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From: ke4zv!gary@gatech.edu (Gary Coffman)
Newsgroups: rec.guns
Subject: Re: Reloading questions: 40 S&W and 45 ACP
Message-ID: <35657@mimsy.umd.edu>
Date: 14 Jun 91 20:05:41 GMT
Sender: magnum@mimsy.umd.edu
Organization: Gannett Technologies Group
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Approved: gun-control@cs.umd.edu


I'm going to tackle this and ask the real experts to straighten out any 
mistakes I might make. I've got a couple of questions of my own.

In article <35456@mimsy.umd.edu> n9020351@unicorn.cc.wwu.edu (James D. Del Vecchio) writes:
#Per mag maximum chamber pressure is about 35,700 cup for 38 super, 40 S&W,
#and 10mm Auto, vs about 19,900 max for 45 ACP.
#
#Why is this and what significance of it?  Are the 45 cases weaker for some
#reason, or is it the construction of the pistols?  Is it possible to make
#a 45 withstand the higher pressures?  Thicker cases maybe?

.45 cases are solid head and plenty strong. The 1911 is the limiting
factor. It's delayed blowback design and lack of full cartridge head
support limit the pressure it can tolerate. High pressure loads will
not drop in pressure enough during the delay time when the slide and
barrel are moving backwards together and it will unlock before pressure
has dropped enough. Also the hotter load will cause stronger recoil
and hammer the gun severely leading to eventual failure. You can shoot
much hotter loads in a Contender where the case is fully supported and
the action remains locked until you reload.

#I've been told the 38 super and 10mm are both louder than the 45, is this
#entirely due to the higher muzzle pressure?

Yes. Now here is where I turn to the experts. A small charge of fast burning 
powder will have higher peak chamber pressure than a larger charge of slower 
burning powder. But, unless the cartridge is way over bore capacity, the 
muzzle pressure as the bullet exits will have dropped so much that the higher
pressure charge is quieter. Is there a formula relating expansion ratio,
peak pressure, muzzle pressure, burning rate, bullet mass, and velocity?

#Is the .40 also louder than the .45, say with the subsonic loads?

That depends on the load. If the velocities and bullet masses are the
same, then yes the .40 should be louder. But the other factors of burning
rate and expansion ratio must be considered. I'm not sure that a load
*has* to be louder in the .40 to deliver the same velocity. Can the 
experts comment?

#According to their charts you can get more velocity using the same powder
#charges and bullet weights in a 4" 40 S&W than you can with a 5" 45.
#Why would this gun be more efficient?

This has to do with something called "bore capacity". The same charge
in a smaller cartridge expanding into a smaller bore will deliver more
pressure to the bullet during the entire time it is in the bore. The
increased pressure over the dwell of the bullet in the barrel more than
offsets the reduced base area of the .40 bullet vs the .45 bullet so the
velocity is greater. There is an ideal point. If the bore is too small,
the cartridge will be over bore capacity and much of the pressure will
be wasted as muzzle blast. If the bore is too large, the cartridge will
be under bore capacity and the pressure will drop too much before the
bullet leaves the barrel. For a severely under bore cartridge, the
bullet can actually slow down in the barrel due to frictional losses being 
higher than remaining pressure.

There is a relationship between exterior ballistics and interior ballistics
that determines ideal cartridges. In rifles, the 7mm bullet has ideal
exterior ballistics, there is an ideal 7mm cartridge that has complimentary
interior ballistics, it's the 7x47 wildcat. A question for the experts,
what is the ideal pistol cartridge based on similar reasoning? Is that
cartridge in commercial production, or has it been wildcated, or does it
remain a theoretical design?

Gary