On 9/5/2013 9:29 PM, pyotr filipivich wrote:
on Thu, 05 Sep 2013 14:19:31 -0500
typed in rec.crafts.metalworking the following:
On 9/5/2013 1:41 PM, pyotr filipivich wrote:
6.35 mm "Pistola Con Caricato

I had to Google that one, pytor.

Fugly little bruiser, isn't it.

That it is.

In the 1920s steel was still becoming steel.
It wouldn't take special alloys for this one.

The .25 doesn't look like much at first glance.
It's a straight walled cartridge, center-fire,
with a semi-rimmed head.

Here are a whole bunch of diameters for you...
http://en.wikipedia.org/wiki/.25_ACP
That should be a fair start.

That is, and gives me cartridge dimensions. Which is fair, but
I'm after the dimensions of the actual "bullet" (the lead pellet) the
round launches. I know the pellet is at least (23.11 - 15.62 =) 7.49
mm - but that is just from the tip of the pellet to the edge of the
cartridge rim.

Warning - bad ASCII art ahead B-)

Brass Cartridge length:
0------------v = 15.62 -.51 mm
__________
| |____
| |_____)
|_________ |

OAL Length -------^ is 23.11 mm

My question is How far left of the "v" (15.62) is the bullet seated? 1
mm? 2 mm? (Bullet over all length is 7.49 mm +????mm)


Here is one...

http://www.ranchdogoutdoors.com/bin/...let_sketch.jpg

http://www.ranchdogoutdoors.com/bin/...amber01_70.jpg

http://www.dave-cushman.net/shot/25_acp.html

That's all I found though...