Re: Intersect circle and irregular shape

glaroc <glaroc@xxxxxxxxx> wrote in message <66c9ae27-c839-41e2-
a9a3-d5dcc7f311a7@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>...
Hi, I am looking for ways to calculate the percentage of the
circumference of a circle enclosed within an intersecting irregular
polygon. For example, if I have a circle centered within a triangle, I
want to know how much of the circumference of the circle is contained
within the triangle. This could be anywhere between 0 and 100. I know
I can do this in a topological GIS, but I really need Matlab for this.
I want to use this to build an edge correction program. I couldn't
find anything in the Matlab file central or anywhere else.

Thank you for any help you may provide,

GL
--------
Here's an outline of how I would tackle your problem, GL. First, find all the
points of intersection between the circle and the various line segments of the
polygon. Then sort such intersection points in accordance with the angles
they make with the circle's center. These points now divide up the circle into
non-overlapping arcs. Find the arc "midpoint" on the circle between each
pair of successive intersection points. Finally, use the function 'inpolygon' to
determine which of these midpoints lie inside the polygon. Each of the entire
circular arcs containing such midpoints must also lie within the polygon.
Then the sum of these "inside" arcs' angles divided by 2*pi times 100 gives
you the percentage you desire. I leave you to deal with the special case when
there are no intersection points.

Here is a procedure for finding the intersection between a circle having
radius r and center (a,b) with a line segment whose endpoints are (c,d) and
(e,f). Any point on the infinite line through these points has the parametric
representation

x = (c+e)/2+(c-e)/2*t
y = (d+f)/2+(d-f)/2*t

with parameter t. If t lies between -1 and +1, then such a point is actually
within the line segment. To also lie on the circle, we must have

(x-a)^2+(y-b)^2 = r^2

By substituting in the parametric expressions in t for x and y in this last
equality, we get a quadratic equation in t which we can use matlab's 'roots'
function to solve. This done as follows:

t = roots([(c-e)^2+(d-f)^2, ...
2*((c-e)*(c+e-2*a)+(d-f)*(d+f-2*b)), ...
(c+e-2*a)^2+(d+f-2*b)^2-4*r^2]);

where the three expressions show the derived coefficients of the above
quadratic. There will be either two, one, or no distinct real roots to this,
according as the circle intersects that many points of the infinite line. As
mentioned above, those that have a real t value between -1 and +1 yield the
points of intersection with the line segment, and their coordinates are
determined by substituting t into the above formulas for x and y.

Roger Stafford

.

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