This weekend i got a chance to shoot my 6.5/284 for the first time out to 788 yards, where i was the shooing the wind was very non-constant, gusting between 5/20mph,i had wind flags set out at 10,500,and 788 yards, and they were never in the same position for any of my shots, and guessing wasn't working to well because i only hit the soda bottle once, if it was a 3ltr there might have been a few more holes [img]images/icons/grin.gif[/img]. Is there any way to accuratly shoot in these conditions?
what i should have asked was how do you guys accuratly judge the windage when its blowing hard at the bench but not at all/very little at the target or vis versa. I know from shooting that day that bullets were most effected by wind at the muzzle rather than at the longer range.
I've been toying with a graph to show you how wind works based on start, stop and overall trajectory distance effected but my brain broke!
Think on this for a bit. (I shoot a 308 Win with 175 Sierra at 2660 fps) If you're shooting 1000 yards and there is a 10 mph full value wind for the first 100 yards and no wind for the remaining distance the amount of correction you'll need is about 1 MOA, same as if you're only shooting 100 yards in a 10 mph full value wind.
If you think you have a grasp on that and the reason answer back this question; the amount of correction you'd need if the 10 mph wind started at the 500 yard mark and stopped at the 600 yard mark with no wind anywhere else on the range with the target still at 1000 yards.
Is the answer 1/5 moa adjustment, if it is then say 0/200 would be 2moa or say 800/900 it would be 1/8moa, from what i saw shooting this weekend more, wind at my position caused more drift and no wind at my position and more wind at target the bullet drifted alot less, so these answers make alot of sense. Thanks alot for clearing this up for me.
The reason is that the wind provides an angular deflection. The earlier the angle change, the more it changes the point of impact. I've done this with my MPM program. It allows you to define a "wind field" which is velocity as various points, <x,y,z>. It then interpolates over the points to find wind at any point.
I'll get back to you on your addition in a little bit, we still need some work I think.
For the 10 mph wind at 500 and stopping at 600 and the target at 1000. You'll need .5 MOA correction as I see it. Here's why, the 100 yards of 10 mph wind will cause 1 MOA of deflection from 500 to 600 yards. This deflection is caused by the projectile adopting an additional sideways motion caused by the wind. This adopted motion does not cease at the 600 yards line when the wind stops, it continues on the same vector, so we have a projectile with a 1 MOA deflection, this equates to 1.047 inches per 100 yards of travel. Add 1.047 inches of deflection for the 600 through 1000 yardsages and your get 5.235 inches of Point Of Impact (POI) to Point Of Aim (POA) error. To figure the amount of correct the shooter needs to correct this ~5" miss we need to turn it into MOA. 5" of error / 10 (for the 1000 yards because MOA is used per 100 yard increments herein)= .5 MOA of correction.
Hopefully one of our resident "egg heads" will double check this and chime in if I'm all wet.