how to find the area of the triangles formed as a result of delaunay triangulation
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Roger Stafford on 3 Mar 2014
If a triangle is in two dimensional space with vertices at (x1,y1), (x2,y2), and (x3,y3), its area is given by:
area = 1/2*abs((x2-x1)*(y3-y1)-(x3-x1)*(y2-y1));
area = 1/2*abs(det([x1,y1,1;x2,y2,1;x3,y3,1]));
If it is in three dimensions with vertices at P1 = [x1,y1,z1], P2 = [x2,y2,z2], P3 = [x3,y3,z3], its area is:
area = 1/2*norm(cross(P2-P1,P3-P1));
These formulas are more accurate for numerical computation than those which depend on the lengths of the three sides of the triangle.
Of course you must adapt these formulas for use with your 160 triangles, either using the necessary for-loop or possibly a vectorized expression.
More Answers (1)
Christopher Rock on 2 Aug 2018
Vectorised code based off Roger Stafford's answer.
function A = triarea(t, p)
% A = TRIAREA(t, p) area of triangles in triangulation
Xt = reshape(p(t, 1), size(t)); % X coordinates of vertices in triangulation
Yt = reshape(p(t, 2), size(t)); % Y coordinates of vertices in triangulation
A = 0.5 * abs((Xt(:, 2) - Xt(:, 1)) .* (Yt(:, 3) - Yt(:, 1)) - ...
(Xt(:, 3) - Xt(:, 1)) .* (Yt(:, 2) - Yt(:, 1)));