Quote:
Originally Posted by neoshaakti
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?
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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.