# Calculate the surface area of each face of a hexahedron and it's outward normal,given coordinates

Basically, I have a hexahedral finite element mesh. I know the coordinates of elements, I used the coordinate transformation into an isoparametric structure and shape(basis) functions to calculate the Jacobian to find the volume.

I also need the surface area of each face separately along with its outward unit normal.

What would be the best possible way to do that? Can the Jacobian be used to do this?

I have tried working with using vector algebra, by finding the cross-product of diagonals of each face to find the direction of the outward unit normal and the area but that doesn't work for an irregular hexahedron( which I think shouldn't be the case). Can this be another general approach?

I can't understand where I am going wrong here. Will be grateful if someone can give me some idea.

• For surface are you can use any of the number of elementary formulas for area of a triangle. For sign of unit normal you take it to point in the opposite way as one going to any other point of the hexahedral. there are standard ways to find normal of a surface z=ax+by. what im trying to say is you can search it here of google – KALLE DA BAWS Sep 12 '18 at 20:46
• Hi, since this can be an arbitrary hexahedron in a finite element mesh. I wanted to use an isoparametric transformation formula. It would not be feasible to calculate using the physical coordinates for all the elements. I know that the area can be found using cross product of the two sides but wanted to know if there is any room for exploration using the isoparametric formulation. – Schneider Sep 12 '18 at 21:19

## 1 Answer

The problem is that, in general, the faces of an irregular hexahedron are not flat. There is not one normal, the surface area is curved in space and difficult to calculate exactly (whatever the additional value of "exact" is in a numerical context). So I think you should anyway be satisfied with some sensible approximations. Here comes a proposal.

Let the vertices of the (in general non-planar) face of an irregular hexahedron be given by $\,\vec{r}_A , \vec{r}_B , \vec{r}_C , \vec{r}_D$ , as shown in the picture. Then form the vertices of the gray ( Varignon) quadrilateral as : $$\vec{r}_1 = \frac{1}{2}\left(\vec{r}_A+\vec{r}_C\right) \quad ; \quad \vec{r}_2 = \frac{1}{2}\left(\vec{r}_B+\vec{r}_D\right) \\ \vec{r}_3 = \frac{1}{2}\left(\vec{r}_A+\vec{r}_B\right) \quad ; \quad \vec{r}_4 = \frac{1}{2}\left(\vec{r}_C+\vec{r}_D\right)$$ It can be easily proved that the quadrilateral $\overline{1324}$ is a paralellogram and hence it is planar. The intersection point of the (yellow) diagonals of the Varignon paralellogram is a good place to define the normal, as a vector perpendicular to $\overline{1324}$ . Last but not least the total area of the face is defined by the sum of five planar areas: the area of the paralellogram $\overline{1324}$ and the four areas of the triangles $\,\overline{A31} , \overline{B23} , \overline{D42} , \overline{C14}$ . Hope this helps.

• Thanks. This looks to be a good way to go forward. Probably can afford this much approximation. – Schneider Sep 17 '18 at 19:22