I took a cerrosafe cast of my blankety blank Smith and Wesson forcing cone, as well as a cast of the muzzle, to see if there was a restriction at the forcing cone.   There wasn’t.   I also took a cast of my pet Ruger 357’s forcing cone, just to compare notes.   The Ruger is cast-friendly, the Smith is not.   Why the difference?

Both guns have slightly choked barrels, about 0.0003″ tighter at the muzzle than near the forcing cone.   The Smith was firelapped years ago when firelapping was fashionable.   I don’t recall firelapping the Ruger.

Both guns have 5 groove rifling with equal width lands and grooves.   I didn’t even know that Ruger made 5 groove rifling.   The Smith’s groove runs around 0.4297″, the Ruger around 0.358″.   No surprises there.

The Ruger’s throats are 0.0005″ over groove, or 0.3585″.   The Smith’s throats are 0.0037″ over groove, or 0.4334″.

The Ruger is fed bullets that are at least throat diameter or up to 0.0015″ larger, depending on the mold.   The Smith is usually fed bullets that are 0.001″ smaller than the throat, because the Smith’s chamber is too tight to swallow a throat diameter bullet, at least not without turning the cases.

I don’t have a precise way to measure the angle, but the Ruger appears to have a 5° forcing cone.   The blankety blank Smith’s was 18°, but I recently recut it to 5°.   Because the cone was oversize, I only cut the inner part of the cone, not the outer, so the result is a compound ange, 18° at the entrance, then 5° further in.

The entrance to the Ruger’s cone is 0.018″ larger than groove, or 0.376″.  The entrance to the blankety blank Smith’s cone is 0.026″ larger than groove, or 0.456″.   It easily swallows a 45acp bullet.

To sum things up, the Ruger’s dimensions are perfect in every respect, while the Smith’s dimensions are flawed in most respects.

Anyone want to buy a Dirty Harry era M29 ?????

How large should the entrance to the forcing cone be, in the perfect world?   I will theorize, and feel free to disagree with me, because this is just armchair theory, that the entrance should be groove diameter plus twice the amount of barrel/cylinder misalignment so that misaligned bullets can enter without shaving lead.   The bullet may obturate to fill that entrance, so we don’t want it to be larger than necessary.   How much misalignment are we talking about?

Back in the early 80’s someone made a test indicator to measure the misalignment on revolvers, and measured a bunch of guns and wrote it up in the IHMSA rag.   Freedom Arms was the best, around 0.001″ – 0.002″, Dan Wessons were around 0.003″, Rugers were around 0.005″, and Smiths were the worst at 0.006″ – 0.008″.   That was a long time ago so I am not 100% sure about those numbers, but I’m sure they are in the ball park because that article made a strong impression on me.


If the Smith has 0.008″ misalignment then the cone entrance should be groove plus 0.0016″ or 0.446″.   Now if you had a Freedom Arms with 0.001″ misalignment, then the cone entrance could be smaller.   Again, that’s just my armchair theory of the moment, and I am only wrong about 75% of the time.

A Taylor throat might be better yet, but I don’t have any experience with Taylors so I can’t comment on them.

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