# How do I determine the curve of the transmission output on a torque-angular velocity graph for a DC motor?

I'm doing a homework problem that involves motor characteristic graphs. For this problem, I have a motor with stall torque of 8 N-cm and a no-load angular velocity of 80 rad/s. The motor powers a 4-wheel drive robot that requires an input of 60 N-cm from the motor to power all the wheels. Additionally, the robot has a transmission with a gear ratio R = 20 and is frictionless.

I need to find the point where the transmission output line intersects the horizontal "wheel requirement" line (= 60 N-cm) so that I can find the velocity of the robot at this point (called the operating point).

The issue I'm having is that I don't know the equation of the transmission line. All I know is that the $t(0) = t_s$ (read as "torque at angular velocity of 0 rad/s = stall torque"), $t(w_n) = 0$ (read as "torque at no-load speed = 0 N-cm"), and that $t(w_i) = 60$ N-cm (read as "torque at angular velocity of interest = 60 N-cm"). Once I know the stall torque and no-load speed of the transmission output line, I can figure out the equation of the line and the angular velocity at the operating point.

• R=20 means for each revolution of the motor the transmission revolves 72 degrees with 20 times the torque (no losses because frictionless). So the new torque-velocity chart will pass 1600cm on torque axis and 20 rad/sec on velocity axis. – Gregory Kornblum Sep 30 '15 at 3:38

The maximum torque is $T=8 Ncm$ (with the gear ratio - $T_r=160 Ncm$) and the no load velocity is $w = 80rad/s$ (with the gear ratio $w_r=2rad/s$). You need to find $T_f=60 Ncm$ and the formula is:
$$T_f = T_r - w_f \cdot \frac{T_r}{w_r}$$
where $w_f$ is your angular velocity where the torque is $60Ncm$