Re: [sherline] Help
Posted by
Alan Marconett KM6VV
on 2000-08-01 14:22:14 UTC
Hi Wm.
The rotary table has 3600 directly readable divisions per revolution.
You have divided 360/5 = 72 deg. per side. Each division is 1/10
degree, so you need 720 divisions. There are 50 divisions per handwheel
rotation, so you divided 720/50 = 14.4 rotations. So 50 x .4 = 20
divisions "extra" per each 14 "full handwheel rotation". Wouldn't you
just start at 0 on the handwheel, rotate 14 full (from where you were),
then 20 additional divisions to get to the next flat? The following
move would count 14 revs from that location, and 20 more divisions.
You'd just have to remember which "division" you were at when you
started each 14 rev! The "divisions" rotate around by +20 each flat.
So at the end, you come back to 0 division (5 * 20 = 100 = 2
"additional" rotations.
(guys, please feel free to check me on this!)
I'm hoping that with the 3700-CNC version of the table that I want to
get, I won't have to remember as much! Maybe a DRO would help. Yeah,
we can add that...
A simple program could rotate the table for you with a stepper drive.
Might not even need the backlash calc's. Looks like another job for
"CAD_CAM_EDM_DRO" (radius).
Thanks for the practical example. I hadn't had to think a rotary table
problem through yet!
Good luck!
Alan (Hope I helped a little)
The rotary table has 3600 directly readable divisions per revolution.
You have divided 360/5 = 72 deg. per side. Each division is 1/10
degree, so you need 720 divisions. There are 50 divisions per handwheel
rotation, so you divided 720/50 = 14.4 rotations. So 50 x .4 = 20
divisions "extra" per each 14 "full handwheel rotation". Wouldn't you
just start at 0 on the handwheel, rotate 14 full (from where you were),
then 20 additional divisions to get to the next flat? The following
move would count 14 revs from that location, and 20 more divisions.
You'd just have to remember which "division" you were at when you
started each 14 rev! The "divisions" rotate around by +20 each flat.
So at the end, you come back to 0 division (5 * 20 = 100 = 2
"additional" rotations.
(guys, please feel free to check me on this!)
I'm hoping that with the 3700-CNC version of the table that I want to
get, I won't have to remember as much! Maybe a DRO would help. Yeah,
we can add that...
A simple program could rotate the table for you with a stepper drive.
Might not even need the backlash calc's. Looks like another job for
"CAD_CAM_EDM_DRO" (radius).
Thanks for the practical example. I hadn't had to think a rotary table
problem through yet!
Good luck!
Alan (Hope I helped a little)
> [sherline] How do I machine a radius?"Wm. Dubin" wrote:
> "Rich " <fxdguy@...>
>
> My next project will be custom shift linkage for my motorcycle. It
> will be made from 1/4 X 3/4 CRS about 8 inches long. It will have a
> hole in each end and I want to put a nice even radius on each end.
> ....................
>
> I want to make pentagrame shaped nuts for a sculpture I am getting ready to
> start. These will be in sizes 1-72, 2-56 and 4-40.......... so the steel rod
> will not be hard to find.
>
> I have a rotary table and tail-stock, so it's simply a question of milling
> the rod with 5 flats............... however, when I do the math, I get the
> results of turning the table's handwheel 14.4 turns per flat.
>
> Neither my math abilities, nor my rotary table knowledge are good enough to
> cope with .4 on the handwheel...
>
> Has anyone a solution that can be simply applied?
>
> Thanks in advance,
>
> Wm. Dubin