Comments

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  Chris Popp March 4, 2010 at 8:04am

  The previous guy is on the right track. I doubt we have something that small, but more info is really needed to come up with an answer. To determine the amount of torque required to lift the load, you need to know the screw pitch so the mechanical advantage and efficiency can be determined. He offered a suggested pitch. Find out your existing pitch and use his formulas to determine the torque and power. Then you can search more accurately.
• Igor Vončina March 1, 2010 at 11:14am

  Hello. I saw your post for gearmotor for lifting application. Calculation for Power of the needed motor is:
  P = F * Pscrew * n
  where: P = needed power
  F = needed force (10 kg = 98,1 N)
  Pscrew = pitch of the screw (let's say 2mm = 0,002 m)
  n = revolutions (250 rpm = 4,17 revs per seconds)
  
  So...F needs to be in N, Pscrew is needed in m, n needs to be in rev/second.
  
  If we use this data, the result is P = 98,1 N * 0,002 m * 4,17 /s = 0,82 Nm/s = 0,82 W
  Now, this is teoretical, because of friction and so on, the actual needed power is around P/0,8 = 0,82 W / 0,8 = 1,03 W
  
  Now to know what motor to use: P=U/I where
  P = power
  U = voltage
  I = current
  
  So, if you need as small as possible, then use some kind of gear mechanism, where the smaller gear is on the motor and bigger gear is on the shaft of the screw.
  
  gear is: n1/n2 = T2/T1 >> this means, if use gear 1:3, you will need 0,3 W motor with 750 rpm, you get the idea.
  
  I hope i helped you i some way.