Question 3 of 7: Two-segment cubic spline with continuous acceleration (20 marks)

Question 3: Two-segment cubic spline with continuous acceleration (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. Three prescribed joint angles and two segment durations. Each segment carries its own local time origin, so the second cubic is written in $t$ measured from the via point, not from the start of the motion.

Find. All eight coefficients $a_{10}\ldots a_{13}$ and $a_{20}\ldots a_{23}$ of the two cubics, together with the via-point velocity and acceleration that the continuity conditions force on the path.

Approach. Two cubics carry eight unknown coefficients, so exactly eight constraints are needed: three prescribed angles (start, via, goal, the via counted once for each segment), rest at both ends of the motion, and matching of velocity and of acceleration where the segments join.

1. Write the two segments and their derivatives. With local time $t$ on each segment, $$\theta_{1}(t)=a_{10}+a_{11}t+a_{12}t^{2}+a_{13}t^{3},\qquad \dot{\theta}_{1}=a_{11}+2a_{12}t+3a_{13}t^{2},\qquad \ddot{\theta}_{1}=2a_{12}+6a_{13}t,$$ and identically for $\theta_{2}(t)$ with the coefficients $a_{20}\ldots a_{23}$. The manipulator starts at rest and finishes at rest; nothing is prescribed about the rate at the via point, which is precisely what makes a spline smoother than two independent cubics stitched together.
2. List the eight constraints. In order, they are $\theta_{1}(0)=\theta_{0}$, $\theta_{1}(t_{f1})=\theta_{v}$, $\theta_{2}(0)=\theta_{v}$, $\theta_{2}(t_{f2})=\theta_{g}$, $\dot{\theta}_{1}(0)=0$, $\dot{\theta}_{2}(t_{f2})=0$, $\dot{\theta}_{1}(t_{f1})=\dot{\theta}_{2}(0)$ and $\ddot{\theta}_{1}(t_{f1})=\ddot{\theta}_{2}(0)$. The last of these is the condition the question names explicitly: it is what removes the torque step a merely velocity-continuous path would demand of the actuator.
3. Read off the four constraints that solve themselves. Evaluating at the local time origins gives $a_{10}=\theta_{0}$, $a_{11}=0$ and $a_{20}=\theta_{v}$ immediately, so $$\boxed{a_{10}=-20^{\circ},\qquad a_{11}=0,\qquad a_{20}=45^{\circ}.}$$ Only five unknowns remain, and they are linear in the data.
4. Solve the remaining system. Because the two durations are equal, $t_{f1}=t_{f2}=t_{f}=4\,\text{s}$, the system collapses to the standard closed form $$a_{12}=\frac{12\theta_{v}-3\theta_{g}-9\theta_{0}}{4t_{f}^{2}},\qquad a_{13}=\frac{-8\theta_{v}+3\theta_{g}+5\theta_{0}}{4t_{f}^{3}},\qquad a_{21}=\frac{3(\theta_{g}-\theta_{0})}{4t_{f}},$$ $$a_{22}=\frac{-12\theta_{v}+6\theta_{g}+6\theta_{0}}{4t_{f}^{2}},\qquad a_{23}=\frac{8\theta_{v}-5\theta_{g}-3\theta_{0}}{4t_{f}^{3}}.$$ Substituting $\theta_{0}=-20^{\circ}$, $\theta_{v}=45^{\circ}$, $\theta_{g}=25^{\circ}$ and $t_{f}=4\,\text{s}$ gives, for the first segment, $$a_{12}=\frac{540+(-75)+180}{64}=\frac{645}{64},\qquad a_{13}=\frac{-360+75-100}{256}=-\frac{385}{256},$$ $$\boxed{a_{12}=10.0781\ \text{deg}\cdot\text{s}^{-2},\qquad a_{13}=-1.5039\ \text{deg}\cdot\text{s}^{-3}.}$$
5. Finish the second segment. The same substitution gives $$a_{21}=\frac{3(25+20)}{16}=\frac{135}{16},\qquad a_{22}=\frac{-540+150-120}{64}=-\frac{255}{32},\qquad a_{23}=\frac{360-125+60}{256}=\frac{295}{256},$$ $$\boxed{a_{21}=8.4375\ \text{deg}\cdot\text{s}^{-1},\qquad a_{22}=-7.9688\ \text{deg}\cdot\text{s}^{-2},\qquad a_{23}=1.1523\ \text{deg}\cdot\text{s}^{-3}.}$$ Notice that $a_{21}$ is not zero: the joint sweeps through the via point at $8.4375\,\text{deg}\cdot\text{s}^{-1}$ rather than stopping there, which is the whole purpose of splining two segments instead of running two separate point-to-point moves.
6. Check every constraint before quoting the answer. At $t=t_{f1}$ the first cubic gives $-20+10.0781(16)-1.5039(64)=45^{\circ}$, its slope is $2(10.0781)(4)+3(-1.5039)(16)=8.4375\,\text{deg}\cdot\text{s}^{-1}=a_{21}$, and its acceleration is $2(10.0781)+6(-1.5039)(4)=-15.9375\,\text{deg}\cdot\text{s}^{-2}=2a_{22}$. At the end of the second segment $\theta_{2}(4)=25^{\circ}$ and $\dot{\theta}_{2}(4)=8.4375-63.75+55.3125=0$. All eight conditions are satisfied exactly.

The path is worth reading as well as tabulating. Because the goal lies below the via point, the second cubic must decelerate hard: it overshoots to $47.44^{\circ}$ at $t=0.610\,\text{s}$ after the via point before falling back to the goal. That overshoot is a genuine property of a cubic spline through a via point, not an arithmetic slip, and it is the reason a designer who needs the tool to stay inside a corridor uses a higher-order or a linear-with-parabolic-blend scheme instead.