# Actuator recommendation for oscillating motion of a 6 kg object

I have a general question without many specific parameters, but I hope someone will be able to point me in the right direction - at least what actuator best serves my purpose.

I am working on a project where I will be moving a 6 kg object in an oscillating manner up and down, from ideally about 0.1 Hz to 20 Hz at different times. The max vertical motion I would hope to be 30 cm or so. I will be controlling the speed with a microcontroller via serial. It would be ideal to have some control of vertical range of motion, though this is not necessary if it proves to be too difficult. Ideally it would also be able to brake quickly.

I'm not quite sure how to go about this - my first thought was a servo, but I assumed that it would be difficult to get it to move up and down as quickly as I would like, and perhaps the startup torque would be a problem on the up stroke. It would otherwise be ideal as I could control the angle, and therefore the vertical range. Perhaps there is a very fast and high torque servo that can handle this?

I then thought of using an industrial sewing machine motor with a clutch to vary the speed, but all of these seem to use a mechanical clutch which would be burdensome to control (would have to use a separate motor to change the speed via a lever). Also, I would have to convert the rotation to linear motion, and I'd rather avoid making more complex parts if possible.

I also thought of a linear actuator, but these seem to be rather slow, and pneumatics are too imprecise.

Any idea as to what kind of actuator to use? I'm aware I am slightly vague in my usage - this is an art project so there is some flexibility as to how it turns out, other than the general parameters I've mentioned. Any suggestions would be appreciated.

• I think you need to re-examine your requirements. Do you realize that a 30cm displacement at 20Hz results in an acceleration of over 240 g's? With a 6kg mass, that's a bit over 14kN without considering the moving pieces of the fixture itself. – DLS3141 Dec 21 '15 at 14:40
• How fast do you need to reach a given operational frequency, and do you need to control the size (length) of the oscillation independent of frequency? Typically a system is easier to drive with a resonant driver and a spring or similar resonant energy handler attaching the driver to the mass. – Carl Witthoft Dec 21 '15 at 14:53
• Carl - Ideally it would be independent, but I can see how that would complicate things a lot. I did think of using a spring, but as I understand it, the spring would have a fixed frequency. If it came down to it, I'd rather have variable frequency with fixed amplitude. – Rem Dav Nan Dec 21 '15 at 17:48
• DLS3141 - thank you for your comment, that helps. I can see that would be an enormous force to conjure. i would be happy to minimise the vertical displacement and lower the high frequency to some extent to have a more manageable system. Unfortunately the object has a fixed weight. – Rem Dav Nan Dec 21 '15 at 17:52
• 5Hz is actually pretty quick, especially if you're looking to move 6kg a distance of 30cm. What kind of art installation is this going to be? – Chuck Dec 22 '15 at 14:26

Vertical position is: $$y = A \sin{(\omega t)}$$

If your frequency is 20Hz, your angular frequency is $2\pi f$, or 125.66 rad/s.

If your max vertical motion is 30cm, then you could say it's 15cm +/- 15cm, so your amplitude is 15cm = 0.15m.

Now your position is given by: $$y = 0.15 * \sin{(125.66 t)} \\$$

So, first derivative, your speed is given by (chain rule): $$\array{ dy/dt = & 0.15* (\cos{(125.66 t)} * 125.66) \\ dy/dt = & 18.85 \cos{(125.66 t)}}$$

So, second derivative, do the chain rule again: $$\array{ d^2y/dt^2 = & 18.85 * (-\sin{(125.66t)} * 125.66) \\ d^2y/dt^2 = &-2368.6 \sin{(125.66t)}}$$

The line above is the second derivative of position, a.k.a. acceleration. As $F = ma$, and the max value of $-\sin{(\mbox{stuff})} = 1$, this gives you a peak acceleration of $2368.6 m/s^2$. If gravitational acceleration is $9.81 m/s^2$ (it is), then you are looking at a peak acceleration of $2368.6/9.81 = 241\mbox{g's}$.

Going back to the $m/s^2$ definition of your peak acceleration, you're looking at a force of $F = (2368.6 m/s^2)(6kg)$, or 14,211 Newtons, or 3,194 lbf.

