What do you put between the chuck jaws and the barrel when chambering.

[Clark]

If the barrel is in the 6-jaw with nothing else, and push up or down at the spider on the left side of the lathe, try to calculate max force with the spider to reach the stress at the end of barrel elasticity:
1) ignore shear force on long beam.
2) concentrate on bending moment:
a) assume tension on one side and compression on the other. Each are is 0.5 in squared centered 0.5" from the center. Assume the barrel sticks out of the chuck by 25". The stresses are then + 100X the spider force and - 100 times the spider force.
b) Stainless yields at ~ 60,000 psi. The max spider load on the cantilever is then 600 pounds.
When I push on the barrel muzzle with my thumb and move the test indicator on a breech spud, the force is ~~ 1 pound.

 

[Machinist John]

True. The way the old timers taught the young apprentice. Let him make a few mistakes and find out on his own.

Betcha you never chambered a barrel this way:

A mock up showing how I chambered my first in about 1963 when I was 16 years old. Still have the little South Bend.

Even Parker did it "wrong" by today's practice.

 

[tobnpr]

I have a question. How do any of us know that our 4 contact points are in the same radial plane perpendicular to the bore at different diameters? The answer is we don't.

I understand the question, but can't wrap my head around why it should matter?
However one gets there, if the indicator reads near zero TIR where we want it, what difference would it make if they're not?

 

[tobnpr]

True. The way the old timers taught the young apprentice. Let him make a few mistakes and find out on his own.

Betcha you never chambered a barrel this way:

A mock up showing how I chambered my first in about 1963 when I was 16 years old. Still have the little South Bend.

Even Parker did it "wrong" by today's practice.

Still have my 9A (4-1/2' bed, 1 hp VFD, V-belt drive, DRO,restored) and chambered some pretty accurate barrels between centers. All chambering has been done with a Sheldon for a long time but i still use that SB for a lot of other work. No toolroom lathe for sure, but tight and accurate and i can hold a thou which is all that's needed.

 

[DaveTooley]

I understand the question, but can't wrap my head around why it should matter?
However one gets there, if the indicator reads near zero TIR where we want it, what difference would it make if they're not?

If the opposing contact points don't line up exactly then a very good argument could be made that you're bending the barrel.

 

[DaveTooley]

True. The way the old timers taught the young apprentice. Let him make a few mistakes and find out on his own.

Betcha you never chambered a barrel this way:

A mock up showing how I chambered my first in about 1963 when I was 16 years old. Still have the little South Bend.

Even Parker did it "wrong" by today's practice.

I've done many that way. More skills involved but not difficult. Cheap imported lathes with larger spindle bores gave us options. Hell I've had lead shot bags sitting on top of the tailstock to keep the inboard side on the ways so I could run a reamer in.

 

[Clark]

If the barrel is in the 6-jaw with nothing else, and push up or down at the spider on the left side of the lathe, try to calculate max force with the spider to reach the stress at the end of barrel elasticity:
1) ignore shear force on long beam.
2) concentrate on bending moment:
a) assume tension on one side and compression on the other. Each are is 0.5 in squared centered 0.5" from the center. Assume the barrel sticks out of the chuck by 25". The stresses are then + 100X the spider force and - 100 times the spider force.
b) Stainless yields at ~ 60,000 psi. The max spider load on the cantilever is then 600 pounds.
When I push on the barrel muzzle with my thumb and move the test indicator on a breech spud, the force is ~~ 1 pound.

My spider has 5/16-24 screws 20" from the back of my 6-jaw jaws. The maximum force would be ~750 pounds before the barrel would be "bent". This would take ~46 inch pounds of torque. The screw is capable of 415 inch pounds without snapping off.

To test:
I have a Shilen select match 6mm barrel in #8 taper [the ones Brownells stocks] that I re tapered to #3, but the shank in the chuck is the same. When I put it in the 6 jaw set, the bore was 0.002" out of parallel with the lathe spindle bore over an inch of spud. I screwed until I got it parallel. It took 8 inch pounds of torque. That is ~ 130 pound of force by the spider screw on the barrel = ~17% of max allowable muzzle deflection.

What does it all mean?
The barrel is not permanently bent. It springs back straight. All elastic deformation, no plastic deformation.