how to find the area of the triangles formed as a result of delaunay triangulation

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.

[M.S. Khan]

How you counted the number of triangles? Please tell me the methods, regards

Answer 1

[Roger Stafford]

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.

[Abinaya]

thank u so much...

[Valerio Ficcadenti]

Hi

How did you manage the polygons that are not closed? Namely, those with vertexes to infinite?

Valerio

Answer 2

[Christopher Rock]

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)));

[SAsa]

What are your inputs t and p ?

[Diego Andrés Alvarez Marín]

T is the output of delaunay()

P = [X Y]; % array with two columns that contains the coordinates of the nodes