; lead 3.09′
time to 40 yards: .1455 sec; lead 7.47′
time to 60 yards: .2594 sec; lead 13.32′
7-1/2 shot with a 3′ MV of 1400 fps and the lead on a 35 mph 90 degree crosser:
time to 20 yards: .0511 sec; lead 2.62′
time to 40 yards: .1265 sec; lead 6.49′
time to 60 yards: .2290 sec; lead 11.76′
So, the bottom line is that more is more. Well, actually more in this
case is less. More velocity equals less lead.
The Technoid writing for Shotgun Report, LLC
(Often in error. Never in doubt.)