## 2 The Mathematical Modeling and Simulation
\right)-\left( \begin{array}{cccc}e_{i}\cos\beta_{i}&e_{i}\sin \beta_{i}&0\end{array} \right)\left( \begin{array}{c}x\\ y\\ z\end{array} \right)+e_{i}^{2}+l_{i}^{2}-e_{i}^{2}\] \[=x^{2}+y^{2}+z^{2}-2 xe_{i}\cos\beta_{i}-2 ye_{i}\sin\beta_{i}+l_{i} ^{2}\] \[ \begin{array}{cccc}l_{i}\mathbf{r}^{T}\mathbf{u}_{i}&=l_{i}\left( \begin{array}{cccc}x&y&z\end{array}\right)\left(\begin{array}{cccc}\sin \theta_{i}\cos\beta&\sin\theta_{i}\sin\beta_{i}&\cos\theta\end{array}\right)^{T} \\ &=l_{i}(x\sin\theta_{i}\cos\beta_{i}+y\sin\theta_{i}\sin\beta_{i}+z\cos\theta_ {i})\\ \end{array} \] \[ \begin{array}{cccc}l_{i}\mathbf{e}_{i}^{T}\mathbf{u}_{i}&=1_{i}(\begin{array} []{cccc}e_{i}\cos\beta_{i}&e_{i}\sin\beta_{i}&0\end{array})\left(\begin{array} []{cccc}\sin\theta_{i}\cos\beta_{i}&\sin\theta_{i}\sin\beta_{i}&\cos\theta_ {i}\end{array}\right)^{T}\\ &=l_{i}(e_{i}\sin\theta_{i}\cos^{2}\beta_{i}+e_{i}\sin\theta_{i}\sin^{2}\beta_ {i})\\ &=e_{i}l_{i}\sin\theta_{i}\end{array} \] \[ \begin{array}{cccc}l_{i}\mathbf{u}_{i}^{T}\mathbf{r}&=l_{i}(\begin{array} []{cccc}\sin\theta_{i}\cos\beta_{i}&\sin\theta_{i}\sin\beta_{i}&\cos\theta_ {i}\end{array})\left(\begin{array}{cccc}x&y&z\end{array}\right)^{T}\\ &=l_{i}(x\sin\theta_{i}\cos\beta_{i}+y\sin\theta_{i}\sin\beta_{i}+z\cos\theta_ {i})\\ \end{array} \] \[ \begin{array}{cccc}l_{i}\mathbf{u}_{i}^{T}\mathbf{e}_{i}&=l_{i}(\begin{array} []{cccc}\sin\theta_{i}\cos\beta_{i}&\sin\theta_{i}\sin\beta_{i}&\cos\theta_{i }\end{array})\left(\begin{array}{cccc}e_{i}\cos\beta_{i}&e_{i}\sin\beta_{i }&0\end{array}\right)^{T}\\ &=l_{i}(e_{i}\sin\theta_{i}\cos^{2}\beta_{i}+e_{i}\sin\theta_{i}\sin^{2}\beta_{i })\\ &=e_{i}l_{i}\sin\theta_{i}\end{array} \]  
Then, Equation (4) can be extended to Equation (5). \[x^{2}+y^{2}+z^{2}-2 xe_{i}\cos\beta_{i}-2 ye_{i}\sin\beta_{i}+l_{i} ^{2}-2l_{i}(x\cos\beta_{i}+y\sin\beta_{i}-e_{i})\sin\theta_{i}-2l_{i}z\cos \theta_{i}=0 \tag{5}\]  
Considering that Equation (5) is the function of (x,y,z), we can get Equation (6):  
\[f_{i}(x,y,z)=x^{2}+y^{2}+z^{2}+l_{i}^{2}-2(l_{i}\sin\theta_{i}+e_{i})\cos\beta _{i}x-2(l_{i}\sin\theta_{i}+e_{i})\sin\beta_{i}y+2l_{i}e_{i}\sin\theta_{i}-2l_ {i}z\cos\theta_{i}=0. \tag{6}\]  
Let t=(x y z), t (k) is the k-th approximation of the solution t,\(f_{i}\)(t) is expanded in vector form by Taylor series to Equation (7). \[\mathbf{f}_{i}(t)\approx\mathbf{f}_{i}(t^{(k)})+\
abla\mathbf{f}_{i}(t^{(k)})^{T}(t-t^{(k)}), \tag{7}\]  
that is:  
\[\mathbf{f}\left(t\right)=\left( \begin{array}{cccc}\
abla f_{1}(t)^{T}\\ \
abla f_{2}(t)^{T}\\ \vdots\\ \
abla f_{n}(t)^{T}\end{array} \right)=\left( \begin{array}{cccc}\frac{\partial f _{1}(t)}{\partial t_{1}}&\frac{\partial f_{1}(t)}{\partial t_{2}}&\dots&\frac {\partial f_{1}(t)}{\partial t_{n}}\\ \frac{\partial f_{2}(t)}{\partial t_{1}}&\frac{\partial f_{2}(t)}{\partial t_{ 2}}&\dots&\frac{\partial f_{2}(t)}{\partial t_{n}}\\ \vdots&\vdots&\dots&\vdots\\ \frac{\partial f_{m}(t)}{\partial t_{1}}&\frac{\partial f_{m}(t)}{\partial t_{ 2}}&\dots&\frac{\partial f_{m}(t)}{\partial t_{n}}\end{array} \right).\]  
By using the Newton iterative method to solve the formula, the solution t can easily be calculated:\(t^{(k+1)}=t^{(k)}+\Delta t^{(k)}\). The procedure of the MATLAB simulation is as follows. (1) The initial values of coordinate point \(t_{0}=\left( \begin{array}{cccc}x_{0}&y_{0}&z_{0}\end{array} \right)^{T}\) are assigned for the Newton iterative method;  
(2) Calculate the initial value of the function according to the initial value \(t_{0}\);  
(3) Calculating the initial Hesse matrix\(\mathbf{f}_{0}(t)\);  
(4) According to the formula \(\mathbf{f}\left(t^{(k)}\right)\Delta t^{(k)}=-\mathbf{f}\left(t^{(k)}\right)\), the value \(\Delta t^{(0)}\) can be calculated;  
(5) According to the formula \(t^{(k+1)}=t^{(k)}+\Delta t^{(k)}\), the value \(t^{(1)}\) can be obtained;  
(6) Judge whether the end condition of the iteration is met. If not, return to step (2) to continue iteration.