Advanced Marksmanship A question on canting

silhouette

Grey man
Full Member
Minuteman
Mar 29, 2012
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This is a genuine question, not a shit stirring one.
If you are shooting at a distance which requires 10 mils elevation and you cant your rifle 1 degree, how far out (left or right) will your point of impact be?
Please show your method of working this out.

If you are shooting at a distance which requires 10 mils elevation, then 10 divided by 360 (1 degree) = .0277 mils.
Therefore if that distance is 1000m, and 1 mil at 100m = 1m, then .0277 x 10 mils = .277 meters, or 277 millimeters.
Is this correct?
 
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No, if the scope is level there is no cant. But there is an offset which remains the same regardless of distance. Or at least that's how I understand it.

If you zero this way, you will induce windage errors at ranges other than the zero, because the horizontal zero changes at different ranges. It's like when you mount a laser on the side of a rifle... you can either deal with the errors at different ranges, or zero with an offset so that the line of sight and bore axis are parallel. For a small cant, the errors will be minor.

To really understand this, imagine mounting the scope to a side rail, zeroing at 100m, then attempting to shoot at 600m.
 
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Where are you getting your formula

Like you, I don't think the linear interpolation of the 10 mils is quite right. I think you'd want to use the sin of the cant angle. If you cant from 0-90. your bore axis traces out a quarter circle. I'd try sin(1 degree) * 10mil = mil offset. For vertical offset, do 10 - 10*cos(1 degree). You get vertical offset because by canting, your elevation doesn't get the full 10 mil correction. If you can't 90 degrees (hold the gun sideways), you get no vertical correction, and a 10 mil windage correction.
 
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