"Math" for Fixing Tailstock Tilt

Now that I've got my Sheldon up and working I'm determined to try to fix the tailstock tilt on my 9A.

I shimmed the base when I first got the lathe- but did it evenly front/back to bring center to the correct height with the spindle retracted and locked (makes little difference whether locked or not, there's good fitment).

With the spindle extended, there's .004 of "drop" towards the headstock over the 2" length. Add the length of a chuck, and a long drill bit, and it can easily exceed 15 thou. It's impossible for me to do second ops like opening up an existing drilled hole without wallowing it out.

If I'm correcting .004 over a length of 2", how can I calculate how much less the rear of the base needs to be shimmed as compared to the front, to level it out?

length divided into shim is tangent

so .004 / 2.000 = .002

Now - use length of tail stock base

Let's say it was 4.5"

4.5 X .002" = .009"

Some somewhat useless info is that this is also 0.11459 degrees

Perfect. Thank you!

Equally useless but easier to calculate. The angle is 2 milliradians.

$0.02 from the sidelines, but you also want to make sure you've accounted for sag. If the spindle is a slightly loose fit in the tailstock bore, some of your .004 drop when extended could be that slop.

Before you go shimming the headstock end of the tailstock up, put the dial indicator on the extended spindle and give it a tug. See how much of the .004 disappears when you put some upward pressure on the spindle.

^^
Addressed in my second sentence above, thanks.
Messed with it very briefly last night. I had originally shimmed .016 all the way around. Left just a strip of that in the front, and replaced rear shim with one of .006. Put me .007 above center using a test bar that I flip end over end.

Haven't gotten back to it yet, but seems I need to reduce the front shim and eliminate the rear entirely.

Tricky, like dialing in the bore on a rifle barrel. Need to true two axes, not just one.