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I am currently building a machine for a school project.I have a pivot point and wish to use a stepper motor to turn the base at an angle. The base is carrying a load of approximately 10kg.

I was wondering whether a stepper motors rated torque (10kgf-cm) is still maintained if you were to use a 1cm in diameter threaded rod that is 30 cm long.

I have viewed other questions on this topic, and I understand that increasing the radius results in decrease of kilograms able to be moved . But there is no explanation of the effects a rod has on torque.

TLRD: Does a threaded screw, 1cm in diameter and 30cm long vertically, cause a reduction in the stepper motors max torque --> 10kgf-cm.

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Torque is distance times force. A torque of 10kgf-cm means that at a radius of 1cm the force exerted in the (tangent) direction of rotation will be 10kgf. At a radius of 10cm, the force is 10kgf-cm / 10cm = 1kgf. So at a radius of 30cm, you have 10kgf-cm / 30cm = 0.33kgf. Note that torque is constant, but the force changes with the radius at which you measure it. Also, the force is always measured in the direction tangent to the radius you're considering.

If you're asking about using the threaded rod as a lead screw, then that's different. You're still converting torque to force, but the conversion will depend on the pitch of the lead screw, and there will be substantial losses to friction.

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  • \$\begingroup\$ Could you please elaborate on your second dot-point please.Why will the conversion depend on the lead screw? And how can I calculate the losses to friction,also can they be mitigated if I use lube. The second dot-point is what I actually want to do. \$\endgroup\$ – user36278 Apr 24 '15 at 2:38
  • \$\begingroup\$ Why is a lead screw different? This is what I actually want to know. \$\endgroup\$ – user36278 Apr 25 '15 at 2:52

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