Posted by BillS on March 10, 2010 at 09:34:20:
In Reply to: Re: Mikton 132 info posted by Tom Ramsey on March 08, 2010 at 18:10:57:
Thanks Tom for clearing up the index constant.
Tom's right about 16/34. that's the correct index to cut 34T. Your gear train has to be an exact match to 16/34 to cut 34T. No approximation allowed.
Your gear train doesn't match any number of teeth - I get 33.88 teeth! IMHO, this confirms you are seeing driver/driven pairs correctly, if the cutter moves out of tooth spaces as it makes 2, 3, etc. revs. And I think you said that the cutter doesn't track for more than 1 rev.
We've been assuming that the index gears as you find on the machine were left there after cutting a spur gear. But could it's last gear have been a helical gear where feed was also be engaged?
Since, as Tom said, the table is limited to 5 deg of swivel, a left-over helical is unlikely unless it was a small angle helical or worm gear.
The only other conclusion I can see is that someone diabolically put a random gearset on the machine before selling it :^( Hey, it could happen!
(Ed:) Something you want to do, Robi. It's important that we find out what is the first DRIVER gear.
Here's what you do: Drop the swing arm out of engagement. Back the cutter away from the workpiece. Turn on the machine (or rotate by hand.) Identify which of the two index gears on fixed shafts is turning. This is your first driver. If it is the 68T, then the index gear set and ratio is as you first described it:
68/84 x 42/72 = 0.47222222
but this can't produce a spur gear (BAD gearset, bad gearset...).
If the 72 is rotating, then Tom absolutely called it right - you should look at the gear set and ratio as:
72/42 x 84/68 = 2.117647
and that ratio doesn't produce an integer number of teeth either.