I want to build an automatic sliding window shutter and need help with part selection and dimensioning.

Some assumptions:

  • window width 1.4 m
  • sliding shutter weight 25 kg
  • max speed 0.07 m/s
  • max acceleration 0.035 m/s^2
  • pulley diameter 0.04m.

Leaving out friction I need a motor with about 0.02 Nm of torque and a rated speed of 33 rpm.

What I would like to use:

Would you suggest other components or a different setup? How do I connect motor and pulleys (clamping set?)? Do I need additional bearings because of the radial load?


(M motor, P pulley, B bearing, = shaft)

If so I have to extend the motor shaft. What would I use for that (clamp collars, couplings?)? What width do I need for the belts? Which belt profile (T, AT, HDT) should I use?


The construction I am aiming for resembles the one which can be seen on page 6 (pdf numbering) here.

  • 1
    $\begingroup$ Hi tauran. At the moment, this question seems a bit too broad -- you're asking about a varied mix of big and small system components. It's also unclear what the mechanical constraints are; would a garage door opening system work? Would the motor from an automotive power window system work? Why or why not? $\endgroup$
    – Ian
    Feb 5, 2016 at 19:14
  • $\begingroup$ This is a question for Engineering or Home Improvement. Not everything that has a motor is a robot. $\endgroup$ Feb 5, 2016 at 21:25

1 Answer 1


"Leaving out friction I need a motor with about 0.02 Nm [of torque] and [a rated speed of] 33 rpm."

No offense, but verifying your numbers based on your spec's:

window width 1.4 m

sliding shutter weight 25 kg

max speed 0.07 m/s

max acceleration 0.035 m/s^2

pulley diameter 0.04m.

The max linear speed is equal to $r\omega$, where $\omega$ is the angular velocity in radians per second. Your pulley diameter is 0.04m, so your radius is 0.02m. This means that:

$$ v = r\omega \\ 0.07 \mbox{m/s} = 0.02\mbox{m}*\omega \\ \omega = \frac{0.07}{0.02} \mbox{rad/s} \\ \omega = \frac{3.5 \mbox{rad/s}}{1}*\frac{60\mbox{s}}{1\mbox{m}}*\frac{1\mbox{rev}}{2\pi\mbox{rad}} \\ \omega = 33.4\mbox{rpm} $$

So, your rated speed checks out. Now, the torque. If you ignore friction, then the only purpose for torque is to accelerate:

$$ \tau = I\alpha \\ $$

where $I$ is the moment of inertia, $\alpha$ is the angular acceleration is $\mbox{rad}/s^2$, and $\tau$ is the motor torque. Again, $a=r\alpha$, like $v = r\omega$, so $\alpha = a/r$. As the weight of the shutter is going to the pulley, you can consider it a point mass at radius $r$, meaning the moment of inertia $I = mr^2$.


$$ \tau = I\alpha \\ \tau = (mr^2)(a/r) \\ \tau = r(ma) \\ \tau = (0.02\mbox{m})(25*0.035\mbox{kg m/s}^2) \\ \tau = 0.0175 \mbox{Nm} \\ $$

BUT - if this is for a vertical shutter, the motor has to support the weight of the shutter while it's accelerating it, so the moment of inertia is now $mgr$, or $25*9.81*0.02 = 4.9\mbox{Nm}$ - this dwarfs your acceleration requirement.

The required torque is small enough that I would suggest double checking your assumption that friction is negligible. That said, you have a rated speed and a rated torque, now you just need a rated power. For rotating systems:

$$ P = \tau\omega \\ $$

Where the previously calculated values torque and speed are $\omega = 3.5 \mbox{rad/s}$ and $\tau = 0.0175 \mbox{Nm}$, so:

$$ P = (3.5)(0.0175) \mbox{Nm/s} \\ P = 61\mbox{mW} \\ $$

So, based on your problem description, nearly any motor you can find will work, with the exception of the very smallest hobby motors. For instance, this motor has a rated speed of 120rpm, a rated torque of 0.6Nm, which is over 30x your requirement, for a rated mechanical power of about 7.5W.

A couple notes of caution, though:

These calculations make the assumptions that there is no friction and that there is no inertia between the motor and the shutter. Regarding bearings, I would probably use bushings, which are basically wear parts but are significantly less expensive and smaller than true bearings for light loads. I would definitely consider using at least one bushing to help minimize bending forces and wear on the output shaft bearing in the motor.

Without a better description (read: picture) of what your intended setup is I can't comment any further than this. In general, though, find the torque, speed, and power rating for the motor and add some performance margin and then just search the Internet. You have to finish the design before anyone here can comment specifically on it.

  • $\begingroup$ Thanks Chuck for the nice and accurate calculation. I hope my update clarifies the intended design. $\endgroup$
    – tauran
    Feb 7, 2016 at 11:12

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