Let me offer more details as to why I wanted the drag
coef. table. I'm serving as an advisor for two
students working on a high-school science project. The
objective of the project is to determine the speed
with which a bullet shot straight up, on a windless
day, will return to the earth. The bullets we chose to
study are the 22 lr, the 9 mm and 9 mm hollow point.
We chose five approaches:
1) call up police, firing ranges, gun shops, bullet
manufacturers, gun manufacturers,...and simply ask
what the value is. No one knows! Really. The kids
received answers ranging from "Oh, about 32" to "well,
e = mc^2, or about 128 mph," to, "the same speed with
which it was shot vertically upward." All nonsense.
About the only reference found that even attempts to
answer the question is Hatcher's Notebook, published
in the '50s. and he essentially gave up.
2) calculate value by using ballistic tables to deduce
the drag coefficient, and then use the 1/2 density
...v^2 equation with the drag force equalling the
weight. Problem is the drag coeff for such an approach
assumes the bullet falls point first, which we've
found, is not the case.
What we have found is that if the bullet is released
in water, regardless of its initial orientation, very
quickly turns and falls sideways. Thus we were looking
for a table providing the drag coef for an object
approximating the shape of a bullet moving sideways.
Not much chance of finding such a table, but I thought
I'd give it a try. I suppose we'll settle for a
cylinder.
As for Reynolds number, it seems that so long as the
flow is turbulent, and well away from Mach one, the
v^2 dependence is valid.
Oh yes, anyone have any idea what a 9 mm bullet will
freely fall at?
I wonder if the bullet will still be spinning when it
hits the ground? Those same experts consulted above
don't believe so. In this I hope, even expect, they're
right. Any thoughts?
Leon
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