# The Kinematic Analysis of a Wind Turbine Climbing Robot Mechanism
## 3 Kinematic Modeling
\[\mathrm{v_{ir}}(\sin\varphi)=\mathrm{v_{Y}}+\left(\mathrm{C_{i}}\cos\lambda \right)\ \mathrm{\omega_{o}} \tag{5}\]  
\[\mathrm{\omega_{i}}\left(\mathrm{r_{i}}\right)-\mathrm{v_{ir}}(\cos\ \varphi)=\mathrm{v_{Z}}+\left(\mathrm{C_{i}}\sin\lambda\right)\ \mathrm{\omega_{o}} \tag{6}\]  
From Equations (5) and (6), Equation (7) can derive the velocity of the roller in the \(i\)th wheel and Equation (9) the wheels' angular velocity:  
\[\mathrm{v_{ir}}=\frac{\mathrm{v_{Y}}+\left(\mathrm{C_{i}}\cos\lambda\right) \ \mathrm{\omega_{o}}}{\sin\lambda} \tag{7}\]  
\[\mathrm{\omega_{i}}\left(\mathrm{r_{i}}\right)=\mathrm{v_{Z}}+\ \mathrm{\omega_{o}}( \mathrm{C_{i}}\sin\lambda\ )+\left(\mathrm{v_{Y}}+\mathrm{\omega_{o}}( \mathrm{C_{i}}\cos\lambda\ )\right)\left(\frac{\cos\varphi}{\sin\varphi}\right) \tag{8}\]  
\[\mathrm{\omega_{i}}\ =\frac{1}{\mathrm{r}}\ \left[\ \ 1\ \ \ \ \frac{\cos\varphi}{\sin\varphi}\ \ \mathrm{C_{i}}\sin\lambda+\left(\mathrm{C_{i}}\cos\lambda\right)\left(\frac{ \cos\varphi}{\sin\varphi}\right)\ \ \right]\ \left[\ \Considering the facts that wheels have the same size, the offset angle of the roller \(\varphi=45^{\circ}\), the transformation matrix is:  
\[\left[ \begin{array}{c}\omega_{1}\\ \omega_{2}\\ \omega_{3}\\ \omega_{4}\end{array} \right]=\frac{1}{\mathbf{r}}\left[ \begin{array}{rr}1&1&- \left(\mathrm{C}_{i}\sin\lambda\,+\frac{\mathrm{C}_{i}\cos\lambda}{\tan\varphi} \right)\\ 1&-1&\left(\mathrm{C}_{i}\sin\lambda\,+\frac{\mathrm{C}_{i}\cos\lambda}{\tan \varphi}\right)\\ 1&-1&-\left(\mathrm{C}_{i}\sin\lambda\,+\frac{\mathrm{C}_{i}\cos\lambda}{\tan \varphi}\right)\\ 1&1&\left(\mathrm{C}_{i}\sin\lambda\,+\frac{\mathrm{C}_{i}\cos\lambda}{\tan \varphi}\right)\end{array} \right]\left[ \begin{array}{c}\
u_{\mathrm{Z}}\\ \
u_{\mathrm{Y}}\\ \omega_{\mathrm{o}}\end{array} \right] \tag{11}\]  
By replacing the parameters of matrix Equation (11), let \(E=\mathrm{C}_{i}\sin\lambda\,\) and \(F=\mathrm{C}_{i}\sin\lambda\,\) will be Equation (12). \[\left[ \begin{array}{c}\omega_{1}\\ \omega_{2}\\ \omega_{3}\\ \omega_{4}\end{array} \right]=\frac{1}{\mathbf{r}}\left[ \begin{array}{rr}1& 1&-(E+F)\\ 1&-1&(E+F)\\ 1&-1&-(E+F)\\ 1&1&(E+F)\end{array} \right]\left[ \begin{array}{c}\
u_{\mathrm{Z}}\\ \
u_{\mathrm{Y}}\\ \omega_{\mathrm{o}}\end{array} \right] \tag{12}\]  
The inverse kinematic solves the individual wheel speed that the velocity of the system can be obtained by implementing \(\omega_{i}\), the wheel angular velocity and \(\
u_{\mathrm{ir}}\), the velocity of passive rollers in the wheel \(i\) can be achieved by Equation (13) and its corresponding matrix from Equation (11). \[\left[ \begin{array}{c}\omega_{i}\\ \
u_{\mathrm{ir}}\end{array} \right]=T\,\left[ \begin{array}{c}\
u_{\mathrm{Z }}\\ \
u_{\mathrm{Y}}\\ \omega_{\mathrm{o}}\end{array} \right] \tag{13}\]  
In contrary, the equation specifying the velocities is the forward kinematic according to Equations (14) and (15). \[\left[ \begin{array}{c}\
u_{\mathrm{Z}}\\ \
u_{\mathrm{Y}}\\ \omega_{\mathrm{o}}\end{array} \right]=T^{+}\left[ \begin{array}{c}\omega_{ i}\\ \
u_{\mathrm{ir}}\end{array} \right] \tag{14}\]  
\[\left[ \begin{array}{c}\
u_{\mathrm{Z}}\\ \
u_{\mathrm{Y}}\\ \omega_{\mathrm{o}}\end{array} \right]=T^{+}\left[ \begin{array}{c}\omega_{1} \\ \omega_{2}\\ \omega_{3}\\ \omega_{4}\end{array} \right] \tag{15}\]  
\(T\) is the transformation matrix of the inverse equation of motion that corresponds to Equation (11). \[T=\frac{1}{\mathbf{r}}\left[ \begin{array}{rr}1&1&-\left(\mathrm{C}_{i}\sin \lambda\,+\frac{\mathrm{C}_{i}\cos\lambda}{\tan\varphi}\right)\\ 1&-1&\left(\mathrm{C}_{i}\sin\lambda\,+\frac{\mathrm{C}_{i}\cos\lambda}{\tan \varphi}\right)\\ 1&-1&-\left(\mathrm{C}_{i}\sin\lambda\,+\frac{\mathrm{C}_{i}\cos\lambda}{\tan \varphi}\right)\\ 1&1&\left(\mathrm{C}_{i}\sin\lambda\,+\frac{\mathrm{C}_{i}\cos\lambda}{\tan \varphi}\right)\end{array} \right] \tag{16}\]  
The expression of its pseudoinverse matrix \(T^{+}\) is generalized by:  
\[T^{+}=\left(T^{\mathrm{T}}T\right)^{-1}T^{T} \tag{17}\]  
\[T^{+}=\frac{\mathrm{r}}{4}\left[