Posted: 7/25/2008 10:27:11 AM EDT
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We've got a discussion going on here at work. If you drop a flat based rifle bullet off a building with zero wind, will it fall base first or pointy end first? Alternately, if you fired it straight up from a rifle, would the rotation make any change on how it would fall back down? An answer with an explanation would be appreciated. 
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Due to the resistance of the air, the bullet you dropped would probably tumble on the way down. In an non atmospheric environment, ie: as on the mooon, the bullet would fall exactly the way you dropped it. If you fired it perfectly straight up into the air, the bullet would eventually lose all forward momentum and stop dead before it fell back to the ground at terminal velocity. Too slow to do much damage to you if it hit you on the head. It would likely tumble due to insufficient rotation to stabailize it. |
Center of gravity is biased toward the rear of the bullet, which is why rounds tumble when they strike flesh or ballistic gelatin - they are attempting to put themselves in the most stable position, which is base-forward. Unless you meant "unstable" as in "it's wobbling slightly," which will be true. |
And given enough energy will keep trying, probably resulting in a base downward tumble (gyration around an unstable, lengthwise, axis would be more accurate) if dropped off a building. 
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The general answer that is true for all bullets: For the dropped bullet: Given a long enough fall and the absence of stabilizing spin the bullet will orient itself such that the center of gravity (CG) is traveling ahead of the center of pressure (CP) relative to the direction of airflow. The closer the CP is to the CG the less stable the bullet will be during it's travel. At some point the CG and CP are not sufficiently separated for stable flight and the bullet will tumble. For a bullet fired "straight up" w/o any wind or other influences: The spin will stabilize the bullet base down throughout it's flight. |
This. |
I actually knew this one. Yay me. |