# Sizing an AC Servo Motor for lifting pallets

I am trying to size a servo motor and my basic physics and math is so rusted that I can't figure out how to go about it? The scenario is as follows: I need to lift an object in the vertical direction against gravity. The object has a mass of 30kg. It is lifted to a height of 1m. It takes 5 sec to lift the object. I want to select a servo motor which is capable of doing this task. I have two options for the mechanism, rack and pinion (R&P) or lead screw (LS). I want the lifting to be done in three sections

1. Acceleration of 250 mm/s^2 for 1 sec. distance covered is 125mm
2. Constant velocity of 250 mm/s for 3 sec. distance covered is 750mm.
3. Deceleration of 250 mm/s^2 for 1 sec. distance covered is 125mm.

First thing is the power of the motor. From basic physics I know that

Weight (force) = mass x g = 30 x 10 = 300N
Work = Force x Distance = 30 x 1 = 300Nm
Power = Work/time = 300/5 = 60W


This is the base Power if I add in some friction and a safety factor I would say that 100W-120W should be enough. But I have serious doubts about this calculation.

1. Is the power calculation same when I have a Rack and Pinion or a Lead Screw?
2. The downwards force due to gravity in case of Lead screw is acting at an angle on the screw threads so should I be resolving that weight component?
3. What happens to the equations in case of downward motion?

I was doing some online research and many factors came up in sizing of servos. For example the Inertia ratio, velocity profile, max torque, RMS torque . I found a forum post which suggested calculating motor RPM based on linear distance and pitch of the lead screw . Then calculating the motor torque using this relationship Torque = 30 x Power / pi x RPM .

But this formula is not accounting for the required velocity and acceleration

1. Also how can I find the inertia ratio for a vertical moving load? Do I need to calculate the inertia of my payload when the payload is moving in linear direction only?

I want to understand the basic principles to the sizing so I can size the motor for a belt or a rack and pinion or any other mechanism. Any help or guidance is appreciated.

• First (unfortunately) you need to clarify what you mean by "servo motor". To an industrial machine builder, it's a motor that's optimized for use in a servomechanism. To a robotics hobbyist DIY type, it's a model airplane servo. They're two entirely different things. Maybe link to a sample of what you mean? Dec 5, 2019 at 21:20
• @TimWescott I hope the title is more clear now. I couldn't find any specific tags though Dec 5, 2019 at 21:34

Is the power calculation same when I have a Rack and Pinion or a Lead Screw?

Yes and no. Either power calculation needs to take the efficiency of the gearing into account, but you can pretty much count on the efficiency of the rack & pinion being much better than the lead screw.

In fact, lead screws (and worm gears) are often used because they have poor efficiency. Such screw-based gears often have an efficiency less than 50%, and if a gear train has an efficiency less than 50%, then when you take torque off of the input shaft the output will lock; this provides a built-in safety mechanism (or, in the case of driving a large inertial load, a built-in mechanism for destroying gearboxes when the output shaft stops and the load just keeps on going).

The downwards force due to gravity in case of Lead screw is acting at an angle on the screw threads so should I be resolving that weight component?

Not if you have the efficiency as a percent. The torque you need to apply to the mechanism is the torque you'd apply to a frictionless mechanism divided by the efficiency. So for your 30kg mass moving at 0.25m/s, the upward force is roughly 300N (taking $$g = 10 \mathrm{m / s^2}$$ because I'm lazy), the power exerted on the mass is then 75W, so the power exerted on a 40% efficient lead screw would be 187.5W. You'd then need to use the distance/turn spec of the screw to turn that into a torque. The calculation for the rack & pinion is the same except that the efficiency will be different.

What happens to the equations in case of downward motion?

Then the gearbox losses help to brake the motion, instead of hindering. This means that the behavior is asymmetrical. I don't have the exact numbers, but for the case of efficiency greater than 50%, you don't need much torque, and for the case of efficiency less than 50%, you actually need to "push" the geartrain to work against friction.

After giving this some thought I think now I understand more clearly what to do. There is three torques associated to the three stages of motion. 1. The accelerated motion a = 250mm/s^2 2. The constant speed motion a = 0 3. The deceleration with a = -250mm/s^2

The first two torques are of interest for me as they will give me maximum and nominal torque. (The deceleration section will depend on the efficiency of my gear train. But in no case will that torque be more than the acceleration torque.)

My load is 300N, to cause an acceleration of 0.25 m/s^2 in the upward direction I need a net force of 0.25x30 = 7.5N. So the total upward force needed is 300 + 7.5 N = 307.5N. Using the gear train efficiency I can then calculate, as pointed out by TimeWescott, the actual force needed to generate this acceleration and then the turn ration to find the torque. This will be by maximum torque.

For constant speed motion the required upward force will be same as the load. The rest of the calculation for nominal torque will be the same.

For the inertia ratio I will find the load inertia from this formula

Torque = (Jload +Jmotor) x alpha Max torque is calculated and linear acceleration (0.25) can be converted to angular acceleration (alpha) for a lead screw. Jmotor can be found in the datasheet. I can then find the Jload and use it to find the inertia ratio of load and motor.