Quote:
Originally Posted by DampRobot
So, theoretically, we should be designing to maximize power out of our battery as well as out motors? IE load the battery such that V^2/R is maximized when you think you'll need the most power as possible (pushing matches).
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I'm not exactly sure what you mean here, as the electrical power output of the battery increases as more current is drawn. Regardless of the power though, if you are in a pushing match, your main concern would be amperage. With DC brushed motors, torque is proportional to current. Gearing slower gives you more torque for the same current, or less current for the same torque.
IMO, it would be better to focus on reducing resistance on the robot. At 300 amps, each extra milli-ohm is .3 volts. Just a few extra milli-ohms to the robot can make the difference between brownout and not. Also relevant here is acceleration in high speed robots. Since increased resistance means less current for a given voltage and current=torque, a few milli-ohms can mean a decent drop in the available torque for acceleration.