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I have this setup:

-Pully with a mass attached.

-Worm gear drive to lower the mass with a constant speed. (Section A-A)

enter image description here

Now if we look at the cross-section A-A, we can see the worm gear drive:

M1 is the torque applied by the mass on the pully and hence on the gear.

M2 is the torque applyed by the worm gear. enter image description here

Now my question is:

If M1 > M2 is it still possible to lower the mass at constant speed ? ...or in other words, is it possible to make the gear (above the worm gear) turn at a constant rotational speed ?

...or will it just block the system due to friction ?

I will rephrase my question: Let's suppose it is a autolocking worm gear, how can I calculate how much M2 needs to be for the mass to lower at a constant speed ?

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    $\begingroup$ For any constant speed operation, the system must be stable and be at dynamic equilibrium. So, M1 must be balanced by the equivalent torque with M2 ( M1 = M2). In other words, if M1 is greater than M2 ( M1 > M2 ) , the mass will accelerate, which in our case, not a constant velocity motion. Also for checking the locking condition, we need more detail about the setup, as dimensions, angles and the material type for friction coefficent etc. $\endgroup$ – F.Bek Jun 20 '17 at 11:58
  • $\begingroup$ @F.Bek I do not know the exact details yet, I am interested in the general principle. I will rephrase my question: Let's suppose it is a autolocking worm gear, how can I calculate how much M2 needs to be for the mass to lower at a constant speed ? $\endgroup$ – james Jun 20 '17 at 12:12
  • $\begingroup$ (Q:) "How much M2 needs to be applied for the mass to lower at a constant speed ?" (A:) M2 can be found by using the relation [ M2=(m*g)*r ] where m: mass, g: gravity, r: radius of red gear/drive. And the M1 must be equal to the M2 to balance the system. Then with these dynamic equilibrium conditions, the mass will go down with constant velocity as worm rotates. $\endgroup$ – F.Bek Jun 20 '17 at 14:03

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