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Posted: 9/11/2007 5:42:39 PM EDT
| If this question does not belong here I appologize. I have been asked this a few times at our high power sillouette matches. What is the maximum distance a bullet of any given caliber will travel. I'm sure it depends on the air temp, baro pressure, wind and other factors, so let's use pilot talk and call it a standard day(59 degrees, 29.95 pressure, zero wind, and sea level. Is there a link to a chart that anyone knows of? Thanks for your time. |
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The basic equation for range for a ballistic arc (which is only approximate in these conditions, as it doesn't account for many smaller variables) is: R = V^2*sin(2r) / g where R = range V = initial velocity r = angle of launch g = acceleration from gravity (9.8 m/s^2) As you can see, this does not take into account environmental variations, air resistance, ballistic coefficient, etc. |
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Maximum Range is obtained with the muzzle of the firearm elevated 28 to 34 degrees. Examples: 5.56 X 45mm M193, 3842 yards 7.62x 51mm M80, 4469 yards .30-06 Springfield 180 grain SPBT, 5670 yards 22 LR, 1588 yards 9 X 19mm, 2130 yards information from NRA Firearms Sourcebook |
Holly shit, I was interested in the answer but not if I have to do that equation! 
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The nice thing about math is that you can use it to figure out things in any situation. Check these out. |
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