Angles at concave polygon

조회 수: 6 (최근 30일)
Mariana
Mariana 2014년 4월 11일
답변: 信实 郑 2022년 2월 26일
Hello, I have a concave polygon you can see in the picture attached. I need to compute all angles between two consecutive vectors. I tried following:
for k = 1 : size(ps, 1) - 2;
point1 = ps(k, :);
vertex = ps(k+1, :);
point2 = ps(k+2, :);
v1 = point1 - vertex;
v2 = point2 - vertex;
theta(k) = dot(v1, v2)/(norm(v1)*norm(v2));
angle(k) = (acos(theta))*180/pi;
end
ps is nx2 matrix containing coordinates of polygon vertices.
In the output angle(k) there are angles allways smaller then 180, but obviously there angles more then 180. I need to somehow distinct which angles are >180 and <180. What do I do wrong? Thank you in advance for your reply

채택된 답변

Image Analyst
Image Analyst 2014년 4월 11일
Try atan2d().

추가 답변 (1개)

信实 郑
信实 郑 2022년 2월 26일
Use function convex to identify convex and concave parts of polygon.As following:
% ps is nx2 matrix containing coordinates of polygon vertices.
c = convex(ps); % Identify convex and concave parts of polygon
for k = 1 : size(ps, 1)-2
point1 = ps(k, :);
vertex = ps(k+1, :);
point2 = ps(k+2, :);
v1 = point1 - vertex;
v2 = point2 - vertex;
theta(k) = dot(v1, v2)/(norm(v1)*norm(v2));
if c(k)>0
angle(k) = (acos(theta(k)))*180/pi;
else
angle(k) = 360 - (acos(theta(k)))*180/pi;
end
end

카테고리

Help CenterFile Exchange에서 Propagation and Channel Models에 대해 자세히 알아보기

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by