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math wiz? sight cant
Okie, unable to find the formula and I've long forgotten my trigonometry. I added a MI fixed front/rear on a rifle, 16", excessive right windage(on rear) to get zero at 50. I didn't notice at the range that the rail was off at the receiver, loosened and aligned it when home. It appeared to have maybe about .050-.062" right cant. The front is back enough to allow a M300 light at 12:00.
Curious of what that would equate to at 50? I hope to check Friday and hope that corrects the issue before I look further.
Oh-"excessive" the rear aperture was close to the right ear. At mechanical center, I was roughly 5" left at 50.
Mark
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Ok, I think you’re saying that you moved the front sight sideways by .050 —.062”. Let’s split the difference and call it 0.056”.
It’s a 16” barrel, but you don’t say what the distance is between front and rear sights. Let’s call it 15”.
50 yards is 1800”.
The corresponding shift would be
(.056 / 15) x 1800” = 6.7”
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Great! yes, 15" would be about right on the radius. I'm guessing on the shift of the sight with the rail being turned slightly. If that holds true, then I shouldn't be far from center on the rear. My back up plan is to remove material from the front sight, right side where it contacts the rail angle.
Thank you!
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Thanks, couldn't seem to locate that earlier, saved now.
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I misunderstood something when I first saw your post. Re-reading with the benefit of this morning’s coffee, I now realize that you had a handguard that was twisted to where its rail was misaligned with the receiver’s rail by .050 — .062”. If so, the offset of the front sight would have been greater than that amount because it’s farther from the axis of rotation, roughly double.
