Would simply cutting a partial cone, then threading the action on, bolt included, until it stops, work? Then you just measure the gap between receiver face and action face, add your desired clearance, and cut the cone that much deeper. Works on any other cone breech I’ve seen, but I’ve never handled a Coup de Grace.
I don't see how A has anything to do with a 1.250" radiused CB. Just an evil engineer.
Aha. Yes, yes it does.Works great, if you have the action in hand.
I have kept a 2" travel indicator on every manual machine I've ownedIf you have a DRO on your lathe touch off both surfaces.
This seems like the simple solution.. or at least that’s the way I would do it.Make a tube that is bored to seat up against the barrel shoulder on one end and big enough for the ball to fit in on the other end. Make the tube longer than dimension A. Then you can use a depth mic to measure from the face of the tube to the ball and do the math to come up with dimension A.
SameI have kept a 2" travel indicator on every manual machine I've owned
I'm sure you've done many more cone breeches than I can ever even imagine doing, so maybe you have some insight into how to reliably gauge that 0.085 distance. There's no square surface to mic against. I think that's why they supply the gauge ball drawing as well. The ball will seat solidly against the cone and can then be measured by conventional mics..822"-.760= .062"
.147"-.062"= .085" at 30'
Am I missing something
Well, the example I showed was for a Panda barrel in 6.5x47, and I did the CAD sketch to make sure that the ball was resting against the breech cone before I did the measurement. I've confirmed it before with some dykem as well, but you are probably correct for the ARC dimensions, hence the 1.25 ball.Move in .085 from the face of the breech end... use compound to cut the cone.
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The problem with the 1 inch ball bearing method - The ball doesn't make contact with the cone, it rests on edge of the hole... so if there's a chamfer on your chamber... the measurement will be off.
I assume that's why ARC suggests a larger ball... but I'm bad at that kind of math.