Obstacle region for a rigid body

Suppose that the world, ${\cal W}= {\mathbb{R}}^2$ or ${\cal W}= {\mathbb{R}}^3$, contains an obstacle region, ${\cal O}\subset {\cal W}$. Assume here that a rigid robot, ${\cal A}\subset {\cal W}$, is defined; the case of multiple links will be handled shortly. Assume that both ${\cal A}$ and ${\cal O}$ are expressed as semi-algebraic models (which includes polygonal and polyhedral models) from Section 3.1. Let $q \in {\cal C}$ denote the configuration of ${\cal A}$, in which $q = (x_t,y_t,\theta)$ for ${\cal W}= {\mathbb{R}}^2$ and $q = (x_t,y_t,z_t,h)$ for ${\cal W}= {\mathbb{R}}^3$ ($h$ represents the unit quaternion).

The obstacle region, ${\cal C}_{obs}\subseteq {\cal C}$, is defined as

$${\cal C}_{obs}= \{ q \in {\cal C}\;\vert\; {\cal A}(q) \cap {\cal O}\not = \emptyset \} ,$$ (4.34)

which is the set of all configurations, $q$, at which ${\cal A}(q)$, the transformed robot, intersects the obstacle region, ${\cal O}$. Since ${\cal O}$ and ${\cal A}(q)$ are closed sets in ${\cal W}$, the obstacle region is a closed set in ${\cal C}$.

The leftover configurations are called the free space, which is defined and denoted as ${\cal C}_{free}= {\cal C}\setminus {\cal C}_{obs}$. Since ${\cal C}$ is a topological space and ${\cal C}_{obs}$ is closed, ${\cal C}_{free}$ must be an open set. This implies that the robot can come arbitrarily close to the obstacles while remaining in ${\cal C}_{free}$. If ${\cal A}$ ``touches'' ${\cal O}$,

$$\operatorname{int}({\cal O}) \cap \operatorname{int}({\cal A}(q)) = \emptyset \mbox { and } {\cal O}\cap {\cal A}(q) \not = \emptyset ,$$ (4.35)

then $q \in {\cal C}_{obs}$ (recall that $\operatorname{int}$ means the interior). The condition above indicates that only their boundaries intersect.

The idea of getting arbitrarily close may be nonsense in practical robotics, but it makes a clean formulation of the motion planning problem. Since ${\cal C}_{free}$ is open, it becomes impossible to formulate some optimization problems, such as finding the shortest path. In this case, the closure, $\operatorname{cl}({\cal C}_{free})$, should instead be used, as described in Section 7.7.