# Energy consumption from acceleration

I have an interesting problem. I have a machine which has to lift 35 kg (this is the total mass of the object and the lifting arm). The lifting is done by a servo motor.

Now I want to calculate the energy required to lift this object with a certain acceleration and deceleration (let's say $1000\:\rm mm/s^2$)

The height of lifting is $h=1000\: \rm mm$.

Now this is a servo machine, which means that for half of the height it will accelerate and for half of the height will decelerate. So at the middle of the lifting the speed will be $v_{max}=1000\:\rm mm/s$ and the total movement will be $t=2\:\rm s$.

I could also say that the average speed would be $v_{\rm avg}=500\:\rm mm/s = 0.5\:\rm m/s$

Energy formula would give me an energy of $\frac12mv^2= 0.5 \times 35\times 0.5 \times 0.5 = 4.375\:\rm J$. If I divide this by the time required (2s) I get $4.375/2 = 2.19\:\rm W$ of Power.

This doesn't make sense because definitely my machine requires more power and consumes more energy. What am I missing here?

• It also takes energy and power to lift the object against the force of gravity. This means that you will need a LOT more power than 2.19 W, because your object has a weight of approximately 350 N. – David White Aug 21 '18 at 15:34

OK, let's plot it. This is speed (m/s), position (m), acceleration (m/s^2), force (N), mechanical power (W) and energy expended (J) during the movement. Since $$\ 1m/s^2$$ acceleration is much lower than gravity g, the servo has to exert force to push the lift up during the entire travel. It will exert a bit more force during acceleration and a bit less during deceleration, but it will not act as a brake (ie, work at negative power). This is an important detail because we don't know if the servo can do regenerative braking and recover energy during braking... or not.

We know the total amount of work the servo will have to do is $$mgh = 343 J$$ no matter the speed or acceleration profile.

However instantaneous power will vary since it is $$Fv$$ and v is zero at the beginning and at the end, and maximum in the middle. At a constant acceleration, v will increase then decrease linearly, so power will do the same. Power is not constant so it is incorrect to estimate it with "energy/time". Likewise you can't calculate kinetic energy by using the average speed since kinetic energy varies during acceleration and deceleration.

The mistake in your calculation is to not account for the fact that the vast majority of the power is used to lift the object, not accelerate it. Look on the bottom graph, kinetic energy is tiny relative to potential energy. Kinetic e is maximum in the middle at maximum speed, then it is converted back into potential energy as the lift slows down.

A real servo will use more electrical power than the mechanical power it produces, of course.

The motor's rotor inertia could be added to the calculation, but it would only matter if the acceleration was higher than g and the motor had to brake against its own inertia during deceleration. Here it won't matter much, since the "brake" is the weight of the object, not the motor. So the motor's inertia simply adds to the lift's kinetic energy and is converted into potential energy at the end (minus transmission losses).

The servos's output mechanical power peaks at 400W, so once we add up losses in transmission, the motor, its controller, etc, it wouldn't be surprising that electrical power would peak at much higher than that. At this power level, it won't use linear drive, rather PWM, so at least the driver should be efficient.

Motor efficiency depends on speed (rpm) and torque. Check this link for an example efficiency map of brushless vs asynchronous motors. These are big motors, the motor in your servo will be less efficient.

Even when it does not move, and produces zero mechanical power, the servo needs torque and thus electrical power to hold the lift against the weight of the object. So electrical power might be close to zero at the beginning if the lift is bottomed out and rests on the ground, but it will draw electrical power to hold the lift in position at the top unless it has a mechanical brake, or other kind of one-way gearbox which will stop the lift from moving unless the motor turns. And in this case... this will also introduce losses.

So I think you should plot electrical power versus time if you have an instrument to measure it ; the shape of that graph will tell you more about how your servo uses power. Also check how much power it uses just to hold the lift depending on how much weight is on it.

• Also, we have bearing friction to consider too, at whatever RPM the motor will turn and what type (if any) gearing it has. +1 for considering motor inertia. – Robert DiGiovanni Dec 15 '19 at 18:43

It takes more energy to speed up an object from $0.5m/s$ to $1m/s$, than from $0m/s$ to $0.5m/s$. So, to calculate the average power required to speed up an object from $0m/s$ to $1m/s$, we have to find the final kinetic energy and divide it by the speeding up time.

