Frame diagram and the transform relating two frames through a graph (20 marks): 22-Mec-B12 Robotics, December 2018, Q3

Question 3: Frame diagram and the transform relating two frames through a graph (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 homogeneous transforms tying four frames $\{U\}$, $\{A\}$, $\{B\}$ and $\{C\}$ together; the rotation blocks are 3-decimal roundings of exact rotations.

Given data
Transform	Rotation	Translation
${}^{U}_{A}T$	$R_Z(30^\circ)$	$[11,\ -1,\ 8]^{T}$
${}^{B}_{A}T$	$R_X(30^\circ)$	$[0,\ 10,\ -20]^{T}$
${}^{C}_{U}T$	$R_X(30^\circ)R_Z(30^\circ)$	$[-3,\ -3,\ 3]^{T}$

Find. (a) the frame diagram, and (b) the transform of $\{C\}$ relative to $\{B\}$.

Figure 3.1 — part (a): the frame diagram. Each solid arrow is one of the given transforms, drawn from the reference frame to the described frame; the dashed arrow is the unknown ${}^{B}_{C}T$. Traversing an arrow backwards means using the inverse.

Approach. Draw the graph, read off a path from $\{B\}$ to $\{C\}$, and chain the transforms along it, inverting whenever the path runs against an arrow.

Build the graph (part a). Each given transform is one edge. ${}^{U}_{A}T$ points $\{U\}\!\rightarrow\!\{A\}$, ${}^{B}_{A}T$ points $\{B\}\!\rightarrow\!\{A\}$ and ${}^{C}_{U}T$ points $\{C\}\!\rightarrow\!\{U\}$. Only one route joins $\{B\}$ to $\{C\}$: $\{B\}\rightarrow\{A\}\rightarrow\{U\}\rightarrow\{C\}$, of which the last two legs are travelled against the arrows.
Write the chain (part b). Following the route and inverting the two reversed edges, $$ {}^{B}_{C}T = {}^{B}_{A}T\ {}^{A}_{U}T\ {}^{U}_{C}T = {}^{B}_{A}T\left({}^{U}_{A}T\right)^{-1}\left({}^{C}_{U}T\right)^{-1}. $$ The mnemonic is that adjacent sub/superscripts cancel: $B\!-\!A$, $A\!-\!U$, $U\!-\!C$ leaves $B\!-\!C$.
Invert the two edges in closed form. Using $[\,R^{T}\ |\ -R^{T}p\,]$ again, $$ {}^{A}_{U}T = \begin{bmatrix} 0.866 & 0.5 & 0 & -9.026 \\ -0.5 & 0.866 & 0 & 6.366 \\ 0 & 0 & 1 & -8 \\ 0 & 0 & 0 & 1\end{bmatrix}, \qquad {}^{U}_{C}T = \begin{bmatrix} 0.866 & 0.433 & 0.25 & 3.147 \\ -0.5 & 0.75 & 0.433 & -0.549 \\ 0 & -0.5 & 0.866 & -4.098 \\ 0 & 0 & 0 & 1\end{bmatrix}. $$ For example the first translation follows from $-R_Z^{T}(30^\circ)[11,-1,8]^{T} = -[9.026,\,-6.366,\,8]^{T}$.
Collapse the last two factors first. Multiplying right to left keeps the intermediate result meaningful — ${}^{A}_{C}T$ is the camera frame seen from $\{A\}$: $$ {}^{A}_{C}T = {}^{A}_{U}T\ {}^{U}_{C}T = \begin{bmatrix} 0.5000 & 0.7500 & 0.4330 & -6.5752 \\ -0.8660 & 0.4330 & 0.2500 & 4.3171 \\ 0 & -0.5000 & 0.8660 & -12.0980 \\ 0 & 0 & 0 & 1 \end{bmatrix}. $$
Premultiply by ${}^{B}_{A}T$ to finish. $$ {}^{B}_{C}T = \boxed{\begin{bmatrix} 0.5000 & 0.7500 & 0.4330 & -6.5752 \\ -0.7500 & 0.6250 & -0.2165 & 19.7876 \\ -0.4330 & -0.2165 & 0.8749 & -28.3183 \\ 0 & 0 & 0 & 1 \end{bmatrix}} $$ so the origin of the camera frame sits at $[-6.5752,\ 19.7876,\ -28.3183]^{T}$ in $\{B\}$.
Check the answer is itself a rigid transform. The rotation block satisfies $R^{T}R = I$ and $\det R = +1$ to within $2\times 10^{-3}$, the residual coming entirely from the 3-decimal rounding of $0.866$ and $0.433$ in the printed data. A larger residual would signal a multiplication slip rather than rounding.

Question 3 — results
Item	Result
(a) Route through the graph	$\{B\}\rightarrow\{A\}\rightarrow\{U\}\rightarrow\{C\}$ (last two legs inverted)
(b) ${}^{B}_{C}R$ — row 1	$[\,0.5000,\ 0.7500,\ 0.4330\,]$
(b) ${}^{B}_{C}R$ — row 2	$[-0.7500,\ 0.6250,\ -0.2165\,]$
(b) ${}^{B}_{C}R$ — row 3	$[-0.4330,\ -0.2165,\ 0.8749\,]$
(b) ${}^{B}P_{C\,\text{ORG}}$	$[-6.5752,\ 19.7876,\ -28.3183]^{T}$

Topic. Frame graphs and transform chains.

Key relations. ${}^{B}_{C}T = {}^{B}_{A}T\,{}^{A}_{U}T\,{}^{U}_{C}T$; travelling an arrow backwards inserts an inverse; adjacent indices cancel.

Why this works. Rigid transforms form a group under multiplication, so any path through the graph gives the same answer. Drawing the graph first turns an index-juggling exercise into a routing problem that can be checked by eye — which is why part (a) is worth marks in its own right.

Common pitfall. Writing the chain in the wrong order. ${}^{B}_{A}T\,{}^{A}_{U}T$ is correct; ${}^{A}_{U}T\,{}^{B}_{A}T$ is dimensionally meaningless even though the matrices multiply. Let the cancelling indices police the order.

Source. Craig, J. J., Introduction to Robotics: Mechanics and Control, 4th ed., Pearson, 2018, §2.9 (transform equations and frame graphs); Niku, S. B., Introduction to Robotics: Analysis, Control, Applications, 3rd ed., Wiley, 2020, Ch. 2 worked examples.

Question 3 of 7: Frame diagram and the transform relating two frames through a graph (20 marks)

Question 3: Frame diagram and the transform relating two frames through a graph (20 marks)