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Torque and holding torque question

  1. Jul 15, 2010 #1
    Hello,

    I am bit lacking in knowledge in this area and would really appreciate some help :)

    I am building a gyroscopic device and need to calculate the starting torque and holding torque that a motor would require to wind up and spin a 2.5 inch flywheel that is .75 of an inch thick and weighs 2lbs
    can be in either ft-lbs or Nm
    Thanks!
     
  2. jcsd
  3. Jul 15, 2010 #2

    rcgldr

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    Holding torque normally refers to a servo's ability to produce a higher torque if not moving (this is because the poles of the motor don't have to cycle through field changes, but it also produces more heat).
     
  4. Jul 15, 2010 #3
    Oh, sorry.... I guess the power the motor would require to maintain the flywheel at target speed. The speed it will be spinning at is 10,000 rpm
     
  5. Jul 15, 2010 #4
    I'd figure out the center of mass of the level arm of the fly-wheel, somewhat in from the circumference so say 1" from the center (assuming the 2.5" is a diameter not a radius), and multiply by the weight: 1" * 2lb to get about 2 in-lb or .125 ft-lb. Then you have startup inertia and friction to deal with, so maybe double that. I doubt that you will find a 10Krpm motor so you probably have drive train ratios and losses to think about too.

    Once it's up to speed it shouldn't take much to keep it there because you will have to have built a very low friction and balanced system to get it to spin that fast to start with...
     
  6. Jul 15, 2010 #5
    Thanks Schipp!
    Thats simple enough :)
    I have a supplier lined up already who has motors that spin well beyond that with little trouble they were just asking for my loading so they don't under size it.

    Thanks again, much appreciated!
     
  7. Jul 16, 2010 #6
    Actually I realized that I typed too quickly... divided by 16 to get ft-lb from in-lb, shoulda been 12 (12 inches to a foot, 16 oz to a lb, duh). So make it 2/12 ft-lb or 32 in-oz. And I would, if I could, double it for inertia etc. Also don't order a 1000 of them until you test my hypotheses....
     
  8. Jul 16, 2010 #7
    lol, no problem :)
    So if I have a 3" flywheel weighing 2 lbs it would be the radius 1.5*2/12 to get 0.25 ft-lbs?
    Ya, I'm only getting the one to start to test on my prototype.

    Thanks!
     
  9. Jul 16, 2010 #8
    My assumption is that the lever arm you are dealing with is at the center of mass along the radius you are trying to spin. If you imagine all the weight of your flywheel being squished into a narrow circle at that distance, the force need to spin it would be the same as having the weight distributed evenly. Since the weight is proportional to the area (actually volume, but the thickness is fixed so we can ignore it...I hope), that distance should be -- I wasn't un-lazy enough to figure this out in my original reply -- the square-root of 1/2 the square of the Radius. In your 3" diameter case then it would be -- ugh, where's my calculator -- 1.5 * 1.5 = 2.25, / 2 = 1.125, sqrrt = 1.06" -- whereas the 2.5" version should actually be at around .88"

    Which brings up another point. If you are making a gyroscope you want as much mass as you can get to be close to the circumference to maximize the effect. This will also boost your drive requirements though.
     
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