The consumption of energy during the second (deceleration) phase of the lift is trickier, because it depends on how exactly the deceleration is done, but we can assume that it will be smaller than the energy consumed during the acceleration phase (in theory, it could be even negative).

Besides the energy required to accelerate the object, we have to take into account the energy required to increase its potential energy and and the energy spent against friction in the arm, which will further increase the average power consumption.

Of course, we also have to take into account the efficiency of the driver circuit (could be built into the servo) and the efficiency of the motor and the drive train.

• Hm... Let me think about it. – DannyRitiu Sep 5 '18 at 5:36
• Let's say the object is on the ground so potential energy $PE=0$ and kinetic energy $KE=0$ After the acc. phase we have $PE=35\times 9.8\times 0.5 = 171.5 J$ and $KE=0.5 \times 35\times 1 = 17.5 J$ so total $E = 189 J$ At the end of the dec. phase we'll have $PE = 35 \times 9.8 \times 1 = 343 J$ and $KE = 0$ Both phases last for $t=1s$, for the first phase we require $189 W$ and for the 2nd $343-189 = 154 W$ Still too low compared to reality. So the question is: what's the factor that drastically changes the values? I somehow doubt that is the friction or motor efficiency. – DannyRitiu Sep 5 '18 at 6:00
• @DannyRitiu So how much power does your machine consume and how do you measure it? Also, does all the power go to the arm or there is something else the machine is doing? – V.F. Sep 5 '18 at 9:12
• Thanks V.F. for the help. I measure it with a scope software of the drive. For the first phase the average power is 120%, for the dec. phase it's 95%. The motor's nominal power is $750W$ So I believe I require cca $900W$ of power for this movement. In reality I have a lighter load $m<35kg$ All the power of the motor goes to the belt pulley. There is no gearbox. – DannyRitiu Sep 6 '18 at 10:13
• There are software to design motors and choose the required power. I'm aware of those. What I am not aware is: what physics phenomena is happening, which makes me loose that much of energy. Obviously the motor heats up, but efficiency of servos is usually >85% – DannyRitiu Sep 6 '18 at 10:23

Would it be possible for you to draw a free body diagram of your system? Realistically, the best way to solve this problem would be to integrate the motor force output as a function of distance. This will give you the work done on the mass by the motor. Does your servo motor control the speed with proportional control? Or does it take a short cut and control the current supplied with proportional control leading to a lower net force on your mass?

We know the net work done is mg$$\Delta h$$, where m is the mass, g is the gravitational force, and h is the change in height. This gives us a nice lower bound for the total energy exerted by ALL of the forces along the length, not just the motor. Consider the integral below. This is the work on your mass, where $$m$$ is your mass, $$F_m$$ is the force from your motor, $$F_g$$ is the force from gravity, and PE is the potential energy. To go further, you can include your model for motor inefficiency in the motor force generation term, eventually bringing the model to voltage and current.

$$\int_{h_0}^{h^f} F_m - F_g dh = PE$$

• I realize that I wrote PE instead of KE. The integral should be equal to the kinetic energy, not the potential energy. – ZackWoodRD Nov 16 '19 at 1:32
• You are not really answering the question as actually no one has until now. There is a surely a big issue with my theory and I am waiting for somebody to point it out. If you think integrating the force and finding the KE could solve my problem, please go further and help me solve your integration. Otherwise you are not really helping. – DannyRitiu Nov 18 '19 at 18:31
• Hi Danny, I have given it a bit more thought and I have a much better answer for you. Rather than integrating the force over distance you should integrate the motor power over time. I think this will give you a much better result. Again, you have not drawn a free body diagram of your system or detailed the control system so you did not answer any of my questions that will drive this discussion further. – ZackWoodRD Nov 19 '19 at 19:32
• Hi Zack, I don't want to calculate the energy consumption of the motor. I want to understand what is the problem why my theoretical calculation doesn't work... If I could draw a free body diagram, it would be easy-peasy, but that's exactly my problem: there are some hidden forces, which make the system very inefficient. And I just want to know which could be those forces? – DannyRitiu Nov 21 '19 at 10:53
• This is my point, there cannot be a hidden force making the system inefficient. Your model is incomplete. That is why I am asking for questions on the control configuration of your motor: if the motor reduces power near the top it may be a less efficient cycle because the motor is using a lot of power holding the mass up against gravity rather than moving it forward. – ZackWoodRD Nov 22 '19 at 5:17