Cutting a Mod 1 worm

I want to cut a worm to drive a Mod 1 gear. This means that the circular pitch is pi,, unless i've miscalculated. So effectively, I need to cut a thread with a pitch of pi mm

Have spent two days setting up, recalculating and recalculating, a spreadsheet to cover every one of the 1278 possibles available on my lathe.

The closest that i can get is 3.165 mm. This gives an error 0.060%, which will give a pitch error of 0.188 mm (0.0074" in old money) This seems to be a lot.

Is this likely to cause any major concerns, or is it tolerable?

Howard

Depending on the diameter of the gear, the worm will only drive a couple of teeth, so I doubt the error would be significant. You could consider choosing a ratio that is a bit smaller than 3.1415 mm, but cant the worm to an angle that makes the thread fit the teeth better?

If the wanted pitch is 3.14159mm and one can achieve 3.165mm, I make that an error of 0.75%? Furthermore the pitch error is 0.0234mm, or less than a thou. I doubt that will be a problem. Either that, or I shouldn't do maths after two glasses of wine.

My lathe will cut 1 Mod threads with an error which think is 0.06%. PM me if you feel you need to use it.

Andrew

Thank You for the advice.

If I must, I shall go ahead. I was given a worm, some time ago, but put it in a safe place. And, now cannot find the safe place!

So another job added to the Round Tuit!

Howard

The question is, to what accuracy do you need to make an approximation to trigger the find-it fairy?

She usually makes an appearance shortly after you have completed the replacement part!

But I DO wish that she would arrive soon.

Howard

Is this likely to cause any major concerns, or is it tolerable?

Does that not depend on the intended application of your worm? You have just shy of a one thou. in pitch error. Probably OK if you are not expecting to use it in an application where the errors will accumulate (such as a rotary table or instrument rack etc).

For example, if you are planning to use the worm to drive a rotary table with 90:1 ratio, one turn of such a table with your worm would have 90.0 x 0.0009 = .084 thou in. positioning error, you can decide if that is significant or not.

If you are using your worm to drive a hacksaw or the suchlike then the error is not material.

Martin, you have read my mind. The wish is to use the Division Plates off my HV6 .

If the worm remains in hiding, it looks a if I shall have to buy one. HPC quote £19.33 presumably plus carriage and VAT

RS offer a single start, steel worm, 18mm dia ( for £9.78, plus VAT ), free carriage and next working day delivery. Total £11.78, so not bad.

Have just ordered, so Cinderella may yet go to the ball!

Howard

The worm arrived, from RS, yesterday at 0800. So now having worm and 90T gear have little excuse, not to start work!

Have to admit being peeved that my BL12-24, with a 3mm Leadscrew cannot cut a pi mm pitch thread given that with the gearbox, a compound gear and three possible drivers or driven, it seems to be capable of 1180 possible combinations!

So far have failed to come up with another gear that could drive, or be driven, to produce the elusive thread pitch.

But Brian Wood gives a train for a 1.5 mm pitch Leadscrew on a mini lathe which will do what i want with an error of "0.001" Not sure if that is %, a thou or a micron!

Anyway, more jobs to go onto the Round Tuit.

It will be ready tomorrow; and you know what they say about tomorrow!

Howard

Edited By Howard Lewis on 09/07/2019 18:19:05

Howard,

The error was listed in the table as mm, so the figure will be 0.001 mm [1 micron]

It is of course a calculated value which depends on the accuracy of the leadscrew on the machine in question. Having said that you could test the gearing on your lathe with a 60 T gear on the leadscrew just to see if you could cut that pitch, it would perhaps prove the point whether or not your lathe is capable of the task.

Regards

Brian

Thank You Brian.

My spreadsheet was form the BL12-24 with its 3mm pitch Leadscrew.

Having a Norton box, there are only three change gears available, apart from the 120/127 compound Idler, the "standard" is 40T and the alternatives are 30T and 32T. To halve the feed rate, I cut a 80T gear for the input to the Norton box.

having now obtained a 1 Mod worm to match the 90T gear, it is a bit academic, but just out of interest, I would like to work out what gear, if any will allow a pi mm pitch thread to be cut.

With the possibility of using the 30 or the 32 as driver, or driven my wish is unobtainable. One day, I shall return to the spreadsheet and start guessing as to what is needed. With 40 ratios available in the Norton box, it ought to be possible to find something that fits the bill, even if it will increase the total of possible combinations to over 3,000!

Really, it would have caused less wear of the brain cells to have followed Brian's chart for the 1.5 mm pitch leadscrew on a mini lathe, and done things the easy way.

Howard

Edited By Howard Lewis on 10/07/2019 16:33:34

A small error in the worm is not a big issue, as it can be compensated for simply by adjusting the centre distance. If it is 1% too large, it just needs to be 1% further out from the gear centre than the theoretical correct depth, or less than 0.3% of the pitch.

The principle is the same as cutting a slightly oversize gear to compensate for increased gear spacing. Obviously there is a limit to how far you can go but fractions of a a percent won't make any difference.

The nature of a worm/wheel pair is that the drive ratio is entirely unaffected by worm pitch errors.

Neil

Hello again Howard,

I can save your brain cells some, but not all, the work.

