Transforming the robot model

Consider transforming a robot model. If ${\cal A}$ is expressed by naming specific points in ${\mathbb{R}}^2$, as in a boundary representation of a polygon, then each point is simply transformed from $a$ to ${h}(a) \in {\cal W}$. In this case, it is straightforward to transform the entire model using $h$. However, there is a slight complication if the robot model is expressed using primitives, such as

$$H_i = \{ a \in {\mathbb{R}}^2 \;\vert\; f_i(a) \leq 0 \} .$$ (3.22)

This differs slightly from (3.2) because the robot is defined in ${\mathbb{R}}^2$ (which is not necessarily ${\cal W}$), and also $a$ is used to denote a point $(x,y) \in {\cal A}$. Under a transformation ${h}$, the primitive is transformed as

$${h}(H_i) = \{ {h}(a) \in {\cal W}\;\vert\; f_i(a) \leq 0 \} .$$ (3.23)

To transform the primitive completely, however, it is better to directly name points in $w \in {\cal W}$, as opposed to ${h}(a) \in {\cal W}$. Using the fact that $a = {h^{-1}}(w)$, this becomes

$${h}(H_i) = \{ w \in {\cal W}\;\vert\; f_i({h^{-1}}(w)) \leq 0 \} ,$$ (3.24)

in which the inverse of ${h}$ appears in the right side because the original point $a \in A$ needs to be recovered to evaluate $f_i$. Therefore, it is important to be careful because either ${h}$ or ${h^{-1}}$ may be required to transform the model. This will be observed in more specific contexts in some coming examples.