## 3 Kinematic Analysis
The coefficients L, M, N, P, and \(Q\) are terms of \(t_{3}\), as in Equation (61). \[\frac{1}{4}(l_{1}C\theta_{8}+l_{2}C\rho_{2}-l_{1}C\theta_{9}-l_{2}C\rho_{3})^ {2}+\left(l_{1}S\theta_{8}+l_{2}S\rho_{2}-l_{1}S\theta_{9}-l_{2}S\rho_{3}\right) ^{2}\]7  
Next, by substituting the trigonometric identities formula \(S\varnothing_{i}=\frac{2t_{i}}{1+t_{i}^{2}}\) and \(C\varnothing_{i}=\frac{1-t_{i}^{2}}{1+t_{i}^{2}}\) where \(t_{i}=tan\frac{\varnothing_{i}}{2}\) into (42), the rearrangement of a second-degree polynomial equation can be obtained as (57). The coefficients \(G\), \(H\), and \(I\) are also in terms of \(t_{3}\)  
\[\frac{1}{4}(l_{1}C\theta_{8}+l_{2}C\rho_{2}-l_{1}C\theta_{9}-l_{2}C\rho_{3})^ {2}+\left(l_{1}S\theta_{8}+l_{2}S\rho_{2}-l_{1}S\theta_{9}-l_{2}S\rho_{3}\right) ^{2}\]8  
By substituting \(t_{1}^{2}=-\frac{I+Ht_{1}}{G}\) from (62) to (61), then, \(t_{1}\) can be obtained as:  
\[\frac{1}{4}(l_{1}C\theta_{8}+l_{2}C\rho_{2}-l_{1}C\theta_{9}-l_{2}C\rho_{3})^ {2}+\left(l_{1}S\theta_{8}+l_{2}S\rho_{2}-l_{1}S\theta_{9}-l_{2}S\rho_{3}\right) ^{2}\]9When substituting \(t_{1}\) from (63) into (49), where \(G0\); then,  
\[ \begin{array}{ll}G^{3}\{Q^{2}G^{4}-QPHG^{3}&-2 QNIG^{3}+IP^{2}G^{3}+QNH^{2}G^{2}+ 3 QMIHG^{2}-PNIHG^{2}\\ &-2 PMI^{2}G^{2}+N^{2}I^{2}G^{2}-QMH^{3}G-4 QLIH^{2}G+PMIH^{2}G\\ &+3 PLHI^{2}G-NMHI^{2}G-2 NLI^{3}G+M^{2}I^{3}G+QLH^{4}-PLIH^{3}\\ &+NL^{2}H^{2}-MLH^{3}+L^{2}I^{4}\}=0\end{array}  \tag{64}\]  
Subsequently, a 16 th-degree polynomial equation was found from (65), in terms of \(t_{3}\):  
\[ \begin{array}{ll}Q^{2}G^{4}-QPHG^{3}-2 QNIG^{3}+IP^{2}G^{3}+QNH^{2}G^{2}+3 QMIHG ^{2}-PNIHG^{2}\\ -2 PMI^{2}G^{2}+N^{2}I^{2}G^{2}-QMH^{3}G-4 QLIH^{2}G+PMIH^{2}G+3 PLHI^{2}G-\\ NMHI^{2}G-2 NLI^{3}G+M^{2}I^{3}G+QLH^{4}-PLIH^{3}+NLI^{2}H^{2}-MLH^{3}+L^{2}I^{4} =0\end{array}  \tag{65}\]  
\(t_{3}\) can be obtained by solving the 16 th-degree polynomial Equation (65) using MATLAB and \(t_{1}\) can be obtained by substituting \(t_{3}\) in (63). \(\varnothing_{1}\) and \(\varnothing_{3}\) can be solved using \(t_{i}=tan\frac{\varnothing_{i}}{2}\) where \(\varnothing_{i}=2 tan^{-1}(t_{i})\) then \(\varnothing_{3}\) can be obtained from (59). After finding \(\varnothing_{1}\), \(\varnothing_{2}\) and \(\varnothing_{3}\) and substituting in (34)-(36), then, \(B_{1}\), \(B_{2}\) and \(B_{3}\) can be solved with knowledge of \(\theta_{7}\), \(\theta_{8}\) and \(\theta_{9}\). The position vector \(O_{7}=\left[O_{7,x}\ O_{7,y}\ O_{7,z}\right]^{T}\), can be calculated by using Equation (66):  
\[O_{7}=\frac{1}{3}[B_{1}+B_{2}+B_{3}] \tag{66}\]  
Then computing orientation \(R_{B}^{A}\) elements \(u_{x}\), \(u_{y}\)_and_\(u_{z}\) By using (23)-(25) as (67)-(69). \[u_{x}=\frac{B_{1,x}-O_{7,x}}{d} \tag{67}\]  
\[u_{y}=\frac{B_{1,y}-O_{7,y}}{d} \tag{68}\]  
\[u_{z}=\frac{B_{1,z}-O_{7,z}}{d} \tag{69}\]  
After calculating \(u_{x}\), \(u_{y}\) and \(u_{z}\) the elements \(v_{x}\), \(v_{y}\) and \(v_{z}\), can be obtained using (23)-(25). \[v_{x}=\frac{2(B_{2,x}-O_{7,x}+\frac{1}{2}du_{x})}{\sqrt{3}d} \tag{70}\]  
\[v_{y}=\frac{-2(\sqrt{3}B_{2,x}-O_{7,y}+\frac{1}{2}du_{y})}{\sqrt{3}d} \tag{71}\]  
\[v_{z}=\frac{2(B_{2,z}-O_{7,z}+\frac{1}{2}du_{z})}{\sqrt{3}d} \tag{72}\]  
Finally, by solving the orthogonal Equations (13)-(15), then, \(w_{x}\), \(w_{y}\)_and_\(w_{z}\) can be obtained.