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Chain Conveyor Speed Calculation

  1. Mar 22, 2017 #1
    Okay this has has been driving me crazy.

    I would like to know how a conveyor speed can be calculated on the basis of chain pitch, sprocket PCD & no. of teeth and output RPM of gearmotor.

    So here is a practical example I'm having to solve at the moment:

    I have to drive a conveyor at a speed of 760mm/min. To drive this, I am using a drive system which consists of a small sprocket mounted on the output shaft of the gearbox. This sprocket is connected to a larger sprocket by means of a chain.

    A shaft from the larger driven sprocket is connected to the main conveyor sprocket on which the conveyor chain is mounted. This chain needs to be driven at 760mm/min or in other words, the part at one station needs to travel to the next station 760mm away in one minute.

    I have enclosed a sketch of this arrangement for reference.
    https://ibb.co/hyhJ5a

    I am going to use a VFD to achieve the required speed. Available specifications:-

    Motor Speed - 915 RPM (6-pole)
    Gearbox Reduction - 73.6 (this I am yet to finalize, Can go for a different reduction ratio if required)
    Output RPM - 12.43
    PCD of Small drive sprocket - 106.14mm
    No. of teeth on small drive sprocket - 13
    PCD of large driven sprocket - 461mm
    No. of teeth on large driven sprocket - 57
    PCD of conveyor sprocket (mounted on same shaft as driven sprocket) - 598mm
    No. of teeth on conveyor sprocket - 74

    Chain Pitch - 1"

    Any help for this would be highly appreciated!

    Thanks,

    KV
     
  2. jcsd
  3. Mar 22, 2017 #2
    If your drive speed is exactly constant, you will not get exactly constant belt speed due to the slight variations introduced by the variable radius of the chain sprocket. For most purposes, this can be ignored. After that, just treat it like a belt drive, taking ratios of the sprocket radii to determine the speed ratios.
     
  4. Mar 22, 2017 #3

    CWatters

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    If I've understood the set up correctly I get quite a different value for the gearbox ratio...

    Looking at the conveyor sprocket. The PCD is 598mm so the circumference = Pi * 598 = 1878mm. (1rpm would move the belt at 1878mm/min which is too much, so we expect the required rpm to be <1rpm).

    The required velocity at the pitch diameter is 760mm/min so the required rpm of the conveyor sprocket is..

    = 760/1878 = 0.405 rpm

    The large driven sprocket is mounted on the same shaft as conveyor sprocket so it must also turn at 0.405 rpm.

    The required rpm of the small sprocket can be calculated from the ratio of the pitch diameters..

    = 0.405 * 461/106.14
    = 1.76rpm

    Using the tooth ratio should also give a similar answer..

    = 0.405 * 57/13
    = 1.78rpm

    If the motor turns at 915rpm I calculate the gearbox ratio required is about

    915/1.78 = 514

    Somewhat different to the 73.6 you have. Perhaps I made a mistake but I can't see it.
     
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