A question on gear cutting

[Cursitor Doom]

This is *not* a hypothetical question, BTW.

One of my lathes I was able to pick up cheap because it had a couple of things missing from it. One of those things was a 16 tooth helical gearwheel that is driven by the leadscrew to provide timing information on when to throw the half-nuts into mesh when screw-cutting. Anyway, the manufacturers (Harrisons) tell me this part is no longer available so I shall have to cut one for myself, basically. What I'd like to know is, can one back-calculate the required dimensions of this gearwheel from the leadscrew it meshes with? I'd have thought so, but welcome informed opinion on the matter. Say for example the leadscrew (Acme thread form if it matters) is 6TPI and the missing gear is known, from old user manuals, to have 16 teeth. The helix angle can be taken from the leadscrew (or could be worked out simply from the pitch of it) and 16 teeth at 6TPI should enable one to work backwards to find the pitch circle diameter - or the base circle diameter at least. If I do this, are other parameters like the pressure angle and MOD values automatically taken care of, or is there still some additional data required to cut the new gear that simply cannot be inferred from the existing machinery?

thanks,

CD.

[Nightjar]

Have you asked the manufacturer whether they still have the drawings? Depending on just how long it has been out of production, they could still have those on file, even if they no longer supply the part.

[Cursitor Doom]

On Thu, 21 Apr 2016 09:29:06 +0100, Nightjar Have you asked the manufacturer whether they still have the drawings?

Ha! Fat chance of that. The gear itself is no longer available, but they just happen to have *one* left as part of a complete new indexing head for 240 quid. They won't sell the gear separately. They kindly offered to cut me a new gear, though, for 160! Bear in mind the equivalent Myford part is just 12 quid. So yes, they have the drawings but there's a negative cash incentive to part with them. Anyway, it's more fun to to work things out; more of a challenge I reckon.

[Cursitor Doom]

I've given it some more thought; I'm almost there...I think.

The key dimension of this gear and the one that really matters is its diameter. Everything else - I think - is known and follows from that. Its diameter will be it's circumference divided by pi. So what will its circumference be? The answer is the same as the length that 16 teeth occupy along the leadscrew it meshes with. But I suspect this is only an approximation. Don't mind me talking to myself here. It often helps to clarify my thinking. But feel free to chip in if you've spotted something I may have missed.

[stvlcnc43]

Could you make a cutter that matches the leadscrew, and then use it to cut another gear?

[Fredxxx]

That could work, but you would need to rotate the gear in sympathy with the cutter.

[Cursitor Doom]

Already done it. That was the easy bit.

That's called "hobbing" but there are other methods. I prefer to cut mine on a milling machine with an indexing attachment. In any case, I just need to know the diameter of the blank to use for the gear. That's the only thing that's holding me back - at the moment....

[Julian Barnes]

I made a pair for the M250 out of acetal using a fly cutter but I had the original gears to work from. I did not worry about helix as the gear is only say 4 mm thick.

Anyway this is the answer - I think

Leadscrew is 4 TPI = 6.35 mm per tooth

mm per tooth / Pi = MOD = 6.35/3.1418 = 2.021268 MOD

DP = 2.54/MOD = 2.54/2.021268 = 12.56637 DP

OD of gear = N+2/DP = 16 (desired tooth count)+2 /12.56637 = 1.432"

Yes I could have stuck with MOD but as your lathe is imperial perhaps you work in imperial. I'm sure the more mathematically able would do it differently. The tooth profile etc only has to be there or thereabouts and it will work fine.

[Cursitor Doom]

A helpful contribution, I'm sure. Just can't help thinking that your end result of just under 1.5" seems rather small. I'll check your working later as I'm a bit tied up right now, but I suspect you've made an error somewhere.

[Cursitor Doom]

No, this can't be right, surely. Whoever heard of a DP of 12.56637?? Doesn't make sense. The whole depth of cut on the leadscrew is about

0.134" which would suggest a DP of 16 (they're always round numbers on stock gears). Tables tell us that the Addendum for a 16DP gear is 0.0625" and the OD will be the PD + 2*A. Getting there.....

[Julian Barnes]

Normally that's right. But you're coming from this arse-about-face remember, so you will get a seemingly abnormal answer. Give it some more thought and you'll see for yourself.