I have a Skywatcher 130/900 and I found that if I do an accurate polar alignment I'll need only one motor to track a celestial body, so I decided to build my motor. Right now I'm facing a problem with the mating of the spur gear that drive the worm gear

enter image description here

Using a caliper I found that the gear has the outer diameter of 50mm, and a tooth is 1mm in height, so the reference diameter is 49mm; the gear has 140 teeth, and so the module (reference_diameter / n_of_teeth) is 0.35.

After this I used Solidworks to model a spur gear that could match it, and printed this

enter image description here

When I tried to check the mating, I found this

enter image description here

The teeth seems right, but they're too short and can't mate.

I used a caliper and counted the teeth to obtain the gear specs, and I think is kinda stupid as approach, but I could not think of something better to mate my gear. Is there a better approach to this?

Then, the idea was to try and remove the gear (removing the screw) and replace it with a 3d printed gear, but it does not seems to work (and I think it could break the telescope mount). Did someone try to do it?

Searching online I found that (obviously) there are some tracking motors and another question came to my mind: how can they mate the gear? Is there some kind of standard? Does it mean that every telescope's mount has some kind of standard gear? If yes, where can I found them?

EDIT: That's the situation: on the right there is the spur gear that I'm trying to mate (or eventually remove, if possible) and in the center there is the worm gear that move the telescope on the RA

enter image description here

EDIT2: Here's the project of an involute gear with Solidworks. (I followed this tutorial) but as you can see, the teeth overlap.

enter image description here enter image description here

As a reminder, I have the following parameters

Pitch diameter (PD): 49mm (caliper measurement)
Number of teeth (Z): 140
Module (PD/Z): 0.35

Am I doing something wrong (or am I missing something)?

  • $\begingroup$ If the outside diameter is 50mm, and the teeth are 1mm, wouldn't the reference diameter be 48mm? $\endgroup$
    – Dr Chuck
    Jan 8, 2017 at 13:14
  • $\begingroup$ @DrChuck No, that would describe the base circle (at the base of the tooth), instead the reference diameter is the one that define the pitch circle (the one that define the distance between teeth). Have a look at this as a reference. So, assuming that the pitch circle is at half the height of a tooth, it gives: 50 - 0.5*2 = 49 $\endgroup$
    – igng
    Jan 8, 2017 at 13:23
  • $\begingroup$ I'd ask the folks over at 3dprinting. I'd assume that even if your printer is capable of the required resolution, that the teeth would be too fragile. - Use a file and increase the dedendum by hand? Make it too big from the start and then handcraft the whole thing? $\endgroup$
    – Mazura
    Jan 8, 2017 at 23:45

2 Answers 2


Gear teeth usually have a special shape which is either an epicycloid or involute curve. Making worm gears is even more complicated because the teeth have a helical curve in addition to having an involute profile.

Making good-quality gears requires specialized tools, knowledge and equipment.

  • $\begingroup$ Yeah, I tried to make an involute one but teeth were way smaller (I'll try to add a picture later). Btw I do not want to make a worm gear, it is already there but at the end of the metal bar there is that metal spur gear and I'm not sure that I can replace just the gear it without breaking the mount $\endgroup$
    – igng
    Jan 5, 2017 at 0:24
  • $\begingroup$ Making a gear with a screw: google.com/… You need to get the diameter exactly right, or you get a mess. $\endgroup$ Jan 5, 2017 at 15:42

Ok, there is a lot of calc's and table references to go through. Punching in the Numbers from above into s Sheet I had done during my Advance diploma days.

Gear Pitch circle diameter = 49.3 ( note this is the OD - Addendum) Number teeth = 140 Module m = 0.352 (same as Addendum) = 49.3/140 Whole depth = 0.759264 (Addendum + Dedendum) Clearance = 0.055264 Dedendum = 0.407264 Outside Øg = 50.0 Root Øg = 48.49

All this information makes up the involute tooth profile for your gear. the rootØ is the absolute bottom of the gear tooth.

I hope some of this makes sense, there are some online references, i got mine from my work books going back a few years. Note: i used metric calcs.

  • $\begingroup$ Welcome to astronomy SE! It is nice that you try to answer the question, but in its current form, it is not too accessible or readible. It would be great if you could edit your answer following astronomy.stackexchange.com/help/how-to-answer $\endgroup$
    – B--rian
    Feb 24, 2021 at 8:10

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