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Incremental Encoders

Discussion in 'Homework Help' started by eulerpi, Jul 15, 2007.

  1. eulerpi

    Thread Starter New Member

    Jul 15, 2007
    Two incremental encoders are set up as a Vernier. Both send out 500 pulses per revolution (no quadrature). Encoder A is geared such that 49 turns of the motor result in one turn of Encoder A.

    Encoder B is similar, only geared 50:1. Initially, the two index (AKA home) signals are synchronized. After the system has been running awhile, Encoder A sends a home signal, 30 Encoder B pulses later (constant direction), Encoder B sends a home signal.

    How far has the MOTOR turned??

    I came up with 120 Rev. but that seems like to many.

    [360deg/rot] / [3deg/count] = 120

    I do not think this is correct but I gave it a shot.

    Thanks in advance for any help !!! :)
  2. hgmjr

    Retired Moderator

    Jan 28, 2005
    See subsequent posting for my self-correction to the above statement.

  3. JoeJester

    AAC Fanatic!

    Apr 26, 2005
    The question ... How far has the motor turned? ... is that relative since the begining or relative since the encoder A index signal?

    120 revolutions is way too small if it's just from receipt of A's index to receipt of B's index.
  4. hgmjr

    Retired Moderator

    Jan 28, 2005
    Ooopppsss. I see where I went off the rails. I had the gear ratio reversed on each of the motors. It is 49 motor revolutions per encoder revolution and 50 motor revolutions per encoder revolution on motor A and B respectively.

    With that cleared up, the question boils down to how many encoder pulses does motor B lose each time motor A generates a home pulse.