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Thread: Starting Torque vs Running Torque in DC Motors

  1. #1
    Associate Engineer
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    Starting Torque vs Running Torque in DC Motors

    Hello,

    I am trying to find some information about how the starting torque changes with relation to the running torque.
    I have an application with low duty cycle that requires the motor to operate more towards the starting torque side than the running torque.

    What I would like to find out is the following:
    1) Is there a dramatic drop in torque from starting to running or is it a gradual decrease?
    2) Is there an equation that relates the two torques?
    3) What are the different variables I should be considering? (speed, current, load ...)

    Motor is a brush DC motor.

    Thanks!

  2. #2
    Technical Fellow
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    Welcome and here's the standard first reply.

    More information please. Aaarrghhhh!

    Volts?
    Amps?
    HP?
    RPM range?
    Opening a six-lane draw bridge or moving a hand puppet on a string?

    Depending on load at start-up, the torque-drop may be negligible or very significant.

  3. #3
    Principle Engineer
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    1) T=K1*I

    2) V0=+LI'+RI +K2*@'

    3) J@"+c@'=T-Tl

    V0 impressed voltage
    I=current
    T= torque delivered by motor
    Tl = load torque
    K1=torque constant
    K2=back emf constant
    J= polar moment of inertia
    @ = rotor angle
    @' = rotor speed
    @"= rotor acceleration
    I' rate of change of current
    L armature inductance
    R= armature resistance
    c = damping coefficient

    These are the equations that govern a DC motor and are readily solvable by eliminating T in eq 1 and 2
    Then you have 2 remaining equations in I and @ and then eliminate @ resulting in a single linear differential equation in I.
    To give you an idea of what happens when you throw the switch, look at eq 2 where the LI' would only allow a smooth riise of current. As I rises, T increases untill it overcomes Tl and accelerates the mass increasing the back emf which slows the rate of riise of current. At steady state T=Tl+c@' and V0=K2@'+RI
    Last edited by zeke; 09-18-2012 at 09:14 AM. Reason: correction for steady state

  4. #4
    Administrator Kelly Bramble's Avatar
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  5. #5
    Associate Engineer
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    Thank you everyone! This is a big help. 12VDC, 7 Amps, 30 RPM nominal, 80 in-lbs nominal torque, 450 in-lbs starting torque, ~100 Watts - moving a 90 pound load with 8" lever from ground straight up, like dead lift.

  6. #6
    Technical Fellow
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    With that kind of load, you may be better off looking at an electro-clutch and leave the motor running. Or if infrequent operational use, start the motor with delay relay, then engage the clutch when it has stabilized.

  7. #7
    Principle Engineer
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    8" lever x 90 lb weight comes to 720 lb-in
    How does this square with your starting torque of 450 lb-in?.
    Also, to what height is the weight lifted?
    What is you power source?

    Since ,as you say , this is low duty cycle, why do you care about the torque characteristics?

    I'm a bit uncomfortable with using such a large lever for the lift.

    If you post a description of the problem, maybe this forum can give you better options.

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