# What motor torque specs that is suited for a horizontal to vertical 90 degree lift?

I have this simple design to lift an access door, but I am looking for the motor specs to do this heavy duty job. This is a simple design, I have added some information of the time, length of the part and weight that I am looking to move.

• the time factor that I am looking for this is, 3sec to 6 sec speed for the lift.
• the door weight is 250lbs
• The door length is 7’6”
• Door access will move in 90 degree in a horizontal to vertical motion.

Everything is made of steel (frame and access door). Looking for a motor and gear box that can handle that work in a daily basis.

If any more info is needed to help me out to find a motor that handle this work, i am available to provide it. • Welcome to Engineering! This is a great start to the question, but engineering is the art of solving problems around constraints, so we'll need you to edit your question with information regarding what exactly those are in this case: there's a lot of context-dependency on what "best" means.
– Wasabi
Mar 7 '20 at 15:44
– John
Mar 7 '20 at 18:03
• Hi, I just change the question on the picture and the title.
– John
Mar 7 '20 at 18:20
• Hi, can you verify if my review is suitable to open the question.
– John
Mar 9 '20 at 0:32
• very few applications would try to life a door like that using motor power. Add a spring or counter-weight to reduce the motor load. Mar 9 '20 at 15:07

So if I understand the question correctly, we are raising a "drawbridge type" door via a motor/gearbox at the pivot.

If we assume that the door is a uniform shape/density then it has a mass of 113kg (250 lbs) acting at a distance of 1.14m (7'6" / 2) from the pivot. This produces a torque of:

$$1.14\mathrm{\:m} \times 113\mathrm{\:kg} \times 9.81\mathrm{\:m/s^2} = 1264\mathrm{\:Nm}$$

So your motor-gearbox combination will need to produce this amount of torque at zero-speed just to hold the door stationary at floor level. As the door closes the torque required to hold it in each position will reduce - to zero at closed (vertical). However in order to actually close the door some additional torque is required to accelerate it.

Although we know the maximum acceptable time period to close through $$90^\circ$$ (6 seconds) the additional torque required will depend on the level of acceleration at each point in the move (the 'move profile'). For ease of calculation lets assume constant acceleration throughout the 6 seconds; so for the door to rotate through $$\pi/2\mathrm{\:rad}$$ (90$$^\circ$$) in 6s then the maximum rotational velocity of the door needs to be $$\pi/6\mathrm{\:rad\:s^{-1}}$$. To achieve that in 6s of constant acceleration needs an acceleration of $$\pi/36\mathrm{\:rad\:s^{-2}}$$.

Achieving that requires a torque calculated based on the moment of inertia, which for our door is:

$$\frac{1}{3} \times 113\mathrm{\:kg} \times 2.28\mathrm{\:m^2} = 196\mathrm{\:kg\:m^2}$$.

Meaning that the additional torque required for acceleration is:

$$196\mathrm{\:kg\:m^2} \times \pi/36\mathrm{\:rad\:s^{-2}} = 17.1\mathrm{\:Nm}$$

This move profile will mean maximum speed is at the end of movement, i.e. the door is slamming pretty hard! So in actual fact we probably want to accelerate faster at the start and then decelerate to stationary at the end. This will require a higher initial acceleration (and therefore more torque). On the other hand the mass of the door being supported is decreasing through the closure, so the available torque for accelerating will be increasing. This means that a move profile with low initial acceleration, then increased acceleration through mid-closure and a short period of deceleration at the end, would probably be the best option.

In practice these type of systems are typically not closed-loop controlled (i.e. apply power to the motor and let it do it's thing), with aspects like damping at the end of the stroke being added mechanically. This makes it difficult to know the exact move profile, but from our calculations it seems likely that around $$1300\mathrm{\:Nm}$$ of torque from our system is probably the minimum we could get away with. In addition we need to be able to achieve a minimum rotational speed of $$\pi/6\mathrm{\:rad\:s^{-1}}$$, or 12 seconds per complete rotation (5 rpm). This means the motor needs to have a minimum power of:

$$\pi/6\mathrm{\:rad\:s^{-1}} \times 1300\mathrm{\:Nm} = 681\mathrm{\:W}$$

As we have ignored lots of stuff like mechanical friction you'd probably want to be looking at something more like $$1\mathrm{\:kW}$$ with a suitable gearbox to achieve the torque/speed specs.

As noted in the comments, springs/counter weights or actuating the door from a different point (e.g. cables connected to the non-pivot end) may also make the motors job considerably easier.