Thread: 2 Speeds
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Unread 26-05-2011, 23:38
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Jared Russell Jared Russell is offline
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Re: 2 Speeds

Yep, your math is pretty much dead on for a theoretical system. For a real world design problem, though, I would also consider two (very important) additional factors:

1) Efficiency. You can expect to chop 10-40% off the top of your total available power due to transmission inefficiencies. The exact number has to do with the number of reduction stages, types of reductions, tolerances, alignment, and the like.

2) Current draw. If you gear so that your motors are just barely traction-limited at stall in low gear, you won't be able to push (or resist pushing) all day - you will be drawing ~120A per motor and will eventually pop your breakers. It won't be an instantaneous thing, but eventually they will trip (temporarily disabling the motor while it cools). If you design for 40A per motor or below, you will be able to push all day without popping a breaker (in theory). Whether this matters or not depends on your application.

So there are definitely justifiable reasons for going lower.

Quote:
Originally Posted by lemiant View Post
When graphing, I found that the value of R that rendered the highest TgFS, would always result in Ds being twice TgFS. I have not yet figured out how to make a rigorous proof of this, if you could help here I would be very grateful.
What you discovered is indeed true. It has to do with the physics of DC motors. The angular speed of the motor decreases linearly with load torque. The total output power of the motor is the product of angular speed (rad/sec) and torque (N*m). (1 N*m/s = 1 Watt). You can always trade off between speed and torque with gearing on the output, but you often** want to make sure that the torque on the motor at the input to the gearing is as close as possible to the "max power" point. As it happens, max power is maximized when the motor is running at 1/2 free speed and 1/2 stall torque (on your triangle graph, imagine graphing x*y for each point on the hypotenuse-that is your power curve).

** often but not always, as the most EFFICIENT point on the power curve (in terms of transforming electrical power into mechanical power) is NOT at the max power point. Different applications will have different design criteria.

Last edited by Jared Russell : 27-05-2011 at 11:19. Reason: fixed typo