Question 2 of 7: Mapping a position vector and a velocity vector between frames (20 marks)

Question 2: Mapping a position vector and a velocity vector 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 transform and two vectors, one referenced to each frame.

Find. (a) the same point referenced to $\{B\}$, and (b) the same velocity referenced to $\{A\}$.

Approach. Invert the given transform in closed form to get ${}^{B}_{A}T$ for part (a); for part (b) recognise that a velocity is a free vector and apply the rotation block alone.

1. Split the transform and confirm it is rigid. Writing $$ {}^{A}_{B}T = \begin{bmatrix} {}^{A}_{B}R & {}^{A}P_{B\,\text{ORG}} \\ 0\ 0\ 0 & 1 \end{bmatrix}, $$ the columns of ${}^{A}_{B}R$ have unit length and zero mutual dot products and $\det {}^{A}_{B}R = +1$, so the array is a genuine rigid-body transform and the closed-form inverse below is legitimate.
2. Invert without a numerical matrix inversion. For any homogeneous transform the inverse is available in closed form, $$ {}^{B}_{A}T = \left({}^{A}_{B}T\right)^{-1} = \begin{bmatrix} {}^{A}_{B}R^{T} & -{}^{A}_{B}R^{T}\,{}^{A}P_{B\,\text{ORG}} \\ 0\ 0\ 0 & 1\end{bmatrix}, $$ because the rotation block is orthogonal. Gaussian elimination on the 4×4 is never needed and is the usual source of arithmetic slips under exam conditions.
3. Apply the inverse to the point. First subtract the origin offset, $$ {}^{A}P - {}^{A}P_{B\,\text{ORG}} = \begin{bmatrix}10\\-4\\3\end{bmatrix}-\begin{bmatrix}5\\-5\sqrt{3}\\5\end{bmatrix} = \begin{bmatrix}5\\ 5\sqrt{3}-4\\ -2\end{bmatrix} = \begin{bmatrix}5\\ 4.6603\\ -2\end{bmatrix}, $$ then rotate the difference into $\{B\}$ with ${}^{A}_{B}R^{T}$ (whose rows are the columns of ${}^{A}_{B}R$): $$ {}^{B}P = \begin{bmatrix} 0.5 & -0.8660 & 0 \\ 0 & 0 & 1 \\ -0.8660 & -0.5 & 0 \end{bmatrix}\begin{bmatrix}5\\ 4.6603\\ -2\end{bmatrix} = \boxed{\begin{bmatrix} -1.5359 \\ -2.0000 \\ -6.6603 \end{bmatrix}}. $$ In exact form this is $[2\sqrt{3}-5,\ -2,\ 2-5\sqrt{3}]^{T}$.
4. Check part (a) by mapping back. Substituting the answer into the original transform, ${}^{A}_{B}T\,{}^{B}P + $ the translation returns $[10,\,-4,\,3]^{T}$ exactly, so the inverse was formed and applied correctly.
5. Treat the velocity as a free vector. A velocity has magnitude and direction but no point of application, so translating the frame origin cannot change it. Only the rotation block acts: $$ {}^{A}V = {}^{A}_{B}R\ {}^{B}V = \begin{bmatrix} 0.5 & 0 & -0.8660 \\ -0.8660 & 0 & -0.5 \\ 0 & 1 & 0 \end{bmatrix}\begin{bmatrix}15\\5\\20\end{bmatrix} = \boxed{\begin{bmatrix} -9.8205 \\ -22.9904 \\ 5.0000 \end{bmatrix}}. $$
6. Check part (b) with the invariant. A pure rotation preserves length: $\|{}^{B}V\| = \sqrt{15^{2}+5^{2}+20^{2}} = 25.4951$ and $\|{}^{A}V\| = \sqrt{9.8205^{2}+22.9904^{2}+5^{2}} = 25.4951$. Had the translation been applied by mistake, the magnitude would have changed and the error would have been caught here.