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Consider a single quadrilaterally-faced hexahedron. If given the co-ordinates of the vertices, $\mathbf{v}_i$, of a face in counter-clockwise orientation, I can compute the corresponding unit outward normal vector using a cross-product:

$\mathbf{n}=\frac{(\mathbf{v}_2-\mathbf{v}_1)\times(\mathbf{v}_4-\mathbf{v}_1)}{\parallel(\mathbf{v}_2-\mathbf{v}_1)\times(\mathbf{v}_4-\mathbf{v}_1)\parallel}$

However, consider now an arbitrarily shaped volume which is discretised into a large number of quadrilaterally-faced hexahedra. Consider in particular a face shared by two adjacent hexahedra. If now given the co-ordinates of the vertices of the face in a certain orientation, the orientation will be clockwise for one hexahedron and counter-clockwise for the other.

If I do not know the orientation beforehand, how can I ensure that the unit normal is pointing outwards (assuming you know the co-ordinates of all vertices of all faces of all hexehedra)?

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Your hexahedron is convex, so you can do the dot product of the normal vector that you found, and a vector from one of the vertices to some point inside the hexahedron. If the dot product is negative, then your normal vector points out.

If you already know a point inside the hexahedron, then you're in luck. Otherwise you would have to find this, which would pose challenges of its own.

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you can computee the dot product between the normal and the vector:

[center of element, center of the current face].

If it's > 0, the normal is outwards to the element.

If not, it's inward and you just have to do : normal = -normal, to change that

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