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by Robotbeat
1950 days ago
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The other part is that there's a straightforward trade-off between specific power and efficiency. Two motors on the same shaft can each be run at half the current, and since power loss due to resistance is: P=I^2*R, your losses due to resistance would halve. (There are, of course, other loss mechanisms.) So it's good to start out with a really high specific power because you can often trade that back for efficiency. |
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