Bonus: Motors & gearboxes

Motors are fast and weak

An electric motor spins fast but pushes gently. A simple model: at standstill it gives its stall torque $\tau_s$, and the faster it spins, the less it can give, down to nothing at its no-load speed $\omega_0$. At motor speed $\omega_m$:

$$\tau_{\max} = \tau_s\left(1 - \frac{\omega_m}{\omega_0}\right)$$

This step's base motor has $\tau_s = 0.1$ N·m and $\omega_0 = 200$ rad/s: about as strong as a fingertip, and as fast as a cordless drill.

Gearboxes trade speed for torque

A gearbox with gear ratio $N$ turns the motor $N$ times for each turn of the joint. The joint gets $N$ times the torque at $1/N$ of the speed, so at joint speed $\omega$ it can have

$$\tau \le N\tau_s\left(1 - \frac{N\omega}{\omega_0}\right)$$

Braking (torque against the motion) is limited to $N\tau_s$ at any speed.

Gear ratio $N$	5	40	160
Torque at rest	0.5 N·m	4 N·m	10 N·m (the joint's limit)
Top speed	40 rad/s	5 rad/s	1.25 rad/s

Too little gearing can't get the arm moving; too much tops out at a crawl. The fastest move is somewhere in between.

The fastest stop

Go flat out, then brake as hard as possible at the last moment. With inertia $I$ (kg·m²), braking torque $\tau_b$ decelerates the base at $\alpha_b = \tau_b / I$. From speed $\omega$, stopping takes (as in step 3)

$$d_{\text{stop}} = \frac{\omega^2}{2\,\alpha_b} = \frac{\omega^2 I}{2\,\tau_b}$$

so brake as soon as $d_{\text{stop}}$ reaches the distance left.

Your task

1. Write stopping_distance(speed, brake_torque, inertia).
2. The program times a 90° move with each ratio in RATIOS and plots the speed. Read the times, then set GEAR_RATIO.
3. The final move must stop within 3° of 90° in less than 1.2× the fastest time any gear ratio allows, and stay there.