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  • Formula to calculate gear train dimensions

    Discussion in 'Calculations' started by Crossup, Sep 20, 2019.

    1. Crossup

      Crossup New Member

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      I'm working on a project for model airplane propulsion.

      I'm trying to design a friction drive gear train which will provide power to coaxial propellers. The chosen solution is to use a drive pinion parallel to the propeller drive shafts.

      This means one driven gear(wheel) will be driven on its OD and the other on the ID. My problem is since the driven gears(wheels) are coaxial and I need different ratios I need to figure the diameter of one driven wheel given the diameter of the other wheel and pinion step. Pinion may need to be stepped so that it has diffenent diameters to provide the desired ratios. So the formula will need to also calculate the pinion size for the unspecified driven wheel.

      My math skills are not up to translating the variables into a formula to solve for them. While I would prefer "hints" as to how to develop such a formula, perhaps its too simple for a "real" engineer to help with beyond just giving me a formula. I suspect, however, that it involves solving simultaneous equations as the results will be a single solution for any given ratio and size of the gear that is initially specified.

      The picture shows the two driven wheels and the black pinion, you can see I have not properly sized the pinion so it does not "fit". Of course the smallest part of the pinion represents the motor shaft the pinion would be on.
       

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      Last edited: Sep 21, 2019
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    3. Crossup

      Crossup New Member

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      To help clarify the details here is an example: I'm wanting the output shafts(which are coaxial) to run at a ratio of 14:15(.9333)(doesnt matter which is which as we can just reverse the gear sets position(front or rear) but the end result needs to be the front prop runs slower than the rear by that 14/15 ratio. I'm going to be using a motor to prop drive ratio of approxmately 1:15 motor to output shaft, so we can choose one gear set to be 4mm pinion to 60mm driven for 1:15. The needed formula would tell us the diameter of pinion and driven wheel for the other gear set given a specified front/rear differential as the 0.9333 is subject to a number of factors which may require tweaking to optimize performance.

      Back to the numbers: so in my example .I specify a ratio of 0.0666(1:15) and a differential speed ratio of .9333 so the other gear set needs to produce a ratio of .0622 rounded off. So the real question is how to calculate the diameters of pinion and driven wheel that produce that ratio(.0622) while allowing them to have the same shaft spacing as the other gearset.
       
    4. Crossup

      Crossup New Member

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      Ok, I figured it out. It does take simultaneous equations.
      I took the two axis centerlines to get a distance I can use for calculating the second pinion/wheel set, which is z=1/2x + 1/2y where x is the pinion dia. and y the driven wheel so 2z=x+y.
      Keep in mind the second pinion /wheel set are different values for x and y than what I pick to start with for the first pinion/wheel.
      . So 2Z is known since we picked the first gear set, in my case x=3 and y=42 so 2z= 47. And the desired ratio, in this case is 1:15 gives us 15=y/x So solve 47= y+x and 15=y/x and the answer is X=2.9375 and Y=44.0625
       
    5. s.weinberg

      s.weinberg Well-Known Member EngineeringClicks Expert

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      Not 100% sure I get what you're after, but, in case I'm right:

      It makes no difference to the speed whether you're driving a wheel internally or externally. It affects only direction (assuming driving by external of pinion, external would reverse direction, internal would be same direction as pinion).

      Your speed ratio between your pinion and any wheel is Dwheel/Dpinion (D being diameter, and with the larger being slower).

      That's your hint.

      As far as a specific answer, will depend on how your pinions vary. However, for a single pinion size, your speed ratio between wheels is just the ratio of diameters. If one is 15/14 times the other, it will move at 14/15 the speed with the same pinion.
       

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