- Oct 5, 2013
Here is a good tutorial on the subject.

http://lancet.mit.edu/motors/motors3.html#tscurve

Unfortunately, we don't have torque/speed/power curves for the BEST motors we are currently using. We have requested this information but don't count on it this year. You can perform your own tests to create the curve. A great exercise and excellent material for your notebook!

---In bestinc@yahoogroups.com, <dmgauntt2002@...> wrote:

A technical question was asked on the BEST Q&A:If the motor is changed to 1/2 speed (127 to 60) does the Torque also go to half torque or stay the same.

This doesn't involve the game specific rules, so I should be able to answer it here.**The quick answer**is: if you cut the applied voltage in half, both the torque and the maximum speed are cut in half.**A more complete answer**: the performance of a DC motor can be described by a torque-speed curve; this graph would show the speed at which the motor runs as a function of the torque applied to the axle. To a good approximation it is a straight line running between two points:

1) Torque="stall torque", speed=0

2) Torque=0, speed=maximum speed

Motors are typically specified in terms of the "stall torque" (in inch-pounds (English) or newton-meters (metric)), and the maximum speed (rotations per minute (RPM), or rotations per second (Hz)).

These specifications depend on the applied voltage. As a rule of thumb, when you double the voltage you will double the stall torque and you will double the maximum speed.

I would guess that when you reduce the argument to the "setMotor" function from 127 to 64 that the voltage will reduce by a factor of two, although I have never made the measurement.

Note that this does *NOT* mean that if you cut the voltage in half that the motor speed is cut in half. For example: suppose a motor's maximum torque at 7 volts is 24 inch-pounds, and the maximum speed is 42 RPM. You are applying 12 inch-pounds to the motor, and 7 volts to the motor coil. If you draw the torque-speed curve, you will set that the motor is now moving at 21 RPM. If you drop the applied voltage to 3.5 volts, the stall torque drops to 12 inch-pounds, and the maximum speed drops to 21 RPM. You are now running the motor at its stall torque, so the speed drops to zero.: the torque applied by the motor is proportional to the electric current through the motor. If the motor is not moving, this will be equal to the voltage applied to the motor divided by the electrical resistance of the motor coils. This is why the stall torque is proportional to the applied voltage.

A yet more complete answer

If the motor is allowed to move, then the motor starts to act like a generator, and produces a "back-voltage" that opposes the applied voltage. This back-voltage is proportional to the motor speed. The current running through the motor is now the applied voltage minus the back voltage, divided by the electrical resistance. If there is no torque applied to the motor, then the motor will speed up until the back voltage is equal to the applied voltage. This is why the maximum speed is proportional to the applied voltage.

In terms of equations, the torque and back voltages areT=a Icoil

where T is torque, Icoil is electrical current through the motor coil, a is the torque constant, b is the back-voltage constant, and F is the motor speed.

Vback=b F

The current is equal toIcoil=(Vapp-Vback)/Rcoil

where Vapp is the applied voltage, Vback is the back voltage, and Rcoil is the motor coil resistance.

When you wrap these togetherT=Tstall*(1-F/Fmax)

This still sweeps under the rug a lot of details having to do with friction, inductance, rotor inertia, etc, but it is a pretty good approximation to the steady-state (non-accelerating) behavior of DC motors.

Tstall=Vapp*(a/Rcoil)

Fmax=Vapp/b

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