All joint-3 angles reaching a given position (20 marks): 22-Mec-B12 Robotics, Undated paper, Q4

Question 4: All joint-3 angles reaching a given position (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 four-revolute arm whose link parameters are all zero except the four listed, and a target position of the origin of frame $\{4\}$ expressed in the base frame.

Given data - nonzero link parameters
Row $i$	$\alpha_{i-1}$	$a_{i-1}$	$d_{i}$	$\theta_{i}$
1	$0$	$0$	$0$	$\theta_{1}$
2	$-90^{\circ}$	$0$	$1$	$\theta_{2}$
3	$45^{\circ}$	$0$	$0$	$\theta_{3}$
4	$0$	$1$	$0$	$\theta_{4}$
target	$^{0}P_{4ORG}=[0.0,\;1.0,\;1.414]^{T}$, all joints limited to $\pm180^{\circ}$

Find. Every value of $\theta_{3}$ inside the joint limits for which the origin of frame $\{4\}$ can occupy the stated position.

Figure 4.1 — the outboard chain of the 4R arm, drawn in the plane containing $\hat{Z}_{2}$ and $\hat{Z}_{3}$. The offset $d_{2}$ and the link $a_{3}$ form a fixed triangle whose closing side is $^{0}P_{4ORG}$; only $\theta_{3}$ changes its length, and the two solution branches (navy and amber) close the same length in different directions.

Approach. Build $^{0}P_{4ORG}$ outward from the tool, then take its squared magnitude. Joint 1 only spins the whole arm about $\hat{Z}_{0}$ and joint 2 only spins the outboard chain about $\hat{Z}_{2}$, so neither can change the distance from the base; the distance is a function of $\theta_{3}$ alone, and one scalar equation delivers the answer.

Place the tool point in frame $\{3\}$. Only $a_{3}=1$ separates frame $\{4\}$ from frame $\{3\}$, so $$^{3}P_{4ORG}=\begin{bmatrix}a_{3}\\0\\0\end{bmatrix} =\begin{bmatrix}1\\0\\0\end{bmatrix}.$$ Joint 4 turns about $\hat{Z}_{4}$ through that very origin and therefore cannot move it - which is why a 4R arm has only three position degrees of freedom and $\theta_{4}$ never enters this question.
Carry it into frame $\{2\}$. The link transform contributes the twist $\alpha_{2}=45^{\circ}$ and the joint rotation $\theta_{3}$, with no offsets, so $$^{2}P_{4ORG}=R_{X}(\alpha_{2})\,R_{Z}(\theta_{3})\begin{bmatrix}1\\0\\0\end{bmatrix} =\begin{bmatrix}\cos\theta_{3}\\ \sin\theta_{3}\cos\alpha_{2}\\ \sin\theta_{3}\sin\alpha_{2}\end{bmatrix}.$$ The link of length $a_{3}$ sweeps a cone of half-angle $45^{\circ}$ about $\hat{Z}_{3}$ as $\theta_{3}$ runs through its range.
Carry it into frame $\{1\}$. Adding the offset $d_{2}=1$ along $\hat{Z}_{2}$ and then the twist $\alpha_{1}=-90^{\circ}$ with the joint rotation $\theta_{2}$, $$^{1}P_{4ORG}=R_{X}(\alpha_{1})\,R_{Z}(\theta_{2}) \left(\;{}^{2}P_{4ORG}+\begin{bmatrix}0\\0\\d_{2}\end{bmatrix}\right).$$ Both operations are rotations, so they leave the length of the bracketed vector untouched - and that observation is the whole solution.
Take the squared distance from the base. Since $^{0}P_{4ORG}=R_{Z}(\theta_{1})\,{}^{1}P_{4ORG}$ is another pure rotation, $$\left|^{0}P_{4ORG}\right|^{2} =\left|\;{}^{2}P_{4ORG}+d_{2}\hat{Z}_{2}\right|^{2} =a_{3}^{2}+d_{2}^{2}+2a_{3}d_{2}\sin\alpha_{2}\sin\theta_{3},$$ because the cross term keeps only the component of $^{2}P_{4ORG}$ along $\hat{Z}_{2}$. With the given numbers, $$\boxed{\left|^{0}P_{4ORG}\right|^{2}=2+\sqrt{2}\,\sin\theta_{3}.}$$ The result depends on $\theta_{3}$ and on nothing else, exactly as anticipated.
Impose the target distance. The printed target has $$\left|^{0}P_{4ORG}\right|^{2}=0^{2}+1^{2}+1.414^{2}=2.9994\approx 3,$$ the small shortfall being nothing more than $1.414$ standing in for $\sqrt{2}$. Setting $2+\sqrt{2}\sin\theta_{3}=3$ gives $$\sin\theta_{3}=\frac{1}{\sqrt{2}}=0.7071.$$
Take every branch the joint limits allow. A sine equation has two solutions per revolution, $\theta_{3}$ and $180^{\circ}-\theta_{3}$, and both lie inside the stated $\pm180^{\circ}$ range: $$\boxed{\theta_{3}=45^{\circ}\quad\text{and}\quad\theta_{3}=135^{\circ}.}$$ Using the printed $1.414$ literally instead of $\sqrt{2}$ moves these to $44.97^{\circ}$ and $135.03^{\circ}$, a difference of three hundredths of a degree that no candidate is expected to report. There are exactly two values; no other $\theta_{3}$ in the range puts the tool origin at the required distance from the base, and for each of them $\theta_{1}$ and $\theta_{2}$ remain free to steer the direction.

