Angles at concave polygon

4 次查看（过去 30 天）
Mariana
Mariana 2014-4-11
回答： 信实 郑 2022-2-26
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

请先登录，再回答此问题。

采纳的回答

Image Analyst
Image Analyst 2014-4-11
Try atan2d().

更多回答（1 个）

信实 郑
信实 郑 2022-2-26
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

类别

在 Help Center 和 File Exchange 中查找有关 Elementary Polygons 的更多信息

标签

Community Treasure Hunt

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

Start Hunting!

Translated by