Thread: Change Gears Again

Guys, the stud gear and gearbox gear are both 40, with an idler in between. I'll double check the spindle/stud gear setup today.

The spindle has a 38 which drives the tumblers which drive another 38. That 38 is keyed to a 40, which drives an idler (70), which drives another 40 on the gearbox.

38 / 38 = 1
40 / 70 = 0.571428571
70 / 40 = 1.75

1 x 0.571428571 x 1.75 = 1

Am I doing it wrong?

PS: And yes, the above setup does produce the imperial pitches as per the chart. At least the few I've tried have worked.

That maths looks ok to me.

Originally Posted by Bryan

The spindle has a 38 which drives the tumblers which drive another 38. That 38 is keyed to a 40, which drives an idler (70), which drives another 40 on the gearbox.

38 / 38 = 1
40 / 70 = 0.571428571
70 / 40 = 1.75

1 x 0.571428571 x 1.75 = 1

Am I doing it wrong?

PS: And yes, the above setup does produce the imperial pitches as per the chart. At least the few I've tried have worked.

I've come in half-way through this thread and might be missing the obvious - but you have a 38 tooth drive which ends up driving another 38 through the tumblers - so far we are at 1:1. The 38 tooth is keyed to a 40 tooth so we have a 38:40 step up ratio. The 40 tooth turns another 40 tooth though an idler gear, the number of teeth on the idler is irrelevant. So don't we have an overall step up ratio of 38:40 from end to end?

Well that's what I thought. But I applied the formula I found somewhere. Maybe that doesn't allow for the coupled gears?

Originally Posted by Gavin Newman

The 38 tooth is keyed to a 40 tooth so we have a 38:40 step up ratio.

I don't quite understand where you're going with this bit? The 40 tooth gear, if it is locked to the 38 tooth by a keyway, is turning at the same speed as the 38 tooth gear. No change in ratio until you move onto the gear it is meshing with?

I agree with you about the 70 tooth idler being irrelevant in this case, because the gears either side of it have the same number of teeth, so all it's doing is reversing the rotation. If the gear on the output side of it was different to the one on the input however, it would be a different story.

Originally Posted by Jekyll and Hyde

I don't quite understand where you're going with this bit? The 40 tooth gear, if it is locked to the 38 tooth by a keyway, is turning at the same speed as the 38 tooth gear. No change in ratio until you move onto the gear it is meshing with?

That's what I meant, my terminology might not be 100% but the compound gear gives a step up ratio between input & output.

The idler gear ensures that the gears either side of it rotate in the same direction.

I ran some simple spreadsheets and determined that the best fit for a single compound gear set is N/60 where N is the selection 25, 26, 43, 46 or 47

The 60 gear gives errors -2.6% +1.9% either side of the metric values displayed in the table.

All you do is introduce the N/60 compound within the 1:1 set described by Bryan.
This will work provided the QC gearbox is actually 1:1 in position B1 (not confirmed).

John.

John
If the 38:40 gears speed up the drive chain to the QC gearbox by 1.0523..
Wouldnt the QC gearbox need to be 1:1.0523... to slow it back down?

Stuart

Stuart,
My understanding from what Bryan reported was that the gear set spindle to QC input is 1:1.

The set is:
- spindle 38 drives a layshaft with 38, this is 1:1,
- on the layshaft is a 40 that drives through an idler 70 to a 40 on the QC input, 1:1 again,
- result is 1:1.

What concerns me is that we have not confirmed that the 1:1 input to the QC actually produces the imperial table.
These gears could be wrong, what if the correct imperial standard set is the 70 gear on the QC input ?

John.

I can confirm that the overall ratio between spindle and leadscrew is 1:1 with B1 selected. I know this by making marks and turning by hand.

I also made marks on the gearbox input shaft. It stopped a bit short of where it started when the spindle was turned exactly once.

Again, the existing setup does produce correct imperial threads, as per the chart.

PS: Again, the existing gears can only go in the positions they are due to differing shaft sizes.

John makes a good point.
Stuart

Originally Posted by Gavin Newman

That's what I meant, my terminology might not be 100% but the compound gear gives a step up ratio between input & output.

The idler gear ensures that the gears either side of it rotate in the same direction.

I still don't get where you're coming from sorry. Unless I'm misunderstanding what is meant by having the 40 tooth 'keyed' to the 38, for there to be a difference in speed (ergo a change in ratio) between the 38 tooth gear and the 40 tooth gear keyed to it, the key would have to slip. The only thing that relates to the 40 tooth gear from the 38 is the speed of the shaft they are both fixed to. Aside from that, the 40 tooth gear is a completely seperate entity, involved only in driving the next shaft in the system.

For the sake of argument, lets say the spindle is turning at 500rpm. Which means both the 38 tooth gears are also turning at 500rpm, due to the 1 to 1 ratio. The 40 tooth gear is keyed to the 38 tooth, so it is also doing 500rpm. The idler gear in this case we agree is essentially unimportant, so a 40 tooth gear doing 500rpm will drive a 40 tooth gear at 500rpm. Where is this step up ratio? I can change the pair of 40 tooth gears for a pair of 80 tooth gears, but the end result is the same, even though your logic suggests there should be a ratio shift of 38:80 - a halving in speed.

*EDIT* I should probably add that I can see the shortcut you are trying to take, but to me it seems to make things overly complicated - as evidenced by the fact you've kind of mixed two approaches together. The step up ratio you mention applies only to the compound gear, not 'end to end' as you suggested earlier. So, with that in mind, 38/40 (compound gear) = 0.95. Starting at the spindle, we have a 38 tooth gear. 38/0.95 = 40. This then drives the final 40 tooth gear via the idler - 40/40 = 1. To my mind, its much simpler to work out the speed of each shaft one at a time.

It would appear that your lathe requires additional gears to cut the Metric pitches displayed.

If this chart is the only one you have and you dont have anything indicating that there should be a Transposeing Compound gear you maybe able tpo cut your pitches with the Stud gear remaining in its position and just changeing the Screw gears listed on the side of the Plate.

Would seem odd though if you wouldnt require some type of Tranposeing Compound.

Originally Posted by Jekyll and Hyde

I still don't get where you're coming from sorry. Unless I'm misunderstanding what is meant by having the 40 tooth 'keyed' to the 38, for there to be a difference in speed (ergo a change in ratio) between the 38 tooth gear and the 40 tooth gear keyed to it, the key would have to slip. The only thing that relates to the 40 tooth gear from the 38 is the speed of the shaft they are both fixed to. Aside from that, the 40 tooth gear is a completely seperate entity, involved only in driving the next shaft in the system..

My Bad - The bit I overlooked was that the gear ratios leading into and out of the compound gear are both 1:1, I was thinking in terms of the case where the compound is used to alter the ratio for inch to metrics. This is one of these areas where the more you think about it the more it does your head in.