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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: 109784" data-attributes="member: 1969"><p>[ QUOTE ]</p><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></p><p>[/ QUOTE ]</p><p></p><p>I'm not arguing that. I just don't understand the equations and I'm a little skeptical that the final equation is so simple. That's all.</p><p></p><p>[ QUOTE ]</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></p><p>[/ QUOTE ]</p><p></p><p>Read my previous post -- different cooridinate systems -- RELATIVE TO THE TARGET, they will be less. The PDF I posted shows why. When I was interested in cant errors, I wanted to know how much I would miss by when I canted accidently. I don't intentionally cant and then correct in my scope for it. What I found, was the relative to the target (which is what matters to me) the shift of the point of impact was much less than the typical formulas give you which were relative to the line of sight. In other words, if your cant formula gives me 57" of windage error in the line of sight coordinate system, that doesn't mean that I'm going to miss the TARGET by 57". It's actually less at the target. Much less. That was my point, for a an accidental cant of less than a degree, I don't worry about it.</p><p></p><p>[ QUOTE ]</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></p><p>[/ QUOTE ]</p><p></p><p>Which is exactly what my original equations give, again, relative to the target. They provided the changes in elevation and azimuth for a rotation around the line of sight. I've shown them to be almost exact, even for large angles.</p><p></p><p>[ QUOTE ]</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></p><p>[/ QUOTE ]</p><p></p><p>You might want to check that drop number, that's about 100" off of what they predict on their website, for a .308 175gn MK. My online program gets about 430" drop at 1000 yards which is close to what they say on their online tables. Am I using the right bullet?</p><p></p><p>[ QUOTE ]</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></p><p>[/ QUOTE ]</p><p></p><p>I agree.</p><p></p><p>[ QUOTE ]</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><p></p><p></p><p>[/ QUOTE ]</p><p></p><p>Again, I'm not disagreeing with you, I was just skeptical. I need to work out the equations myself to see it.</p></blockquote><p></p>
[QUOTE="JBM, post: 109784, member: 1969"] [ QUOTE ] 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." [/ QUOTE ] I'm not arguing that. I just don't understand the equations and I'm a little skeptical that the final equation is so simple. That's all. [ QUOTE ] 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. [/ QUOTE ] Read my previous post -- different cooridinate systems -- RELATIVE TO THE TARGET, they will be less. The PDF I posted shows why. When I was interested in cant errors, I wanted to know how much I would miss by when I canted accidently. I don't intentionally cant and then correct in my scope for it. What I found, was the relative to the target (which is what matters to me) the shift of the point of impact was much less than the typical formulas give you which were relative to the line of sight. In other words, if your cant formula gives me 57" of windage error in the line of sight coordinate system, that doesn't mean that I'm going to miss the TARGET by 57". It's actually less at the target. Much less. That was my point, for a an accidental cant of less than a degree, I don't worry about it. [ QUOTE ] 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. [/ QUOTE ] Which is exactly what my original equations give, again, relative to the target. They provided the changes in elevation and azimuth for a rotation around the line of sight. I've shown them to be almost exact, even for large angles. [ QUOTE ] 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 [/ QUOTE ] You might want to check that drop number, that's about 100" off of what they predict on their website, for a .308 175gn MK. My online program gets about 430" drop at 1000 yards which is close to what they say on their online tables. Am I using the right bullet? [ QUOTE ] 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. [/ QUOTE ] I agree. [ QUOTE ] 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 ] Again, I'm not disagreeing with you, I was just skeptical. I need to work out the equations myself to see it. [/QUOTE]
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