Question 3 of 7: Cubic joint trajectory for a rest-to-rest move

Question 3: Cubic joint trajectory for a rest-to-rest move (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A single revolute joint that starts and finishes at rest, with the start angle, goal angle and travel time all specified.

Find. The four cubic coefficients, and the position, velocity and acceleration profiles they produce.

Approach. Four boundary conditions determine the four coefficients uniquely; impose them in the order $\theta(0),\dot\theta(0),\theta(t_{f}),\dot\theta(t_{f})$ so that the first two fall out immediately and only a 2×2 system remains.

1. Write the derivatives. From $\theta(t)=a_{0}+a_{1}t+a_{2}t^{2}+a_{3}t^{3}$, $$\dot\theta(t)=a_{1}+2a_{2}t+3a_{3}t^{2},\qquad \ddot\theta(t)=2a_{2}+6a_{3}t.$$
2. Apply the two initial conditions. $\theta(0)=\theta_{0}$ gives $a_{0}=\theta_{0}=-5^\circ$ at once, and $\dot\theta(0)=0$ gives $a_{1}=0$. The cubic therefore has no linear term, which is the signature of a start from rest.
3. Apply the two final conditions. With $a_{0}$ and $a_{1}$ known, $$\theta(t_{f})=\theta_{0}+a_{2}t_{f}^{2}+a_{3}t_{f}^{3}=\theta_{f},\qquad \dot\theta(t_{f})=2a_{2}t_{f}+3a_{3}t_{f}^{2}=0.$$ Solving the pair gives the standard rest-to-rest result $$a_{2}=\frac{3(\theta_{f}-\theta_{0})}{t_{f}^{2}},\qquad a_{3}=-\frac{2(\theta_{f}-\theta_{0})}{t_{f}^{3}}.$$
4. Substitute the numbers. The travel is $\theta_{f}-\theta_{0}=80-(-5)=85^\circ$ over $t_{f}=4\ \text{s}$, so $$a_{2}=\frac{3(85)}{16}=15.9375\ \text{deg/s}^{2},\qquad a_{3}=-\frac{2(85)}{64}=-2.65625\ \text{deg/s}^{3}.$$ Collecting all four, $$\boxed{\theta(t)=-5+15.9375\,t^{2}-2.65625\,t^{3}\ \ [\text{deg}]}$$
5. Verify the endpoints. $\theta(4)=-5+15.9375(16)-2.65625(64)=-5+255-170=80^\circ$ and $\dot\theta(4)=2(15.9375)(4)+3(-2.65625)(16)=127.5-127.5=0$, so both goal conditions are met exactly.
6. Extract the profile features that the plot must show. The velocity $\dot\theta=31.875\,t-7.96875\,t^{2}$ is a downward parabola peaking at $t=t_{f}/2=2\ \text{s}$, where $$\dot\theta_{\max}=\frac{3(\theta_{f}-\theta_{0})}{2t_{f}}=31.875\ \text{deg/s},$$ and the acceleration $\ddot\theta=31.875-15.9375\,t$ is a straight line running from $+31.875\ \text{deg/s}^{2}$ at $t=0$ to $-31.875\ \text{deg/s}^{2}$ at $t=4\ \text{s}$, crossing zero at the mid-point. At $t=2\ \text{s}$ the position is $37.5^\circ$, the exact average of the two endpoints — a cubic rest-to-rest profile is antisymmetric about its mid-point.
7. Convert to radians for a controller. Multiplying by $\pi/180$, $a_{0}=-0.08727\ \text{rad}$, $a_{1}=0$, $a_{2}=0.27816\ \text{rad/s}^{2}$ and $a_{3}=-0.04636\ \text{rad/s}^{3}$; the peak rate is $0.5564\ \text{rad/s}$.

The profile is smooth in position and velocity but its acceleration jumps discontinuously at $t=0$ and $t=t_{f}$, which excites structural modes on a real arm. Where that matters the same boundary-value method is applied to a quintic with two extra conditions $\ddot\theta(0)=\ddot\theta(t_{f})=0$.