Change Gears Again

[Bryan]

Sorry I know this is a bit of a pain in the bum topic, and I feel like I should be able to figure it out but I'm stumped. And since it seems to be a slow day ...

Now that I have a metric machine in the stable I want to get metric screwcutting organised. Has anyone seen a chart like this?

The part at the top is straightforward imperial threads, with the operator left to fill in the blanks. The coloured numbers on the left I believe are metric thread pitches. They line up in columns with the gearbox letters. And the colours link the pitches with the gearbox numbers at the top.

Say you wanted a one mm pitch. You find the 1 in the table, see that it lines up with the B, and that it's red. Red corresponds to 5, therefore B5 is the ratio for a 1mm pitch.

The baffling bit is the column on the far left. I can only imagine it's about the change gears; that the tooth count on the left should somehow produce all the pitches on that row. But I cannot for the life of me make sense of those numbers as change gears.

Before I go into the numbers, I'd like to know if there are any flaws in my logic so far? Is there some other way this chart can be read?

Edit: Have reproduced the chart as a pdf and hopefully made it easier to read.

[Vernonv]

I reckon they are change gears also. Do you have any gears that match those numbers?
Do you have a picture of your gear train?

[Bryan]

Vernon, the only gears that came with the lathe were 2 x 40s and a 70 idler. I don't believe there was ever a 127 because it simply wouldn't fit.

[Stustoys]

So the 3 in 43 isnt green then?

I think the numbers are the driver gear.

my e.g goes like this
Take 0.7= 26 c yellow

1.125 is also C yellow so the only difference is the 25 and 43
43 is 17 teeth bigger than 26
17 about 65% of 26
0.7 + about 65% = 1.157

Sort of about right

Of course I've been wrong before, more than once ;)
Stuart

[Bryan]

Stuart, yes the 3 in 43 is green, and I have no explanation for that. Got a theory?

I think I can follow your logic. It might be a question of how close is close enough?

The way I thought about it was this:
Leadscrew = 8TPI.
8TPI = B1.
B1 = 1:1.
Change gear ratio = metric pitch at B1 / imperial pitch at B1 (in mm).
Change gear ratio = 2.25 / 3.175 = 0.7087

I've used a spreadsheet to multiply .7087 by every whole number from 2 to 127 inclusive, and there's not a single whole number among the answers. The closest ones are 24:17.0088 and 127:90.0049. They may be close enough but they don't sound like realistic change gears to me. Something's not right.

[Gavin Newman]

Some lathes use a 63 tooth gear instead of a 127 tooth for the imperial/metric conversion (given that 63 * 2 is 126 which is close to 127) due to the size.

[Stustoys]

Gavin beat me to it.
As I understand it, all thread cut on the "other" leadscrew are approximations, some are just closer than others.
Not theory on the green 3 sorry
I dont follow your maths really, you could have picked 2.5 instead of 2.25, its B1 also.

Stuart

[Bryan]

Gavin beat me to it.
As I understand it, all thread cut on the "other" leadscrew are approximations, some are just closer than others.

So the pairing would be 63:60? That gives 1.05. Lead of 3.175 / 1.05 = 3.0238. I'm guessing a 1 percent-ish error wouldn't matter a damn for normal purposes. So just replacing my idler with a 63/60 would give me some metric threads. But it doesn't explain the chart. I guess I can live with that if I have to.



I dont follow your maths really, you could have picked 2.5 instead of 2.25, its B1 also.

Stuart

Yes, I see. I don't have the answer yet. Looks like a busy weekend ahead with family commitments, so I don't know when I'll get a clear head for this. Wish I was more competent with maths. Thanks for the input so far.

[electrosteam]

Bryan,
It would assist greatly if we knew the gear set to give the imperial table.

If you cannot easily determine them, look at just the gearbox.
It is easy to mark the input gear, select a ratio (say A1), and observe the leadscrew rotation to determine the gear ratio for that selection.
Only do it once as the rest of the box is just multipliers or divisions from that value.
Once you know the gearbox overal ratio,and using the leadscrew value, you can calculate the ratio spindle to gearbox input, and that in turn will tell you the gears required.

