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<blockquote data-quote="Brent" data-source="post: 41007" data-attributes="member: 99"><p>Phil,</p><p>I'm going to take my compass and draw the first circle around my POA (bull) that's radius is 1 unit. This is the the bore distance from the scope height like you said above.</p><p></p><p>As we cant, we rotate around the LOS always being fixed on the bull, and the bore is ALWAYS maintains this distance from the scope, 1 unit away in any cant position.</p><p></p><p>Likewise, the bore line (BL) always intersects the target 2 units away from our LOS, on the opposite side the bore is offset to. (normal fire - bore offset 1 unit away at 6 o'clock, BL intersects target 2 units away at 12 o'clock, cant the rifle 180 degrees and it reverses) </p><p></p><p>That said, we draw another circle with our compass, this time 2 units away from our LOS to represent the BL intersection point on our target at any given cant angle.</p><p></p><p>We KNOW, that the bullet WILL leave the bore and travel the BL angle, no matter what the cant angle. The only thing that causes it to deviate is ONE thing: GRAVITY (wind is discounted here). Picture this shooting downhill 90 deg if that helps.</p><p></p><p>Now, we look at our target and see the BL circle 2 units away from our LOS. </p><p></p><p>We absolutely KNOW the bullets POI WILL BE dirrectly below the point where the BL intersects the target with ANY given cant angle. i.e. directly below that point on the circle 2 units away (3 units total in your previous example I believe).</p><p></p><p>Here's where we run into discrepancies, I think. </p><p></p><p>There is NOT 3 units of drop from the BL, there is 2 units of drop, and 1 unit of scope height though, but not DROP.</p><p></p><p>Dave's circle around the LOS accounted for both, sight height and BL drop.</p><p></p><p>Sight height is 1 unit, in our example, so in the 90 degree canted position, if this 1 unit was all that was dialed into the scope, the bullet would fall directly below our LOS, as the BL and LOS coinside perfectly. </p><p></p><p>Also, if nothing was dialed in at all, the bullet's POI would be 1 unit "away" from our LOS for a 90 degree canted position, and drop would be "2" units below "that".</p><p></p><p>Now, we KNOW that we have 2 more MOA dialed into the scope, not just the 1 MOA for scope height compensation (yeah, I'm calling it MOA here, I'm tired of units and my brain thinks in MOA damnit <img src="http://images/icons/grin.gif" alt="" class="fr-fic fr-dii fr-draggable " style="" /> <img src="http://images/icons/wink.gif" alt="" class="fr-fic fr-dii fr-draggable " style="" />) So, we now have, as a result of this additional 2 MOA, 2 MOA worth of deflection from cant at a 90 degree angle, also with the 2 MOA drop that gravity dictates. (there's that nicely formed 2 MOA box you taught me about at 90 degrees to verify your zero point, remember Dave <img src="http://images/icons/smile.gif" alt="" class="fr-fic fr-dii fr-draggable " style="" />) Drop is equal to horizontal deflection at 100 yards when fired at 90 degrees, forming a group at a low corner of an imaginary box diaganal of your POA.</p><p></p><p>That all said....... The circle 2 MOA around your LOS, which represents your BL in a given canted position, drop is ALWAYS 2 MOA PLUMB below that point in which the BL is intersecting the target (in our example at 100 yards). </p><p></p><p>Therefore, centering our compass at the point 2 MOA PLUMB below our LOS, we set our compass at 2 MOA, which should reach the LOS (bull) at the high point and draw a circle around this point. Refer to Dave's lower circle in his picture, that's it there, your POI at any given cant angle. </p><p></p><p></p><p>POP QUIZ. <img src="http://images/icons/grin.gif" alt="" class="fr-fic fr-dii fr-draggable " style="" /> </p><p></p><p>What happens to POI when we add more MOA to the vertical turret? I figure the BL "circle" is larger, so the amount of deflection MOA is increased also, and multiplied by the range it is set to rezero at?</p><p></p><p>Now, what's the formula to figure the deflection MOA per degree due to cant angle for a given range? </p><p></p><p>I'm talking to Jim right now about the odd numbers in RSI SL. Zero MOA deflection at 100 yards is not correct, and the first one with the very high deflection numbers is not either.</p><p></p><p>That's an informative Link, Dave. </p><p></p><p>One thing:</p><p></p><p><strong>Notice in section 2.1 of the referenced web article the figure B representation. This can be thought of as a "near" zero representation if you discount the projectile path, examine only the Line-Of-Sight line. Cant the rifle 90 degrees to the side and the deflection goes to zero -0-. The "drop" still exists and the round will strike below the POA, "BUT it will be DIRECTLY BELOW the POA". </strong></p><p></p><p>They don't consider you have 2 MOA dialed into the scope to compensate for that drop though, and "that's" where the deflection from cant comes from, they cannot discount it, unless the 2 MOA is taken "off" the turret, it exists and must be accounted for in deflection. Make sense?</p><p></p><p>Ric, You are RIGHT. <img src="http://images/icons/smile.gif" alt="" class="fr-fic fr-dii fr-draggable " style="" /> The bore line pointing left or right of LOS is the key here. The more you dial, and the farther we shoot, the worse it all gets.</p></blockquote><p></p>
