## 3 Natural Coordinate of Point in Tetrahedral Element  
A three-dimensional (3D) hexahedral finite element is commonly used for modeling three-dimensional bodies of general shape. A 3D hexahedral finite element can have a different number of nodes, typically ranging from 8 to 21. For modeling complex, irregular shapes like dams, tetrahedral finite elements with intermediate nodes are often used (Figure 5) [41]. An arbitrary position of a measuring point (M\({}_{\text{P}}\)) in a tetrahedral element with intermediate nodes is shown in Figure 5.  
Geometry interpolation functions [42] have the following form:  
\[\mathbf{x}= \begin{cases}\mathbf{x}\\ \mathbf{y}\\ \mathbf{z}\end{cases} =\mathbf{H}\mathbf{X} \tag{1}\]  
where the variables are defined as follows:  
\(\mathbf{x}\)--is the coordinate vector of the point in the element,  
\(\mathbf{H}\)--is the interpolation matrix,  
\(\mathbf{X}\)--represents vector of node coordinates. The vector of the node coordinates is given by Equation (2):  
\[\mathbf{X}^{\text{T}}=[X_{1}\,Y_{1}\,Z_{1}\,X_{2}\,Y_{2}\,Z_{2}\,\dots\dots\,X_ {N}\,Y_{N}\,Z_{N}] \tag{2}\]  
The interpolation matrix is as follows:  
\[\mathbf{H}= \begin{bmatrix}\text{h}_{1}&0&0&\text{h}_{2}&0&0&\dots\dots\, \text{h}_{N}&0&0\\ 0&\text{h}_{1}&0&0&\text{h}_{2}&0&\dots\dots\,\text{0}&\text{h}_{N}&0\\ 0&0&\text{h}_{1}&0&0&\text{h}_{2}&\dots\dots\,\text{0}&0&\text{h}_{N}\\ \end{bmatrix}  \tag{3}\]  
The interpolation functions for the tetrahedral finite element are given in Table 1 [42]. Midnodes of the tetra element are not used to calculate the local coordinates of the measuring point. The coordinates of the measuring point M\({}_{\text{P}}\) (\(x,y,z\)) can be expressed using the coordinates of the element in which it is located, according to Equation (1), as follows:  
\[ \begin{array}{l}x=(1-r-s-t)X_{1}+rX_{2}+sX_{3}+tX_{4}\\ y=(1-r-s-t)Y_{1}+rY_{2}+sY_{3}+tY_{4}\\ z=(1-r-s-t)Z_{1}+rZ_{2}+sZ_{3}+tZ_{4}\end{array}  \tag{4}\]  
with the following definitions:  
\(X_{1},Y_{1},Z_{1};\,X_{2},Y_{2},Z_{2};\,X_{3},Y_{3},Z_{3};\,X_{4},Y_{4},Z_{4}\)--the coordinates of the nodes of the tetrahedral element; \(r,s,t\)--the local coordinates of the measuring point M\({}_{\text{P}}\). Figure 5: Tetrahedral element with 10 nodes and arbitrary position of measuring point. By solving the system of Equations (4), natural coordinates are obtained:  
\[r=\frac{\left\{ \begin{array}{l}x[Z_{1}(Y_{3}-Y_{4})+Z_{3}(Y_{4}-Y_{1})+Z_{4}(Y_{ 1}-Y_{3})]+X_{1}[Y_{3}(Z_{4}-z)+y(Z_{3}-Z_{4})+Y_{4}(z-Z_{3})]+\\ X_{3}[Y_{4}(Z_{1}-z)+y(Z_{4}-Z_{1})+Y_{1}(z-Z_{4})]+X_{4}[Y_{3}(z-Z_{1})+y(Z_{1} -Z_{3})+Y_{1}(Z_{3}-z)]\\ \end{array} \right\}}{\left\{ \begin{array}{l}X_{2}[Z_{1}(Y_{3}-Y_{4})+Z_{3}(Y_{ 4}-Y_{1})+Z_{4}(Y_{1}-Y_{3})]+X_{1}[Y_{3}(Z_{4}-Z_{2})+Y_{2}(Z_{3}-Z_{4})+Y_{4} (Z_{2}-Z_{3})]+\\ X_{3}[Y_{4}(Z_{1}-Z_{2})+Y_{1}(Z_{2}-Z_{4})+Y_{2}(Z_{4}-Z_{1})]+X_{4}[Y_{3}(Z_{ 2}-Z_{1})+Y_{2}(Z_{1}-Z_{3})+Y_{1}(Z_{3}-Z_{2})]\\ \end{array} \right\}} \tag{5}\]  
\[s=\frac{\left\{ \begin{array}{l}x[Z_{1}(Y_{4}-Y_{2})+Z_{2}(Y_{1}-Y_{4})+Z_{4}( Y_{2}-Y_{1})]+X_{1}[Y_{2}(z-Z_{4})+y(Z_{4}-Z_{2})+Y_{4}(Z_{2}-z)]+\\ X_{2}[Y_{4}(z-Z_{1})+y(Z_{1}-Z_{4})+Y_{1}(Z_{4}-z)]+X_{4}[Y_{2}(Z_{1}-z)+y(Z_{2} -Z_{1})+Y_{1}(z-Z_{2})]\\ \end{array} \right\}}{\left\{ \begin{array}{l}X_{2}[Z_{1}(Y_{3}-Y_{4})+Z_{3}(Y _{4}-Y_{1})+Z_{4}(Y_{1}-Y_{3})]+X_{1}[Y_{3}(Z_{4}-Z_{2})+Y_{2}(Z_{3}-Z_{4})+Y_ {4}(Z_{2}-Z_{3})]+\\ X_{3}[Y_{4}(Z_{1}-Z_{2})+Y_{1}(Z_{2}-Z_{4})+Y_{2}(Z_{4}-Z_{1})]+X_{4}[Y_{3}(Z_{ 2}-Z_{1})+Y_{2}(Z_{1}-Z_{3})+Y_{1}(Z_{3}-Z_{2})]\\ \end{array} \right\}} \tag{6}\]  
\[t=\frac{-\left\{ \begin{array}{l}[X_{3}(Y_{1}-Y_{2})+X_{1}(Y_{2}-Y_{3})+X_{2 }(Y_{3}-Y_{1})][X_{2}(Z_{1}-z)+X_{1}(z-Z_{2})+x(Z_{2}-Z_{1})]-\\ [X_{2}(Y_{1}-y)+X_{1}(y-Y_{2})+x(Y_{2}-Y_{1})][X_{3}(Z_{1}-Z_{2})+X_{1}(Z_{2} -Z_{3})+X_{2}(Z_{3}-Z_{1})]\\ \end{array} \right\}}{\left\{