5
votes
How can one prove the duality of Voronoi and Delaunay?
The duality between Voronoi cells and vertices of the triangulation is pretty clear: each vertex of the Delaunay triangulation is a site in the Voronoi diagram which gets associated with its Voronoi ...
5
votes
Accepted
Relationship between number of nodes, elements and sides in a triangular 2D mesh
Yes there is a relationship, the Euler characteristic:
For a 2-dimensional orientable manifold with boundaries embedded in $\mathbb{R}^3$, the Euler characteristic is
$\chi = V - E + F = 2 - 2g - b$
...
4
votes
Accepted
Is the marching triangles algorithm guaranteed to terminate?
According to the following paper, the algorithm creates cracks like you have (see figure 1 and surrounding discussion).
Fournier, Marc. "Surface Reconstruction: An Improved Marching Triangle
...
3
votes
Accepted
How to remove triangles in a hollow hemisphere shape?
Rather than relying on a delaunay triangulation, you could consider making a structured mesh directly, as you know how you are constructing the points and the ...
3
votes
Delaunay triangulation for datasets with four or more co-circular points
You can use exact predicates to detect co-circular points, and use symbolic perturbation to consistently decide which triangles to generate.
Regarding exact predicates:
If your point coordinates ...
2
votes
How to remove triangles in a hollow hemisphere shape?
I agree with Mikael's suggestion to use a structured mesh if you can get away with it.
But there is a more fundamental reason why the Delaunay function from matlab isn't doing what you want and I ...
1
vote
Can TETGEN generate triangulation of a 2D point set?
TetGen is specifically designed to perform 3-D meshes -> tetrahedralization. I do not see a reliable and direct way to use it explicitly as a 2-D mesher. It is well pointed out that Triangle has a "...
1
vote
Iterating through a 3D triangle
To iterate over the triangle points you could follow this algorithm:
find the longest edge and use it as direction for the inner loop, in the following it is assumed that the longest edge is (p1,p2) ...
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