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Rifles, Reloading, Optics, Equipment
Rifles, Bullets, Barrels & Ballistics
Canting - the right answer
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<blockquote data-quote="TiroFijo" data-source="post: 109778" data-attributes="member: 974"><p>JBM, I believe the basic formulas I posted are widely used, perhaps with a different twist or additional refinements. When Gustavo, our dutch friend Jeroen (the author of the article in the link I posted) and I were trying to find out more about canting, I wrote to Bill McDonald of Sierra, and he told me: "I believe your equations are precisely correct. Drop is the radius of the circle as measured from the extended bore line."</p><p></p><p>Of course I don't want this or any discussion to turn into a "magister dixit" kind of argument, I'm just sharing what we have found and the formula we came out to correct at any range, cant angle, and zero range. </p><p></p><p>The values you posted for horizontal deflection seem too small, by a factor of ten or so, compared to everything I've ever read on this topic.</p><p></p><p>In all cases the important thing is to know the angle of the LOS to the bore line, and its projection at any given distance. </p><p></p><p>For example, if we take a 175 SMK @ 2680 fps, std. cond., SH = 1.75", the Sierra Infinity program predicts a drop of 541.13" at 1000 yds</p><p></p><p>To zero at 1000 yds, we have to compensate for SH + drop = .75" + 541.13" = 542.88"</p><p></p><p>this is: 542.88"/10.47 = 51.85 MOA = 0.864º</p><p></p><p>the angle between LOS and bore line is 0.864º</p><p></p><p>the two lines intersect at: 1.75" / sin 0.864º = 116.05" = 3.224 yds</p><p></p><p>With any cant angle to the right, the POI is going to move to the LEFT, and UP, from the muzzle up to about 3 yds; and from that point on is going to move RIGHT and DOWN. </p><p></p><p>this angle subtends at 1000 yds: (1000 - 3.224) * sin 0.864º = 15.030 yds = 541.13" (this is the drop at this distance)</p><p></p><p>so the POI movement, in relation to cant angle, is:</p><p></p><p>horizontal projection: X = drop*sin ß</p><p></p><p>vertical projection: Y = drop*(1 - cos ß)</p></blockquote><p></p>
[QUOTE="TiroFijo, post: 109778, member: 974"] JBM, I believe the basic formulas I posted are widely used, perhaps with a different twist or additional refinements. When Gustavo, our dutch friend Jeroen (the author of the article in the link I posted) and I were trying to find out more about canting, I wrote to Bill McDonald of Sierra, and he told me: "I believe your equations are precisely correct. Drop is the radius of the circle as measured from the extended bore line." Of course I don't want this or any discussion to turn into a "magister dixit" kind of argument, I'm just sharing what we have found and the formula we came out to correct at any range, cant angle, and zero range. The values you posted for horizontal deflection seem too small, by a factor of ten or so, compared to everything I've ever read on this topic. In all cases the important thing is to know the angle of the LOS to the bore line, and its projection at any given distance. For example, if we take a 175 SMK @ 2680 fps, std. cond., SH = 1.75", the Sierra Infinity program predicts a drop of 541.13" at 1000 yds To zero at 1000 yds, we have to compensate for SH + drop = .75" + 541.13" = 542.88" this is: 542.88"/10.47 = 51.85 MOA = 0.864º the angle between LOS and bore line is 0.864º the two lines intersect at: 1.75" / sin 0.864º = 116.05" = 3.224 yds With any cant angle to the right, the POI is going to move to the LEFT, and UP, from the muzzle up to about 3 yds; and from that point on is going to move RIGHT and DOWN. this angle subtends at 1000 yds: (1000 - 3.224) * sin 0.864º = 15.030 yds = 541.13" (this is the drop at this distance) so the POI movement, in relation to cant angle, is: horizontal projection: X = drop*sin ß vertical projection: Y = drop*(1 - cos ß) [/QUOTE]
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Canting - the right answer
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