3
$\begingroup$

I have some doubts about a quick-release mechanism that I'm working on and I'd like to run it by you.

I have hollow tube in which there's a spring-loaded rod. Part of it is made wider and there's a metal round pin that restrains the entire mechanism as shown below: enter image description here

The spring rate is 1.23 N/mm its free length is 135mm but it is compressed to 35 mm.

I'm using a servo quick release mechanism similar to this: enter image description here

It is driven by a servo with a stall torque of 9.8kg/cm

My question is, would the lateral force the rod is exerting on the pin be a problem for the motor? E.g. the friction between the pin and the hole of cylinder being so great that the motor won't be able to move? Is there a way to mitigate this?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

The force which is needed to pull the pin out, will depend on the friction of the pin and the neighbouring material, and the surface at which the friction works. Let's assume a friction factor $\mu$ of 1, though it's probably less for plastics. That means that theoretically, just as much force is needed to remove the pin, as the force put on the pin by the spring, which is 123N(or 12.5kg) in your case, assuming the springrate is constant. If the servo has an arm of 2 cm, then the maximum force the servo will be putting on the pin, is half it's stall torque, so 4.9kg. That wouldn't be enough to remove the pin. However, if the friction factor, or the arms length is lower, it may be enough.

But I'd rather be concerned that the pin doesn't get damaged after a few times, as your servo pulls it out of the tube under force. That used to be the one lifelimiting factor in the BB guns I had as a kid. Those used the same construction as you show here.

I'd advice you to use a solenoid to pull the pin out of it's hole, it does so with enough force, and much faster than a servo, greatly reducing the stress on the pin. You can easily find a solenoid strong enough to pull the pin out.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Very detailed response! Thank you very much Bart! I had a solenoid in my original design, but I dismissed it, because I wasn't able to find a DC solenoid strong enough. for the job (my project is to run off a 7.6V LiPo battery). What I gather from your explanation, the solenoid should have a force of 125N (about 12kgf) Any recommendations? $\endgroup$
    – JackS
    Commented Jan 17, 2018 at 23:03
  • $\begingroup$ @JackS You're welcome. You can use a lever to reduce the pull needed from the solenoid. The force in my calculation is based on assumptions, so reality may very likely differ. If friction factor is lower, less force is needed to remove the pin. Look for a solenoid with rather much windings, and designed to work on 6V or less. If you only need to power the solenoid briefly, you can quite heavily overload it with 12V without problems. I'd look for a solenoid with a core, that can pull 10-15kg to have some margin. Remember for your circuit, that solenoids are very inductive. $\endgroup$
    – Bart
    Commented Jan 18, 2018 at 18:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.