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How to simply accomplish moving a swinging arm back and forth 90 deg via switch?


#1

Hi,

Update: what about a linear actuator? Which I know nothing about!

I want a 3 foot arm running parallel to the ground to move 90 degrees back and forth, like a windshield wiper - but not quite. One press of a button moves the arm 90 degrees - then the arm stops and stays rigidly in place. Another press of the button moves the arm back where it stops. Speed of this movement should be slow, say 2-3 RPMs. The load at the end of the 3 foot arm is 5-10#. I know nothing about math, physics, or motors or coding. I am looking for the simplest option, please.

Which motor would do this? AC? DC? How much torque would I need? Battery run or plug in? Gear box - don’t really know what that is - would the motor mount at the fulcrum? Guess I would need a counterweight too - or do I make the fulcrum a hinge and attach a rod say in the middle of the arm like the mechanism of an automatic door opener. Trying to keep this under 50 bucks. Also fire safety is a big deal.

Is there a switch that will reverse the motor? Do I need a servo and some computer programming or can this be done more simply? Simple is preferred. I really don’t want programming.
Thank you very much.


#2

Hello.

We do not have anything pre-made like that, and it will probably be difficult to get something like for the budget you have. Some of the questions you have are things you will have to figure out as they will be specific to your project, for example your torque requirements, or whether to use a battery or plug-in. If you want to try and figure out your torque requirements, you might find this blog post helpful.

By the way, your questions were pretty broad, so I moved your post to Robotics and electronics discussion.

Grant


#3

So the weight. I will assume that you have structure that this will be mounted on, so I will not talk to that. I am also assuming that by “parallel to the ground” you mean the arm moves like it’s sitting on a table top. A windshield wiper would be moving perpendicular to the ground.
Any object changing speed has an acceleration acting on it. A ball dropped has gravity 32ft/sec^2 acting on it. Force = massacceleration F=ma. For fun lets use 1g acceleration on your mass. That make the number kind of easy. 10#'s at 3ft =30 ft-lbs torque, or 3012 = 360 in-lbs. But maybe 1g is too fast. If you were do just drop a weight, the formula for velocity is V=Vi +aT . Vi Velocity initial is 0 so V=aT. For 1g after falling 0.1 sec: 32ft/s^2 * 0.1 s = 3.2 ft per sec’ So even though we are going in a circle a 1g acceleration gets to 3.2ft/sec. Remember dia*pi = circumference. So 6ft * 3.1415 is 18.8 ft around. 5.8 sec per revolution. you wanted 20 to 30 seconds per revolution. You don’t need anything near 1g… How about 1/10 that… you would accelerate for 0.25sec to get to about 0.8ft/sec or 23.5 sec/rev. So about 1/10th gravity, so we could make a guess that 36 in-lbs (1/10 above) would be a reasonable torque. Again this is for the mass not changing in height to the ground. (Rotating flat on table)

In this case I would look for a surplus 2rpm 12volt gear motor producing 75 in-lbs. It’s DC so reversible.

Hope this can help.


#4

Thank you q silver!