## 3 Analysis of Kinematics
### Inverse Position Kinematics  
For the inverse position kinematics, the actuated joint variables \(q_{i}(i=1,2,3)\) are solved for a specific position on the moving platform. In terms of Figure 3, the vector-loop closure equation for each arm is:  
\[\left(\overrightarrow{OP}+\overrightarrow{PB_{i}}-\overrightarrow{OA_{i}} \right)^{\mathrm{T}}\left(\overrightarrow{OP}+\overrightarrow{PB_{i}}- \overrightarrow{OA_{i}}\right)=l^{2} \tag{3}\]  
Substituting (1) and (2) into (3) yields:  
\[\left(\overrightarrow{u}_{i}^{\mathrm{T}\rightarrow}\overrightarrow{u}_{i} \right)q_{i}^{2}+\left(-2\overrightarrow{u}_{i}^{\mathrm{T}}\overrightarrow{ OP}-2\overrightarrow{u}_{i}^{\mathrm{T}}\overrightarrow{PB_{i}}\right)q_{i}+ \left(2\overrightarrow{PB_{i}}^{\mathrm{T}}\overrightarrow{OP}+ \overrightarrow{PB_{i}}^{\mathrm{T}}\overrightarrow{PB_{i}}+\overrightarrow {OP}^{\mathrm{T}}\overrightarrow{OP}-l^{2}\right)=0 \tag{4}\]  
where \(l\) is the length of the kinematic chain. From Equation (4), the solution for \(q_{i}\) is:  
\[q_{i}=\frac{-b_{i}\pm\sqrt{\left(b_{i}^{2}-4a_{i}c_{i}\right)}}{2a_{i}}\quad( i=1,2,3) \tag{5}\]  
where  
\[\left\{\begin{array}{l}a_{i}=\overrightarrow{u}_{i}^{\mathrm{T}\rightarrow} \overrightarrow{u}_{i}\\ b_{i}=-2\overrightarrow{u}_{i}^{\mathrm{T}}\overrightarrow{OP}-2 \overrightarrow{u}_{i}^{\mathrm{T}}\overrightarrow{PB_{i}}\\ c_{i}=2\overrightarrow{PB_{i}}^{\mathrm{T}\rightarrow}\overrightarrow{OP}+ \overrightarrow{PB_{i}}^{\mathrm{T}}\overrightarrow{PB_{i}}+\overrightarrow {OP}^{\mathrm{T}}\overrightarrow{OP}-l^{2}\end{array}\right. \tag{6}\]  
\begin{table}
\begin{tabular}{c c c} \hline \hline
**Parameters** & **Description** & **Value** \\ \hline \(r\) & The length between \(P\left(x_{p0},y_{p0},z_{p0}\right)\) and \(B_{i}(i=1,2,3)\) & 60 mm \\ \hline \(l\) & The length of kinematic chain & 670 mm \\ \hline \(q_{\mathrm{max}}\) & The maximum of the actuated joint variables & 500 mm \\ \(q_{i}(i=1,2,3)\)(the maximum stroke) & & \\ \hline \hline \end{tabular}
\end{table}
Table 1: Geometry parameters for the PH-LDR.  
Figure 3: Coordinate frames of the PH-LDR.  
In Equation (5), there are two solutions for each actuator. However, only the negative square root solution satisfies the current assembly for the mechanism in which the three actuators are inclined inward from top to bottom. Thus, the inverse position kinematic equation for \(i\)th arm is:  
\[q_{i}=\frac{-b_{i}+\sqrt{\left(b_{i}^{2}-4a_{i}c_{i}\right)}}{2a_{i}}\quad(i=1,2,3) \tag{7}\]  
Substituting (1), (2) and (7) into (4) yields:  
\[\left\{\begin{array}{l}q_{1}^{2}-2q_{1}r+r^{2}-l^{2}-2q_{1}x+2 rx+x^{2}+y^{2} +z^{2}=0\\ q_{2}^{2}-2q_{2}r+r^{2}-l^{2}+q_{2}x-rx-\sqrt{3}q_{2}y+\sqrt{3}ry+x^{2}+y^{2}+ z^{2}=0\\ q_{3}^{2}-2q_{3}r+r^{2}-l^{2}+q_{3}x-rx+\sqrt{3}q_{3}y-\sqrt{3}ry+x^{2}+y^{2}+ z^{2}=0\end{array}\right. \tag{8}\]  
As \({}^{\mathcal{O}}P_{P}=\left\{x\,yz\right\}^{\mathrm{T}}\) is given, the set of solutions for \(q_{i}\) is:  
\[\left\{\begin{array}{l}q_{1}=x+r+\left(l^{2}-y^{2}-z^{2}\right)^{\frac{1}{2} }\\ q_{2}=\frac{1}{2}[2r-x+\sqrt{3}y+\left(4l^{2}-3x^{2}-2\sqrt{3}xy-y^{2}-4z^{2} \right)^{\frac{1}{2}}]\\ q_{3}=\frac{1}{2}[2r-x-\sqrt{3}y+\left(4l^{2}-3x^{2}+2\sqrt{3}xy-y^{2}-4z^{2} \right)^{\frac{1}{2}}]\end{array}\right. \tag{9}\]