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Help with my Non-Standard Gears.

  1. Jan 11, 2013 #1
    This may be simple to some, but this is my first experience with designing gears, so bear with me...

    I'm trying to "cut" a spur gear to match the profile of the pinion gear I designed. This gear won't carry any load, it's basically for esthetics, but I want good contact between the two gears.

    I have an 8 tooth pinion & 60 tooth spur gear.
    Here is my pinion gear:
    untitled_zps3f730a44.jpg

    I am working with one tooth profile and trying to simulate it moving out of the mating gear. Seems simple enough...

    This is the shape of one tooth, and the line/arc segments are something I drew that approximates what I should end up with (I used the assembly and adjusted the points until it was close).
    Untitled_1_zpse920e356.jpg

    I could just use those line segments, but I really want to figure this out mathematically.

    This is what I figured should be right, but obliviously it's not...
    -I drew a circle around the main gear with a radius of the distance between my gear shafts.
    -Keeping the center of the pinion gear tooth coincident with that circle I rotated it 5 degrees (I plan on making this increment smaller once I get the calculation figured out).
    -Then I rotated the tooth around the center of the spur gear at the increment that it would rotate if the pinion rotated that 5 degrees.
    That was .6666666667 degrees. My math:
    60 tooth spur - 8 tooth pinion
    1 rotation = 7.5 rotations
    360 degrees = (360 * 7.5) 2700 degrees
    2700 / 5 (sketch increment) = 540
    360 / 540 = .666666667

    I'm sure I’m over simplifying this because here is my result:
    Untitled_2_zpse7eda389.jpg
    The red line is the line/arc that represents where I should be.

    I know it's really cluttered, but I figure once I get this figured out I can just use this file as a template.

    So obliviously .66666667 is too much of an increment.

    Just playing with the numbers I found .59 is really close to what I need. But I don't know how to get there mathematically.
    Untitled_3_zpscdeb415c.jpg

    Does anyone know if I'm just completely over simplifying this, or am I doing something wrong in the math????

    Please help!!

    Thanks,
    Bill

    Not sure if it will work but I also attached a .gif depicting what I'm trying to accomplish.
    generation.gif
     
  2. jcsd
  3. Jan 13, 2013 #2

    mfb

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    Staff: Mentor

    We had a thread about non-standard gears (even more exotic) a while ago in this forum, but I cannot find it any more :(. One user presented a software to design those gears in that thread. Maybe you can find it with the search function ("non-standard gears" and "gear software" and 1-2 other queries did not work).
     
  4. Jan 16, 2013 #3
    If anyone has a link to this discussion, I would greatly appreciate them posting it. I've done extensive searching and can't seem to find it. :grumpy:
     
  5. Jan 16, 2013 #4

    mfb

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    Staff: Mentor

  6. Jan 16, 2013 #5
    The first two figures in this paper are about asymmetric gears. It might be helpful.
     
  7. Jan 17, 2013 #6
    Awesome mfb, thanks for that. I emailed the developer, hopefully something will come of it.
     
  8. Jan 23, 2013 #7
    Well I've figured it out. So for anyone else trying to draw odd gears here it is. And it's rather simple...

    The Movement of the Pinion Tooth Around the Main Gear is this:
    360 / Nm = Tm (Nm=Number of Teeth on Main Gear, Tm=Angle Between Teeth on Main Gear)
    Tm / X = M (X=Increment Factor, M=Increment Around Main Gear)

    And for the Pinion:
    360 / Np = Tp (Np=Number of Teeth on Pinion Gear, Tp=Angle Between Teeth on Pinion Gear)
    Tp + Tm = Y
    Y / X = P (X=Increment Factor, M=Increment Around Pinion Gear)

    So from my original post:
    Main Gear is 60t, Pinion is 8t
    360 / 60 = 6 / 12 = .5 (I drew 20 teeth and linked them, and with the increment factor set at 12, the 20th tooth was just outside the main gear max OD, so I had plenty of steps inside to cut a good profile)
    360 / 8 = 45 + 6 = 51 / 12 = 4.25

    As I drew it, each tooth is rotated around the center of the main gear .5 deg.

    Then the C/L of each tooth is rotated 4.25 deg. from the last.

    That's it plain n simple. I've already made a few gears using this and it works every time.
     
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