# Precision Analysis and Error Compensation of a Telescope Truss Structure Based on Robotics
## 2 Structural Analysis of the Telescope Manipulator Based on Stiffness Performance Index
### Optimization Principle
\(T(q)\) is the pose matrix formed by the kinematic model that corresponds to the end of the robotic arm. Before optimizing the stiffness coefficient, the initial pose of the robotics must be solved. First, we determined the position vector \(P\) and direction vector \(N\) of the end of the robotic arm according to the relative positions of the PM and SM; then, the initial pose of the end of the robotic arm is  
\[T_{0}=\left[ \begin{array}{cccc}U_{X}&V_{X}&N_{X}&P_{X}\\ U_{Y}&V_{Y}&N_{Y}&P_{Y}\\ U_{Z}&V_{Z}&N_{Z}&P_{Z}\\ 0&0&0&1\end{array} \right] \tag{8}\] \begin{table}
\begin{tabular}{c c c c c c c c} \hline  & **J1** & **J2** & **J3** & **J4** & **J5** & **J6** & **x** \\ \hline \(\theta_{i}\) (\(rad\)) & \(\theta_{1}\) & \(\theta_{2}\) & \(\theta_{3}\) & \(\theta_{4}\) & \(\theta_{5}\) & \(\theta_{6}\) & 0 \\ \(\alpha_{i}\) (\(deg\)) & \(-90^{\circ}\) & \(0^{\circ}\) & \(-90^{\circ}\) & \(90^{\circ}\) & \(-90^{\circ}\) & \(0^{\circ}\) & 0 \\ \(a_{i}\) (\(mm\)) & 525 & 790 & 250 & 1190 & 100 & 0 & 150 \\ \(d_{i}\) (\(rad\)) & 0 & 132 & 132 & 0 & 0 & 0 & 0 \\ \hline \end{tabular}
\end{table} 
Table 1: Denavit–Hartenberg (DHm) parameters of the robot in the research case. where \(P=\left[ \begin{array}{cc}P_{X}&P_{Y}&P_{Z}\end{array} \right]^{T}\), \(N=\left[ \begin{array}{cc}N_{X}&N_{Y}&N_{Z}\end{array} \right]^{T}\), and \(U=\left[ \begin{array}{cc}U_{X}&U_{Y}&U_{Z}\end{array} \right]^{T}\) is an arbitrary unit vector in the plane perpendicular to \(N\), and \(V=\left[ \begin{array}{cc}V_{X}&V_{Y}&V_{Z}\end{array} \right]^{T}\) is the cross product of \(\left[ \begin{array}{cc}N_{X}&N_{Y}&N_{Z}\end{array} \right]^{T}\) and \(\left[ \begin{array}{cc}U_{X}&U_{Y}&U_{Z}\end{array} \right]^{T}\). All the inverse kinematic solutions are solved according to the initial pose of the robot end pose, and the maximum stiffness performance index value is selected to be \(q_{0}\) as the initial joint angle vector. Then, the inverse kinematic problem (IKP) method based on Jacobian matrix iteration can solve the joint angle of each feasible pose and the maximum stiffness coefficient is selected in the range of \([\theta_{min},\theta_{max}]\). The solution process is as follows: \begin{tabular}{l l} (1) & Initial \(q=q_{0}\) \\ (2) & \(\{p,Q\}=pk(q)\) \\ (3) & \(\Delta p=p_{d}-p\), \(\Delta\alpha=\alpha_{d}\alpha^{T}\) \\ (4) & \(\Delta\delta=\left[\begin{array}{c}\Delta p\\ vec(\Delta\alpha)\end{array}\right]\) \\ (5) & \(\Delta q=J(q)^{-1}\Delta\delta\) \\ (6) & \(q=q+\Delta q\) \\ (7) & If \(q\) is reachable, go next; Else, go back to (2) \\ (8) & Output \(k_{stif}=\frac{1}{\sqrt{\det(C_{u})}}\) \\ \end{tabular}   
where \(pk(q)\) represents the positive kinematics solution, \(p\) is the real-time position vector of the robotic arm end in the iterative process, \(\alpha\) is the rotation matrix representing the real-time posture of the robotic arm end in the iterative process, \(p_{d}\) represents the end target position vector, and \(\alpha_{d}\) represents the rotation matrix of the end target pose. \(\Delta q\) represents a small change in the angle vector, while \(\Delta\delta\) represents a small change in the end posture, and is defined as  
\[\Delta\delta=\left[ \begin{array}{c}\Delta p\\ vec(\Delta\alpha)\end{array} \right], \tag{9}\]  
where \(\Delta p\) denotes a small change in position, \(\Delta\alpha\) is a small change in the rotation matrix, and \(vec(\Delta\alpha)\) is the axial vector of \(\Delta\alpha\):  
\[vec(\Delta\alpha)=\frac{1}{2}\left[ \begin{array}{c}\Delta\alpha_{32}- \Delta\alpha_{23}\\ \Delta\alpha_{13}-\Delta\alpha_{31}\\ \Delta\alpha_{21}-\Delta\alpha_{12}\end{array} \right]. \tag{10}\]