4.2.2 Tetrahedron Barycentric Coordinates

Let P be a point inside the tetrahedral element as shown in Fig. <4.7>. It divides the tetrahedron in four sub-tetrahedrons

$$V_1 = V_{P234}, V_2 = V_{P134}, V_3 = V_{P124}, V_4 = V_{P123}.$$

The barycentric coordinates of the point P are given by

$$\lambda^e_1(\vec{r}) = \frac{V_1}{V_e} = \frac{1}{4} + (\vec{r} -... ...r}_5)}{J} = \frac{1}{4} - (\vec{r} - \vec{r}_B)\cdot\frac{F_1 \vec{n}_1}{3V_e}$$ (4.66)

$$\begin{split}\lambda^e_2(\vec{r}) = \frac{V_2}{V_e} & = \frac... ...vec{r} - \vec{r}_B)\cdot\frac{F_2 \vec{n}_2}{3V_e} \end{split}$$ (4.67)

$$\lambda^e_3(\vec{r}) = \frac{V_3}{V_e} = \frac{1}{4} + (\vec{r} -... ...r}_1)}{J} = \frac{1}{4} - (\vec{r} - \vec{r}_B)\cdot\frac{F_3 \vec{n}_3}{3V_e}$$ (4.68)

$$\lambda^e_4(\vec{r}) = \frac{V_4}{V_e} = \frac{1}{4} + (\vec{r} -... ...}_2)}{J} = \frac{1}{4} - (\vec{r} - \vec{r}_B)\cdot\frac{F_4 \vec{n}_4}{3V_e},$$ (4.69)

where $F_i (i\in[1;4])$ is the area of the triangular face of the tetrahedron opposite to the vertex $i$

$$F_1 = F_{234}, F_2 = F_{134}, F_3 = F_{124}, F_4 = F_{123}.$$

The vector $\vec{n}_i (i\in[1;4])$ is normal to its according face, has the length $1$ and points outwards. The position vector is written as

$$\vec{r} = \vec{r}(x,y,z) = x \vec{e}_x + y \vec{e}_y + z \vec{e}_z$$ (4.70)

and $\vec{r}_i$ is the position of the vertex $i$ .

Analogously to the two-dimensional case it can be shown that the barycentric coordinates are equal to the linear element shape functions in (4.65). Thus the same notation $\lambda_i^e$ is used.

The gradient of the barycentric coordinates is a constant vector

$$\vec{\nabla}\lambda^e_1(\vec{r}) = \frac{(\vec{r}_4\times\vec{r}_5)}{J} = - \frac{F_1 \vec{n}_1}{3V_e}$$ (4.71)

$$\vec{\nabla}\lambda^e_2(\vec{r}) = \frac{(\vec{r}_2\times\vec{r}_3)}{J} = - \frac{F_2 \vec{n}_2}{3V_e}$$ (4.72)

$$\vec{\nabla}\lambda^e_3(\vec{r}) = \frac{(\vec{r}_3\times\vec{r}_1)}{J} = - \frac{F_3 \vec{n}_3}{3V_e}$$ (4.73)

$$\vec{\nabla}\lambda^e_4(\vec{r}) = \frac{(\vec{r}_1\times\vec{r}_2)}{J} = - \frac{F_4 \vec{n}_4}{3V_e}.$$ (4.74)

Analogously to the two-dimensional case it is valid

$$\lambda^e_i(x_j,y_j,z_j) = \delta_{ij} = \left\{ \begin{array}{lc} 1 & i = j 0 & i \neq j \end{array} \right..$$ (4.75)

The barycentric coordinate $\lambda^e_i$ is constant on a plane parallel to the element facet opposite to the $i$ -th node and it is zero on this opposite facet, which ensures the inter-element continuity of the element interpolation function (4.58).

Only three of the four linear element form functions are independent

$$\sum^4_{i=1}\lambda^e_i = 1.$$ (4.76)

For points inside the element and on the element facets

$$0 \leq \lambda^e_1 \leq 1, 0 \leq \lambda^e_2 \leq 1, 0 \leq \lambda^e_3 \leq 1, 0 \leq \lambda^e_4 \leq 1$$ (4.77)

is satisfied. Similar to the two-dimensional case some of the barycentric coordinates of a point outside the element can be negative or greater than 1.

Otherwise, the barycentric coordinates can be used to represent the coordinates of each point inside the tetrahedral element

$$\vec{r} = \sum_{i=1}^4\lambda^e_i\vec{r}_i.$$ (4.78)

Using (4.76), equation (4.78) leads to

$$\begin{split}x & = (x_1 - x_4)\lambda^e_1 + (x_2 - x_4)\lambd... ...2 - z_4)\lambda^e_2 + (z_3 - z_4)\lambda^e_3 + z_4, \end{split}$$ (4.79)

which gives the coordinate transformation as shown in Fig. <4.8>.

Figure 4.8: Three-dimensional domain transformation.