Suppose I have a robot of a given mass, and I'm choosing between 2 different wheels and 2 different motors to put on it. For each wheel I have the diameter, and for each motor I know the stall torque and free speed. How would I figure out which motor and wheel combination will make the robot move the fastest?
My calculations show to use the big motor with the big wheels, but the small motor with the small wheels goes faster than the big motor with the small wheels. I am not sure my calculations are correct, I need to know the correct way to work this problem.
Best Answer
Regardless of the size of the wheels, and ignoring air resistance, if the motor is making $P = T(\omega)\;\omega$ power then the acceleration is
$$ a= \frac{T(\omega)\; \omega}{m v} $$
The motor speed is $\omega = \frac{v}{r} $ where $r$ is the wheel radius. If the torque at $\omega=0$ is $T_0$ and the motor speed at $T=0$ is $\omega_0$ then the torque function is
$$ T(\omega) = T_0 \left( 1- \frac{\omega}{\omega_0}\right) $$
The time it takes to reach a certain speed $v$ is
$$ t = \int_0^v \frac{1}{a}\;{\rm d}v $$ $$ t = \int_0^v \frac { m v } { T_0 \left( 1- \frac{v}{\omega_0\,r}\right) \frac{v}{r} } \; {\rm d} v $$ $$ t = \frac{m \omega_0 r^2}{T_0} \ln\left(\frac{\omega_0 r}{\omega_0 r - u} \right) $$
or
$$ v(t) = v_0 \left( 1 - \boldsymbol{e}^{-\frac{T_0}{r} \frac{t}{m v_0} } \right) $$ where $v_0 = \omega_0\,r$ is the theoretical top speed.
So to get to $99$% of top speed you need
$$ t_{99} = \frac{m\, r^2\, \omega_0} {T_0}\; 2\ln(10) $$
From here you plug in your values and see which one has the highest top speed and which one has the highest acceleration (least time).