# Good question. Here's the idea...

March 17 2002 at 5:44 PM

AR&P

Response to Steve in NC, when you multiply the BC by 8000 ...

The "8000" is really 8000 yards. 8000 yards is (appoximately) the downrange point where a projectile with a BC of 1.0 (i.e., the Ingalls ballistic reference projectile) will have lost 63% (i.e., 1 - 1/e) of its MV. By dividing the range by 8000yds, you're effectively normalizing everything to the "natural" or "Naperian" range of the ballistic standard projectile.

This "Naperian range" (a term I made up from its relationship to the constant "e" = the Naperian logarithm base) is proportional to BC. So, for example, a wadcutter pellet with a BC of 0.01 (implying a rate of velocity loss -- a.k.a. retardation factor -- 100x faster than the standard) is down to 37% (1/e) of its MV at 8000 x .01 = 80yds, while a CPL (BC ~ .025 = retardation factor 40x faster than the reference slug) can make it to 8000 x .025 = 200yds before slowing down that much.

The corresponding constant for energy loss is half as big (4000yds instead of 8000yds) because the rate of energy loss is exactly twice as fast as velocity due to the fact that kinetic energy is proportional to velocity squared.

Ultimately, there's nothing magic about 8000yds, except that when Ingalls picked his arbitrary standard projectile (I think it was something like a two-pound, 1" caliber pointy-nosed thing), that's just how its (subsonic) weight-to-drag ratio worked out.

Does that help -- or just make it worse?

Cheers,
Steve

 This message has been edited by pneuguy on Mar 17, 2002 5:50 PM

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