Question 5 of 7: Inverse orientation for a spherical wrist — Z–Y–Z Euler angles (20 marks)

Question 5: Inverse orientation for a spherical wrist — Z–Y–Z Euler angles (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. The symbolic Z–Y–Z Euler form of the wrist rotation and its required numerical value.

Find. Every triple $(q_{4},q_{5},q_{6})$ reproducing that rotation.

Approach. Extract $q_{5}$ from the single-term element $r_{33}=c_{5}$, take both signs of $s_{5}$, then read $q_{4}$ and $q_{6}$ from the third column and third row respectively.

1. Confirm the given matrix is a legitimate rotation. Before inverting anything, check $RR^{T}=I$ and $\det R=+1$. Both hold exactly for the given surds, so the orientation is achievable and the wrist equations will have real solutions. A matrix that failed this test would signal a transcription error and no amount of algebra would rescue it.
2. Isolate the middle angle from the one element that stands alone. In the Z–Y–Z form the element $r_{33}$ is simply $c_{5}$, so$$\cos q_{5}=\frac{\sqrt{2}}{2}\;\Longrightarrow\;\boxed{q_{5}=\pm 45^\circ}$$Both signs are admissible because the cosine cannot distinguish them, and this is precisely where the two solution branches are born. Each choice of sign then determines $q_{4}$ and $q_{6}$ uniquely.
3. Take the first branch, $q_{5}=+45^\circ$, so $s_{5}=+\tfrac{\sqrt{2}}{2}$. The third column supplies $q_{4}$ and the third row supplies $q_{6}$:$$c_{4}=\frac{r_{13}}{s_{5}}=\frac{-\sqrt{6}/4}{\sqrt{2}/2}=-\frac{\sqrt{3}}{2},\qquad s_{4}=\frac{r_{23}}{s_{5}}=\frac{-\sqrt{2}/4}{\sqrt{2}/2}=-\frac{1}{2},$$$$c_{6}=\frac{-r_{31}}{s_{5}}=0,\qquad s_{6}=\frac{r_{32}}{s_{5}}=\frac{\sqrt{2}/2}{\sqrt{2}/2}=1.$$Both sine and cosine are known for each angle, so $\mathrm{Atan2}$ resolves the quadrant without ambiguity:$$\boxed{q_{4}=210^\circ\ (\equiv-150^\circ),\qquad q_{5}=45^\circ,\qquad q_{6}=90^\circ}$$
4. Take the second branch, $q_{5}=-45^\circ$, so $s_{5}=-\tfrac{\sqrt{2}}{2}$. Every ratio flips sign:$$c_{4}=\frac{-\sqrt{6}/4}{-\sqrt{2}/2}=\frac{\sqrt{3}}{2},\quad s_{4}=\frac{-\sqrt{2}/4}{-\sqrt{2}/2}=\frac{1}{2},\quad c_{6}=0,\quad s_{6}=-1,$$giving the companion solution$$\boxed{q_{4}=30^\circ,\qquad q_{5}=-45^\circ,\qquad q_{6}=-90^\circ\ (\equiv 270^\circ)}$$The two branches are related by the standard Euler-angle identity $(q_{4},q_{5},q_{6})\rightarrow(q_{4}+180^\circ,\,-q_{5},\,q_{6}+180^\circ)$, which is a useful independent confirmation that nothing has been missed.
5. Substitute both triples back into the symbolic matrix. Reconstructing $R_{z}(q_{4})R_{y}(q_{5})R_{z}(q_{6})$ from each solution reproduces all nine given elements exactly. Checking element $r_{11}$ by hand for branch 1, for instance,$$r_{11}=c_{4}c_{5}c_{6}-s_{4}s_{6}=\left(-\tfrac{\sqrt{3}}{2}\right)\left(\tfrac{\sqrt{2}}{2}\right)(0)-\left(-\tfrac{1}{2}\right)(1)=\tfrac{1}{2},$$which matches. Verifying one off-diagonal element by hand is worth the thirty seconds it costs, because it tests the branch assignment rather than merely repeating the division.
6. State the degenerate case for completeness. The solution above divides by $s_{5}$, so it fails when $q_{5}=0$ or $180^\circ$. There the first and third wrist axes become collinear, the wrist loses a degree of freedom, and only the sum $q_{4}+q_{6}$ (or the difference) is determined — a one-parameter family of solutions rather than two isolated ones. That is the classic wrist singularity of a Z–Y–Z spherical wrist, and it does not arise here because $q_{5}=\pm45^\circ$ is comfortably away from it.