Linear movement: Power calculation

Hello Mechatronitians!

April 2021

A lot of theory… but no practice. Does this ring a bell? In engineering studies or technical training, a lot of weight is put on teaching students complex formulae. In real life, however, quite a few of us wonder whether it is possible to acquire in-depth knowledge about a very specific field that is focused on the day-to-day activities of engineers and helps them solve their actual problems.

If this is your case, then today is your lucky day: we are going to share a little secret with you, a simple formula that will help you calculate the power required for a given application using a linear motor. It is so simple that you won’t need your calculator, just a mental calculation. But before we begin, let us give you some theoretical background to ensure you can easily understand everything.

First let us talk about force, specifically about mechanical power. Force is the amount of energy required to accelerate, move or stop a mass which is moving or idle.

P=ʃ(VxF)

The higher the velocity and force, the higher the power. Reducing speed for a certain force obtains a higher power, and vice versa. A practical example is an elevator; whenever a motor cannot lift a certain mass, we have two options: reduce speed using a geared motor, or use a larger gear ratio.

The engine’s available force does not change after a reduction (only the torque does), as it does not depend on the specific mechanical solution: it is constant and depends only on the motor.

P net = P motor – P wasted

Going back to the previous example of the elevator with force issues, another possibility would be to improve the performance of the components, for example by replacing a crown worm gear (µ = 40%) with a helical gear box (µ = 80%). In this case, load capacity would be increased by 100%.

Having cleared up this matter, let us see the first formula to calculate the power:

Where:

M – mass to move, kg

V – speed of the movement, mm/s

100000 is a constant, the result of converting the physical units

µ – performance of the whole system (from 0.1 to 0.9)

With the example of the elevator, where we need to lift 500 kg at a speed of 50 mm/s. If we choose a trapezoidal spindle shaft with 0.3µ, the required power would be:

P = 500 x 50 / 100000 x 0.3 = 0.83 kW

If the requested speed is 500 mm/s, the required power would be:

P = 500 x 500 / 100000 x 0.3 = 8.3 kW

10 times higher, since we have multiplied the speed by 10.

Some tips for you: if the application has several transmission parts, performance values must be multiplied by one another and the result added to the formula. If, for instance, we use a trapezoidal spindle µ = 0.4 with a worm gear box µ = 0.4, the resulting performance is µ = 0.12. This is the value which should be used in the formula.

This figure is both impressive and astonishing: a performance of 0.12 means that 88% of the force/energy is lost and becomes residual heat. These are losses of power that cannot be recovered, whether in force, electricity or any other type of energy. When designing a new machine, great care must be taken in deciding the performance of each single component. Therefore, only components with the highest performance should be used. The ideal value is above µ = 0.6 or 0.65, where only one third of the performance would be lost.

Machine wear can be reduced when working with more efficient components. The same goes for the dimensions of the cabinet and the cable cross section. If we all cooperate, we can make a more efficient, and therefore cleaner, world.

We recommend putting this formula into practice and reviewing the components you are using! There’s room for optimisation, that’s for sure.

If you would like to know more, please get in touch with us on our Web SINADRIVES or leave a comment below. We would be delighted to help you.