A quick sanity check on the pictured pose confirms the geometry. Putting $\Theta=[0,0,90^{\circ},0]^{T}$ into the chain gives $^{0}P_{4ORG}=[0,\,1.707,\,-0.707]^{T}$, whose squared length is $2+\sqrt{2}=3.414$ - precisely what the boxed law predicts at $\sin\theta_{3}=1$, the farthest the tool origin can ever get from the base. The nearest it can get is $\sqrt{2-\sqrt{2}}=0.765$ at $\theta_{3}=-90^{\circ}$.

Check: the printed target vector, component by component. The magnitude condition above is the only thing $\theta_{3}$ controls, and it is what the question asks for. Reading the three printed components literally, however, demands $^{0}P_{4ORG}\cdot\hat{Z}_{0}=1.414$, and $\theta_{1}$ spins the arm about $\hat{Z}_{0}$ without changing that component: at $\theta_{3}=45^{\circ}$ or $135^{\circ}$ its largest attainable magnitude is $\sqrt{1-\tfrac{1}{2}\sin^{2}\theta_{3}}=0.866$. The stated direction is therefore outside this arm's reach as the parameters are printed, although the stated distance is comfortably inside it; the same target with its components permuted so that the zero falls on $\hat{Z}_{0}$ is reachable, and gives the same two answers. A transposed component ordering is the most likely explanation. Nothing above depends on the printed direction: $\theta_{3}=45^{\circ}$ and $135^{\circ}$ follow from the distance alone.

Question 4 - results
Quantity	Expression	Value
squared reach law	$a_{3}^{2}+d_{2}^{2}+2a_{3}d_{2}\sin\alpha_{2}\sin\theta_{3}$	$2+\sqrt{2}\sin\theta_{3}$
required $\left\|^{0}P_{4ORG}\right\|^{2}$	$0+1^{2}+1.414^{2}$	$2.9994\approx3$
required $\sin\theta_{3}$	$(3-2)/\sqrt{2}$	$0.7071$
solution branch 1	$\arcsin(0.7071)$	$\theta_{3}=45^{\circ}$
solution branch 2	$180^{\circ}-45^{\circ}$	$\theta_{3}=135^{\circ}$
reach envelope of the tool origin	$\sqrt{2\pm\sqrt{2}}$	$0.765\le\left\|^{0}P_{4ORG}\right\|\le1.848$

Concept & Theory
Topic. Inverse position kinematics by an invariant: using the magnitude of a position vector to decouple one joint variable.
Key relations. Modified Denavit-Hartenberg link transform $^{i-1}_{\ i}T=R_{X}(\alpha_{i-1})D_{X}(a_{i-1})R_{Z}(\theta_{i})D_{Z}(d_{i})$; rotations preserve length, so $|Rv|=|v|$; and a revolute joint whose axis passes through a point cannot move that point.
Why this works. Every joint outboard of the base contributes either a rotation or a fixed offset. Rotations about $\hat{Z}_{0}$ and $\hat{Z}_{2}$ leave $|^{0}P_{4ORG}|$ unchanged, so the squared distance can only involve $\theta_{3}$, and expanding $|v+d_{2}\hat{Z}_{2}|^{2}$ produces a single sine term. Reducing a three-unknown inverse-kinematic problem to one scalar equation this way is the standard first move in closed-form inverse kinematics; the same trick isolates the elbow angle of an articulated arm and the extension of a SCARA.
Common pitfall. Reporting only the principal arcsine. The question says “find all values”, and the supplementary angle is just as valid because the joint range is $\pm180^{\circ}$ - forgetting it costs half the marks. A second trap is expecting $\theta_{4}$ to matter: it turns about an axis through the point being located, so it changes the tool orientation and nothing else.
Source. Craig, Introduction to Robotics, 4th ed., §3.4-3.6 (link parameters and transforms) and Ch. 4 (§4.4 algebraic solution by reduction); Spong et al., 2nd ed., §3.2 and §3.3.

Question 4 of 7: All joint-3 angles reaching a given position (20 marks)

Question 4: All joint-3 angles reaching a given position (20 marks)