Quote:
Originally Posted by Ether
@Eric: In other words, when you're cruising along at your desired 3 m/s, you need some headroom to increase the voltage.
Suppose you leave 2 volts headroom, so you're operating at 10 volts at 3 m/s (I have no idea if 2v headroom is enough or too much, this is just for sake of discussion).
CIM free speed at 10v is 4425 rpm. To go 3 m/s at that motor rpm with 12.5" dia wheels, you'd need a gear ratio of 24.5:1, not 34:1.
Of course, at free speed a motor is producing no torque output. So let's say 90% of 10v free speed, or 3983 rpm CIM speed (again, I do not know if that is sufficient1). Now you need a gear ratio of 22.1:1.
1At 10v and 3983 rpm, each CIM produces about 28.6 ozin torque. Run this through the 22.1:1 geartrain and assume 10% torque loss in the drivetrain and you've got about 3 ft-lb torque at each wheel. For a 12.5: dia wheel this translates to about 5.7 lb motive force per wheel, or 11.4 lb total for both wheels. Is that enough to move your 100 lb robot+payload at 3 m/s up a 5% incline on a gravel surface?
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@Ether
Suppose all of these calculations are correct and work for the robot. How do I determine if the battery will last 45min?
PS: I'm an electrical engineer intern so I don't really know a lot about the mechanical side of things so this might be an easy question for you to answer, but when you say 11.4 lb total motive force does that mean that the force would be able to move 11.4 lbs? so therefore not enough to move my 100lb robot+payload?