# Torque and power for elevator with counterweight

I need help calculating required torque and power for an elevator to select appropriate motor and gearbox.

The mass (weight of elevator plus max load) is 2000kg

The speed of the elevator is 10 m/s.

The radius of the pulley is 0,5 m

I want to use a counterweight - mass of counterweight is not specified, what is appropriate here? I dont get it how to count when I have a counterweight in the system.

• Hi, I suggest you to consult an engineer, rather asking it here. The calculations are not that complex but there are other safety concerns.
– user14407
Dec 27, 2018 at 13:30
• Regarding your second question, you have to find the tension in the cable and set a free body diagram of the whole system. but it's very ideal, you have to count for friction as well.
– user14407
Dec 27, 2018 at 13:32
• Counter weight is a tradeoff between acceleration speed (more weight = more innerta) and energy cost of moving at constant speed once acceleration is done. So if counter weight is equal to elevator load then the constant speed is nearly free. Now what you choose depends on what you are optimizing. Dec 27, 2018 at 13:47
• @SamFarjamirad thank you for tour answer. I´m asking for a school project not actually constructing an elevator, so if you feel like sharing the calcultations and help me how to think it would be helpful, I found different ways of calculating and it makes me a bit confused.
– Jdog
Dec 27, 2018 at 14:47
• Well, do you know anything about required acceleration? Or at least can you choose the counterweight freely? Otherwise we have to do some trial and error to optimise the problem.
– user14407
Dec 27, 2018 at 16:02

Let's say we ignore the friction for simplicity now, and assume mass of counter weight 600kg, this should be the weight of elevator plus cable and depending on the traffic pattern some extra weight.

More important than the speed is what acceleration you need. Say we choose 4m/s otherwise at constant speed of 10m/s we just need much smaller tension of holding the difference weight of 2000- 600=1400kg. So the tension would be only 13729N and torque would be 6860N.m, but we will see the torque needed to accelerate the elevator is much larger.

The tension force in the cable to accelerate the elevator and its counter weight is $$F_{net} = T_{cable} - mg = m\alpha$$ $$and \ F_{net}= \ m\alpha = 2600* \alpha = 2600*4 = 10400N$$ So:

$$T_{cable}= 10400 + mg =10400 + 1400 *9.8 = 24120N$$

now we can calculate the torque $$\tau = f*r = 24120 * 0.5 = 12060 N.m$$ So we see the torque needed to accelerate the elevator is nearly two times that which is needed to move it at 10m/s. The real scenario is each manufacturer uses their own proprietary pattern of relays and controls that adjust the acceleration in a curve , ramping up, for comfort and minimizing wear and tear.

• thank you so much! So, the motor need to be able to give torque of 22460 Nm? (or the torque given after gearing). In calculating the power needed, I startet out by thinking P=mgv (mass, gravity, speed), is that correct? And if so, do I use the mass of the difference between elevator and counterweight? Another thing, what about moment of inertia, do I somewhere need to take that into consideration? Thank you!
– Jdog
Dec 27, 2018 at 17:52
• @Jdog, if you review my answer it addresses all you questions. only I edited the acceleration to a lower amount for typical 2-3 story elevators. we need to calculate torque for acceleration. Dec 27, 2018 at 17:59