Editing Normal (geometry) (section)

===Transforming normals===
{{hatnote|in this section we only use the upper <math>3 \times 3</math> matrix, as translation is irrelevant to the calculation}}

When applying a transform to a surface it is often useful to derive normals for the resulting surface from the original normals.

Specifically, given a 3×3 transformation matrix <math>\mathbf{M},</math> we can determine the matrix <math>\mathbf{W}</math> that transforms a vector <math>\mathbf{n}</math> perpendicular to the tangent plane <math>\mathbf{t}</math> into a vector <math>\mathbf{n}^{\prime}</math> perpendicular to the transformed tangent plane <math>\mathbf{Mt},</math> by the following logic:

Write '''n&prime;''' as <math>\mathbf{Wn}.</math> We must find <math>\mathbf{W}.</math>
<math display=block>\begin{alignat}{5}
W\mathbb n \text{ is perpendicular to } M\mathbb t \quad \,
&\text{ if and only if } \quad 0 = (W \mathbb n) \cdot (M \mathbb t) \\
&\text{ if and only if } \quad 0 = (W \mathbb{n})^\mathrm{T} (M \mathbb{t}) \\
&\text{ if and only if } \quad 0 = \left(\mathbb{n}^\mathrm{T} W^\mathrm{T}\right) (M \mathbb{t}) \\
&\text{ if and only if } \quad 0 = \mathbb{n}^\mathrm{T} \left(W^\mathrm{T} M\right) \mathbb{t} \\
\end{alignat}</math>

Choosing <math>\mathbf{W}</math> such that  <math>W^\mathrm{T} M = I,</math> or <math>W = (M^{-1})^\mathrm{T},</math> will satisfy the above equation, giving a <math>W \mathbb n</math> perpendicular to <math>M \mathbb t,</math> or an <math>\mathbf{n}^{\prime}</math> perpendicular to <math>\mathbf{t}^{\prime},</math> as required.

Therefore, one should use the inverse transpose of the linear transformation when transforming surface normals. The inverse transpose is equal to the original matrix if the matrix is orthonormal, that is, purely rotational with no scaling or shearing.