2
\$\begingroup\$

I have a round door knob, which allows opening the door.

To release the door bolt, one must pull the door knob to make it translate horizontally for about 10 centimeters.

Once you stop pulling, a spring bring it back to its normal position.

I've tried measuring how much force is needed to pull the knob. I've to figure that out using a weight I could measure, tying a metal cable to my door knob, and trying to make the weight pull it. Here is a scheme of what I've tried.

enter image description here

I don't need gram precision, I've calculated the weight necessary is 1.3kg, being about 13N.

Knowing that, I've bought a motor which has a couple of 800g.cm. I've bound its main axis to a 10cm diameter pulley, to which I've attached a metal cable. Then, I've loosely attached another pulley to the door knob, and I've passed the metal cable through it. The metal cable then is attached to a fix point on the wall. Here is a scheme.

enter image description here

Currently, when I plug in my motor, it just tries to turn and blocks immediately, without even pulling a little on the door knob.

What could I try to demultiply the force of my motor so I could open that door knob ?

\$\endgroup\$
2
  • \$\begingroup\$ Your motor is only going to give 160g pull at radius of 5cm. Which is not far off being 1/10 of what you need. You need a radius of something closer to 0.5cm. Can you try without the pulley? Now i see the second diagram, I see you have a 2:1 advantage there which will help, but you still need to make the pulley diameter smaller. \$\endgroup\$
    – user1844
    Oct 26, 2015 at 20:42
  • \$\begingroup\$ I smell some XY problem here. What is it that you are really trying to achieve? probably some modification of the door mechanism is easier... \$\endgroup\$
    – PlasmaHH
    Oct 26, 2015 at 20:59

2 Answers 2

3
\$\begingroup\$

Your motor can pull 800g, or 8N with a radius 1cm pulley, but you used a 5cm (diam 10cm) one, which means you are down to 1/5 of that value, being 8/5N. By using the pulley scheme that you are, the needed force is halved to 13/2N, much more than the available force.

You can either trim down the winch to about 1cm, or apply more tiers of pulleys (yet with every pulley you also add drag, so have some force extra...).

With your current pulley diameter you would need a 5-tier pulley system, so i would advocate the smaller winch. (possibly just use the motor axle, it will take more turns to winch in 10cm of wire, but it will turn more quickly if its not as near to its operational limits)

\$\endgroup\$
1
\$\begingroup\$

I would strongly suggest that a far better solution is to get a motor with some type of decently strong gear box on it. This will allow a multiplier of the available torque of the motor. This will be much more compact and simple than the elaborate cable and pulley system that will be required to use your current motor. In all likelihood the motor with the appropriate gear box will draw less current than your current motor does. And you can simply spool the cable around the output shaft of the gear box of the working bend radius of the cable can go that low.

Even better, although actuation time will be longer, is to get a motor that uses a worm gear drive. These can even hold their position when the motor is unpowered if the gear ratio is high enough.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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