Rotation

The robot, ${\cal A}$, can be rotated counterclockwise by some angle $\theta \in [0,2 \pi)$ by mapping every $(x,y) \in {\cal A}$ as

$$(x,y) \mapsto (x \cos\theta - y \sin\theta,\; x \sin\theta + y \cos\theta) .$$ (3.30)

Using a $2 \times 2$ rotation matrix,

$$R(\theta) = \begin{pmatrix}\cos\theta & -\sin\theta \sin\theta & \cos\theta \end{pmatrix},$$ (3.31)

the transformation can be written as

$$\begin{pmatrix}x \cos\theta - y \sin\theta x \sin\theta + y \cos\theta \end{pmatrix} = R(\theta) \begin{pmatrix}x y \end{pmatrix}.$$ (3.32)

Using the notation of Section 3.2.1, $R(\theta)$ becomes $h(q)$, for which $q = \theta$. For linear transformations, such as the one defined by (3.32), recall that the column vectors represent the basis vectors of the new coordinate frame. The column vectors of $R(\theta)$ are unit vectors, and their inner product (or dot product) is zero, indicating that they are orthogonal. Suppose that the $x$ and $y$ coordinate axes, which represent the body frame, are ``painted'' on ${\cal A}$. The columns of $R(\theta)$ can be derived by considering the resulting directions of the $x$- and $y$-axes, respectively, after performing a counterclockwise rotation by the angle $\theta$. This interpretation generalizes nicely for higher dimensional rotation matrices.

Note that the rotation is performed about the origin. Thus, when defining the model of ${\cal A}$, the origin should be placed at the intended axis of rotation. Using the semi-algebraic model, the entire robot model can be rotated by transforming each primitive, yielding ${\cal A}(\theta)$. The inverse rotation, $R(-\theta)$, must be applied to each primitive.