Don't worry about odd ratios, for example my Bantam has a 7:1 in the gearbox somewhere and 49 teeth gears for some metric threads.

The imperial table is identical to numerous lathes copied off the imperial South Bend 9A.
My Sheraton 9A with an 8 tpi leadscrew, for example, would produce an identical set of values with a 40 Driver and a 56 Follower gears.

The reverser yoke seems odd, the two gears engaging on the spindle must be the same size to get equal thread pitch Forward and Reverse.
I count the teeth from the photo as: 20, 25, 40, 40, 70.
Is it possible the two 40s should be on the reverser yoke ?

John.

[Stustoys]

The reverser yoke seems odd, the two gears engaging on the spindle must be the same size to get equal thread pitch Forward and Reverse.
I count the teeth from the photo as: 20, 25, 40, 40, 70.
Is it possible the two 40s should be on the reverser yoke ?

John.
You sure about that John?
One rotation of the spindle is XX teeth, doesn't matter what size and number of gears you put that through they all turn XX teeth. So only the gear on the spindle and the gear on the input shaft of the gearbox matter.

Unless you have a double gear like a 127/120(I think thats my double gear, or is it called a compound gear?)

As the spindle gear looks pretty fixed on Bryans lathe, I assume its the gear on the gearbox input shaft the table it meaning.

Stuart

[Bryan]

John, attached to post 1 is a PDF chart which shows all the imperial thread pitches. As a starting point I was trying to simplify matters by assuming QC ratio B1, which is 1:1. I know this because the thread produced is the same pitch as the leadscrew, with change gears of 40:40. If I can get the right conversion ratio at B1, I figure the rest should fall into place. The gears I have, including the tumblers, are in the only positions they can go.

[electrosteam]

Stuart,
You are correct, all the gears in the photo are only idlers between the spindle and the QC input gear.
I spent so much time analysing my Sheraton, I just followed that arrangement with a layshaft after the tumber reverse allowing a selection of driver gears.

Bryan,
My Sheraton has a QC ration in A1 of 1:2.8 and for B1, 1:1.4.
The Hercus will be similar, but not necessarily identical.
Are you really sure B1 is 1:1 ?

Once you know the actual QC ratio, you can calculate the compound intermediate to produce the metric table (after a few tries because of the approximation).

John.

[Bryan]

I thought the way to know what gears I needed would be to understand the thread chart provided on my lathe. That may be true, but I can't understand the chart. The numbers that we think are tooth counts do not work with my existing gears in any combination that produces that chart. So other gears must be needed. If there's a way to calculate those missing gears I'm not smart enough to figure it out.

However, by once again messing around with spreadsheets I've concluded there are numerous ways to achieve common metric pitches on my lathe. And I can do it by only adding two or three gears. No transposing gears needed. (Transposing gears I understand to mean a keyed pairing of either 127/120 or 63/60.) So I think I'll settle for that.

Thanks to all respondents.
:fixed:

Edit: John you posted while I was composing. I think my 1:1 logic is sound. If you want to keep chasing this bunny I'll watch. :)

[Stustoys]

Bryan.
I might be confused as to what you are thinking here but.
The 1:1 ratio of B1 is for the complete gear chain from spindle to leadscrew. Not just the gearbox.

Stuart

p.s. what are the tooth counts on the spindle and gearbox input?

[electrosteam]

Stuart,
Exactly - there is probably a reduction from the spindle to the QC input, then a speed-up within the QC for the A1 position.

My Sheraton:
40 Driver to 56 Follower gear reduction 56/40 = 1.4 from spindle to QC input,
QC position A1 speed-up of 2.8 input to leadscrew,
result is leadscrew speed is double the spindle speed,
equals 4 tpi with an 8 tpi leadscrew.

Once we have the QC ratio, we can calculate what the gear set should be.
Don't assume that the gears as set in the photo give the correct results until confirmed by Bryan.

John.

[Bryan]

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.

[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.

[Vernonv]

That maths looks ok to me.

[Gavin Newman]

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?

[Bryan]

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

[Jekyll and Hyde]

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.

[Gavin Newman]

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.

[electrosteam]

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.