The basic gearing calculation is really not too difficult. For this example we do need to incorporate an approximation for pi/2 in the gearing, which is neatly given by 55/35 to an error of roughly 1 in 2000

The full sum therefore becomes required Pitch P = 55/35 x R x Leadscrew pitch [ R is the combined value of change wheel ratio and in your case a gearbox ratio]

Putting in the known value for P = 3.142, we can rearrange the formula to find the value of R

R = 3.142 x 35/55 x 1/ Leadscrew pitch [1/3 in this case] which gives a value for R of 0.6664

Without knowing what change wheels you have available, nor the gearbox ratios that you can call upon, it surely can't be too taxing to find values for these components in the calculation, which multiplied together amount to 2/3

Regards

Brian

Hello Brian!

Thank You. Secretly, I had been hoping that you add your words of wisdom!

I would be most grateful for your help.

I should have taken the easy way and followed the advice in your book, "Gearing of lathes for Screwcuttimg" and used the mini lathe with nits 1.5mm pitch Leadscrew. Because I am so much more conversant with it, I was looking to use my Engineers Tool Room BL12 -24.( This is a dual dialled version of a Chester Craftsman, or metric Warco BH600 ) It has Norton gearbox giving 40 combinations, fed via 120/127T Compound gear. I have 30T, 32T and 40T gears which can either drive the 120/127 or be driven as the input to the Norton box. So, by my calculations there are 1180 possible results. None of which provide a pitch of pi mm!

To my shame, I have cheque book engineered a solution by buying a 90T gear from Arc Euro, and a 1 Mod worm from RS. So the immediate problem has gone, but the curiosity as to how one achieves the desired result, still remains. You never know, I may want to cut another 1 Mod worm in the future!

So far my guesses have been futile!

The compulsion was a wish to be able to use the Division Plates from my HV6 on a small Dividing Head.

Probably, what I should have done was to cut a worm, (possibly 3/8 BSF or 1/2 UNF) and then a custom gear to match.

If you would kindly PM your E mail address, I will send over my spreadsheet, in the hope that you can suggest a way out for a simpleton like me.. My brain cells have just about seized on their spindles, by now!.

Presumably, this will entail cutting another 1.25 Mod changewheel to supplement the existing three.

Howard

Hello Howard,

How very flattering, I'll do what I can to help.

I am not familiar with your BL12-24 lathe, but I suspect it will have a 40T gear off the spindle permanently included in any change wheel set up. I don't know how comprehensive the Norton gearbox is either, so a picture or two to enlighten me on those aspects will be helpful; plus of course your spreadsheet.And a picture to show how your changewheels are arranged on the banjo would be helpful.

Do you have any feel for the ratios the gearbox provides? One way of establishing that is to take a look inside and count the teeth on the input gear, then count the teeth on the cone of gears that can be selected. I can work out what that gives as ratios if you would prefer not to.

I hope it won't be necessary for you to make other change wheels, we'll see.

My email is wood_y(at)btinternet(dot)com all in lower case, no spaces and note the underscore between d and y

Regards

Brian

.

It just wears out [or, for positive thinkers, beds in] more quickly

MichaelG.

Posted by Howard Lewis 04/07/2019 21:09:34

I want to cut a worm to drive a Mod 1 gear. This means that the circular pitch is pi,, unless i've miscalculated.
So effectively, I need to cut a thread with a pitch of pi mm
Have spent two days setting up, recalculating and recalculating, a spreadsheet
to cover every one of the 1278 possibles available on my lathe.
The closest that i can get is 3.165 mm. This gives an error 0.060%, which will
give a pitch error of 0.188 mm (0.0074" in old money) This seems to be a lot.
Is this likely to cause any major concerns, or is it tolerable?

Howard

Looking though Ivan Law's book gears and gear cutting on page 97 and the adjacent pages
provides some explanations on the pitch of worms.
It would not be of any use to be able to cut a worm exactly the pitch to suit the CP of the gear
(assuming that this is to mesh with a spur gear) since the worm has to be set over at an angle
in order to mesh with the gear.He does make the point that if the helix angle is low about 3 deg the
difference can be ignored.
However by adjusting the helix angle in this case to 6.9 deg the figures work out to the 3.165 mm
or .1246 inches .
As described in the book dividing the circular pitch by the cosine of the helix angle of the
worm to obtain the new pitch CP 3.14 mm or .1237 inch divided by cos 6.9 deg = .1246 in
or 3.1649 mm ,at these figures the worm would be .328 inch PCD , .407 inch OD .
It just depends if you can live with the 7 deg set over and the root diameter of about .240 inch .

I don't screw cut worms anymore , i mill them the finish is much better along with a cnc system
the pitch can be adjusted over many teeth and along with the control with rotation any size or
pitch can be achieved ,the same basic parameters need to be observed in respect of lead
helix angle and pcd.
Here is a 1 module worm and hobbed gear at 90 to 1.

John

Thank You all for your wisdom and advice.

The next time that I am faced with this problem, I shall know that by offsetting the worm, I can cut one that will suffice

As always, on this Forum, huge amounts of experience and help!

Howard.

Discussing this problem with other Forum members, have realised that my spreadsheet contains an assumption that may not be valid, ie that Feedshaft and Leadcsrew rotate at the same speed for the same gearbox setting.

Need to calculate the apparent ratios, for Threads, as well as Feeds. They may be different.

You learn something every day!

Beginning to look as if there are some different possible combinations.

More work on EXCEL tomorrow!

Howard.