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Rifles, Reloading, Optics, Equipment
Rifles, Bullets, Barrels & Ballistics
Canting - the right answer
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<blockquote data-quote="Buffalobob" data-source="post: 109775" data-attributes="member: 8"><p>Gustavo and TiroFijo</p><p></p><p></p><p>Y(R)=H(R)*cos(cant)-DROP(R)</p><p></p><p>And</p><p></p><p>vertical projection: Y = drop*(1 - cos ß)</p><p></p><p></p><p>are using left hand and right hand reference frames</p><p></p><p>This will introduce a small error in your calculation (or more appropriately in your ability to communicate smoothly with each other) at a latter date if you refine your equations to include other interactions.</p><p></p><p>Using your convention, I would suggest that you modify your equations to add a function which we can call F of theta { F(theta)}. This function will account for such things as right hand twist barrels and vertical gravity velocity vectors. One of the underlying assumptions/simplifications you have used is a horizontal line of sight.</p><p></p><p>A 30MOA angle between the line of sight and line of departure of the bullet will result in a gravity induced velocity difference at a 1000yds of about 0.3 fps when the gun is in the zero cant position and in the 180 degree cant position. This introduces an error of about 0.1 inches in your answer.</p><p></p><p>The modified equation would look something like this</p><p></p><p></p><p>Y(R)=H(R)*cos(cant)-DROP(R) + F(theta)</p><p></p><p>Where F(theta) = {k1(0.5 gt^2) + k2(spin)}sin(theta)</p><p></p><p>K1(0.5gt^2) can be readily calculated from the JBM's ballistic calculator or it can be simply had by approximation and parameterization. </p><p>K2(spin) must be determined experimentally for each gun and bullet, however, because the effect itself is small a constant approximation may be adequate. It all goes back to my original question on your original post. What is it that you really want. Accuracy of the calculation only should be tailored to your needs.</p><p></p><p>Here is a quote from Kregg Slack and you will notice he has five degrees (not minutes) of angle between scope and barrel.</p><p></p><p></p><p>"this here is the unlucky dog that i shot at 3125yds with the 338 lapua. 300 gr seirra bullet time of flight was a little over 8-sec it took 325 min of elvation to get there at 2750 fps at 5000 ft elvation. the bases on this rifle were solids made to shoot only at that range.but i had 60 min or so in the scope for fine tuneing. fine cross hair 25x50x50mm scope from dick thomas."</p><p></p><p></p><p></p><p>Now then, if you are really bored and have time on your hands you can further eliminate certain other simplifying assumptions and the calculation will be exponentially more difficult. On of your assumptions is that at zero cant the scope/ line of sight is centered over the bore/ line of departure. There are many applications where the scope is offset to one side of the bore. I have such a gun.</p><p></p><p>Just some thoughts to keep the neurons firing.</p></blockquote><p></p>
[QUOTE="Buffalobob, post: 109775, member: 8"] Gustavo and TiroFijo Y(R)=H(R)*cos(cant)-DROP(R) And vertical projection: Y = drop*(1 - cos ß) are using left hand and right hand reference frames This will introduce a small error in your calculation (or more appropriately in your ability to communicate smoothly with each other) at a latter date if you refine your equations to include other interactions. Using your convention, I would suggest that you modify your equations to add a function which we can call F of theta { F(theta)}. This function will account for such things as right hand twist barrels and vertical gravity velocity vectors. One of the underlying assumptions/simplifications you have used is a horizontal line of sight. A 30MOA angle between the line of sight and line of departure of the bullet will result in a gravity induced velocity difference at a 1000yds of about 0.3 fps when the gun is in the zero cant position and in the 180 degree cant position. This introduces an error of about 0.1 inches in your answer. The modified equation would look something like this Y(R)=H(R)*cos(cant)-DROP(R) + F(theta) Where F(theta) = {k1(0.5 gt^2) + k2(spin)}sin(theta) K1(0.5gt^2) can be readily calculated from the JBM’s ballistic calculator or it can be simply had by approximation and parameterization. K2(spin) must be determined experimentally for each gun and bullet, however, because the effect itself is small a constant approximation may be adequate. It all goes back to my original question on your original post. What is it that you really want. Accuracy of the calculation only should be tailored to your needs. Here is a quote from Kregg Slack and you will notice he has five degrees (not minutes) of angle between scope and barrel. “this here is the unlucky dog that i shot at 3125yds with the 338 lapua. 300 gr seirra bullet time of flight was a little over 8-sec it took 325 min of elvation to get there at 2750 fps at 5000 ft elvation. the bases on this rifle were solids made to shoot only at that range.but i had 60 min or so in the scope for fine tuneing. fine cross hair 25x50x50mm scope from dick thomas.” Now then, if you are really bored and have time on your hands you can further eliminate certain other simplifying assumptions and the calculation will be exponentially more difficult. On of your assumptions is that at zero cant the scope/ line of sight is centered over the bore/ line of departure. There are many applications where the scope is offset to one side of the bore. I have such a gun. Just some thoughts to keep the neurons firing. [/QUOTE]
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Canting - the right answer
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