a CIM motor has a torque of 343.4oz-in at stall

What does the measure of ib-in exactly mean?

Torque is measured in (units of force)-(units of distance), and the order is replacable. The measurement is simply the stall torque, and oz-in is just the units they chose to use. You could convert from there to ft-lbs, N-m, or whatever your heart desires. I hope this helps your question.

kinda, but how exactly does torque factor in to making your robot?

how would I know my robot was “at 343.4oz-in” and therefore causing a stall?

343.4oz-it simply means that at the output shaft of the motor you could attach a bar 1 inch long and it would take 343.3 oz of pressure applied to the end of that bar to stop the motor from turning. also you can divide that number by 12 and get the ft-oz and then divide it by 16 to get the ft-lbs of torque (1.78 ft-lbs). also remember gearing will change this value so if your motor has a 30 tooth gear and you meshed it with a 60 tooth gear the 2nd gear would have 686.6 in-oz of torque at stall. also remember that the motor will be drawing way more than 40 amps at stall so the electrical system wont let the motor stall anyways.

hope that helps

Forest

To answer this it might help if we use an example. I’ll choose an example that is relatively easy to implement, and might be used in a FIRST robotics competition robot if it matched the needs of a team’s game strategy.

Suppose your robot weighs 135 lb (with a battery in it) and has four wheels that share its weight equally. So each wheel supports 135/4 = 33.75 lb.

Now let’s assume your wheels have a friction coefficient (on standard FRC carpet) of 1.0, which is typical of the 6 inch diameter AndyMark FIRST kit wheel. So each wheel can develop a traction (pushing) force of up to 33.75 lb; any more force would cause it to slip on the carpet. The maximum useful wheel torque is therefore the product of the wheel radius and the slip force, 3 in x 33.75 lb = 101.25 lb-in. As sumadin and fimmel pointed out earlier, this torque could also be expressed in ounce-inches (oz-in): 101.25 lb-in x 16 oz/lb = 1620 oz-in.

Now let’s make one more assumption: we select a speed reduction ratio from the CIM motor to the wheel of 12 to 1. This ratio is typical of gearboxes that have been included in past FRC kits of parts. Using this ratio, we can calculate that the CIM motor torque required to make your wheels slip is 1620/12 = 135 oz-in. [Notice I have assumed that each wheel is driven by its own CIM motor, and I have ignored frictional losses in the gearing.]

We know that the CIM motor can develop that torque because it is less than the rated stall torque. The CIM motor can develop that torque while drawing 135/343.4 = 0.393 or 39.3% of its rated stall current. The rated stall current is found on the CIM motor data sheet and its value is 133 Ampere. So the current required to spin this robots wheels would be 39.3% x 133 Ampere = 52.3 Ampere. [This current would be drawn by each of four CIM motors, just before they began to spin freely.]

The robot’s top speed would be given by the CIM motor’s free speed x the wheel’s circumference, divided by the gear ratio: 5310 rev/minute x (1 minute/ 60 seconds) x (6 inches * PI) / 12 = 139 inches per second, or 11.6 feet per second. The robot will not actually reach this speed, because friction and other power losses will limit top motor speed to something less than the free speed. Factors that influence the choice of robot speed have been discussed in another thread recently.

Of course a practical drivetrain design needs to consider many other factors that I neglected in the simple example above. You might want to use wheels with a different friction coefficient, or use a different gear ratio. You might want to use a different number of driven wheels. You might not have equal sharing of the robot’s weight among the wheels. Any of these would complicate the analysis above, but the principles remain the same.

Chief Delphi has a very good White Papers collection. Many good references on drive train design can be found there.

Also keep in mind you don’t want to design for stall torque. Most things don’t like to operate at their limits, case in point the CIM motor… it becomes more a electro->thermal energy conversion device as opposed to an electro->mechanical energy conversion device as its torque output approaches stall.

Remember, this heating is nothing to fool around with: A CIM motor at stall eats approximately 133A… since we run CIMs at 12VDC that would make the motor a 1,596watt space heater… thats a more heat than even your biggest plug-in portable space heaters.

I posted the CIM pdf a little while ago, i’m sure you’ll find the torque curves and points of interest charts useful: http://rapidshare.com/files/51500157/CIM.pdf

I also have the big (3") CIM empirical data graphs and tables somewhere… just have to figure out where I left them :o

Also, if you are still having trouble understanding torques, I have a physics refrence (ebook) i can direct you to.

Have fun…

-q

Please remember that electrics have nothing to do with stall. Although the motor is capable of 133 amps at stall, the battery, when fully charged, can actually supply well in excess of 400 amps. The main breaker and 40 amp motor breakers won’t trip for for several seconds in this condition.

Yes in addition… if you DO get a motor to go to stall, it will most likely fry! We tried that for our arm with a mabuchi and all of the magic black smoke came out of it!

Ok we learned that when all the magic smoke comes out of a motor/wire it doesn’t work and more. So therefore you must try to keep all the magical smoke in!!! When it runs out of it… it dies! However when a CIM stalls it doesn’t die very easy

for the record torque has the units Nxm not N*m. N*M is a J (joule) which be this equation is a measure of work.

Well… unless my physics professor lied to me/was confused (I asked him last week in class) they are the same thing and you could theoretically express torque in Joules (it would just be stupid to do so).

I tell my class each year that N*m for torque and Energy may be the same units multiplied together, but when it is for Energy we can use the derived unit Joule. How is the Joule derived? It comes from the scalar product of Force and Distance. Torque is the Cross Product of Force and Distance. Probably why “BanksKid” used Nxm as being different from N*m. The point is somewhat moot since units are just labels, there is no vector quality to them; which is why we have unit vectors, bold print, arrows on top of symbols to help denote these subtlities.