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Ι need help calculating the power usage for a linear axis motor. it must move up and down at a maximum speed of 0.5 m/s a load of 50 kg

What I did so far:

The load (i.e. weight) acting on the motor is:

$$F = mg = 50 * 9.81 = 490.5 [N]$$

where:

  • F-force [N]: weight of the mass

  • m - mass [kg] mass of the arms + mass of the belt + estimate of the mass of the moving parts of the axis

  • g - gravitational acceleration [m/s2]

Calculation of the power required to be performed by the engine

$$P = F * v = 490.5 * 0.5 = 245.25 W$$

where:

  • P: work [W]

  • F: force [N]

  • v: linear speed [m/s]

Given an engine efficiency of 0.8 it will be obtained that the required power is:

$$P_{t,η} = \frac{P}{ η} = \frac{245.25 }{ 0.8} = 307 W$$

And when a safety factor of Ν=2 is taken into account it will be obtained that the required power is:

$$P_{t, s} = P_{t,η} \cdot N = 614 [W]$$

The problem that I have is how to take into calculations the inertia or the acceleration I need? What other formulas i need?

I searched the web and did not find what I need.

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    $\begingroup$ You need to know the time or distance of the motion. $\endgroup$
    – r13
    May 18, 2021 at 13:48

2 Answers 2

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TL;DR: Given that you took into account the efficiency $\eta$, and added on top of that the safety factor $N$, in most cases you will be selecting a motor capable of performing the task you describe.


Account for acceleration

You'd only need for acceleration and the mass moment of inertia, if you had significantly high values for acceleration.

For example, if you wanted to reach the maximum velocity of 0.5 m/s in 0.01 sec (i.e. you'd have an acceleration of a = 50 $\frac{m}{s^2}$), then the force acting on the motor (during the acceleration stage) would be equal to:

$$F = m\cdot (g + a) = 2990.5 [N] $$

NOTE: this acceleration value is extreme, and its only for illustration purposes.

In that case, you might want to consider also, the added torque required to accelerate the rotational masses of your system (e.g. the motor shaft etc).

However, as I said, the safety factor, gives you a safe margin, so I don't thing you need to overcomplicate things if you've found an appropriate motor (you might get away with 0.5[kW] motor if you are not too bother about the acceleration and the friction losses on your setup are small).


why you can't find a more detailed analysis on the web

you mention that you couldn't find a detailed analysis/equations for this problem.

The main reason is that the parameters that will affect the problem (beyond the weight and the velocity), will have high fluctuations and be affected by the implementation and the assembly.

More specifically, -IMHO- the other parameter that will affect significantly the calculations is the friction losses on your setup, which will be depended on:

  • alignment of assembly (small misalignment of sub millimeter can have a detrimental effect on the friction),
  • condition of the bearings
  • temperature
  • dust on the assembly
  • etc.

These are not easy to account for in a mathematical model, so the engineering way out is to use a safety factor (exactly how you did it).

A final note is that, -IMHO-, N=2 is a bit high for this purpose, given that the motor also has some margins of safety but then again, it won't hurt to have it, if you can control it.

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  • $\begingroup$ You gave an answer in newtons but the question asks for how much power, which is measured in watts. $\endgroup$
    – DrZ214
    Jun 13 at 16:12
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For linear acceleration of a mass, the power required will depend on how quickly you want to accelerate it.

This is because the Kinetic Energy formula is $E_k = \frac{1}{2} m v^2$. Note the v squared. So going from 0 to 1 m/s will cost less energy than going from 1 to 2 m/s, even if it accelerates at the same rate both times.

So I'll just make up more details of your example to show you how to do the math. You have a mass of 50 kg and a final velocity of 0.5 m/s, and of course it starts at zero velocity.

I'll choose an a of 0.5 m/s$^2$. That means it takes 1 second to accelerate to the final velocity.

The final KE of this is $0.5*50*0.5^2$ = 6.25 J. Therefore you had to do 6.25 joules of work over a time of 1 second. The unit watt is just 1 joule per second. Therefore you need 6.25 watts to do this acceleration...

However, this is only taking into account the linear force assuming no resistance of any kind besides inertia. Your problem said it must move up, in other words, against gravity. So we must take into account the change in potential energy.

$E_p = mgh$. First we have to find h. There is a kinematic formula to determine distance traveled during constant acceleration, $x_f = x_i + v_i * t + \frac{1}{2} * a * t^2$. Starting position and starting velocity are zero, so it simplifies and distance is 0.25 m. So EP = 122.625 J.

So over a time of 1 second, you changed the EP from 0 to 122.625 joules, and therefore requires 122.625 watts.

The final power is the sum: 128.875 watts. Use whatever efficiency and safety factors on top of this that you think is best.


It is interesting here that this is less power to move it normally with no acceleration. Maybe it is too coincidental, but I can do one more example to show you how different things are when choosing a different a.

Let's make a ten times bigger. a = 5.0 m/s$^2$. Final velocity is still 0.5 m/s, mass is still 50 kg, but now time to reach that final velocity is 0.1 s.

Distance moved upwards during the acceleration phase is $0.5 * 5.0 * 0.1^2$ = 0.025 m, so potential energy at that height is 12.2625 J, but this time that had to be done over a time of 0.1 seconds. So the power is 122.625 W. This is the same as before, for now.

Now consider the change in KE. The final velocity is the same, so KE still is 6.25 joules. But it was done over 0.1 seconds, so the power is actually 62.5 W.

The sum is 185.125 watts. Still less than your requirement for the linear phase.


Just to be thorough, I want to calculate the power for the linear phase, in a different way than you did. We can use the change in potential energy again, $E_p = mgh$. And note that is linear this time, not squared, so if it's constant velocity then it doesn't matter where you start and end.

A 50 kg mass moving up at 0.5 m/s, so ever second it changes its PE by $50 * 9.81 * 0.5$ = 245.25 joules. Inputting that much energy per second is equivalent to 245.25 watts. You already calculated this with force times velocity, which works here, but I prefer changes in PE when going against gravity.

Anyway, since 245 watts is bigger than the power needed for a quick 0.1 second acceleration to final velocity, I think you have nothing to worry about here. You will probably see the motor jump to steady velocity faster than your eye can detect.

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