Mathmatical assistance please

I am trying to calculate the centre distance of a worm and worm gear and al though I have looked lots of formulae on line I am, as usual, confused!
The worm gear has 40 teeth and the OD at the largest point of the gear is 4 .375?. The worm OD is 1.167? and it is a three start worm, if that is important.
Could somebody explain in words of one syllable how I do this calculation a nd what answer I should get?
I need the easy to understand method of calculation as I know I will need i t again in the not too distant future.
As usual, thanks in advance.
John H
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I am trying to calculate the centre distance of a worm and worm gear and although I have looked lots of formulae on line I am, as usual, confused!...
John H
Gears contact each other at their pitch diameters. http://www.fi.edu/time/Journey/Time/Lessons/printgeomgears.html
http://www.salemcompany.com/cgi-bin/Store/pdfs/gearsClsect4.pdf
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But it's slightly more complex (sorry, don't have the answer) in the case of a worm wheel which wraps around the worm
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wrote:

The effective diameter of a gear is its pitch diameter.
The DP (diametral pitch) of the gear is <number of teeth>/<pitch diameter>.
Normally the pitch diameter is halfway down the tooth, so:-
<outside diameter>=(<number of teeth+2)/DP
If it's specifically a worm wheel (with the teeth slightly dished), rather than an abused gear, one generally uses:-
<outside diameter>=(<number of teeth+3)/DP
with the tooth height still being 1/DP
for the worm,
Pitch diameter= outside diameter - 1/DP
So, we have
4.375=(40+3)/DP
DP=(40+3)/4.375C/4.375=9.8285714285714285714285714285714
Might be good to check that it isn't DP !!!
if the gear really is 10DP, we get:- <gear pitch diameter>= 40/10=4"
If it's the calculated number, we get:- <gear pitch diameter>= 40/9.82857142857142857142857142857144.069767441860465116279069767442
with 10DP, we get:- <Worm pitch diameter> = 1.167-1/10 = 1.067 (there's a suspicion that might be exactly 1")
with 9.8285714285714285714285714285714DP, we get:- <worm pitch diameter>-1.167- 1/9.8285714285714285714285714285714 <worm pitch diameter)=1.0652558139534883720930232558139
So, for 10DP we get
spacing = <worm pitch diameter>/2 + <gear pitch diameter>/2 spacing = 1.067/2 + 4/2 = 2.5335"
For 9.8285714285714285714285714285714 DP we get:-
spacing= 1.065255/2 + 4.069767/2 = 2.567511
You're going to have to make some guesses and assumptions, because a shaft spacing of 2.5" looks very reasonable, but the actual numbers point to a shaft spacing of:- up to 2 9/16"
I don't know if that's helped or made things worse :-(
regards Mark Rand RTFM
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"Mark Rand" wrote in message

Impressive bit of maths there Mark!
I would be tempted to mount the worm and wheel on some suitable adjustable mount, and tweak for a nice running fit, then measure. Rather like I understand clockmakers use a depthing set up.
The chuck of the lathe could carry the worm on a suitable arbor, and the wormwheel could be mounted in the toolpost again on a suitable arbor. (or vice versa) using the cross slide dials for measurement.
AWEM
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Hut that's for a different reason, and is to locate the transition point between the hypocycloidal and the epicycloidal curves. There won't be the same effect with involute teeth because all that the happens is that there is a change in the effective pressure angle, a dodge sometimes used to cut different numbers of teeth on the same notional pitch cirlce diameter.
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"gareth" wrote in message

...'twas but an illustration of the set up in the absence of pretty pictures !!!!!!
AWEM
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On Monday, February 11, 2013 11:32:01 PM UTC, John H wrote:

although I have looked lots of formulae on line I am, as usual, confused! T he worm gear has 40 teeth and the OD at the largest point of the gear is 4. 375?. The worm OD is 1.167? and it is a three start worm, if that is im portant. Could somebody explain in words of one syllable how I do this calc ulation and what answer I should get? I need the easy to understand method of calculation as I know I will need it again in the not too distant future . As usual, thanks in advance. John H
Thank you Mark, your calculations have sorted out my understanding of the f ormulae that I have read.
As I said, I will be doing this type calculation again. Hopefully now I sha ll be able to, but with a few less decimal places!!
John H
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wrote:

although I have looked lots of formulae on line I am, as usual, confused! The worm gear has 40 teeth and the OD at the largest point of the gear is 4.375. The worm OD is 1.167 and it is a three start worm, if that is important. Could somebody explain in words of one syllable how I do this calculation and what answer I should get? I need the easy to understand method of calculation as I know I will need it again in the not too distant future. As usual, thanks in advance. John H

Andrew's suggestion of trying it for fit and measuring is the optimal solution if it's at all possible. I discovered this when I made a load of gears to replace the clapped out ones in a gearbox and then found that mine were a few thou oversize!
Regards Mark Rand RTFM
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On Wed, 13 Feb 2013 19:35:51 +0000, Mark Rand

although I have looked lots of formulae on line I am, as usual, confused! The worm gear has 40 teeth and the OD at the largest point of the gear is 4.375. The worm OD is 1.167 and it is a three start worm, if that is important. Could somebody explain in words of one syllable how I do this calculation and what answer I should get? I need the easy to understand method of calculation as I know I will need it again in the not too distant future. As usual, thanks in advance. John H

I needed some small custom worm and worm gear sets and decided to use a tap as the hob. I made a few different sets, measured the parts, calculated center distances, and still had to adjust more than I thought I would need to. Measuring over the gears with the worm nested in the worm gear gave me much better results, so I think I goofed a little on the math. The best part though was how easy it is to make small worm gears with a high spiral tap held in a collet in the lathe. In fact, it was so much fun that I made extra sets of different sizes that I didn't need and now just sit around as show and tell parts on my desk. Eric
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Is that hobbing with the workpiece driven synchronously, or free hobbing which would require a spiral tap to ensure continuous engagement?
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On Fri, 15 Feb 2013 06:56:52 -0000, "gareth"

Free hobbing with a high spiral tap. Super easy. And I always got the number of teeth I expected. Eric
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On Monday, 11 February 2013 23:32:01 UTC, John H wrote:

although I have looked lots of formulae on line I am, as usual, confused!

4.375?.

t.

and what answer I should get?

it again in the not too distant future.

What is a 'high' spiral tap, please?
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On Sat, 16 Feb 2013 07:12:00 -0800 (PST), " snipped-for-privacy@yahoo.co.uk"

A high spiral tap should really be called a low angle helix flute tap. In my opinion. But anyway, instead of having straight flutes like most taps the flutes wrap around the tap. If you were to try to hob a worm gear with a regular tap, using the tap to turn the work, it wouldn't work too well because the tap would not be constantly engaging the work. A high spiral tap will always have teeth engaged in the work. And the purpose of the taps is to eject chips out of the hole, not push them ahead like a spiral point or let them pack the flutes like a hand tap. See this link: http://www.yamawa.com/product/tap.html Eric
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