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Rifles, Reloading, Optics, Equipment
Rifles, Bullets, Barrels & Ballistics
30 Cal 260's
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<blockquote data-quote="groper" data-source="post: 474026" data-attributes="member: 12550"><p>Rich,</p><p></p><p>Ok ive modelled your projectile in 3D CAD as per your dimensions... it came out at 262grains. You didnt give me all the dimensions so i made a few small assumptions such as a jacket thickness of 0.020" etc</p><p><img src="http://i439.photobucket.com/albums/qq120/gravitygroper/260SXR.jpg?t=1297750498" alt="" class="fr-fic fr-dii fr-draggable " style="" /></p><p></p><p><img src="http://i439.photobucket.com/albums/qq120/gravitygroper/260SXRrendered.jpg?t=1297750498" alt="" class="fr-fic fr-dii fr-draggable " style="" /></p><p></p><p><img src="http://i439.photobucket.com/albums/qq120/gravitygroper/260SXRexploded.jpg?t=1297751337" alt="" class="fr-fic fr-dii fr-draggable " style="" /></p><p></p><p>These are the principal moments of inertia for your complete projectile, taking into account the aluminium tip, core material and jacket in their relative locations- sorry i live in a metric country <img src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" class="smilie smilie--sprite smilie--sprite1" alt=":)" title="Smile :)" loading="lazy" data-shortname=":)" /></p><p></p><p>Ix (axial) = 1.1166 g.cm^2</p><p>Iy (transverse) = 18.926 g.cm^2</p><p>Ky = (radii of gyration) = 10.540mm</p><p></p><p>Now the only other variable we need to know is the Cma or pitching moment derivative. There is only 2 ways to get an accurate number for this. 1 is live testing in wind tunnels or ballistics range with doppler radar etc, the other is with advanced CFD modelling- neither of which i can help you with. However, we can make some reasonably good assumptions based on the McDrag work done by Robert McCoy. Using his formulas and stability equations, we can assume a Cma of about 4.6 for your projectile aerodynamic geometry. - This is a higher than usual number due to the extreme length of your projectile.</p><p></p><p>So using the above numbers, i calculate a stability factor as follows, assuming velocity of mach 2.5 (~2850fps) and std atmosphere and temp at sea level.</p><p></p><p>Sg,</p><p>from a 1:8 twist = 1.61</p><p>from a 1:9 twist = 1.27</p><p>from a 1:10 twist = 1.02 - will most likely tumble.</p><p></p><p>So the ideal twist rate at sea level is 1:8.3 for an Sg of 1.49.</p><p></p><p>Take it upto 5000ft, and your ideal twist rate is 1:9 for a 1.47 stability factor.</p><p></p><p>Take it upto 9000ft, and a 10 twist will work just fine <img src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" class="smilie smilie--sprite smilie--sprite1" alt=":)" title="Smile :)" loading="lazy" data-shortname=":)" /></p><p></p><p>So a 1:9 will work ok with an Sg of 1.27. Its a little lower than ideal and you may notice some slightly poorer accuracy at short range under 300yds, but it should be ok other than that.</p></blockquote><p></p>
[QUOTE="groper, post: 474026, member: 12550"] Rich, Ok ive modelled your projectile in 3D CAD as per your dimensions... it came out at 262grains. You didnt give me all the dimensions so i made a few small assumptions such as a jacket thickness of 0.020" etc [IMG]http://i439.photobucket.com/albums/qq120/gravitygroper/260SXR.jpg?t=1297750498[/IMG] [IMG]http://i439.photobucket.com/albums/qq120/gravitygroper/260SXRrendered.jpg?t=1297750498[/IMG] [IMG]http://i439.photobucket.com/albums/qq120/gravitygroper/260SXRexploded.jpg?t=1297751337[/IMG] These are the principal moments of inertia for your complete projectile, taking into account the aluminium tip, core material and jacket in their relative locations- sorry i live in a metric country :) Ix (axial) = 1.1166 g.cm^2 Iy (transverse) = 18.926 g.cm^2 Ky = (radii of gyration) = 10.540mm Now the only other variable we need to know is the Cma or pitching moment derivative. There is only 2 ways to get an accurate number for this. 1 is live testing in wind tunnels or ballistics range with doppler radar etc, the other is with advanced CFD modelling- neither of which i can help you with. However, we can make some reasonably good assumptions based on the McDrag work done by Robert McCoy. Using his formulas and stability equations, we can assume a Cma of about 4.6 for your projectile aerodynamic geometry. - This is a higher than usual number due to the extreme length of your projectile. So using the above numbers, i calculate a stability factor as follows, assuming velocity of mach 2.5 (~2850fps) and std atmosphere and temp at sea level. Sg, from a 1:8 twist = 1.61 from a 1:9 twist = 1.27 from a 1:10 twist = 1.02 - will most likely tumble. So the ideal twist rate at sea level is 1:8.3 for an Sg of 1.49. Take it upto 5000ft, and your ideal twist rate is 1:9 for a 1.47 stability factor. Take it upto 9000ft, and a 10 twist will work just fine :) So a 1:9 will work ok with an Sg of 1.27. Its a little lower than ideal and you may notice some slightly poorer accuracy at short range under 300yds, but it should be ok other than that. [/QUOTE]
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