[Stustoys]

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

[electrosteam]

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.

[Bryan]

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.

[Stustoys]

:doh:John makes a good point.
Stuart

[Jekyll and Hyde]

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.

[pipeclay]

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.

[Gavin Newman]

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.

[Jekyll and Hyde]

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.

Agreed, I was triple guessing myself trying to see if I'd missed something really obvious and was just making an idiot out of myself. :2tsup:

[Gavin Newman]

Agreed, I was triple guessing myself trying to see if I'd missed something really obvious and was just making an idiot out of myself. :2tsup:

I'm good at that :C:C:C. I think I'll leave this one to the experts and go out and work on the race car - it's gotta be easier.

[Jekyll and Hyde]

I'm good at that :C:C:C. I think I'll leave this one to the experts and go out and work on the race car - it's gotta be easier.

Good idea that, I've got a whole front end to straighten out and weld back on. Panel work is painful, especially when you're dealing with the sins of a previous owner (inch thick bog and bad lap joints anyone?)

[Bryan]

So it now sounds like we have agreement that the ratio between spindle and gearbox should in theory be 1:1. Then how can my observation be explained that the gearbox shaft turns less than once for each spindle turn? Both those things can't be true. (I will be triple checking that tomorrow.)

[Gavin Newman]

So it now sounds like we have agreement that the ratio between spindle and gearbox should in theory be 1:1. Then how can my observation be explained that the gearbox shaft turns less than once for each spindle turn? Both those things can't be true. (I will be triple checking that tomorrow.)

That's what I don't understand - can we have more photos of the gear train. There must be something in the gear train we are missing????

[Jekyll and Hyde]

So it now sounds like we have agreement that the ratio between spindle and gearbox should in theory be 1:1. Then how can my observation be explained that the gearbox shaft turns less than once for each spindle turn? Both those things can't be true. (I will be triple checking that tomorrow.)

Looking at the photo of your gearbox, I think you may not have given all the info we needed.

When you move the lever at the top up or down, does the drive go through both 'tumblers' to get to the compound gear? As far as I can tell, when you select one position, one tumbler will engage to the spindle and the other to the compound gear. If you select the other, the gear previously engaged to the spindle will engage the compound gear, and the one engaged to the compound gear will now engage the spindle? If this is correct, you need to count the teeth on the two tumblers as well, because they are changing the ratio between the spindle and the compound gear....

[Jekyll and Hyde]

Scratch that. I just counted the teeth on your tumblers, and did some numbers, still works out to be 1:1 between spindle and compound. Seems that it doesn't matter how many idlers you have, they still don't do anything. Also worked out I'm wrong about how they operate anyway, since that's the FWD/REV lever I'm on about...

Maybe recount the teeth on the spindle gear? It seems bigger than the 38 tooth on the compound, but so do the teeth....

*edit* If you had a 40 tooth spindle gear, everything else the same, you'd get a 0.95 reduction over the whole geartrain. And this is driving me nuts, I want answers!

[rusty steel]

Hello Bryan,
I checked the chart details (metric) on my lathe by setting my digital vernier calipers up in the tool holder so that the "rod" came up against the headstock. I then turned the chuck ten turns for each combination of gear setting. Perhaps you could do this and see how the results compare to your chart.. I suspect that you may need more change gears.
Regards,
Russell

[Stustoys]

Then how can my observation be explained that the gearbox shaft turns less than once for each spindle turn?
Nothing that I can think of. Either your teeth counts are out or maybe there was some slop in the gears on the first turn?
Or we are still missing something ;)

Stuart

[Bryan]

I have an explanation for the anomoly. I was expecting to find a difference so when I thought I found one I didn't look any harder. Must have been a backlash error. I should have repeated the test several times. That's what I just did and it's definitely 1:1. My apologies for the red herring.

[Bryan]

Attached are some solutions, including the default gearset(!).
Highlighted are the pitches I expect to find useful.

[Stustoys]

I think there is something wrong with your table.
For e.g.
In the top table
A1 = 6.35 = 4 tpi Ok
B1 = 3.175 = 8tpi Ok
A6 = 9.525 = 2.66666666666666667tpi ????????

or am I missing something yet again?

