I am building a machine that has to lift a load of approx 80 kg including the material and lifting mechanism's weight. The velocity requirement is quite low (around 4.5 cm/s or 0.045 m/s). The whole lifting assembly will run in a guiding channel and will have rolling contact with the guide. The beam shown in blue will be the one to be lifted. Also attached to it will be another assembly. The lifting approach i have chosen for this purpose is telescope lift (also called continuous lift), both assemblies will be lifted using a pulley system and steel wires being pulled by a motor sitting at the base frame. Also during the uplift,there will be a continuous decrease in pulling force due to outflow of material (about 25 kg). I don't want to use a speed control circuit. A friend of mine has suggested to use a very high torque output and over powered motor to cater for variation in load.

My question is that what type of motor is most suited for this job. enter image description here

  • $\begingroup$ Nice rendering. $\endgroup$ Apr 18, 2016 at 14:01

2 Answers 2


An AC motor will turn at a constant speed with 1% to 5% slip provided the load is geared to be less than the listed maximum torque of the motor. An AC gear motor should bring it into the gear ratio range of your pulley system.

The approximate power required by your system can be calculated with the gravitational potential energy equation.

PE = mgh
PE = 80kg * 9.81m/s^2 * 0.045 m
PE = 35.3 joules
Power = 35.3 joules/s = 35.3 Watts
Power = 35.3/ (746W/hp) ~ 1/20hp
Remember that cheap gear boxes are probably only 50% efficient.

Here are some motors that may fit your application:
Dayton 1/10hp 60rpm AC Gearmotor
Dayton 1/10hp 30rpm AC Gearmotor

I'm not sure how much travel you need, but it may be worth looking at prebuilt linear actuators instead of building your own pulley system.


To calculate the needed torque, you could use the principle of virtual labor.

I would write this as a comment, if I had enough reputation.

  • 1
    $\begingroup$ If you could expand a bit on what you mean by editing your answer, this could be a perfectly acceptable answer. $\endgroup$
    – Wasabi
    Apr 18, 2016 at 11:13

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