Determining if a point is inside a polygon or along the outer edge
9 views (last 30 days)
Show older comments
I have a 2D polygon in 3 dimensions. Some of the points are inside of this polygon, and some define the outer edge of it. How can I detect which points are inside the polygon, and which define the outside of it?
My initial idea is to create a vector that is defined by a set of 2 points, then see if that vector intersects a line segment in the positive and negative direction. If it only intersects on one side, then the point is on the outside of the polygon. If it intersects a line segment in the positive and negative direction, then the point is inside the polygon. But, in order for that to be any use, I need to figure out a condition that tells me if the line defined by that vector intersects any of the line segments defined by the other points.
Any ideas?
4 Comments
Matt J
on 16 Jul 2013
Edited: Matt J
on 16 Jul 2013
No matter what you do, you will need to define some tolerances. No set of 2D points are ever exactly coplanar in 3D when floating point calculation errors are involved. Likewise, no 3 points that are all supposed to lie at the edge of a polygon can be relied upon to be exactly colinear.
Accepted Answer
Matt J
on 16 Jul 2013
Edited: Matt J
on 16 Jul 2013
[A,b,Aeq,beq]=vert2lcon(points);
vertices = lcon2vert(A,b,Aeq,beq);
All points that are not vertices are considered by the code as lying inside of it or along an edge, up to the tolerance parameters of the code.
If you really have to distinguish between interior points and edge points. the points satisfying
A*points<=b-tolInterior
can be considered interior points up to a tolInterior tolerance parameter set by you..
8 Comments
Matt J
on 17 Jul 2013
Edited: Matt J
on 17 Jul 2013
Glad it's working Joshua, but I'm starting to think that it's better, since you know a prior that it's a 2D polygon, to take advantage of that and pre-map the coordinates into 2D space, like Jos and cyclist were suggesting.
At that point you could still use vert2lcon to derive inequalities for the polygon from the 2D coordinates, but it would be more stable and you wouldn't have to fuss with tolerances.
More Answers (0)
See Also
Categories
Find more on Computational Geometry in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!