# Worm Drive Calculation

I have this setup:

-Pully with a mass attached.

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

• 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. – F.Bek Jun 20 '17 at 11:58
• @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 ? – james Jun 20 '17 at 12:12
• (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. – F.Bek Jun 20 '17 at 14:03