I've watched a few videos explaining how actuators work and it seems to me that a regular, non-blocking, actuator would freely move if not powered. But theory and practice are rarely the same thing. I've never owned an actuator, and nobody on the entire internet have tried to push on an actuator that is not powered, hence this question.

To give a bit more clarity, I'll describe my application and what I want to use the actuator for. There's this door that I often forget to close. I want the door to close after some time, if open.

I don't need the actuator to ever open the door, just to close it, if open for more than, say 5 minutes. Edit: The closing mechanism has to be on the "pull to close" side of the door. I want to be able to manually open and close the door, as that will still be the primary mode of operating the door. The logic part will be done using an arduino and some sort of sensor that figures whether the door is open or not.

An actuator seems to fix my problem, but only if the shaft can freely move when not powered. So, can a cheap actuator like this be moved freely when not powered? If not, what would my options be?

  • $\begingroup$ A standard spring door closer, with a catch to keep the door open, and open the catch with your Arduino? $\endgroup$
    – TimWescott
    Sep 30, 2019 at 19:20
  • $\begingroup$ Come to think of it, screen door closers are pneumatically damped, and you can adjust the damping. Put a solenoid valve between the cylinder and the damping adjust and connect the valve to the Arduino. With the valve closed the door will creep closed, but probably slowly enough you won't cane. With it opened it'll close at whatever rate you set. $\endgroup$
    – TimWescott
    Sep 30, 2019 at 19:23
  • $\begingroup$ This depends entirely on the mechanism. There is no one way to do a linear actuator. For example a pneumatic piston can easily be de-energized and re-energized if needed. You can do it easily with a rack and pinion or a screw mechanism too just disengage the gear (this is after all how most manual lathes work). But really your over complicating things. Just take a standard door closer mechanism and shorten the hand so that it can move sligltly over the critical. This will cause it to stay open, then just move the door back to other side of critical and it will close. $\endgroup$
    – joojaa
    Oct 2, 2019 at 15:02
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    $\begingroup$ You can even do it without the arduino just using a spring and delay mechanism so you dont even need an arduino. Let alone electrical power $\endgroup$
    – joojaa
    Oct 2, 2019 at 15:05
  • $\begingroup$ @joojaa So many questions... What do you mean by "then just move the door back to other side [..] and it will close"? The whole point of the mechanism is to move the door to close. Sorry, I don't get it. Also, what is a delay mechanism that doesn't need power? I can imagine how it would work in theory, but I have no clue where I would buy one. $\endgroup$
    – Andrei
    Oct 2, 2019 at 15:24

1 Answer 1


Your primary question, if a linear actuator can be moved with the power off, will be no, unless you find a mechanism specifically created for that purpose.

These actuators are effectively a pinion gear design with a very high reduction ratio. You'll note in the link you provided that heavy lift mechanisms such as a motorcycle lift and a medical bed are presented for this type of use. They work well because you can position the bed/lift top where you wish and stop without additional locking mechanisms, although you'll find most have a safety latch of some sort.

For your application, closing a door, this type of linear actuator will be poorly suited.

I can envision a more mechanically complex arrangement which would use a linear actuator. Consider the manual release for a garage door opener. The chain and carrier for the opener can be considered the linear actuator.

If I disengage the manual release, then activate the button or remote, the door will remain closed. I can open and close manually the garage door without interference from the motorized mechanism, as long as I do not engage the latch system that is part of this garage door device.

If the linear actuator in fully-open mode has a pusher block attached, and remains in open mode, your door will operate normally. A sensor tells your system that the door is closed after the appropriate period.

If not, the device actuates the actuator and pushes the door closed, until the sensor informs the device that it is closed, at which point, the actuator returns to the original location, ready to push again on command.

There may be other approaches using clutches, cables, etc., but this is the one that popped into my alleged mind as an option.

EDIT: similar method to my suggestion, based on new information regarding the placement.

Consider that the actuator is on the inside of the door, as specified. When idle (door closed), the actuator is retracted, to a point where the end of the ram is not in contact with the door.

At the end of the ram is a magnet of sufficient strength, matching placement on the door of a metal panel.

After the door is ajar for the specified time, the ram/actuator will travel to the limit, "collecting" the door in the process. If that aspect is undesirable, a limit switch could be placed on the actuator, triggering motor reversal. Simple microswitches such as those used in 3D printers are inexpensive and easily found online.

When the actuator contacts the door, either at full travel or with the switch activated, the magnet will pull the door closed as the ram retracts. When the door is closed, the additional space between the ram and the door will disengage the door, returning it to normal service.

If magnets are not desirable, a more complex design using the limit switch concept to trigger a latch and socket mechanism could be implemented. Additionally, a low-to-medium force detent mechanism might work, but that would require full extension to engage. Think of a horizontal plate on the door with a large diameter hole in the surface. The end of the ram would have a sphere that has to deflect or be deflected in order to engage the hole. Once engaged, reversal will pull the door closed until the retraction force pops the sphere free from the hole. Less reliable, less fussy to build, but still an option.

  • $\begingroup$ I see what you mean. It's just that the mechanism must be on the "pull to close" side of the door. This complicates things because I would need some sort of catch and release mechanism for the tip of the actuator. Regardless, my primary question is answered, so thanks. $\endgroup$
    – Andrei
    Sep 30, 2019 at 17:38
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    $\begingroup$ Please edit your question with that constraint about which side of the door you need the mechanism to be. $\endgroup$
    – TimWescott
    Sep 30, 2019 at 19:17
  • $\begingroup$ added new suggestion based on placement update $\endgroup$
    – fred_dot_u
    Oct 2, 2019 at 14:26

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