How can I compute the gradient for a tetrahedral mesh (3D)? For triangular mesh, I got an answer from the following post Calculating the gradient of a triangular mesh
How can I get a similar formula for 3D mesh?
Thanks in advance.
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Assuming a value is assigned to each vertex of the mesh and we use purely linear interpolation, then there will be a constant gradient vector within each tetrahedron.
Linear interpolation can be expressed using barycentric coordinates, like $$ f(x,y,z) = f_1 w_1(x,y,z) + f_2 w_2(x,y,z) + f_3 w_3(x,y,z) + f_4 w_4(x,y,z) $$ where $f_1 \ldots f_4$ are the values of the function at the four vertices, and $w_1 \ldots w_4$ are the barycentric weights for each vertex. Then, finding the gradient of $f$ reduces to finding the gradients of all of the weights.
This can be worked out geometrically by noting that each $w_i$ is 1 at the $i$th vertex, falling off to 0 at the plane formed by the other three vertices. The gradient vector will therefore be normal to that plane, pointing back towards the $i$th vertex, with a magnitude equal to 1 / the distance from the plane to the vertex.
Once you've calculated those barycentric gradients, you can multiply them by $f_1 \ldots f_4$ and sum them up to arrive at the gradient of $f$ overall.
This reasoning works for triangles too, by the way, only replace "plane" with "line".
answered Aug 30 '20 at 6:31
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$\begingroup$ Is there a similar way to compute the Laplacian operator for 3D mesh? $\endgroup$ – Bis Aug 31 '20 at 1:16
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$\begingroup$ Laplacians are tricky to even define, because it's a second-derivative operator, so with linear interpolation it would just be zero. Getting a useful answer out of it requires defining some higher-order interpolation scheme first. You can see some approaches to this in Laplace–Beltrami: The Swiss Army Knife of Geometry Processing $\endgroup$ – Nathan Reed Aug 31 '20 at 2:33
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