Building Robots at School

September 15, 2008

Torque… it all comes down to Torque.

Filed under: robots,teaching,Tech Ed — dtengineering @ 10:07 pm

One of the most useful concepts in building competitive robots is the concept of torque.  Oddly, for a concept this simple but powerful, it isn’t officially covered in the BC curriculum until Physics 12… and then often towards the end of the year.  That is why I always talk about torque near the beginning of any robotics project, even with my grade 9’s.

Torque is the amount of “twist” about a pivot point, and is determined by two factors:  the amount of force being exerted and the distance that force is away from the pivot point.  Consider example “A”, a “teeter-totter”.  These things used to exist in all sorts of kid’s playgrounds, and were always a great opportunity to see if you could actually launch your little brother into the air.  Playgrounds might be safer now that they are gone, but a whole generation is deprived of the deep understanding of physics that comes from having the person on the other end jump off while you are six feet in the air.  There is nothing like landing flat on your butt on packed dirt to leave a lasting impression of the importance of torque in your brain.

But I digress.  The formula for torque is simply “Force x Distance”, or, in physics symbology, “τ=F*D”.  Thus in example “A”, the teeter totter is balanced… the torque on one side 20Nm clockwise about the pivot point, is countered by the 20Nm torque in the other direction.  This should be pretty obvious, but what isn’t always obvious is that you could replace the force and distance on one side with the driveshaft of a motor.  So long as the motor could produce 20Nm of torque, the system would be balanced.

Torque also describes the amount of “pushing force” you can get from your robot wheels.  In example “B”, a wheel is driven by a 20Nm motor.  Since the wheel has a radius of 2m, you will get a pushing force of 10N.  That’s not bad… the Newton is the metric unit of force, and… roughly translated, one Newton is roughly equivalent to the gravitational force exerted by a 100g weight near the surface of the Earth.  So this robot would push with a force equivalent to “one kg” (10N≈1kg).  If you don’t quite comprehend the difference between a Newton and a kg, go bug a science teacher until they explain it to you.  They will be delighted that you asked.  Anyways, back to that wheel… what would happen to the amount of pushing force if you used a 1m diameter wheel?  What would be the “trade off” for the extra pushing force?  And what idiot came up with an example using a 2m radius wheel?  How big would that robot have to be?

Finally, torque also describes the situation where you are attempting to use a robot arm to lift an object, as shown in example “C”.  By understanding torque it is easier to understand how a counterbalance (an opposing force on the end opposite the load) can make it easier for your robot to lift heavy loads.  Note that the counterbalance doesn’t have to be a weight… you could attach elastic bands between the tower and the “short” end of the robot arm.  What isn’t immediately apparent, however, is that torque will also determine whether or not this robot will flip forward as the arm lifts the load (you don’t have enough information to calculate that here, but it fairly easy to do… just consider the front wheel of the robot to be the pivot point.)

One thing to keep in mind is that our good friends south of the border, who long ago fought a bitter war to gain independence from the British Empire, are ironically about the only nation left on the planet still using the old British “Imperial” system of measurement.  So if you happen to be talking to Americans, you’ll find that they use terms such as “foot-pounds” and “ounce-inches” to describe torque.  It’s the same concept, of course, but in different units.

These are all simple torque calculations (we haven’t discussed the weight of the arm, for instance, or how to deal with loads that are at an angle other than 90° to the level arm), but as students develop a better understanding of torque they can apply it to almost every aspect of robot design…. but without at least a basic understanding of torque, it is almost impossible to design a robot.

P.S.  A great question was asked at the end of my torque lesson the other class, “But Mr. Brett, isn’t the formula for work (F*d) the same as the formula for torque (F*d)?”  The answer is both yes, and no… the formulas are the same, but in work, the “d” refers to the distance an object moves, which is more correctly referred to as Δd, “the change in distance”.  (Alternatively one may refer to it as Force*Displacement, if one simply insists on using F*d).


1 Comment »

  1. […] If you can’t get more kits then spend a couple weeks working on design problems, studying torque and other important concepts and you’ll find that only the truly hardcore roboteers will […]

    Pingback by Registering for VEX in BC: 2011/2012 Version « Building Robots at School — September 17, 2011 @ 9:37 am | Reply

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

Blog at

%d bloggers like this: