Transforming a position and a velocity between frames (20 marks): 22-Mec-B12 Robotics, May 2016, Q1

Question 1: Transforming a position and a velocity between frames (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. One homogeneous transformation and two three-component vectors, each referenced to a different frame.

Given data
Quantity	Value	Referenced to
Position $P$	$(-5,\;3,\;4)$	frame $\{A\}$
Velocity $V$	$(10,\;20,\;-15)$	frame $\{B\}$
Rotation in $^{A}_{B}T$	$R_{z}(30^\circ)$	$\{B\}$ seen from $\{A\}$
Translation in $^{A}_{B}T$	$(11,\;-3,\;9)$	origin of $\{B\}$ in $\{A\}$

Find. (a) the same physical point expressed in frame $\{B\}$; (b) the same physical velocity expressed in frame $\{A\}$.

Figure 1.1 — the given transformation places {B} 30° rotated about z and displaced (11, −3, 9) from {A}. A bound point carries the translation; a free vector does not.

Approach. Recognise that the given transform describes $\{B\}$ relative to $\{A\}$, so part (a) needs its inverse, while part (b) needs only the rotation block because a velocity is a free vector.

Read the transformation and confirm it is a rigid-body motion. The upper-left block is $R={}^{A}_{B}R=\begin{bmatrix}\tfrac{\sqrt{3}}{2} & -0.5 & 0\\ 0.5 & \tfrac{\sqrt{3}}{2} & 0\\ 0 & 0 & 1\end{bmatrix}=R_{z}(30^\circ)$, a pure rotation of 30° about the common $z$ axis, and the fourth column holds the translation $^{A}P_{Borg}=(11,\,-3,\,9)$. Checking $RR^{T}=I$ and $\det R=+1$ costs one line and rules out a transcription error before any arithmetic is committed.
Establish which way the given transform points. By the exam's own nomenclature, $^{A}_{B}T$ describes $\{B\}$ relative to $\{A\}$, so as an operator it converts B-coordinates into A-coordinates:$$^{A}P = {}^{A}_{B}T\;{}^{B}P.$$Part (a) asks for $^{B}P$ from $^{A}P$, which is the opposite direction, so the inverse transform is required.
Invert the homogeneous transform analytically, not numerically. For any rigid-body transform the inverse is obtained by transposing the rotation and re-referencing the translation:$$^{B}_{A}T=\left({}^{A}_{B}T\right)^{-1}=\begin{bmatrix} R^{T} & -R^{T}\,{}^{A}P_{Borg}\\ 0\;0\;0 & 1\end{bmatrix}.$$Here $R^{T}=R_{z}(-30^\circ)$, and the new translation column is$$-R^{T}\begin{bmatrix}11\\-3\\9\end{bmatrix}=-\begin{bmatrix}\tfrac{11\sqrt{3}}{2}-1.5\\[2pt] -5.5-\tfrac{3\sqrt{3}}{2}\\[2pt] 9\end{bmatrix}=\begin{bmatrix}-8.026\\ 8.098\\ -9.000\end{bmatrix}.$$Never form this by a general 4×4 matrix inversion in an exam — the closed form is faster and cannot drift.
Apply the inverse to the position vector. Rotating the point first and then adding the re-referenced translation gives$$^{B}P=R^{T}\,{}^{A}P+\left(-R^{T}\,{}^{A}P_{Borg}\right)=\begin{bmatrix}-2.830\\ 5.098\\ 4.000\end{bmatrix}+\begin{bmatrix}-8.026\\ 8.098\\ -9.000\end{bmatrix}.$$The radicals collapse neatly, which is a useful sign that the arithmetic is clean:$$\boxed{^{B}P=\begin{bmatrix}3-8\sqrt{3}\\ 8+3\sqrt{3}\\ -5\end{bmatrix}=\begin{bmatrix}-10.856\\ 13.196\\ -5.000\end{bmatrix}}$$
Classify the velocity vector before transforming it. A velocity is a free vector: it is the difference of two position vectors taken at the same instant, and in that difference the translation column cancels identically. Only the rotation block survives, so$$^{A}V = {}^{A}_{B}R\;{}^{B}V,$$with no translation term. Adding $(11,-3,9)$ here is the classic error on this question and it is worth a sentence in the answer book to show the omission is deliberate.
Rotate the velocity into frame $\{A\}$. This time the un-transposed rotation is wanted, because the vector is being carried from $\{B\}$ into $\{A\}$:$$^{A}V=\begin{bmatrix}\tfrac{\sqrt{3}}{2} & -0.5 & 0\\ 0.5 & \tfrac{\sqrt{3}}{2} & 0\\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}10\\ 20\\ -15\end{bmatrix}=\begin{bmatrix}5\sqrt{3}-10\\ 5+10\sqrt{3}\\ -15\end{bmatrix}$$$$\boxed{^{A}V=\begin{bmatrix}-1.340\\ 22.321\\ -15.000\end{bmatrix}\ \text{units s}^{-1}}$$
Check the result. A pure rotation preserves length, so $\lVert{}^{A}V\rVert$ must equal $\lVert{}^{B}V\rVert$. Both evaluate to $\sqrt{725}=26.926$, and the $z$ component is unchanged at $-15$ because the rotation is about $z$. Both observations follow from the structure of the problem rather than from the arithmetic, so they are genuine independent checks.

Question 1 — results
Part	Quantity	Result
(a)	$^{B}P$	$(3-8\sqrt{3},\;8+3\sqrt{3},\;-5)=(-10.856,\;13.196,\;-5.000)$
(b)	$^{A}V$	$(5\sqrt{3}-10,\;5+10\sqrt{3},\;-15)=(-1.340,\;22.321,\;-15.000)$
check	$\lVert V\rVert$	$26.926$ in both frames (rotation preserves length)

Topic. Homogeneous transformations; bound versus free vectors.
Key relations. $^{A}P={}^{A}_{B}T\,{}^{B}P$; $\left({}^{A}_{B}T\right)^{-1}=\begin{bmatrix}R^{T} & -R^{T}p\\ 0 & 1\end{bmatrix}$; free vectors transform with $R$ alone.
Why this works. A homogeneous transform packs a rotation and a translation into one 4×4 operator acting on points written in homogeneous coordinates $(x,y,z,1)$. A direction, velocity or force is written $(x,y,z,0)$, and that trailing zero is exactly what annihilates the translation column — the distinction is built into the algebra, not bolted on afterwards.
Common pitfall. Adding the translation to the velocity, or using $R$ where $R^{T}$ is needed. The two errors often cancel visually (the answer still “looks like a vector”), so state the direction of the given transform in words before touching the numbers.
Source. Craig, Introduction to Robotics, 4th ed., Ch. 2 (§2.4 mappings, §2.8 inverting a transform); Spong et al., Robot Modeling and Control, 2nd ed., §2.6.

Question 1 of 7: Transforming a position and a velocity between frames (20 marks)

Question 1: Transforming a position and a velocity between frames (20 marks)