huntsman
Member
I have two guns in .45acp, but nary a 1911.
The area of a circle is pi X the radius (1/2 the diameter)squared, so the area of the 9mm is 3.1416 X .177 squared (.0316). Thus, 9mm = .0989 sq.inch. .45 cal diameter = .1597 sq. inch. So the area (size of the hole) of the .45 is 1.6X the area of the 9mm.
Think about that.
I remember back in 2011 the 1911 graced the cover of every gun magazine, and people raved about them and how great they were and combat proven, people went out and bought them, in piles.
yes, this may be true. But, keep in mind, E=M(V squared), so when deriving the kinetic energy that an object has, the velocity component of the equation has a larger effect on the product. But, 0 times anything is 0, so the mass of the projectile is relevant.
I'm NOT trying to pick a fight here, just point out that there are more variables in play than simply the diameter of tissue displaced by said object. We can do all the math and physics we like, but real-world testing into things like meat products (say a side of pig) or a block of known-density ballistic gel will show things that may be difficult to 'calculate'.
Polar Express is absolutely correct. Basic high school-level physics.
That would be a big NO!Bottom line, .45 is obsolete as a defensive round, which is the reason (or a least the stated reason) that most people buy handguns. You are giving up 50% of the capacity, increasing weight, and increasing cost vs. a plastic 9mm, for virtually nothing in return.
I'm sure the 1911 will still be around for shooters long after all of us are dead though.
I have two guns in .45acp, but nary a 1911.
I have a 1911 45. and a East German Mak and a XDs 9 mm and I like them all.Have you personally shifted away from the .45 and/or 1911?