### Author Topic: Torque from pulling wires help  (Read 1144 times)

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#### Hal_Emmerich

• Jr. Member
• Posts: 25
##### Torque from pulling wires help
« on: November 21, 2010, 10:38:22 PM »
Hello, I am posting about the SMA actuator robotic arm (seen here) that I have received so much help with. I am having some final problems figuring out torque and power consumption, and any help is appreciated.

I am using 0.006" diameter nitinol wires and about 24 inches in length, and here are their specs:

Ohms/inch = 1.4
Pull force = 0.71 pounds
Approx. current for 1 second contraction = 410 mA

When the wires contract, they pull the lower arm up from a point approx. 1 inch out from the elbow (which is my axis of rotation). The forearm is nearly straight compared to the upper arm, and the wires are inside the upper arm and pulling the forearm at about a 20 degree angle.

Here is my question: how do I calculate the amount of torque one wire can produce from pulling at this point on my arm?

I was hoping to use the moment arm tutorial, but I can't figure out how to apply it to find out the torque each of my wires can produce and with that the load they can lift when pulling that close to the rotational axis.

My arm will weigh about 3 lbs total, with 1.5 lbs each for the upper and lower arm halves. The hand weight is part of the lower arm (forearm) weight (its only about 0.05 lbs), and I was hoping to lift about 5 lbs with it.

I know this seems like a problem  I should solve on my own, but I always get these calculations wrong. Any help is greatly appreciated. Thank you!
« Last Edit: November 21, 2010, 10:46:23 PM by Hal_Emmerich »

#### waltr

• Supreme Robot
• Posts: 1,944
##### Re: Torque from pulling wires help
« Reply #1 on: November 22, 2010, 09:07:02 AM »
The moment arm tutorial does have the info you need to calculate the torque. Also try reading Wiki on torque. http://en.wikipedia.org/wiki/Torque

So a little help. 0.71 lbs force X 1 inch from the pivot is 0.71 lb-in of torque. (see on-line converters if you need this in other units).
http://www.unitconversion.org/moment-of-force/newton-meters-to-pound-force-foots-conversion.html
http://en.wikipedia.org/wiki/Pound-force

The Imperial Units can be very confusing so try to use the metric units instead.
0.71 lbf = 3.15 Newton, 1in = 25mm = 0.025m
torque = 0.0788 N-m

The remained of your problem doesn't have enough information.
How long is the arm that weighs 3 lbs? Is the weight evenly distributed? (I assume so if 1.5 lbs  are on each half).
Convert to metric: 1.5lb weight = 0.68kg. Force (Newton) = 0.68kg * 9.8m/s^2 = 6.68 N.

So lets assume that the arm is 12 inches (0.3m) long. The torque required to lift just the arm is 6.68 N X 0.3m (center of mass) = 2 N-m of torque.
This is way too large for the wire to lift by a factor of 2.5.......and is not even taking into consideration the additional 5 lb payload. Now this was just guessing on the length of the arm. You can either or both shorten the arm and/or reduce the weight of the arm but you still have the payload.

You could also increase the torque produced by using multiple strands of the muscle wire, 3 strands would just do.

#### Hal_Emmerich

• Jr. Member
• Posts: 25
##### Re: Torque from pulling wires help
« Reply #2 on: November 22, 2010, 05:13:00 PM »
Thank you very much, waltr! Your calculations help me immensely! And yes, the weight is evenly distributed, ao your calculations are perfect for me.

I have to do a presentation on this stuff soon, and since my physics knowledge has always been poor, I am very grateful to have your data for reference and guidance.

Thank you again!

#### waltr

• Supreme Robot
• Posts: 1,944