Mapping a point and a free vector between frames: 22-Mec-B12 Robotics, December 2019, Q1

Question 1: Mapping a point and a free 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 describing frame {B} relative to frame {A}, one point expressed in {A} and one velocity expressed in {B}.

Given data
Quantity	Symbol	Value
Point, referenced to {A}	$^{A}P$	$[10,\,-4,\,3]^{T}$
Velocity, referenced to {B}	$^{B}V$	$[15,\,5,\,20]^{T}$
Rotation block of $^{A}_{B}T$	$^{A}_{B}R$	$\begin{bmatrix}0.5 & 0 & -0.8660\\ -0.8660 & 0 & -0.5\\ 0 & 1 & 0\end{bmatrix}$
Translation block of $^{A}_{B}T$	$^{A}P_{BORG}$	$[5,\,-5\sqrt3,\,5]^{T}=[5,\,-8.6603,\,5]^{T}$

Find. The same point written in frame {B}, and the same velocity written in frame {A}.

Frame {B} is described relative to frame {A}. A point carries the translation as well as the rotation; a free vector such as a velocity carries only the rotation.

Approach. Invert $^{A}_{B}T$ in closed form to map the point the “wrong way” along the arrow, then rotate the velocity with the rotation block alone, because a velocity is a free vector and has no point of application.

Confirm the rotation block really is a rotation. The three columns of $^{A}_{B}R$ are the axes $\hat X_B,\hat Y_B,\hat Z_B$ written in {A}, so each must be a unit vector and the three must be mutually orthogonal. $\|[0.5,-0.8660,0]\|=\sqrt{0.25+0.75}=1$, $\|[0,0,1]\|=1$, $\|[-0.8660,-0.5,0]\|=1$, and every pairwise dot product vanishes ($0.5(-0.8660)+(-0.8660)(-0.5)=0$). Finally $\det{}^{A}_{B}R=+1$, so the matrix is a proper rotation (the columns are in fact a rotation of $-60^\circ$ about a skewed axis, but the numeric test is what matters).
Write the closed-form inverse. For any homogeneous transform the inverse is not a general 4×4 matrix inversion; it is $$^{B}_{A}T=\begin{bmatrix}^{A}_{B}R^{T} & -{}^{A}_{B}R^{T}\,{}^{A}P_{BORG}\\ 0\;0\;0 & 1\end{bmatrix}.$$ Transposing gives $$^{A}_{B}R^{T}=\begin{bmatrix}0.5 & -0.8660 & 0\\ 0 & 0 & 1\\ -0.8660 & -0.5 & 0\end{bmatrix}.$$
Form the inverse translation. With $^{A}P_{BORG}=[5,\,-5\sqrt3,\,5]^{T}$, $$^{A}_{B}R^{T}\,{}^{A}P_{BORG}=\begin{bmatrix}0.5(5)+(-0.8660)(-8.6603)\\ 5\\ (-0.8660)(5)+(-0.5)(-8.6603)\end{bmatrix}=\begin{bmatrix}10\\ 5\\ 0\end{bmatrix},$$ so $\boxed{^{B}P_{AORG}=-{}^{A}_{B}R^{T}\,{}^{A}P_{BORG}=[-10,\,-5,\,0]^{T}}$. The exact cancellation in the third row is a useful arithmetic check: $-\tfrac{\sqrt3}{2}(5)+\tfrac{1}{2}(5\sqrt3)=0$.
(a) Map the point. Applying the inverse to $^{A}P$, $$^{B}P={}^{A}_{B}R^{T}\,{}^{A}P+{}^{B}P_{AORG}=\begin{bmatrix}8.4641\\ 3\\ -6.6603\end{bmatrix}+\begin{bmatrix}-10\\ -5\\ 0\end{bmatrix},$$ giving $$\boxed{^{B}P=[-1.536,\;-2.000,\;-6.660]^{T}}.$$
Check the point by pushing it back. $^{A}_{B}T\,[-1.536,-2.000,-6.660,1]^{T}=[10,-4,3,1]^{T}$, which reproduces the given $^{A}P$ exactly, so the inverse was formed correctly.
(b) Rotate the free vector. A velocity has magnitude and direction but no location, so the translation column must be ignored: $$^{A}V={}^{A}_{B}R\;{}^{B}V=\begin{bmatrix}0.5(15)+0(5)-0.8660(20)\\ -0.8660(15)+0(5)-0.5(20)\\ 0(15)+1(5)+0(20)\end{bmatrix},$$ hence $$\boxed{^{A}V=[-9.821,\;-22.990,\;5.000]^{T}}.$$
Check the velocity by its magnitude. A rotation preserves length, so $\|{}^{A}V\|$ must equal $\|{}^{B}V\|=\sqrt{15^2+5^2+20^2}=\sqrt{650}=25.495$. Indeed $\sqrt{9.821^2+22.990^2+5.000^2}=25.495$. Had the translation been included by mistake the answer would have been $[-4.821,-31.651,10.000]^{T}$, whose norm is $33.6$ — the failed length check is exactly how that error is caught.

Final results
Quantity	Symbol	Value
Position vector referenced to {B}	$^{B}P$	$[-1.536,\;-2.000,\;-6.660]^{T}$
Velocity vector referenced to {A}	$^{A}V$	$[-9.821,\;-22.990,\;5.000]^{T}$
Origin of {A} seen from {B} (intermediate)	$^{B}P_{AORG}$	$[-10,\;-5,\;0]^{T}$
Magnitude check on the velocity	$\\|V\\|$	$25.495$ (unchanged)

Concept & Theory

Topic. Homogeneous transforms; the distinction between a bound point and a free vector.

Key relations. $^{A}P={}^{A}_{B}T\,{}^{B}P$; the closed-form inverse $^{B}_{A}T=[\,R^{T}\,|\,-R^{T}p\,]$; for a free vector $^{A}V={}^{A}_{B}R\,{}^{B}V$; a rotation satisfies $R^{T}R=I$ and $\det R=+1$.

Why this works. A homogeneous transform is simultaneously a description of a frame, a mapping of coordinates and an operator. Its rotation block is orthonormal, so its inverse is its transpose and the inverse translation is just the original translation carried back through that transpose — no Gaussian elimination is ever needed, and the closed form is numerically exact. The point/vector distinction is the homogeneous fourth coordinate: a point carries a 1 and therefore picks up the translation column, while a free vector such as a velocity, an angular rate or a force direction carries a 0 and does not.

Common pitfall. Students routinely apply the full transform to the velocity, which silently adds the frame offset and changes the vector length. A second frequent slip is inverting the 4×4 matrix numerically and then trusting a result that is subtly non-orthonormal; always use the transpose form and re-check the length of any rotated vector.

Source. Craig, Introduction to Robotics, 4th ed., §2.3–2.9 (mappings, operators and the inverse of a transform); Niku, 3rd ed., ch. 2.

Question 1 of 7: Mapping a point and a free vector between frames

Question 1: Mapping a point and a free vector between frames (20 marks)