# Calculating worm gear TPI and diameter



## Runner (Feb 23, 2012)

Hi all,

This is only an exercise. I have spare gear wheels for my 9" Southbend Type C Lathe. I want to cut a worm gear that would mesh with a gear wheel and I am unsure of the diameter and the TPI that the worm gear would need to be. For ease of presentation lets say that the gear wheel has 44 teeth, has a major diameter of 2.55" and a root diameter of 2.3", it has straight cut teeth and 0.25 " thick. Using schoolboy arithmetic the mean diameter of the gear wheel is 2.425". The circumference is (22/7 X 2.425) 7.62" and that the pitch of the teeth is (7.62/44) 0.173". The TPI therefore (1/0.173) is 5.78. This would be the same for all the gear wheels irrespective of size. I can cut either 5.5 TPI or 6 TPI, which would be better suited to mesh with the gear wheel? That is the easy part. The diameter of the worm gear is beyond my ability to determine. The parameter to consider on the worm gear is angle of the helix (I don't know if I am using the right terminology) with a small TPI it is quite large, so with a straight cut gear wheel which is relatively thick (0.25") there is going to be some interference between the worm gear and the gear wheel. The way to reduce this is interference is to have the diameter of the worm gear large, effectively reducing the angle of the helix and/or widening the gap between the teeth of the worm gear. How do I calculate the diameter of the worm gear that would mesh reasonably well with the gear wheel?

Thanks in advance.

Brian


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## steamer (Feb 23, 2012)

The other alternative would be to mount the worm gear at the helix angle with respect to the gear.

Sounds like a SB version of Geoge Thomas's lathe mounted dividing head.

What is the diametral pitch of a South bend change gear?.....I want to say 18....but I'm not sure.

Dave


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## mklotz (Feb 23, 2012)

OD of gear [2.35 in] ? 2.55
Number of teeth [45] ? 44

Diametral Pitch = 18.0392
Module = 1.4080
Number of teeth = 44
Outside Diameter = 2.5500 in = 64.7700 mm
Pitch Diameter = 2.4391 in = 61.9539 mm
Addendum = 0.0554 in = 1.4080 mm
Dedendum = 0.0685 in = 1.7405 mm
Whole Depth = 0.1240 in = 3.1485 mm
Circular Pitch = 0.1742 in = 4.4235 mm
Tooth Thickness = 0.0836 in = 2.1233 mm

B & S cutter number used to cut this gear = 3

Yup, looks like it's 18.

Brian,

Note in the calculations above that the circular pitch of 0.1742 in = 5,74 tpi supports your calculations.


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## steamer (Feb 23, 2012)

And might I mention Marv has some really cool software on his web site.... ;D

Dave


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## Mainer (Feb 24, 2012)

I'm pretty hazy on this. I don't think wormgears use standard diametral pitch measurement. They use another system that includes pi as a factor. Maybe somebody else can give the details.


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## Runner (Feb 24, 2012)

Thanks for your responses.

The problem is more acute than I first realised, the thickness of the gear wheel is 0.4375" not 0.25". I decided to try and draw the meshing arrangement and see if that produced anything close to a solution. Using 6TPI worm gear derived above. The footprint of a straight cut gear wheel tooth is 0.4375" X 0.0349". If this is to sit in the helix of the worm gear, the worm gear cannot have a symmetrical thread form, i.e. the root width must be larger than the crest width. Hope I am making sense. Let's assume a 1:2 ratio (crest/root). It gets complicated here because the thread form is not square more like an ACME. The root width could be 0.12975". The helix angle would be the tan of 0.04325/0.4375 = 0.0989. Don't have a set of tables, but it's about 5.5 degs. But doesn't matter we are going to use the tan of the angle to determine the diameter. The find the diameter of the wormgear Tan 5.5 degs = 0.173/dia. Therefore dia = 0.173/0.0989 = 1.5". This is much larger than I had hoped. That's assuming a fully engaged thread, but as Dave has intimated it is only to drive a dividing head, so meshing could be partial. 

If I have given the impression that I have solved my own problem and readers need not provide further response, then this is not the case. I have used what little I know to only give a ball park indication, my solution contains approximations and the math maybe incorrect.

Brian


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