Stuart

[Bryan]

Stuart you are correct. Collect $200. The numbers should get smaller as they go across, not bigger. Because per inch is a division. I remembered to correct going down, but not across. Thank you. I will start again. :B

[electrosteam]

Bryan,
The spreadsheets you posted with combinations of gears to discover 'near enough' metric threads is the method I have used in the past with great success.
You just need to record your own tables of gear sets versus metric threads.
(once you and Stuart get them correct)

The original question, I believe, was to understand the table affixed to the lathe.
I determined earlier that the only single gear that could reasonably produce the metric table was N/60 with N being selected from 25, 26, 43, 46, 47.

The set would hypothetically be a reduction:
- 38 spindle mesh via one or more idlers to 38 on layshaft on reverser yoke,
- yoke layshaft keyed to a 40,
- 40 mesh with 60 on layshaft on the banjo,
- banjo layshaft keyed to N,
- N mesh with 40 on QC input.
But, I am not at all sure you could fit a 25/60 compound on the banjo and have physical space to mesh the 25 with the 40 on the QC input.

Are you in a position to judge the fit as suggested above ?
What is the bush diameter that is used on the QC input shaft to space the gear outwards one thickness ?
What gear specification has been used, 20 DP 14.5 degrees, or other ?
With some help from Pipeclay on gear OD sizes, we should be able to determine if the set proposed above will work.

John.

[Bryan]

The original question, I believe, was to understand the table affixed to the lathe.

John, yes I was interested in solving the original table. But it's dawned on me that a lot of threads in that table I would never use. So it's an academic exercise rather than a practical one. The tangible outcome I need is to be able to cut common metric threads, ie 1, 1.25, 1.5, 1.75, 2, 2.5, 3 mm. The ones that are hard to find are 1.25 and 1.75.



The set would hypothetically be a reduction:
- 38 spindle mesh via one or more idlers to 38 on layshaft on reverser yoke,
- yoke layshaft keyed to a 40,
- 40 mesh with 60 on layshaft on the banjo,
- banjo layshaft keyed to N,
- N mesh with 40 on QC input.


I've, um 'XL'd' your suggestion as I understand it. Please see if I've done it right. It gets closer with some ranges than others. How close is close enough? I really don't know, but for some reason I have my sights set on 1% as a maximum error. If that means swapping two gears instead of one I think that's worthwhile. So again this is tending away from the concept of the original table as I understand it.



But, I am not at all sure you could fit a 25/60 compound on the banjo and have physical space to mesh the 25 with the 40 on the QC input.


Yes that does sound tight. At the moment I'm working on the idea of leaving the idler alone and just swapping stud & screw gears. If I can get the pitches I want that way in realistic sizes it will save some hassle. At least it seems simpler to my mind. If that doesn't work out I'll revisit the idea of adding a compound gear. Thanks, I appreciate your efforts.



What gear specification has been used, 20 DP 14.5 degrees, or other ?


16DP 20 PA. At least they're standard, unlike other gears on this machine.

[Bryan]

Something like this:

[electrosteam]

Bryan,
I did try to post with a copy of my spreadsheet(just love computers !), but yours is the same as mine.
Note that for the nominated metric threads, the errors are plus/minus.
I tried 61 and 59 teeth with the obvious systematic error one way or the other,
You are on to a solution now, leave it with you.

One thing, however, intrigues me; the presence of the 70 gear.
Is it possible that the 70 gear is part of the metric set ?
If 70 gear was on the QC input, the compound would have to be adjusted.
Instead of N/60 it would be N/(60x40/70) = N/34.2, say 34 teeth.

I have attached my spreadsheet showing the result.
Some better, some worse.
But, you can set it on the banjo/quadrant no problem.
The N/34 with N meshed to 70 could be the original design intent.

John.

[Bryan]

John, all I can tell you about the 70 is that it's intended to be an idler. It has no keyway or dowel hole or any way to drive anything.

That's a neat way to show the errors. I copied that on mine and got no more than .54%. Pretty happy with that. I also expanded the highlighting. Those two tables will cover diameters from 5-36 in coarse and 4-100 in fine. Oughta do me. :U