# What type of actuator should I use? [closed]

I need to find out if There is a way to get at least 60 Hz of linear Motion with at least 5 mm of stroke that I intend to make linear persistence of vision device(not rotating one)It must be small and light as possible. ( maybe 50 mm long and 10-15 mm diameter or around these) (less than 500 grams) The Load will be around 50 grams. There are voice coils that is very expensive, can I use solenoids for instance or what do you recommend?

Thanks

• No I am a PhD Student in electronics Engineering department. You're right I edited the question. Thanks – user52709 Apr 20 '16 at 17:29

Max displacement: +/- 5mm. Frequency: 60Hz = 377 rad/s.

Now, some math:

$$x = 0.005 \sin{(377 t)} \\ v = (377)(0.005) \cos{(377 t)} \\ a = -(377^2)(0.005) \sin{377 t)} \\$$

Then, the max values are wherever the particular trig function is equal to 1, so:

$$x_{\mbox{max}} = 0.005 \mbox{m}\\ v_{\mbox{max}} = 1.9 \mbox{m/s}\\ a_{\mbox{max}} = 710 \mbox{m/s}^2 \approx 72 \mbox{g}\\$$

Now, $F = ma$, so the peak force you are applying is:

$$F = ma \\ F = (0.3)(710) \\ F = 213 \mbox{N} \\$$

Peak power is force times speed. $$P = (213)(1.9) \\ P = 405 W \\$$

Per your guidelines, you want (I'm assuming) a cylinder that's 0.05m long and 0.01m in diameter. This means that the volume is (approximately)

$$V = Ah \\ V = \pi r^2 h \\ V = \pi (0.005)^2 (0.05) \\ V = 0.00000393 \mbox{m}^3 \\ V = 0.00393 \mbox{liters} \\ V = 3927 \mbox{mm}^3 \\$$

Now, you can try looking at power density and see that you want 405W of peak power in a space of 0.00393 liters, or a power density of about $405/0.00393 = \boxed{103 \mbox{W/l}}$, or about the same power density as a Ferrari engine.

I'm not saying it's not possible, but good luck.

If you are a PhD student at a university, you should consider asking your committee members to guide you to another grad student for assistance with this.

Typically I would vote to close an open-ended design/shopping question like this, but I figured I'd go through the steps to show that what you want isn't really attainable.

• chances are that parts of the actuator also move which adds to the total mass and thus to the power required. – Bending Unit 22 Apr 20 '16 at 18:04
• btw. To separate your $ab$, you can use a \cdot: $a\cdot b$ or a ~: $a~b$ for example. – Bending Unit 22 Apr 20 '16 at 18:25
• @BendingUnit22 - Thanks for the tip. Internal masses are shockingly significant, especially in the case where a reduction gear is used. In my job I found that, to a drive motor, the inertia of the motor shaft and gearbox was the same as the 35 tonne piece of equipment the motor moved. – Chuck Apr 20 '16 at 19:49
• Thanks Chuck ,that was very helpful, so i can lower my load to 50 grams that will need around 20 Watts of power. Should i use solenoids with spring or sth else you know it is better. – user52709 Apr 21 '16 at 13:14
• You've posted this as an answer to your original question, but you should edit the original question to update the payload weight. FYI, though, you're not applying the equations I posted correctly. At 50 grams, it's 67 watts. My hope was that the equations I gave you will allow you to look (for yourself) for a product that meets your specifications. Asking us to search for products for you is a shopping question which means your question will be closed. – Chuck Apr 21 '16 at 14:04