Question 2 of 7: Frame diagram and composition of transformations (20 marks)

Question 2: Frame diagram and composition of transformations (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 of the six possible links between the four frames $\{U\}$, $\{A\}$, $\{B\}$ and $\{C\}$.

Find. (a) the frame diagram; (b) $^{B}_{C}T$, the description of $\{C\}$ relative to $\{B\}$.

Approach. Read a route from $\{B\}$ to $\{C\}$ off the frame diagram, invert the two links that are traversed against their arrows, and multiply in that order.

1. Draw the frame diagram (part a). Represent each of the four coordinate systems as a node and each given transformation as an arrow drawn from the reference frame to the frame it describes. The three given links are therefore $\{U\}\rightarrow\{A\}$, $\{B\}\rightarrow\{A\}$ and $\{C\}\rightarrow\{U\}$, as plotted in Figure 2.1. Drawing the graph before any algebra is not decoration: it shows at a glance that $\{A\}$ is a sink with two arrows entering it, which is precisely why the chain to $\{C\}$ cannot be written without inverting something.
2. Read the route and note which links run backwards. Travelling $\{B\}\rightarrow\{A\}\rightarrow\{U\}\rightarrow\{C\}$ uses $^{B}_{A}T$ forwards, then $^{U}_{A}T$ backwards, then $^{C}_{U}T$ backwards. Adjacent superscripts and subscripts must cancel:$$^{B}_{C}T = {}^{B}_{A}T\;\;{}^{A}_{U}T\;\;{}^{U}_{C}T= {}^{B}_{A}T\;\left({}^{U}_{A}T\right)^{-1}\left({}^{C}_{U}T\right)^{-1}.$$The inner indices pair off $A$ with $A$ and $U$ with $U$, leaving $B$ above and $C$ below, which is the required result. This index bookkeeping is the fastest available check that the product has been written in the correct order.
3. Invert the first backwards link. Using $T^{-1}=\begin{bmatrix}R^{T} & -R^{T}p\\ 0 & 1\end{bmatrix}$ with $R=R_{x}(-90^\circ)$ and $p=(11,-1,8)$:$$^{A}_{U}T=\left({}^{U}_{A}T\right)^{-1}=\begin{bmatrix}1&0&0&-11\\ 0&0&1&-8\\ 0&-1&0&-1\\ 0&0&0&1\end{bmatrix}.$$The all-integer result is a strong hint that the transcription is right, since the rotation here is an exact quarter turn.
4. Compose the first two links. Multiplying gives the description of $\{U\}$ relative to $\{B\}$:$$^{B}_{U}T={}^{B}_{A}T\;{}^{A}_{U}T=\begin{bmatrix}1&0&0&-11\\[2pt] 0&\tfrac{1}{2}&\tfrac{\sqrt{3}}{2}&\tfrac{21}{2}-4\sqrt{3}\\[2pt] 0&-\tfrac{\sqrt{3}}{2}&\tfrac{1}{2}&-24-\tfrac{\sqrt{3}}{2}\\[2pt] 0&0&0&1\end{bmatrix}=\begin{bmatrix}1&0&0&-11.000\\ 0&0.500&0.866&3.572\\ 0&-0.866&0.500&-24.866\\ 0&0&0&1\end{bmatrix}.$$Every rotation in this question is about the $x$ axis, so the composite rotation must also be a pure $x$ rotation — here $R_{x}(30^\circ-90^\circ)=R_{x}(-60^\circ)$, which the numbers confirm.
5. Invert the second backwards link and complete the product. With $R=R_{x}(45^\circ)$ and $p=(-3,-3,3)$,$$^{U}_{C}T=\left({}^{C}_{U}T\right)^{-1}=\begin{bmatrix}1&0&0&3\\[2pt] 0&\tfrac{\sqrt{2}}{2}&\tfrac{\sqrt{2}}{2}&0\\[2pt] 0&-\tfrac{\sqrt{2}}{2}&\tfrac{\sqrt{2}}{2}&-3\sqrt{2}\\[2pt] 0&0&0&1\end{bmatrix},$$and therefore $^{B}_{C}T={}^{B}_{U}T\;{}^{U}_{C}T$ evaluates to$$\boxed{^{B}_{C}T=\begin{bmatrix}1 & 0 & 0 & -8.000\\0 & -0.259 & 0.966 & -0.102\\0 & -0.966 & -0.259 & -26.987\\0 & 0 & 0 & 1\end{bmatrix}}$$
6. Verify the composite rotation against the sum of the individual turns. All three given rotations are about $x$, and rotations about a common axis simply add, so the answer must be $R_{x}(30^\circ-90^\circ-45^\circ)=R_{x}(-105^\circ)$. Indeed $\cos(-105^\circ)=\tfrac{\sqrt{2}-\sqrt{6}}{4}=-0.259$ and $\sin(-105^\circ)=-\tfrac{\sqrt{2}+\sqrt{6}}{4}=-0.966$, matching the boxed matrix exactly. The translation column checks out as the exact surds $\left(-8,\;\tfrac{21}{2}-4\sqrt{3}-\tfrac{3\sqrt{6}}{2},\;-24-\tfrac{3\sqrt{2}}{2}-\tfrac{\sqrt{3}}{2}\right)$. The $x$ translation is a whole number because no $x$ rotation can ever move the $x$ components out of alignment.