How do I calculate a normal vector based on multiple triangles sharing a vertex?

Question

If I have a mesh of triangles, how does one go about calculating the normals at each given vertex?

I understand how to find the normal of a single triangle. If I have triangles sharing vertices, I can partially find the answer by finding each triangle's respective normal, normalizing it, adding it to the total, and then normalizing the end result. However, this obviously does not take into account proper weighting of each normal (many tiny triangles can throw off the answer when linked with a large triangle, for example).

Answer 1

This blog post outlines three different methods and gives a visual example of why the standard and simple method (area weighted average of the normals of all the faces joining at the vertex) might sometimes give poor results.

Answer 2

I think a good method should be using a weighted average but using angles instead of area as weights. This is in my opinion a better answer because the normal you are computing is a "local" feature so you don't really care about how big is the triangle that is contributing... you need a sort of "local" measure of the contribution and the angle between the two sides of the triangle on the specified vertex is such a local measure.

Using this approach a lot of small (thin) triangles doesn't give you an unbalanced answer.

Using angles is the same as using an area-weighted average if you localize the computation by using the intersection of the triangles with a small sphere centered in the vertex.

Answer 3

The weighted average appears to be the best approach.

But be aware that, depending on your application, sharp corners could still give you problems. In that case, you can compute multiple vertex normals by averaging surface normals whose cross product is less than some threshold (i.e., closer to being parallel).

Search for Offset triangular mesh using the multiple normal vectors of a vertex by SJ Kim, et. al., for more details about this method.

Answer 4

You can give more weight to big triangles by multiplying the normal by the area of the triangle.

Answer 5

Check out this paper: Discrete Differential-Geometry Operators for Triangulated 2-Manifolds.

In particular, the "Discrete Mean Curvature Normal Operator" (Section 3.5, Equation 7) gives a robust normal that is independent of tessellation, unlike the methods in the blog post cited by another answer here.

Answer 6

Obviously you need to use a weighted average to get a correct normal, but using the triangles area won't give you what you need since the area of each triangle has no relationship with the % weight that triangles normal represents for a given vertex.

If you base it on the angle between the two sides coming into the vertex, you should get the correct weight for every triangle coming into it. It might be convenient if you could convert it to 2d somehow so you could go off of a 360 degree base for your weights, but most likely just using the angle itself as your weight multiplier for calculating it in 3d space and then adding up all the normals produced that way and normalizing the final result should produce the correct answer.