[QUOTE="Brent, post: 41007, member: 99"] Phil, I'm going to take my compass and draw the first circle around my POA (bull) that's radius is 1 unit. This is the the bore distance from the scope height like you said above. As we cant, we rotate around the LOS always being fixed on the bull, and the bore is ALWAYS maintains this distance from the scope, 1 unit away in any cant position. Likewise, the bore line (BL) always intersects the target 2 units away from our LOS, on the opposite side the bore is offset to. (normal fire - bore offset 1 unit away at 6 o'clock, BL intersects target 2 units away at 12 o'clock, cant the rifle 180 degrees and it reverses) That said, we draw another circle with our compass, this time 2 units away from our LOS to represent the BL intersection point on our target at any given cant angle. We KNOW, that the bullet WILL leave the bore and travel the BL angle, no matter what the cant angle. The only thing that causes it to deviate is ONE thing: GRAVITY (wind is discounted here). Picture this shooting downhill 90 deg if that helps. Now, we look at our target and see the BL circle 2 units away from our LOS. We absolutely KNOW the bullets POI WILL BE dirrectly below the point where the BL intersects the target with ANY given cant angle. i.e. directly below that point on the circle 2 units away (3 units total in your previous example I believe). Here's where we run into discrepancies, I think. There is NOT 3 units of drop from the BL, there is 2 units of drop, and 1 unit of scope height though, but not DROP. Dave's circle around the LOS accounted for both, sight height and BL drop. Sight height is 1 unit, in our example, so in the 90 degree canted position, if this 1 unit was all that was dialed into the scope, the bullet would fall directly below our LOS, as the BL and LOS coinside perfectly. Also, if nothing was dialed in at all, the bullet's POI would be 1 unit "away" from our LOS for a 90 degree canted position, and drop would be "2" units below "that". Now, we KNOW that we have 2 more MOA dialed into the scope, not just the 1 MOA for scope height compensation (yeah, I'm calling it MOA here, I'm tired of units and my brain thinks in MOA damnit [img]images/icons/grin.gif[/img] [img]images/icons/wink.gif[/img]) So, we now have, as a result of this additional 2 MOA, 2 MOA worth of deflection from cant at a 90 degree angle, also with the 2 MOA drop that gravity dictates. (there's that nicely formed 2 MOA box you taught me about at 90 degrees to verify your zero point, remember Dave [img]images/icons/smile.gif[/img]) Drop is equal to horizontal deflection at 100 yards when fired at 90 degrees, forming a group at a low corner of an imaginary box diaganal of your POA. That all said....... The circle 2 MOA around your LOS, which represents your BL in a given canted position, drop is ALWAYS 2 MOA PLUMB below that point in which the BL is intersecting the target (in our example at 100 yards). Therefore, centering our compass at the point 2 MOA PLUMB below our LOS, we set our compass at 2 MOA, which should reach the LOS (bull) at the high point and draw a circle around this point. Refer to Dave's lower circle in his picture, that's it there, your POI at any given cant angle. POP QUIZ. [img]images/icons/grin.gif[/img] What happens to POI when we add more MOA to the vertical turret? I figure the BL "circle" is larger, so the amount of deflection MOA is increased also, and multiplied by the range it is set to rezero at? Now, what's the formula to figure the deflection MOA per degree due to cant angle for a given range? I'm talking to Jim right now about the odd numbers in RSI SL. Zero MOA deflection at 100 yards is not correct, and the first one with the very high deflection numbers is not either. That's an informative Link, Dave. One thing: [B]Notice in section 2.1 of the referenced web article the figure B representation. This can be thought of as a "near" zero representation if you discount the projectile path, examine only the Line-Of-Sight line. Cant the rifle 90 degrees to the side and the deflection goes to zero -0-. The "drop" still exists and the round will strike below the POA, "BUT it will be DIRECTLY BELOW the POA". [/B] They don't consider you have 2 MOA dialed into the scope to compensate for that drop though, and "that's" where the deflection from cant comes from, they cannot discount it, unless the 2 MOA is taken "off" the turret, it exists and must be accounted for in deflection. Make sense? Ric, You are RIGHT. [img]images/icons/smile.gif[/img] The bore line pointing left or right of LOS is the key here. The more you dial, and the farther we shoot, the worse it all gets. [/QUOTE]
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