Frame graph and a compound transform: 22-Mec-B12 Robotics, December 2019, Q2

Question 2: Frame graph and a compound transform (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 a universe frame {U} and three body frames {A}, {B}, {C} together. In the notation defined on page 2 of the paper the leading superscript is the reference frame and the leading subscript is the frame being described.

Given data
Quantity	Symbol	Value
Frame {A} described relative to {U}	$^{U}_{A}T$	$R_{Z}(30^\circ)$, $p=[11,\,-1,\,8]^{T}$
Frame {A} described relative to {B}	$^{B}_{A}T$	$R_{X}(30^\circ)$, $p=[0,\,10,\,-20]^{T}$
Frame {U} described relative to {C}	$^{C}_{U}T$	$R$ as printed, $p=[-3,\,-3,\,3]^{T}$
Rounding of the printed sines/cosines	—	3 decimals ($0.866,\,0.433,\,0.25$)

Find. (a) the frame graph, and (b) the transform $^{B}_{C}T$ that describes frame {C} relative to frame {B}.

Part (a): the frame diagram. Three arrows are given and the fourth, from {B} to {C}, is the one required. Two of the given arrows must be traversed backwards, which is what makes two inversions necessary.

Approach. Draw the four frames as nodes and each given transform as an arrow from its reference frame to its described frame. Any path from {B} to {C} then gives the answer as an ordered product, with an inverse wherever the path runs against an arrow.

(a) Read the arrows off the notation. $^{U}_{A}T$ is an arrow {U}→{A}, $^{B}_{A}T$ is an arrow {B}→{A}, and $^{C}_{U}T$ is an arrow {C}→{U}. The graph above is the answer to part (a); the required $^{B}_{C}T$ is the missing arrow {B}→{C}.
Choose a path and write the product. Walking {B}→{A} forwards, then {A}→{U} backwards along $^{U}_{A}T$, then {U}→{C} backwards along $^{C}_{U}T$ gives $$^{B}_{C}T={}^{B}_{A}T\;{}^{A}_{U}T\;{}^{U}_{C}T={}^{B}_{A}T\,\left({}^{U}_{A}T\right)^{-1}\left({}^{C}_{U}T\right)^{-1}.$$ Subscript–superscript cancellation ($B\!\to\!A$, $A\!\to\!U$, $U\!\to\!C$) is the bookkeeping check that the order is right.
Invert $^{U}_{A}T$. Its rotation is $R_{Z}(30^\circ)$, so $R^{T}=R_{Z}(-30^\circ)$ and $-R^{T}p=-[0.866(11)+0.5(-1),\,-0.5(11)+0.866(-1),\,8]^{T}$, giving $$^{A}_{U}T=\begin{bmatrix}0.866 & 0.5 & 0 & -9.026\\ -0.5 & 0.866 & 0 & 6.366\\ 0 & 0 & 1 & -8\\ 0 & 0 & 0 & 1\end{bmatrix}.$$
Invert $^{C}_{U}T$. The same rule with the printed rotation gives $-R^{T}p=[3.147,\,-0.549,\,-4.098]^{T}$ and $$^{U}_{C}T=\begin{bmatrix}0.866 & 0.433 & 0.25 & 3.147\\ -0.5 & 0.75 & 0.433 & -0.549\\ 0 & -0.5 & 0.866 & -4.098\\ 0 & 0 & 0 & 1\end{bmatrix}.$$
Multiply the three transforms in order. Carrying out $^{B}_{A}T\,{}^{A}_{U}T\,{}^{U}_{C}T$ column by column, $$\boxed{^{B}_{C}T=\begin{bmatrix}0.500 & 0.750 & 0.433 & -6.575\\ -0.750 & 0.625 & -0.217 & 19.788\\ -0.433 & -0.217 & 0.875 & -28.318\\ 0 & 0 & 0 & 1\end{bmatrix}}$$
Check the answer is still a rigid-body transform. The rotation block satisfies $R^{T}R=I$ to within $1.2\times10^{-4}$ and $\det R=1.000$. That residual is not an arithmetic slip: the paper prints $\sin30^\circ$ and its products only to three decimals ($0.866$, $0.433$, $0.25$), and multiplying three such matrices accumulates exactly this order of rounding error. Reporting the result to three decimals is therefore the honest precision.
Sanity-check the translation. $^{B}P_{CORG}=[-6.575,\,19.788,\,-28.318]^{T}$ has magnitude $34.9$, which is the right order of magnitude given that {A} sits $20$ units below {B} and {U} is a further $8$–$11$ units away in each direction.

Final results
Quantity	Symbol	Value
Frame diagram (part a)	—	{U}→{A}, {B}→{A}, {C}→{U} given; {B}→{C} required
Path used	—	$^{B}_{C}T={}^{B}_{A}T\,({}^{U}_{A}T)^{-1}({}^{C}_{U}T)^{-1}$
Rotation of {C} in {B}	$^{B}_{C}R$	$\begin{bmatrix}0.500 & 0.750 & 0.433\\ -0.750 & 0.625 & -0.217\\ -0.433 & -0.217 & 0.875\end{bmatrix}$
Origin of {C} seen from {B}	$^{B}P_{CORG}$	$[-6.575,\;19.788,\;-28.318]^{T}$
Orthonormality residual (rounding only)	$\\|R^{T}R-I\\|_{\max}$	$1.2\times10^{-4}$

Concept & Theory

Topic. Compound transforms and the frame (transform) graph.

Key relations. $^{A}_{C}T={}^{A}_{B}T\,{}^{B}_{C}T$ (subscript–superscript cancellation); $\left({}^{A}_{B}T\right)^{-1}={}^{B}_{A}T=[\,R^{T}\,|\,-R^{T}p\,]$; a closed loop of transforms multiplies to the identity.

Why this works. The frame graph turns an algebra problem into a path-finding problem. Because transform composition is associative, any path from {B} to {C} yields the same answer, and traversing an arrow backwards simply inserts that arrow’s inverse. Drawing the graph first prevents the single most common error in this topic, which is multiplying the given matrices in whatever order they appear on the page. The cancellation rule is a mechanical guarantee: if the inner indices do not match, the product is meaningless.

Common pitfall. Reversing an arrow without inverting the matrix, and inverting with a general matrix routine instead of the transpose form. Also watch the exam’s own notation: here the leading superscript is the reference frame, so $^{C}_{U}T$ describes {U} relative to {C} — not the other way round.

Source. Craig, Introduction to Robotics, 4th ed., §2.6–2.10 and Example 2.5 (transform equations and frame graphs); Spong et al., 2nd ed., ch. 2.

Question 2 of 7: Frame graph and a compound transform

Question 2: Frame graph and a compound transform (20 marks)