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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

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

The barycentric coordinates of the point P are given by

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

\begin{displaymath}\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}\end{displaymath} (4.67)

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

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

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

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

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

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

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

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

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

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

Analogously to the two-dimensional case it is valid

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

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

For points inside the element and on the element facets

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

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

Using (4.76), equation (4.78) leads to

\begin{displaymath}\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}\end{displaymath} (4.79)

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

Figure 4.8: Three-dimensional domain transformation.
\includegraphics[width=14cm]{figures/scalarfem3d/domtrans.eps}


next up previous contents
Next: 4.2.3 Assembling Up: 4.2 Three-Dimensional Scalar Finite Previous: 4.2.1 Linear Shape Functions   Contents

A. Nentchev: Numerical Analysis and Simulation in Microelectronics by Vector Finite Elements