HelixToolkit.Wpf.MeshBuilder.FindCornerNormal C# (CSharp) Method

        private Vector3D FindCornerNormal(Point3D p, double eps)
        {
            var sum = new Vector3D();
            int count = 0;
            var addedNormals = new HashSet<Vector3D>();
            for (int i = 0; i < this.triangleIndices.Count; i += 3)
            {
                int i0 = i;
                int i1 = i + 1;
                int i2 = i + 2;
                var p0 = this.positions[this.triangleIndices[i0]];
                var p1 = this.positions[this.triangleIndices[i1]];
                var p2 = this.positions[this.triangleIndices[i2]];

                // check if any of the vertices are on the corner
                double d0 = (p - p0).LengthSquared;
                double d1 = (p - p1).LengthSquared;
                double d2 = (p - p2).LengthSquared;
                double mind = Math.Min(d0, Math.Min(d1, d2));
                if (mind > eps)
                {
                    continue;
                }

                // calculate the triangle normal and check if this face is already added
                var normal = Vector3D.CrossProduct(p1 - p0, p2 - p0);
                normal.Normalize();

                // todo: need to use the epsilon value to compare the normals?
                if (addedNormals.Contains(normal))
                {
                    continue;
                }

                // todo: this does not work yet
                // double dp = 1;
                // foreach (var n in addedNormals)
                // {
                // dp = Math.Abs(Vector3D.DotProduct(n, normal) - 1);
                // if (dp < eps)
                // continue;
                // }
                // if (dp < eps)
                // {
                // continue;
                // }
                count++;
                sum += normal;
                addedNormals.Add(normal);
            }

            if (count == 0)
            {
                return new Vector3D();
            }

            return sum * (1.0 / count);
        }