Posted by Bill Andrews on 19/10/2016 09:48:54:
Good morning all.
Further to my original question on 10/12 Perhaps I should have added my own calculation which didn't work no matter what I did with it. leadscrew tpi 12×5=60 :-2=30
tpi required 26×5=130 :-2=60 60/30 on the gear arm…Where did I go wrong??
best rgds. Bill.
YOu made a slip up when you divided 130 by 2 and came up with 60. It should be 65.
So your overall ratio between the headstock spindle and the leadscrew needs to be 30/65.
Normally, on a "normal" lathe, you would put a 30T on the head stock spindle and a 65T on the leadscrew with whatever idlers needed in between to simply fill the gaps and away you would go.
But because you have a 40T permanently attached to the spindle, you need to calculate a compound train beginning with 40T that has an overall ratio of 30/65.
It's tough one. But if you multiply 30 by a factor of 44, you get 1320. (Which is divisible by 40, our first gear)
If you multiply 65 by a factor of 44, you get 2860.
This means that 30/65 = 1320/2860
1320 and 2850 are nice round numbers that can be broken down into change gear sized factors.
1320 = 40 x33 (Your set as posted on the chart above has a 33T homemade included I believe)
2860 = 52 x 55 (Your set also has a homemade 52T ISTR)
So: 12/26 = 30/65 = 1320/2860 = 40 x 33 / 52 x 55
So you need set up your gear train with the 40 tooth headstock gear (or the 40T fixed gear it drives) driving a 52T gear that is compounded on the same stud as a 33T. This then drives, through whatever idlers of any size that fit, a 55T on the lead screw.
So the gear train would go 40 – 52/33 – 55 plus idlers as needed.
The 55 and the 52 could be swapped with each other and you still get the same correct ratio.
So the penciled in chart you posted is correct, except one of the 55s read as 52.
Edited By Hopper on 20/10/2016 03:48:57
Edited By Hopper on 20/10/2016 03:51:57