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Effacer les filtres

intersection points between random polygon and a circle

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KalMandy
KalMandy le 9 Nov 2016
Hi, I have a random polygon (with known coordinates of the vertices) and a circle(with know center coordinates and center). I want to find the intersection points in between them. I have used linecirc function, but the problem is it also gives the intersection points which does not lie on the polygon. I have developed the following program, but sometimes it does not provide all the intersection points. Please help me to find out where I went wrong. Thanks
for i=1:length(V_x) %closed polygon with 7 vertices
V_x(8)=V_x(1);
V_y(8)=V_y(1);
m(i)=(V_y(i)-V_y(i+1))/(V_x(i)-V_x(i+1));
b(i)=-m(i)*V_x(i)+V_y(i);
[xout,yout] = linecirc(m(i),b(i),C_x,C_y,R);
crossX(i,1)=xout(1,1);
crossX(i,2)=xout(1,2);
crossY(i,1)=yout(1,1);
crossY(i,2)=yout(1,2);
DistCircleVertex(i)=distance(C_x,V_x(i),C_y,V_y(i));
EdgeLength(i)=distance(V_x(i),V_x(i+1),V_y(i),V_y(i+1));
DistPoint1Vertex1(i)=distance(V_x(i),crossX(i,1),V_y(i),crossY(i,1));
DistPoint1Vertex2(i)=distance(V_x(i+1),crossX(i,1),V_y(i+1),crossY(i,1));
DistPoint2Vertex1(i)=distance(V_x(i),crossX(i,2),V_y(i),crossY(i,2));
DistPoint2Vertex2(i)=distance(V_x(i+1),crossX(i,2),V_y(i+1),crossY(i,2));
%checking if intersection points lies on the polygon
if EdgeLength(i)==DistPoint1Vertex1(i)+DistPoint1Vertex2(i)& EdgeLength(i)==DistPoint2Vertex1(i)+DistPoint2Vertex2(i)
InpointX(i,1)=crossX(i,1);
InpointY(i,1)=crossY(i,1);
InpointX(i,2)=crossX(i,2);
InpointY(i,2)=crossY(i,2);
else if EdgeLength(i)==DistPoint1Vertex1(i)+DistPoint1Vertex2(i)
InpointX(i,1)=crossX(i,1);
InpointY(i,1)=crossY(i,1);
InpointX(i,2)=NaN;
InpointY(i,2)=NaN;
else if EdgeLength(i)==DistPoint2Vertex1(i)+DistPoint2Vertex2(i)
InpointX(i,1)=NaN;
InpointY(i,1)=NaN;
InpointX(i,2)=crossX(i,2);
InpointY(i,2)=crossY(i,2);
else
InpointX(i,1)=NaN;
InpointY(i,1)=NaN;
InpointX(i,2)=NaN;
InpointY(i,2)=NaN;
end
end
end
end

Réponse acceptée

Matt J
Matt J le 9 Nov 2016
Modifié(e) : Matt J le 9 Nov 2016
I have used linecirc function, but the problem is it also gives the intersection points which does not lie on the polygon.
Why not just discard the points that do not lie on the polygon? You can test this with inpolygon or with one of several File Exchange submissions, e.g.,
  3 commentaires
KalMandy
KalMandy le 9 Nov 2016
Hi Matt, I figured out the way to use inpolygon for my case. Thank you very much!
Khouloud Eledlebi
Khouloud Eledlebi le 3 Août 2017
Hello KalMandy, I am working on a project and I got stuck at how to find the intersection between polygon and circle. Can you please send me your final version of the code which you have adjusted earlier please. I would really appreciate your help very much

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