Thanks for the explanation, but I still can’t make sense of what the pdf is showing …..I’m sure that if we were face to face sketching on a whiteboard it’d be immediately apparent! [img]/ubbthreads/images/graemlins/smile.gif[/img]
Meanwhile, I notice my diagrams weren’t clear to you either:
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I'm not quite sure what you're sketching. Your first couple of sketches make it seem like you're using a triangle with a 10 degree angle to calculate the windage. This isn't really physical. Think about what happens when we cant a firearm. You rotate about the centerline of the scope/sights. For a clockwise cant, this causes the the barrel to swing to the right -- higher azimuth angle and lower elevation angle.
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....that’s exactly what the pics were trying to show!! [img]/ubbthreads/images/graemlins/smile.gif[/img]….... LOS fixed with the applied elevation rotating about it by 10deg
...I'm not sure that it's important whether we regard the rifle as having rotated 'under' the LOS
....or the line of departure having rotated above the LOS
... the endstate is the same:
-LOS fixed on Tgt.
-Elevation (intended to be applied vertically) now being applied at 10 degrees from the vertical.
Hopefully these sketches are clearer:
Correcting my earlier QE=TE+AS confusion:
We now have the new QE (quadrant elevation) …..8.8mils
The TE (tangent elevation) calculated and applied is unchanged (because the target range and bullet drop are unchanged)
Therefore we have, in effect, reduced our AS (angle of sight) by 0.28mils
Therefore we will miss low by the distance subtended by 0.28mils at 914m
(as well as miss right by the distance subtended laterally by the new bearing!)
I think I’ll ‘draw stumps’ there! ...keen to receive feedback, but I'm done sketching!…..if nothing else, I’ve improved my PowerPoint skills! [img]/ubbthreads/images/graemlins/smile.gif[/img]
Thanks to all for a good brain workout!