Lathe runout question

[dml66]

I have a LMS 7 x 16, much of my work is with 3jaw chuck, I tried it in the 3 mounting positions with a ground shaft chucked up,, when I found the best spot,, <.0004 I marked the chuck and mounting spindle/plate so the it returns to the same place..

I did the same with my ER32 spindle flange-mounted chuck, the marked spots on the flange and the chuck lasted quite a while, then the assembly decided one more counter-clockwise position provided the lowest TIR . Of course, I'd marked mine with an engraver. I still check it from time to time, it's not uncommon to get different TIR readings simply by twisting the chuck left or right before tightening the 3 screws, or letting it float.

 

[tobnpr]

If your spindle is off center relative to the axis of rotation, any part machined will still be round... or have less runout...

Round, true- but not necessarily of consistent OD.

 

[extropic]

Except for what has been contributed by @682bear and myself, this thread is loaded with misleading/defective information. I suggest those who don't understand should disregard the entire thread to preclude becoming interminably misinformed.

Better yet, @dml66 mount a piece of raw stock in the ER chuck, in the lathe spindle and turn a diameter, any diameter, to 100% cleanup. Without removing the chuck or workpiece, check the turned diameter for runout. . . . Get it?

 

[dml66]

Ok... picture turning a crankshaft journal... the end in the chuck may be offset several inches, but the journal that is being turned would end up round.

Same concept... if your runout is caused by the chuck not holding the work pefectly concentric with the axis of rotation. What you machine will be concentric... and round, barring all other negative factors.

However, if the part is removed from the chuck, turned, then reinstalled, your machined diameter will now show some runout... theoretically, if it is turned 180 degrees, it should show two times the runout of the chuck.

-Bear

Staying in the theoretical lane:

Would it be safe to say that if the chuck offset (measured as TIR) is greater than the amount of material to be removed, the goal cannot be met?

Lets say for example, the workpiece is already round, even perfectly round, I need to reduce it's diameter by 0.001". Chucked, the workpiece exhibits 0.002" TIR, I think it's not possible to accomplish the goal. If the chucked workpiece has 0.001" TIR then I think the goal could be met.

 

[682bear]

Yes, I would agree with that.

-Bear