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Thread: A task on synchronization of two gears (2 x AC drives; 1 x encoder).

  1. #11
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    Guys, what about understandable answer on my #7 post?

    regards.

    Chelny

  2. #12
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    180212-1225 EST

    Chelny:

    I believe some responders have not understood what you want to do.

    I believe what you want to do is to adjust rotation of a pair of gears, that are not yet engaged (teeth not touching), in such a manner that while rotating you can bring the gears together in mesh (no clanking, no teeth hitting on top of one another, and no tooth damage).

    To do this means you need position information from both shafts.

    1. Use a position encoder on each shaft, or
    2. Use photocells or prox detectors on each gear to determine tooth relationship, or
    3. Use a vision system looking at the tooth relationship.

    .

  3. #13
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    Quote Originally Posted by gar View Post
    180212-1225 EST

    Chelny:

    I believe some responders have not understood what you want to do.

    I believe what you want to do is to adjust rotation of a pair of gears, that are not yet engaged (teeth not touching), in such a manner that while rotating you can bring the gears together in mesh (no clanking, no teeth hitting on top of one another, and no tooth damage).

    To do this means you need position information from both shafts.

    1. Use a position encoder on each shaft, or
    2. Use photocells or prox detectors on each gear to determine tooth relationship, or
    3. Use a vision system looking at the tooth relationship.

    .
    Sorry, I didn't get that you were going to be meshing two gears together that are already turning.

    So I agree with gar, although I think the effectiveness of #2 is likely untenable without some precision in the smaller drive, which then comes back to #1. It's one thing to count teeth with a sensor, it's another thing to know WHERE in the rotation those teeth are and for that, you will need an absolute encoder feedback on both shafts.

    The vision system idea has merit though. I've never done it, but I've seen demos of it working.
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  4. #14
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    Jraef:

    I have built test equipment for checking the teeth in encoders used in automotive ABS systems. Using the sensor that goes with the encoder gear in the differemtial I was able to get good measurement information within a tooth to tooth cycle. These were generally dv/dt sensors (magnetic and coil) and thus have a phase shift. A Hall sensor would be better in this application. One can definitly get good tooth position information with a magnetic sensor where the speed is approximately constant.

    .

  5. #15
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    Chelny:

    It would be useful if you ask more questions, and provided more information about your gears.

    From 30 year ago memory I believe the following information is about correct for magnet-coil pickup:

    1. At 3 MPH we needed a minimum of about 1 V. Output voltage is proportional to speed. Thus, one uses zero crossing information. 3 MPH = 4.4 feet per second, because 30 MPH is 44 ft/sec.
    2. Tire diameter about 30", and thus circumference = 94" = 7.9 feet.
    3. Thus, axle RPM at 3 MPH = about 1/2 RPS.
    4. Tone ring (gear) had about 100 teeth on a diameter of about 6". Thus, tooth pitch was about 0.25".
    5. 100 teeth at 1/2 RPS is about 50 Hz.
    6. Air gap tooth tip to sensor possibly 0.020".
    7. Output somewhat like a sine wave.
    8. Very reliable except for magnetic metal chips, but you don't want those in a differential anyway.

    .

  6. #16
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    Quote Originally Posted by Chelny View Post
    Guys, what about understandable answer on my #7 post?

    regards.

    Chelny
    If this is still a work in progress (ie.) being designed to suit your specific needs. . . this synchronization dilemma can be addressed far more than what is presented in the drawing. Torque, speed and positioning of the gears and even holding the gears in place when machine is in idle.

    Accurate positioning and easy control by using pulse signals from a controller along with servo to provide feedback signal from the output shaft of the smaller gear.

    Check out stepper motors applications with some high torque in a compact body for ease of integration into your machine.

    Both servo and stepper motor can be mounted at the input and output shaft.

    I've use smaller versions of these in small-scale robotics.

  7. #17
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    As an aside, I assume the bigger gear is the input?
    Since both shafts are marked AC drive, and probably having its own motor. . . an auxiliary planetary gear (not shown in drawing) is where the work is being performed. The two gears are there to perform one common task?

    Some drawings, when it comes to controllable speed motors--the VFD controller is sometimes labeled VFD Drive referring to the controller itself not the motor.

    Is this common in Russia?

  8. #18
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    Quote Originally Posted by myspark View Post
    If this is still a work in progress (ie.) being designed to suit your specific needs. . . this synchronization dilemma can be addressed far more than what is presented in the drawing. Torque, speed and positioning of the gears and even holding the gears in place when machine is in idle.

    Accurate positioning and easy control by using pulse signals from a controller along with servo to provide feedback signal from the output shaft of the smaller gear.

    Check out stepper motors applications with some high torque in a compact body for ease of integration into your machine.

    Both servo and stepper motor can be mounted at the input and output shaft.

    I've used smaller versions of these in small-scale robotics.
    I
    Last edited by myspark; 02-15-18 at 06:39 PM. Reason: missing word

  9. #19
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    it can be done
    engagement speed is a factor
    assume land and tooth the same on both
    both gears turn same speed

    the land has an angle that sweeps across g2
    as the land ang on gear 1 approaches perpendicular to gear 2
    start moving gear 2 torwards gear 1
    time gear 1 land to be perpen to gear 2 when they meet
    gear 1 ang should sweep at least 2 gear 2 teeth
    land vs gear position are arbitrary

    I need to calc
    engagement speed as a function g1 swept ang on g2
    if it misses the first g2 tooth it will snag the next

    need to do math
    too burned out now lol



  10. #20
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    think about this (assuming he is linearly and radially engaging the gears)

    g1 dia = 15, roots = 60, teeth = 40
    g2 = 10, 40, 27
    typically the root is 1.5 x tooth area per rev
    let's say 60% root and 40% tooth
    and a gear ratio of 1.5:1
    assume tooth to tooth vel for g1 is 100
    and root a bit less (smaller dia) 98
    both gears ~ the same since
    vg1 = D1 x Pi x v1 rev/sec
    vg2 = 2/3 x D x Pi x 3/2 x v1
    same

    set the encoder up so it always starts to engage g2 when a root is pointing/perpendicular towards/at g2
    so consider it stationary

    your odds are
    g1 is always 1 or 100% root since it is positioned by the encoder
    g2 root 60% and tooth 40%
    that you hit a root 60%
    that you hit a tooth 40%
    if you hit a tooth good
    if you hit a root we need to adjust, but how do we know? we have no feedback

    you need to make sure that by the time the next root on g1 is 'in position' we have a tooth
    in line on g2

    I assume g2 is driving the load
    and g1 is xmting power to it
    so let us run g2 a bit slower if required
    g1 root to root 98/60 = 1.63, same as g2
    g2 tooth to tooth 100/60 = 1.67, same as g2

    if it is synch'ed will continue to strike at the same point, root-root
    g1's next root will be in position again 1.63 later
    but if g2 is run a bit slower it will have a tooth in position
    root to land delta 1.67-1.63 = 0.04 (using vel, easily converted to time, deg, etc)

    so if g2 is run 0.04/1.67 x 100% = 2.4% slower than g1
    if it doesn't engage the first tooth it will get the second
    speed of linear engagement = tooth depth/0.04 in this case



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