Mass, inertia & F = ma

Newton's second law, turning

Push a mass $m$ (kg) with a force $F$ (N) and it accelerates at $a = F/m$: that's $F = m\,a$. Turning works the same way. A torque $\tau$ (N·m) on a joint gives it an angular acceleration $\alpha$ (rad/s²):

$$\tau = I\,\alpha$$

$I$ is the moment of inertia (kg·m²), the turning version of mass. The further the mass sits from the axis, the bigger $I$, so a stretched-out arm is harder to spin than a tucked-in one.

Measure it

Apply a known torque for $T$ seconds and watch the speed. If the angular velocity $\omega$ goes from $\omega_0$ to $\omega_1$, then $\alpha = (\omega_1 - \omega_0)/T$ and $I = \tau/\alpha$.

Why robots can't stop instantly

A spinning joint carries angular momentum $I\omega$, and only torque takes it away. The base motor gives at most 10 N·m, so stopping takes time: braking at $\alpha$ from speed $\omega$ takes $\omega/\alpha$ seconds and $\omega^2/(2\alpha)$ radians of travel. That's why the test push has to be followed by an equal push backwards.

The fastest move that stops

To turn a distance $D$ (rad) and stop: full torque forwards for a time $t_1$, then the same torque backwards for $t_1$. Speeding up from rest for $t_1$ covers $\tfrac12\alpha t_1^2$, and braking covers the same again, so $\alpha t_1^2 = D$:

$$t_1 = \sqrt{\frac{D}{\alpha}} = \sqrt{\frac{D\,I}{\tau}}$$

Nothing is measured during the move: it is open loop, only as good as your $I$. A 5 % error in $I$ puts the base 4.5° off.

Speed-limited moves like this one come back in Trajectory Generation, step 3.

Your task

The shoulder, elbow and wrist are held for you, so only the base turns.

1. The program pushes the base with TEST_TORQUE for TEST_TIME. From the speeds before (qd0[0]) and after (qd1[0]), set BASE_INERTIA (within 5 %).
2. Write turn_base(distance, torque) with two calls to push_base.
3. Choose MOVE_TORQUE (up to 10 N·m). The base must stop within 3° of 90°, slower than 0.05 rad/s, and the move must take less than 1.5× the time at full torque.