I'm planning on using a lead screw for my robotic arm project. My question was that if I for example use a 3Nm torque motor and couple it to a lead screw (four-start lead screw), does the torque remain the same? Or does it increase?

How do I calculate the torque that the lead screw will give out? I know that a lead screw converts rotary motion into linear motion.

Lead Screw Driven Robot Arm

Thank you in advance for any help

  • $\begingroup$ Torque is torque. If the leadscrew is coupled directly to the motor it will still be 3 Nm. Do you mean axial force the lead screw will provide (linear force)? The lead screw (and nut) converts your torque into axial force. $\endgroup$
    – jko
    Feb 16, 2022 at 20:17
  • $\begingroup$ Thank you for your reply! Yes I believe I am talking about the linear force my bad. From the diagram I have added to the question, my question is that will the motor that is driving the lead screw still need a torque of 10Nm? $\endgroup$
    – user36991
    Feb 16, 2022 at 20:45
  • $\begingroup$ It's called a worm gear. $\endgroup$
    – DKNguyen
    Feb 16, 2022 at 23:53
  • $\begingroup$ @jko, torque is not torque. Power works this way, but torque is affected by simple machines like screws or gears. $\endgroup$
    – Tiger Guy
    Feb 17, 2022 at 17:17
  • $\begingroup$ @TigerGuy yes but if it's coupled directly to the motor it will have the same torque $\endgroup$
    – jko
    Feb 18, 2022 at 12:42

1 Answer 1


It is all just leverage, mechanical advantage. You can solve all of it through ratio of movements on each side.

Assuming 3 N/m and screw advances about 8 mm per 1 turn and its radius is 4 mm.

Forst find linear force, tangential at screw's side in this case

3 N/m / 0.004 m

Gives 750 N at the screw's side, tangential. To get axial force, find ratio of movement:

750 / ( 0.004 * 2 * 3.14) * 0.008

Gives 239 N axial force of the screw. Then you also draw another lever at your arm, but it ratio of shoulders is close to 1, so it can be ignored. If it is far from 1:1, calculate it too. If hand part is 3 times as long as the muscle part, then you get 80 N at the hand part.

And yes, worm gear or screw drive give large mechanical advantage that is used for making slow and strong movements.

  • $\begingroup$ @user36991 its a typo, i've fixed it $\endgroup$ Feb 16, 2022 at 23:27
  • $\begingroup$ Ah ok thanks!!!, so basically the 3Nm torque from the motor (rotational motion) is converted into 239N axial force (linear motion. Is that correct? $\endgroup$
    – user36991
    Feb 16, 2022 at 23:41
  • $\begingroup$ @user36991 yes. For the sizes I mentioned $\endgroup$ Feb 16, 2022 at 23:51
  • $\begingroup$ Ok great, also how do I calculate the axial force that I will be needing for my arm. Is it all the mass*gravity for each of the masses on that link? $\endgroup$
    – user36991
    Feb 17, 2022 at 0:04
  • $\begingroup$ @user36991 force * length should be equal on both sides of the lever $\endgroup$ Feb 17, 2022 at 0:19

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