Now, if you are interested in the power required to do this, that's force times speed. Because you get peak force when speed is zero and vice versa, let's just multiply the equations together and work from there, as peak power is going to happen somewhere in between peak force and peak speed. And again, since these are sine and cosine, we can drop the negative because it's absolute values we're concerned with:

$$\array{ P = & F*(dy/dt) \\ P = & (6*2368.6 \sin{(125.66t)} N)(18.85 \cos{(125.66 t)} m/s) \\ }$$

Rearrange, and then use the double-angle formula $\sin{(2\theta)} = 2\cos{\theta}\sin{\theta}$:

$$\array{ P = & 267,894 \sin{(125.66t)}\cos{(125.66t)} \\ P = & 133,947 \sin{(2 * (125.66 t)} (Nm/s)\\ }$$

And again, here the peak value of $\sin{(\mbox{stuff})} = 1$, so the peak power you're going to require is $P = 133,947 Nm/s$, or

$$\boxed{P = 134 \mbox{kW}}\\$$

If you think of the peak power as $P_p = V_p * I_p$, then you could consider RMS power as $P_{\mbox{rms}} = \frac{V_p}{\sqrt{2}}\frac{I_p}{\sqrt{2}}$, or $P_{\mbox{rms}} = P_p/2$.

So, going this route, if your home electrical service is 120VAC, then you're looking at drawing:

$$I = P/V = (134,000/2)/120\\ I = 558A$$

To put this into perspective, your home probably has "100 amp service", meaning that if you added all of the circuits in your house together you could only get 100A combined - you're trying to draw all the power available to a block. The breakers in your house are probably rated for 15A each, though, so this would blow the circuit well before you ever got close to what you're asking for. Also the motor would burn up. And the clutch.

--EDIT--

After getting back to my computer, I set up a spreadsheet with the above equations to run through some different scenarios. I found a math error this way - I had left off mass in the power equation $P = Fv$, or $P = mav$. What I calculated was just $P = av$, which is incorrect. It threw my numbers off by a factor of , which was 6 in this case, so where I thought you were drawing 93A at 120VAC you are actually drawing 558A.

Anyways, if you lower your frequency from 20Hz to 5Hz, then your power requirement drops from 67kW, or 90HP - about the output power of an economy class sedan, to 1kW, or about 1.4HP. This would have you drawing about 8.7A at 120VAC, which is entirely within the realm of possibilities for actual implementation.

Now, with a reasonable set of specifications, you can go hunting for applicable actuators. At this point, your "general question without many specific parameters" is too vague to give much guidance beyond some broad categories. For instance, I would hope you have considered how the following plays into your decision:

1. Cost
2. Positional accuracy
3. Complexity
4. Noise

Those are the things I would consider, in approximately that order of priority. Once you set a budget for what you are (and are not) willing to spend, then that should weed out a lot of actuators. Beyond that, your positional accuracy should then seal the deal. You said you didn't want to use pneumatics because "they are too imprecise", but you have not stated how precise you want your positioning to be.

I have actually used this controller before for some relatively fine position control, but that controller is about \$900, and you still need to buy the cylinder (probably about \$100) and a suitable air compressor (maybe \$150-\$200), and pneumatic hose and fittings.

You could use hydraulics, but that requires a hydraulic power unit (~\$400), a position regulator (~\$100), and then the cylinder (~\$100) and the hydraulic hose and fittings, which are likely to be considerably more expensive than pneumatic hose and pneumatic fittings. You could use electric linear actuators, but those are typically much slower. At 5Hz, for a +/- 15cm stroke, you're looking at a top speed of 4.7 m/s. This "high speed" linear actuator moves at 4.5 inches per second, or about 0.11m/s, which would correspond to a frequency of about 0.1Hz for you - one cycle every ten seconds instead of five cycles every second. All of the above are linear actuators. You can go with a motor (probably would be my choice), but as you mentioned this will require a cam/lever/crank system to convert the rotary motion to linear motion. Electric motors rated for 1.5HP or 1kW should be pretty easy to find, I would guess for about \$150 or so, but you'd still need a motor controller, which could easily run \$250-\$500, and then, if you care about position, you would need a position encoder for the motor.

