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Rifles, Reloading, Optics, Equipment
Rifles, Bullets, Barrels & Ballistics
Canting - the right answer
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<blockquote data-quote="JBM" data-source="post: 109672" data-attributes="member: 1969"><p>[ QUOTE ]</p><p></p><p>...only 6" of horizontal deflection at 1000yds for a 10deg cant seems an extraordinarily low value.</p><p></p><p></p><p>[/ QUOTE ]</p><p></p><p>Granted, I haven't run out to the range to check since last night (I wish!), but it makes sense. The 10 degree rotation adds about 0.5 moa of windage. That's a little over 0.5" for every hundred yards.</p><p></p><p>It agrees with my online calculator which does a velocity vector rotation, not an approximation.</p><p></p><p>[ QUOTE ]</p><p></p><p>Working this out the 'other' way (taking bullet path -294 as meaning bullet drop is approx -328") gives a horizontal deflection of 57" (ie approx 10 times your value!) and a vertical change of -5"</p><p></p><p></p><p>[/ QUOTE ]</p><p></p><p>That does seem excessive to me. In my example, the elevation angle is 3.35 moa -- not much, the elevation angle after canting is going to be somewhat less (but greater than zero) so at most (with a 90 degree cant), you're only going to get about 3.35 moa change or about 50" at 1000 yards. With a 10 degree cant the change in azimuth is very small, about 0.05 moa.</p><p></p><p>Think about it this way, if the elevation angle (angle between the line of sight and the bore) was zero, and the sight height was zero you wouldn't get any change. Yet the other forumulas appear to me to show a change in drop and windage. My formula does not. Would somebody run that case with the other formulas? I probably just don't understand them.</p><p></p><p>[ QUOTE ]</p><p></p><p>...I haven't tried to rework your calculations. Have you only worked on 1 deg rather than 10? ....or quoted the vertical value as the windage value (5" being almost the same as 6") .......or are these differing calculation approaches giving values that are different by a factor of 10?! /ubbthreads/images/graemlins/smile.gif</p><p></p><p></p><p>[/ QUOTE ]</p><p></p><p>Pretty sure. That's why I verified it with my online programs using a different method.</p><p></p><p>[ QUOTE ]</p><p></p><p>...if 10 deg cant only gives a 6" deflection at 1000 yds; it's not really worth worrying about a more realistic degree or 2! </p><p></p><p>[/ QUOTE ]</p><p></p><p>I guess it depends on what you're trying to hit!</p><p></p><p>buffalobob asked about a 180 degree cant. Yes my formula will work before I simplified it for angles less than 90. I removed a term of y/|y| (y is the y component of the velocity unit vector) which would include a minus sign for large angles. It would then give an elevation angle of -3.35 moa and azimuth of 0.0 moa which is what you would expect. </p><p></p><p>If I have some time, I'll include this on my website with some graphics.</p></blockquote><p></p>
[QUOTE="JBM, post: 109672, member: 1969"] [ QUOTE ] ...only 6" of horizontal deflection at 1000yds for a 10deg cant seems an extraordinarily low value. [/ QUOTE ] Granted, I haven't run out to the range to check since last night (I wish!), but it makes sense. The 10 degree rotation adds about 0.5 moa of windage. That's a little over 0.5" for every hundred yards. It agrees with my online calculator which does a velocity vector rotation, not an approximation. [ QUOTE ] Working this out the 'other' way (taking bullet path -294 as meaning bullet drop is approx -328") gives a horizontal deflection of 57" (ie approx 10 times your value!) and a vertical change of -5" [/ QUOTE ] That does seem excessive to me. In my example, the elevation angle is 3.35 moa -- not much, the elevation angle after canting is going to be somewhat less (but greater than zero) so at most (with a 90 degree cant), you're only going to get about 3.35 moa change or about 50" at 1000 yards. With a 10 degree cant the change in azimuth is very small, about 0.05 moa. Think about it this way, if the elevation angle (angle between the line of sight and the bore) was zero, and the sight height was zero you wouldn't get any change. Yet the other forumulas appear to me to show a change in drop and windage. My formula does not. Would somebody run that case with the other formulas? I probably just don't understand them. [ QUOTE ] ...I haven't tried to rework your calculations. Have you only worked on 1 deg rather than 10? ....or quoted the vertical value as the windage value (5" being almost the same as 6") .......or are these differing calculation approaches giving values that are different by a factor of 10?! [img]/ubbthreads/images/graemlins/smile.gif[/img] [/ QUOTE ] Pretty sure. That's why I verified it with my online programs using a different method. [ QUOTE ] ...if 10 deg cant only gives a 6" deflection at 1000 yds; it's not really worth worrying about a more realistic degree or 2! [/ QUOTE ] I guess it depends on what you're trying to hit! buffalobob asked about a 180 degree cant. Yes my formula will work before I simplified it for angles less than 90. I removed a term of y/|y| (y is the y component of the velocity unit vector) which would include a minus sign for large angles. It would then give an elevation angle of -3.35 moa and azimuth of 0.0 moa which is what you would expect. If I have some time, I'll include this on my website with some graphics. [/QUOTE]
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Canting - the right answer
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