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    Default Cutting Gears

    Hi all, I am going to attempt to cut some small gears but do not know which cutter to use here is what I have.
    40Tooth 18mm round.
    Thanks Pat
    Attached Thumbnails Attached Thumbnails pl-570-gear.jpg  

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    What's the pressure angle? 14.5 degrees? 20 degrees? Eyeball gauge suggests it's 20 degrees, but you better know for sure before you start cutting anything.

    Given that the dimensions are metric, the only way that drawing makes sense as a standard pitch is if 18mm is the pitch diameter, not the outside diameter, and if it's 0.45 module. If it's a bastardized drawing of an inch part in metric units, it might be 64 DP rather than 0.45 module, but I'd think that even a sloppy draftsman would have not written 18.0mm in that case.

    So you would select an involute gear cutter of 0.45 module and the appropriate pressure angle. Typically, those parameters give you a set of eight numbered cutters. To cut 40 teeth, you want the #6 [edit: original was #3] cutter, which is appropriate for cutting 35-54 teeth.

    [Added in edit] Actually, I am making a major assumption that this is an involute tooth form, which is almost universal. If these gears are for clock mechanisms, they might well be cycloidal tooth form. Better know that before cutting anything...

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    Excuse me, the Module cutter sets are numbered backwards to the Imperial set; please check that Module Cutter #6 would be correct. Most module gears are 20 degree PA, although they don't have to be, this is the most common. Example: KBC,.75 NO.6 20 DEGREE MODULE GEAR CUTTER,1-202-010-6,KBC Tools & Machinery

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    Ah, thank you Video_Man, I was not aware of that the metric cutter sets use #8 for the rack, where the DP sets use #1 for the rack.

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    Quote Originally Posted by sfriedberg View Post
    What's the pressure angle? 14.5 degrees? 20 degrees? Eyeball gauge suggests it's 20 degrees, but you better know for sure before you start cutting anything.

    Given that the dimensions are metric, the only way that drawing makes sense as a standard pitch is if 18mm is the pitch diameter, not the outside diameter, and if it's 0.45 module. If it's a bastardized drawing of an inch part in metric units, it might be 64 DP rather than 0.45 module, but I'd think that even a sloppy draftsman would have not written 18.0mm in that case.

    So you would select an involute gear cutter of 0.45 module and the appropriate pressure angle. Typically, those parameters give you a set of eight numbered cutters. To cut 40 teeth, you want the #6 [edit: original was #3] cutter, which is appropriate for cutting 35-54 teeth.

    [Added in edit] Actually, I am making a major assumption that this is an involute tooth form, which is almost universal. If these gears are for clock mechanisms, they might well be cycloidal tooth form. Better know that before cutting anything...
    This was a diagram done in CAD from someone this is what was on there OD is 18mm and the PA is 20 degress and is an involute tooth form.
    Thanks Pat

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    Video Man, I am new to this what does the .75 mean??
    Pat

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    In the link Video_Man gave, the cutter is for 0.75 module. If you follow the link, it actually says "Module Pitch 0.75" in the little table on the catalog page. 0.75 module is much bigger than the teeth on your drawing, so that's just an example of cutter #6 being the one you'd use for tooth counts of 35 through 54.

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    Quote Originally Posted by pjf610 View Post
    OD is 18mm and the PA is 20 degress and is an involute tooth form.
    Well, either OD = 18mm is a mistake (for pitch diameter = 18mm) or this gear calls for a non-standard, custom gear cutter.

    It is good that you know the pressure angle and tooth form!

    Is this for a turntable?

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    Quote Originally Posted by sfriedberg View Post
    Well, either OD = 18mm is a mistake (for pitch diameter = 18mm) or this gear calls for a non-standard, custom gear cutter.

    It is good that you know the pressure angle and tooth form!

    Is this for a turntable?
    Yes the gear motor for a pioneer PL-570

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    Quote Originally Posted by pjf610 View Post
    Video Man, I am new to this what does the .75 mean??
    Pat
    That's the module of the gear, which is a measurement roughly similar to diametral pitch in Imperial gears. (Think of it as being "how many threads per mm." in a tap.) And... the example is not the actual cutter you need, it is to show the number of the cutter for the number of teeth it can cut. I was making the point of the way module cutters are numbered, because the kind response of sfriedberg might have needed a second look on that issue.

    Oddly, metric cutter sets are numbered inversely to the numbers of Imperial sets, that is, 1-8 in Imperial starts at #1, cuts a rack and many small teeth, and ends at #8 with a few big teeth, and module sets start with few big teeth and work to the #8 cutter, which cuts a rack and small teeth.

    To buy a metric cutter, you need to know the module, pressure angle, and number of teeth to be cut. As sfriedberg accurately points out, you seem to have a non-standard gear or a defective drawing.

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    Your post and drawing both show a gear with an 18mm outer diameter and 40 teeth. If (and I did say IF) your gear has standard form teeth and not stub sized teeth, which are smaller, then we can add 2 to the tooth count for a gear with 18mm as the pitch circle diameter. So the formulae for such a gear would give either a metric module:

    M = 18 / 42 = 0.42857...

    That, obviously is not a round number nor very close to one. So it is not a standard module metric gear.

    18 mm = 0.70866" and the DP, which is usually used for English measure gears would be given by:

    DP = 42 / 0.70866 = 59.26666...

    Again, that is not a round number. Therefore it is not a standard DP, English measure gear.

    If it does have stub teeth, then the most common factor to add to the tooth count is 1.6. For your gear, this gives us:

    M = 18 / 41.6 = 0.43269...

    or

    DP = 41.6 / 0.7066 = 58.70222...

    Once again, neither of these are round numbers.

    My conclusion is that your gear is a non-standard pitch. This means that you are not going to find a commercially available cutter and will have to MAKE ONE.

    I welcome anyone who wishes to check my math.

    As a suggestion, I have seen parts for this Pioneer turntable for sale on the web and I would suggest contacting some of these sellers and asking about your gear. You may pay more for shipping than for the gear itself. Or you may be able to just buy a "for parts" turntable and make one good one from the two. Then you can sell the excess parts and probably make a few dollars on the deal.



    Quote Originally Posted by pjf610 View Post
    Hi all, I am going to attempt to cut some small gears but do not know which cutter to use here is what I have.
    40Tooth 18mm round.
    Thanks Pat

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    Without knowing the centre to centre distance and dimensions of the other gear its playing in the dark


    Peter

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    Hi all, I would like to thanks everyone for their help.

    Pat


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