**comp.graphics.algorithms**

## Subject: **Re: medial axis for convex polygons**

news:1146582083.258229.163740@u72g2000cwu.googlegroups.com...

> ... But If we take two pieces of lines [A,B] and [C,D] which dont

> intersect, the subset of points equidistant from [A,B] and [C,D] is not

> only a piece of the bisector of (A,B) and (C,D), there are some

> parabolics arcs ... so it could be possible to have some parabolic arcs

> in the medial axis of a convex polygon ...

You are trying to abstract the problem outside of the

realm of polygons. The segments [A,B] and [C,D] that

you choose have endpoints that are vertices of the

convex polygon. It does not matter that the set of

points equidistant from A and [C,D] is a parabolic arc.

The reason it does not matter is that the vertex A itself

is on the medial axis because it is equidistant (of distance

zero) from two edges of the polygon (the edges that share

the vertex A).

Think about the construction this way. At a vertex A,

a portion of the medial axis starts at A and lies along the

bisector of the two edges sharing A. Initially, each point

on the bisector is equidistant from exactly two edges

(those sharing A). Eventually, you reach a point B that

is equidistant from three or more edges. This is a "branch

point" of the axis. Construct all such portions of the medial

axis generated by pairs of edges, each pair sharing a vertex

of the convex polygon.

Now the branch points are connected by line segments that

lie on the bisectors of pairs of edges, each pair not sharing

a polygon vertex.

In all steps, you are always constructing segments whose

points are equidistance from pairs of edges.

--

Dave Eberly

http://www.geometrictools.com

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