4.2 Scalar treatment

Combining Equations 1, 2 and 3, we obtain Equation 5.

$$\mathbf{n}=\left( \mathbf{P}_{3}-\mathbf{P}_{2}\right) \times \left( \mathbf{P}_{2}-\mathbf{P}_{1}\right)$$ (5)

Expanding this into the individual components, and using the definition of vector cross-product, we obtain the three Equations 6 to 8.

$$n_{x} = \left( P_{3y}-P_{2y}\right) \left( P_{2z}-P_{1z}\right) -\left( P_{3z}-P_{2z}\right) \left( P_{2y}-P_{1y}\right)$$ (6)

$$n_{y} = \left( P_{3z}-P_{2z}\right) \left( P_{2x}-P_{1x}\right) -\left( P_{3x}-P_{2x}\right) \left( P_{2z}-P_{1z}\right)$$ (7)

$$n_{z} = \left( P_{3x}-P_{2x}\right) \left( P_{2y}-P_{1y}\right) -\left( P_{3y}-P_{2y}\right) \left( P_{2x}-P_{1x}\right)$$ (8)

In order to normalise these values, we find the scalar value , using the values for , and found in Equations 6 to 8.

$$n=\left\vert \mathbf{n}\right\vert =\sqrt{n_{x}^{2}+n_{y}^{2}+n_{z}^{2}}$$ (9)

We now use this normalisation factor to find the normalised versions of Equations 6 to 8.

$$\hat{n}_{x} = \frac{1}{n}n_{x}$$

$$= \frac{1}{n}\left( P_{3y}-P_{2y}\right) \left( P_{2z}-P_{1z}\right) -\frac{1}{n}\left( P_{3z}-P_{2z}\right) \left( P_{2y}-P_{1y}\right)$$ (10)

$$\hat{n}_{y} = \frac{1}{n}n_{y}$$

$$= \frac{1}{n}\left( P_{3z}-P_{2z}\right) \left( P_{2x}-P_{1x}\right) -\frac{1}{n}\left( P_{3x}-P_{2x}\right) \left( P_{2z}-P_{1z}\right)$$ (11)

$$\hat{n}_{z} = \frac{1}{n}n_{z}$$

$$= \frac{1}{n}\left( P_{3x}-P_{2x}\right) \left( P_{2y}-P_{1y}\right) -\frac{1}{n}\left( P_{3y}-P_{2y}\right) \left( P_{2x}-P_{1x}\right)$$ (12)

The values , and given in Equations 10 to 12 are now the , and coordinates of the normalised surface normal to the triangle with vertices , and .