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The Triangle 

We first consider the triangular shaped two-dimensional element. We will use the formulation of Sherwin [20] to describe the element.
 
 
 
Figure 2.1: The tensor coordinates of the triangle.
 

The natural coordinate systems will be given in terms of the ordered-pair ${\bf x} = (x,y)$. The coordinate system of the reference elements will be given in terms of (r,s). Finally the local tensor space for each element will be given in terms of (a,b) coordinates.

The reference triangle, shown in figure 2.1, is described

It is mapped to a straight-sided physical triangle with the following mapping:

\begin{displaymath}{\bf x} = -\frac{(r+s)}{2} {\bf v^1} + \frac{(1+r)}{2} {\bf v^2} + \frac{(1+s)}{2} {\bf v^3} \end{displaymath}

where the ${\bf v^1}$,${\bf v^2}$, and ${\bf v^3}$ are the physical coordinates of the vertices of the triangle labelled in counter-clockwise manner. 

The Jacobian for this mapping is:

\begin{displaymath}\frac{\partial(x,y)}{\partial(r,s)}= \vert({\bf v^2} - {\bf v^1}) \times ({\bf v^3}-{\bf v^1})\vert\end{displaymath}

The tensor (square) element is the set of points:

\begin{displaymath}Q_{tensor}^2 = \{(a,b) \vert -1 \leq a,b \leq 1\}\end{displaymath}

The tensor element is mapped to the reference triangle by:

The Jacobian for this mapping is:

\begin{displaymath}\frac{\partial(r,s)}{\partial(a,b)}= \frac{(1-b)}{2}\end{displaymath}

We notice that this mapping is singular at b=1. We will demonstrate that this singularity will not present any numerical problems in chapter 4.


next up previous contents
Next: The Quadrilateral Up: The Elements Previous: The Elements 
T. Warburton

10/24/1998