Gradient of the Barycentric Coordinate in 2D

I always forget this fact, so hopefully typing this out will help.

Let T be a triangle with vertices v_1, v_2, v_3, then there exists unique linear functions The derivation is simple. The direction of the gradient is towards the highest growth, and as the barycentric coordinates are linear functions, it’s clear that -n_k is the correct direction. The scaling is to simply note the area as ls says:

when you say “unit normal” which unit normal are you referring to?

  1. It’s probably best with an example. If the triangle has vertices (0, 0), (1, 0), (0, 1), then the unit outward normal for the edge (0, 0) — (1, 0) is (0, -1) (e.g. the vector which points directly south).

    For the edge (0, 0) — (0, 1), it is (-1, 0).

    For the edge (1, 0) — (0, 1), it is (1/sqrt(2), 1/sqrt(2)) (the square roots are there to normalize the vector).

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