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Need help selecting a motor

Discussion in 'Electronic Design' started by Ed, Aug 19, 2005.

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  1. Ed

    Ed Guest

    I am building what is essentially a motorozed clothesline. The length
    of the clothesline is 5 ft long (~10 feet of rope).

    I intend to connect a DC motor (reversible) to the axle of one of the
    spindles to make the clothesline move forward and back. The
    clothesline will not have more than one item on it and the items will
    weight less than 1 pound. The item on the clothesline should be able
    to move from one end to another in about 4 seconds. The diameter of
    the spindle is 2.25 inches. As such, the RPM needs to be about 127 RPM
    by my calculations. The motor need to be small and light (less than a
    pound, hopefully just a few ounces) and hopefuly less than $20US, and
    can be up to 12VDC.

    I tried this with a motor I removed from a tape deck, and it did not
    have nearly enough torque to get the job done.

    My questions are:

    1) Does a motor exist that fits my requirements? What torque do I
    need?
    2) Where can I buy just one?
    3) How can I best connect the spindle (which has a .25 hole in it) to
    the motor shaft?
    4) Is there any type of motor I should use or stay away from (stepper,
    etc)?

    Thanks in advance for your consideration.

    Ed
     
  2. Pooh Bear

    Pooh Bear Guest

    Have you tried improvising using say an electric screwdriver motor ? It's
    more likely to have the kind of torque you need.

    Graham
     
  3. Sounds like a job for a gearbox motor. Perhaps an old windshield
    wiper motor.
     
  4. Sam

    Sam Guest

    Try a motor out of a power window from a auto wrecker. Go for the passenger
    side as it has had less wear and tear.

    Cheap and with the torque and speed you need.

    Cheers.
     
  5. Ed

    Ed Guest

  6. redbelly

    redbelly Guest

    If you use different gear sizes on the motor and spindle shaft, then
    other RPM values could work as well.

    Mark
     
  7. The windshield (UK = windscreen) wiper and window motors would be too
    slow to meet your 4 sec requirement. (Unless, of course, you increased
    the diameter of the driving pulley.) The screwdriver motor suggested
    by Graham should be fine though. Mine run fast and with good torque.

    Your rpm calculation is just a little high. The required speed would
    be (4*15*12)/(pi*d*n), where d = diam in inches and n = rpm.
    So n = 720/(pi*d) = 720/7.068 = 102 rpm

    I'd try these methods to connect the cord:

    1. Tension it with a small spring (trial and error) and see if
    friction proves adequate.

    2. Extend the driving spindle if necessary and wrap 17 turns of the
    cord neatly around it. With the clothes line at its mid-point (i.e.
    with your imaginary wet shirt or whatever hanging 2.5' away), you
    should then have 8 loops symmetrically on either side of the middle
    loop. Secure the middle loop to the spindle at one central point.
    (Several ways of doing that. A simple staple would do.) If you then
    turn the spindle 8.5 turns in one direction, the line will move
    (8.5*2.25*pi)/12 = 5 feet, and all the loops will now be to one side
    of the fixed point. For robustness, I'd make that 19 not 17 loops, so
    there would be less strain on the fixing. If the cord is say 1/6 inch
    (4mm) in diameter, the loops would take 19/6 = 3.2", so a 4-5" spindle
    should be OK. I used the same approach in my curtain opener, albeit
    with only 7 loops on a 1.75" spindle:
    http://www.terrypin.dial.pipex.com/Images/CurtainControlMechanics.jpg

    The method of connecting the motor to the spindle will depend on what
    motor you use. In the case of my ex-screwdriver geared motor, it came
    with a removable double-ended 'bit'. I made the spindle from doweling,
    drilled an undersized hole, and epoxied the bit into it.

    Improvisation is the name of the game <g>.
     
  8. redbelly

    redbelly Guest

    I get 127 rpm, same as Ed. It seems you are using 4 feet for the
    length of rope, not 5.

    The linear speed is:

    [ Length of rope / time to travel rope ]
    = [ wheel circumference x revolution rate ]

    In terms of inches and seconds (converting from feet and minutes):

    (5 x 12 inches) / 4 sec = (pi x 2.25 inches) x (RPM / 60 sec/min)

    or

    RPM = (5 x 12 / 4) / (pi x 2.25 / 60)
    = 15 / 0.1178...
    = 127

    Mark

    (Not that it matters all that much :)
     
  9. You're right, my mistake, sorry. 4 *seconds*, 5 *feet*!
     
  10. Ed

    Ed Guest

    Thanks to everyone for the excellent suggestions. I'll get this thing
    to work one way or another using your advice.

    Thanks!
     
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