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Long Range Hunting & Shooting
Effect of Bullet Spin on Terminal Performance
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<blockquote data-quote="GeorgeS" data-source="post: 1695042" data-attributes="member: 107925"><p>I'll chime in here without getting too physics-y, on two counts: first, the force involved, and second, the rotational decay. </p><p></p><p>In the first case, the bullet is held together (stopped from flying apart) by the material strength of the bullet - lead, jacket, copper, whatever. For a given caliber, the force required to hold a bullet together increases as the square of the rotational speed - a bullet rotating at 40% greater rate than another requires twice the force. That outward force bears on the expansion rate of the bullet as it deforms and "mushrooms." Higher spin, more rapid or greater total mushrooming, all other things held equal. Conversely, the more rapid the mushrooming, the greater the braking force on the rotation rate as tissue drags on the increasing diameter of the bullet. So the wound channel profile differs for different spin rates. </p><p></p><p>Second, while it is true that a bullet is decaying in spin rate in air slowly, and its maximum rotation rate appears to be that of the barrel twist times the velocity, you can see that the ratio of rotation to velocity increases with distance up to the point of contact with the animal, in effect increasing in spin rate as measured in rotations per linear foot; a 1:10 twist might be a 1:8 at impact. At the point of contact, the bullet decelerates dramatically - but the rotational rate does not. Accordingly, while a bullet leaving the muzzle might spin one time in 10 inches, inside an animal that would change, because of the rapid linear deceleration compared to the slow rotational rate deceleration, to a full rotation in perhaps 2 or 3 inches, with mushrooming affecting the rate of decrease of rotational speed. </p><p></p><p>If the bullet opens like a Black Talon, then, the aggregate tissue damage from the "buzzsaw" effect would be far more pronounced at a higher rotational speed. You can see this from the wound cavity geometry in clear gel. The spiral track of the cavity might show a half-dozen spins within a distance of 16 inches, yet no barrel has a twist rate of one in 3 inches. </p><p></p><p>The net: higher destructive force with a greater spin rate, all else held equal; and expansion brings a measure of inequality. The above does not account for tumbling - only for stable transit of a bullet.</p></blockquote><p></p>
[QUOTE="GeorgeS, post: 1695042, member: 107925"] I'll chime in here without getting too physics-y, on two counts: first, the force involved, and second, the rotational decay. In the first case, the bullet is held together (stopped from flying apart) by the material strength of the bullet - lead, jacket, copper, whatever. For a given caliber, the force required to hold a bullet together increases as the square of the rotational speed - a bullet rotating at 40% greater rate than another requires twice the force. That outward force bears on the expansion rate of the bullet as it deforms and "mushrooms." Higher spin, more rapid or greater total mushrooming, all other things held equal. Conversely, the more rapid the mushrooming, the greater the braking force on the rotation rate as tissue drags on the increasing diameter of the bullet. So the wound channel profile differs for different spin rates. Second, while it is true that a bullet is decaying in spin rate in air slowly, and its maximum rotation rate appears to be that of the barrel twist times the velocity, you can see that the ratio of rotation to velocity increases with distance up to the point of contact with the animal, in effect increasing in spin rate as measured in rotations per linear foot; a 1:10 twist might be a 1:8 at impact. At the point of contact, the bullet decelerates dramatically - but the rotational rate does not. Accordingly, while a bullet leaving the muzzle might spin one time in 10 inches, inside an animal that would change, because of the rapid linear deceleration compared to the slow rotational rate deceleration, to a full rotation in perhaps 2 or 3 inches, with mushrooming affecting the rate of decrease of rotational speed. If the bullet opens like a Black Talon, then, the aggregate tissue damage from the "buzzsaw" effect would be far more pronounced at a higher rotational speed. You can see this from the wound cavity geometry in clear gel. The spiral track of the cavity might show a half-dozen spins within a distance of 16 inches, yet no barrel has a twist rate of one in 3 inches. The net: higher destructive force with a greater spin rate, all else held equal; and expansion brings a measure of inequality. The above does not account for tumbling - only for stable transit of a bullet. [/QUOTE]
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