The Home Machinist!

I did not do the math when I made the drawing.I was thinking the 3/8-10 acme would be moving in and out any way so put a pulley on lt and haft the problem is solved. Make that a 4 inch pulley and a 1/2 inch drive pulley with the o ring drive belt. The finished part is a little over 1/2 inch long making the total travel of the 3/8 -10 a little over 5 turns thats about 40 turns on the drive pulley. Is that the right way to do the math? I was thinking the drive belt would have to move with the 4 inch pulley to keep them in line. Do you think the o ring profile would fellow the larger pulley on it's own? Maybe a gear box would be a better way. Jim

I think it would, particularly if you used 1-1/2 right-hand turns around the handwheel shaft. The offset thickness would ensure the advance of the driven pulley would draw the belt along the shaft.

As for the math, the ratio of the circumferences will determine the ratio of turns. So, for your 4" pulley and 1/2" shaft:

0.5" shaft = pi * 5" = 1.571"
4" pulley = pi *4" = 12.566"

12.566/ 1.571 = 7.999:1

That's approximately 8:1, so yes, 40 turns on the handle makes 5 turns on the shaft.

A 1/4" shaft would give you 16:1, but might be a bit too flexible.

My first though this morning was that was good thinking for 8-1 but not for 30-1 Then I read your post. I was using grade school math I had forgot that Pi thing. I will go back to the drawing board. Maybe increase the size of the big pulley say to 5.5 inches make it a wheel use the o ring as tread running on a 3/8 inch drive shaft. That should be 20-1 I think?

"grade school" math works. The 'Pi things' cancel so the ratio of the two circumferences is the same as the ratio of two diameters.