# comp.graphics.algorithms

## Subject: Re: medial axis for convex polygons

wrote in message
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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