## 3 Spherical Area Coordinates (SACs)
### Local SACs
#### 3.2.2 Transformation Laws between SACs and Cartesian Coordinates  
Because Cartesian coordinates have been widely used in many areas of science and engineering, the transformation between Cartesian coordinates and SACs must be established. However, Equation (3) is too complicated for obtaining the transformation; thus, we need a simpler relationship between the spherical triangle area and its vertices. In spherical trigonometry, the following identity can be derived from sine and cosine laws [32]:  
\[S=2\tan^{-1}\left(\frac{\mathbf{p}_{1}\cdot(\mathbf{p}_{2}\times\mathbf{p}_{3})}{1+\mathbf{p} _{1}\cdot\mathbf{p}_{2}+\mathbf{p}_{2}\cdot\mathbf{p}_{3}+\mathbf{p}_{3}\cdot\mathbf{p}_{1}} \right). \tag{6}\]  
From Equation (3), (6), we obtain  
\[\begin{split}\lambda_{1}&=\frac{S_{1}}{S}=\frac{2} {S}\tan^{-1}\left(\frac{\mathbf{p}\cdot(\mathbf{p}_{2}\times\mathbf{p}_{3})}{1+\mathbf{p} \cdot\mathbf{p}_{2}+\mathbf{p}_{2}\cdot\mathbf{p}_{3}+\mathbf{p}_{3}\cdot\mathbf{p}}\right),\\ \lambda_{2}&=\frac{S_{2}}{S}=\frac{2}{S}\tan^{-1} \left(\frac{\mathbf{p}\cdot(\mathbf{p}_{3}\times\mathbf{p}_{1})}{1+\mathbf{p}\cdot\mathbf{p}_{3}+ \mathbf{p}_{3}\cdot\mathbf{p}_{1}+\mathbf{p}_{1}\cdot\mathbf{p}}\right),\\ \lambda_{3}&=\frac{S_{3}}{S}=\frac{2}{S}\tan^{-1} \left(\frac{\mathbf{p}\cdot(\mathbf{p}_{1}\times\mathbf{p}_{2})}{1+\mathbf{p}\cdot\mathbf{p}_{1}+ \mathbf{p}_{1}\cdot\mathbf{p}_{2}+\mathbf{p}_{2}\cdot\mathbf{p}}\right).\end{split} \tag{7}\]  
If the vectors in Equation (7) are given in the Cartesian coordinate system, then it determines the transformation rules from Cartesian coordinates to SAC.  
Given a point \(P\) at \(\mathbf{u}=(\lambda_{1},\lambda_{2},\lambda_{3})\), if the Cartesian coordinates of the vertices \(\mathbf{p}_{1},\mathbf{p}_{2},\mathbf{p}_{3}\) are known, then Equation (7) can be used to obtain  
\[\tan\frac{S\lambda_{1}}{2}(1+\mathbf{p}_{2}\cdot\mathbf{p}_{3}) =\left[(\mathbf{p}_{2}\times\mathbf{p}_{3})-\tan\frac{S\lambda_{1}}{2}( \mathbf{p}_{2}+\mathbf{p}_{3})\right]\cdot\mathbf{p}, \tag{8}\] \[\tan\frac{S\lambda_{2}}{2}(1+\mathbf{p}_{3}\cdot\mathbf{p}_{1}) =\left[(\mathbf{p}_{3}\times\mathbf{p}_{1})-\tan\frac{S\lambda_{2}}{2}( \mathbf{p}_{3}+\mathbf{p}_{1})\right]\cdot\mathbf{p},\] \[\tan\frac{S\lambda_{3}}{2}(1+\mathbf{p}_{1}\cdot\mathbf{p}_{2}) =\left[(\mathbf{p}_{1}\times\mathbf{p}_{2})-\tan\frac{S\lambda_{3}}{2}( \mathbf{p}_{1}+\mathbf{p}_{2})\right]\cdot\mathbf{p}.\]  
Equation (8) can be written in the matrix form:  
\[\mathbf{p}^{T}=A^{-1}\mathbf{k}, \tag{9}\]  
where \(\mathbf{p}=(x,y,z)\), and matrix \(A\) and vector \(\mathbf{k}\) are given by  
\[A=\begin{bmatrix}(\mathbf{p}_{2}\times\mathbf{p}_{3})-\tan\frac{S\lambda_{1}}{2}(\mathbf{ p}_{2}+\mathbf{p}_{3})\\ (\mathbf{p}_{3}\times\mathbf{p}_{1})-\tan\frac{S\lambda_{2}}{2}(\mathbf{p}_{3}+\mathbf{p}_{1}) \\ (\mathbf{p}_{1}\times\mathbf{p}_{2})-\tan\frac{S\lambda_{3}}{2}(\mathbf{p}_{1}+\mathbf{p}_{2}) \end{bmatrix},\qquad\mathbf{k}=\begin{bmatrix}\tan\frac{S\lambda_{1}}{2}(1+\mathbf{p}_ {2}\cdot\mathbf{p}_{3})\\ \tan\frac{S\lambda_{2}}{2}(1+\mathbf{p}_{3}\cdot\mathbf{p}_{1})\\ \tan\frac{S\lambda_{3}}{2}(1+\mathbf{p}_{1}\cdot\mathbf{p}_{2})\end{bmatrix}. \tag{10}\]  
Finally, we obtain transformation laws between the Cartesian coordinates and SACs.