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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: 109833" data-attributes="member: 1969"><p>[ QUOTE ]</p><p>JBM, taking the same conditions as in the example (first post)</p><p></p><p>100 yds ZERO</p><p></p><p>500(x) = 03.27"</p><p>500(y) = 47.30"</p><p></p><p>1000 yds zero</p><p></p><p>500(x) = 030.80"</p><p>500(y) = 108.84"</p><p></p><p>drop (500) = 356.7" </p><p></p><p>[/ QUOTE ]</p><p></p><p>For a 100 yard zero, I have 3.500 moa of elevation.</p><p></p><p>Plugging this into my formulas for a 10 degree cant, I get a new elevation and azimuth of:</p><p></p><p>elevation = 3.446 moa</p><p>azimuth = 0.608 moa</p><p></p><p>For a 500 yard target (I'm assuming that's what 500(x) means):</p><p></p><p>0.608 moa at 500 yards = 3.183"</p><p></p><p>For a change of elevation of 3.500 - 3.446 = 0.054 moa, this gives a change in drop of</p><p></p><p>0.283" or a drop of 49.5 - 0.283 = 49.78"</p><p></p><p>[My 500 yard drop value is -49.5"]</p><p></p><p>My online calculator gives: -49.7" drop and 2.9" of windage, but the bullet starts</p><p>left of the line of sight by 0.3" so total deflection is 3.2". Which agrees.</p><p></p><p>This is the same 10 degree cant I used as my first example that had about a 6.1" deflection at 1000 yards (but I think some parameters were slightly different). But since the range is twice as far I would expect twice the deflection. You can see this from Brown Dog's nice PowerPoints. More elevation means more azimuth when canted. My first example had a zero range of 100 yards -- very small elevation.</p><p></p><p>So they agree at 500 yards. I don't know what your change in drop was, but I'm guessing it was close.</p><p></p><p>For 1000 yard zero, I get 34.61 moa of elevation. Plugging this and the 10 degree cant into my formulas, I get the following adjusted elevation and azimuth:</p><p></p><p>elevation = 34.084 moa</p><p>azimuth = 6.010 moa</p><p></p><p>At 500 yards, the windage value is comes out to 31.47". You have 30.8". My online calculator gives me 31.2" of windage, but again, it starts at -0.3" to the left (due to the cant), so total deflection is 31.5". My online calculator only prints to one significant figure, but it's certainly close enough. Your formula is off by about 3/4".</p><p></p><p>My online calculator gives me 113.5 inches of drop at 500 yards (remember it's zeroed at 1000 yards). </p><p></p><p>For a change of elevation change of 34.61 - 34.084 = 0.526 moa, this gives a change in drop of </p><p></p><p>2.75" or a drop of 113.5 - 2.75 = 110.75". My online calculator gives 110.7". You have 108.84", but I don't know what the uncorrected drop is so the difference may vary well be atmospheric models, etc.</p><p></p><p>The reason that I compare to my online calculator is that it does the velocity vector transformation by cant and los angles. It doesn't use an approximation or my formulas I derived above -- it's a different method and independent check.</p><p></p><p>So I still think my formula is right. I don't know if yours is an approximation or not because I haven't seen a derivation (I'm not really sure it would matter that much, we agree pretty well). I do know that Rinker's book says that for longer ranges, you'll see more lateral deflection than multiplying by the sin() will give you, which tells me that it is an approximation.</p><p></p><p>JBM</p></blockquote><p></p>
[QUOTE="JBM, post: 109833, member: 1969"] [ QUOTE ] JBM, taking the same conditions as in the example (first post) 100 yds ZERO 500(x) = 03.27" 500(y) = 47.30" 1000 yds zero 500(x) = 030.80" 500(y) = 108.84" drop (500) = 356.7" [/ QUOTE ] For a 100 yard zero, I have 3.500 moa of elevation. Plugging this into my formulas for a 10 degree cant, I get a new elevation and azimuth of: elevation = 3.446 moa azimuth = 0.608 moa For a 500 yard target (I'm assuming that's what 500(x) means): 0.608 moa at 500 yards = 3.183" For a change of elevation of 3.500 - 3.446 = 0.054 moa, this gives a change in drop of 0.283" or a drop of 49.5 - 0.283 = 49.78" [My 500 yard drop value is -49.5"] My online calculator gives: -49.7" drop and 2.9" of windage, but the bullet starts left of the line of sight by 0.3" so total deflection is 3.2". Which agrees. This is the same 10 degree cant I used as my first example that had about a 6.1" deflection at 1000 yards (but I think some parameters were slightly different). But since the range is twice as far I would expect twice the deflection. You can see this from Brown Dog's nice PowerPoints. More elevation means more azimuth when canted. My first example had a zero range of 100 yards -- very small elevation. So they agree at 500 yards. I don't know what your change in drop was, but I'm guessing it was close. For 1000 yard zero, I get 34.61 moa of elevation. Plugging this and the 10 degree cant into my formulas, I get the following adjusted elevation and azimuth: elevation = 34.084 moa azimuth = 6.010 moa At 500 yards, the windage value is comes out to 31.47". You have 30.8". My online calculator gives me 31.2" of windage, but again, it starts at -0.3" to the left (due to the cant), so total deflection is 31.5". My online calculator only prints to one significant figure, but it's certainly close enough. Your formula is off by about 3/4". My online calculator gives me 113.5 inches of drop at 500 yards (remember it's zeroed at 1000 yards). For a change of elevation change of 34.61 - 34.084 = 0.526 moa, this gives a change in drop of 2.75" or a drop of 113.5 - 2.75 = 110.75". My online calculator gives 110.7". You have 108.84", but I don't know what the uncorrected drop is so the difference may vary well be atmospheric models, etc. The reason that I compare to my online calculator is that it does the velocity vector transformation by cant and los angles. It doesn't use an approximation or my formulas I derived above -- it's a different method and independent check. So I still think my formula is right. I don't know if yours is an approximation or not because I haven't seen a derivation (I'm not really sure it would matter that much, we agree pretty well). I do know that Rinker's book says that for longer ranges, you'll see more lateral deflection than multiplying by the sin() will give you, which tells me that it is an approximation. JBM [/QUOTE]
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