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Ball Drive Ratio
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Thread: Ball Drive Ratio

  1. #1
    Junior SCH Member
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    Ball Drive Ratio

    I've calculated this too many times and I have to be doing something wrong. The ratio I've read is 4.44:1. Measuring the balls at 1.0625" and the minumum diameter of the output shaft at 0.940" and assuming no slip, the balls will rotate against the outer race and rotate the output only 3.26 times. What am I missing?

  2. #2
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    What you are measuring is not the gear ratio. To quote the VS 57 manual:

    "The planetary drive system, incorporated between the input and output shafts, is the second point of rpm step-up. The ratio of the system is a constant 1 to 4.44 and does not vary under any operating conditions."

    http://www.studebaker-info.org/VS57/vs57s21.jpg

  3. #3
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    It seems the areas of your measurement were only a little off.

    http://www.studebaker-info.org/VS57/vs57s61.jpg

    This link gives the areas of the ratio:

    Inner contact surface of the ball races versus outer contact surface of the output shaft.

  4. #4
    Junior SCH Member
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    I may not have stated it correctly, but your links confirm my findings. The inner contact surface of the ball race is the contact surface of the output shaft plus the diameter of two balls. Otherwise, they don't touch. Interpeting the text in other ways makes the drive ratio even lower. The maximum ratio will be from the smallest diameter of the sun (output shaft) as compared to the largest diameter of the ring (ball race). The satellites (balls) must contact both the sun and outer orbit (ring). I have seen a formula for ratios of planetary systems that is the ring/sun+1, but in setting this up and physically rotating the system, the extra revolution is not present.

  5. #5
    Junior SCH Member
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    Physical measurements
    Contact diameter of output shaft 0.940"
    Contact diameter of ball race 3.065"

    3.065 vs 0.940 = 3.26

  6. #6
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    jim, I do not doubt your measurements.

    Also I do not doubt the engineers at Paxton, and C.C. at Paradise Wheels. When they tell me it's 4.41 I have to go along with them.

    Z. Ray

    '65 K code

  7. #7
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    Well I am stumped. I got out there and finished disassembling mine and got the same measurement of .940" Dia. and adding the 2 ball diameters gets 3.065" Dia. Sorry - I was hoping there was something you missed.

    You rotating it manually and counting the number of turns verifys your findings.

    It makes me wonder about the maximum number of RPM's it can handle and calculating them now.

  8. #8
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    I'm wondering the same things. lol

    I did finally figure out how the extra 1 rev gets into the formula. I forgot you get 1 "free" because of the 1 rev of the ball driver during the orbit of the balls. So now I'm up to 4.26:1 which I was able to confirm by adding some tension to the races. Evidently, my mockup outside of the SC case was too loose to properly rotate the output shaft. So, at least it's a small difference now. But for the life of me, I can't get that last 4% unless it's in the gap between the races not driving the ball on its full diamter. Maybe that results in some mechanical advantage? Sort of like flying Northeast out of New York and arcing around to Southeast is the shortest route to London.

    While I know in my mind that Paxton HAS to be right, it makes me crazy to not be able to come up with their numbers.
    Last edited by jlmccuan; 10-06-2009 at 03:39 AM.

  9. #9
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    OK, if I take a used race and look at the contact area, the ball doesn't touch at the edges. Measuring the approximate middle of the contact path across to the middle of the contact patch on the opposite side of the race the "contact diameter" is close to 2.9". Very hard to measure this accurately. If, in fact, this diameter of contact is what is at play here, our formula must factor in this advantage. Running the numbers:

    Working diameter of the ring would be 3.065"
    Contact diameter 2.90"

    Ratio of 3.065 vs 2.90 = 1.056:1

    Diameter of ring 3.065"
    Diameter of sun 0.940"

    Ratio of 3.065 vs 0.940 = 3.261:1

    3.261 * 1.056 = 3.44

    Add "free" revolution from driven input
    3.44 +1 = 4.44

    YEAH, I can go to bed now.
    Figure 20 minutes to get shower and in bed and it's appropriately 4:44AM

  10. #10
    Senior SCH Member
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    THANK YOU! I was hoping for something like that but was unable to get the math to add up properly. I lost the sleep measuring and taking apart my unit the rest of the way. I had to finish taking it apart anyway.

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