how to find the area of the triangles formed as a result of delaunay triangulation
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I have applied delaunay triangulation to my image. i have got 160 triangles. i need to find area of each triangle. Kindly help me with code. only after i can proceed with my prooj.i am new to matlab.
1 件のコメント
M.S. Khan
2020 年 8 月 24 日
How you counted the number of triangles? Please tell me the methods, regards
採用された回答
Roger Stafford
2014 年 3 月 3 日
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));
or
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.
2 件のコメント
Valerio Ficcadenti
2022 年 8 月 14 日
Hi
How did you manage the polygons that are not closed? Namely, those with vertexes to infinite?
Valerio
その他の回答 (1 件)
Christopher Rock
2018 年 8 月 2 日
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)));
2 件のコメント
Diego Andrés Alvarez Marín
2021 年 4 月 23 日
T is the output of delaunay()
P = [X Y]; % array with two columns that contains the coordinates of the nodes
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