It all gets pretty costly pretty quickly, which I know is a frustrating answer, but that's engineering.

So, to recap: first, set a budget. As in most things, this should significantly narrow the scope of what's feasible for you.

• Thank you for the illuminating answer Chuck. I really had no idea it would be so near impossible to do - if nothing else, I have a better understanding of what goes into analysing the process. I'm on 240v but obviously that's not going to help :) – Rem Dav Nan Dec 21 '15 at 20:40
• @RemDavNan - It's not a terribly difficult process, but most of engineering isn't. It's all just knowing which equations to apply to which problems. I would comment that the frequency is the killer for you. I'm on my phone at the moment so I'm not going to run the numbers, but if you could get the frequency from 20Hz to 2Hz then that order of magnitude reduction could cut your power by 1,000 - from 22kW to 22W, which is very reasonable. Power is proportional to speed cubed. – Chuck Dec 21 '15 at 23:26
• Thank you Chuck - I can definitely reduce the high frequency of this - 20hz was just the desired upper range. I can also reduce the amplitude as well - this is an art project and not an engineering project so there is a fair amount of wiggle room. Any recommendations for motor type or mechanism, or will pretty much anything with the right amount of torque do? – Rem Dav Nan Dec 22 '15 at 7:47
• Thanks again Chuck - all very helpful. indeed, all of the linear actuators I've seen are quite slow, though it would be an easy solution. But I am leaning towards the use of a motor. That pneumatics position controller is very cool :) And very precise it seems. Though $900 is definitely above my limit for now. My earlier comment about precision was based on standard bi directional cylinders without a controller. I've actually used an inverter/variable drive before for someone else's project, so I think I can manage that with a motor. As you say, budget first. Sound advice. – Rem Dav Nan Dec 22 '15 at 20:51 For the lower frequency, the easiest setup I can think of would be a motor (connected via a variable frequency drive) to the object, through a gearbox to reduce the frequency (common VFDs will leave your motor between 0-60 Hz). This then turns a large crank. Connect the bottom end of your object via spring to the ground to ensure stability. So long as your wheel is 30cm in diameter, your object will displace the same fixed 30cm each time. The gear box will make sure you have enough force to lift the 6 kg object by increasing the torque. Based on Chuck's Dynamic analysis at 2 Hz, you will need to be able to have (6kg * g + 142 N) * 15 cm = 30 N * m of torque. Your typical motor will have way less than this. But, your gearbox should be a 30:1 reducer to make your object oscillate at 2 Hz. In this case, you only need 1 N * m. A reliable supplier of motors has a 1/3 hp motor that can handle this loading easily. I'd also recommend investing in a large flywheel to ensure things run smoothly. • If the wheel diameter is 30cm, then the peak static torque generated by the load would be the weight (6kg * 9.81m/s2) multiplied by the radius, 15cm. Also of note, if the motor output is the required static torque, then the system will only ever be static. That is,$\tau_{\mbox{net}} = I\alpha$, where$\tau_{\mbox{net}}$is the net torque,$I$is the moment of inertia, and$\alpha\$ is angular acceleration. If motor torque equals static torque, then net torque is zero and thus acceleration is also zero. – Chuck Dec 22 '15 at 18:25
• Good point - I'll edit for clarity. – Mark Dec 22 '15 at 19:16
• Got it, this is really useful information. The general method makes perfect sense to me, with use of gear reduction and cranks. What I'm a bit blurry about is how it all fits together - for example, are there standard sizes for cranks, flywheels and mechanical parts that allow you to fit them to the motor shaft, or do they have to be custom made? – Rem Dav Nan Dec 22 '15 at 20:42
• I'd start at Grainger - they have some parts that are available for some of the other stuff they sell (rather than trying to open an old chainsaw or lawnmower to get their crankshaft). The real problem with the crankshaft is the bearings. You need them to stay lubricated during the operation, and a bad polish job on them will make everything off. However, if you just plan on going at this for a few days, any old weld shop could make a smooth enough surface that the machine won't die for that short of operation. – Mark Dec 23 '15 at 15:55