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<blockquote data-quote="Jay Kyle" data-source="post: 8617" data-attributes="member: 347"><p>ok, I'll take a stab at this. First lets be sure of the parameters:</p><p></p><p>A. Both barrels from the same blank</p><p>B. Same twist rate</p><p>C. Different length barrels</p><p>D. Same muzzle exit velocity</p><p>E. Same (theoretically identical) bullets</p><p>F. Fired at the same location and time</p><p></p><p>In other words all external factors being equal except for barrel length.</p><p></p><p>Second, let set some definitions.</p><p>Let's draw a line straight through the center of the bullet from tip to stern and label this line as D1, then spin the bullet and label the axis of rotation as D2.</p><p>The average difference between D1 and D2 we'll call the degree of wobble (W)</p><p></p><p>Now the solution. When the bullets leave the respective muzzles - with velocities equal - and spin rates the same, as well as no rotational instability, i.e. D1 = D2 or W = 0, then the flight will be identical. </p><p></p><p>If however due to lack of concentricity, misalignment of the round to the chamber, or some other similar factor, Wobble is introduced to the bullet as it travels down the barrel, then it is plausible that a longer barrel will reduce the degree of wobble and hence the bullet will exit with a lower wobble, than will occur in the shorter barrel. </p><p></p><p>It follows that the lower the wobble at the muzzle, the sooner the bullet goes to 'sleep'.</p><p></p><p>The only other factor that would come into play as I see it would be the gas pressure at the muzzle as the bullet leaves, the greater the pressure, the more the bullet is upset by that pressure. Hence the shorter barrel will likely have a higher exit pressure contributing to the wobble.</p><p></p><p>Is that as clear as mud?</p><p></p><p>AB</p></blockquote><p></p>
[QUOTE="Jay Kyle, post: 8617, member: 347"] ok, I'll take a stab at this. First lets be sure of the parameters: A. Both barrels from the same blank B. Same twist rate C. Different length barrels D. Same muzzle exit velocity E. Same (theoretically identical) bullets F. Fired at the same location and time In other words all external factors being equal except for barrel length. Second, let set some definitions. Let's draw a line straight through the center of the bullet from tip to stern and label this line as D1, then spin the bullet and label the axis of rotation as D2. The average difference between D1 and D2 we'll call the degree of wobble (W) Now the solution. When the bullets leave the respective muzzles - with velocities equal - and spin rates the same, as well as no rotational instability, i.e. D1 = D2 or W = 0, then the flight will be identical. If however due to lack of concentricity, misalignment of the round to the chamber, or some other similar factor, Wobble is introduced to the bullet as it travels down the barrel, then it is plausible that a longer barrel will reduce the degree of wobble and hence the bullet will exit with a lower wobble, than will occur in the shorter barrel. It follows that the lower the wobble at the muzzle, the sooner the bullet goes to 'sleep'. The only other factor that would come into play as I see it would be the gas pressure at the muzzle as the bullet leaves, the greater the pressure, the more the bullet is upset by that pressure. Hence the shorter barrel will likely have a higher exit pressure contributing to the wobble. Is that as clear as mud? AB [/QUOTE]
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