Cant with Burris Rings? Fifty Driver, Goodgrouper anybody......

[goodgrouper]

SOrry Johnny,
I just read this post for the first time tonight and I can see I'm too late, there have already been some great and precise replies.

[Ackman]

that those Burris inserts aren't MOA, they're thousandths - .005" and .010" for the 30mm, and MOA will vary with the distance between rings.

[jb1000br]

[ QUOTE ]
that those Burris inserts aren't MOA, they're thousandths - .005" and .010" for the 30mm, and MOA will vary with the distance between rings.

[/ QUOTE ]


BINGO!!! [img]/ubbthreads/images/graemlins/grin.gif[/img]

JB

[Mysticplayer]

Forgot to mention my simple rule of figuring out how much shims I need. For most scope spacing on reg. length actions (7" to 8"), 1 thou of shim is 1" at the target.

so if I need 20min, I use 20 thou worth of shims. Close enough and you can fine tune either with the knobs or rotating the shims in the rings.

That's right, you are not fixed at the amounts marked on the shims. By rotating them from 'full' value, you can get any amount in between max and min.

Jerry

[jsali]

Someone critique my math here. If I am figuring right the formula should play out as follows:

Difference in elevation of rings (inches) / Spacing of rings (inches) X 3600 (inches in 100 yards).

This should give you the vertical displacement off of center at 100 yards (in inches). Divide this answer by 1.047 (inches per MOA at 100 yds) and this should give you the vertical displacement in MOA. It seems right if I remember my math correctly.

[abinok]

johnny,
3600 is from the miliradian value 1mil=36"@ 1000 yds, and 3.6" at 100yds. There are 3.438moa in a mil. Don't worry about the 1.04719.... this formula yields true moa values.

[jsali]

That makes sense. 1 mil = 1/1000 the distance. I was just referring back to some old math formulas. They seemed like they would apply.