## 3 Kinematics Modeling of Self-Insulating Mechanical Arm Joints
### Static Modeling  
After completing the kinematic modeling, this manuscript conducts static modeling of the mechanical arm to verify its behavior when the end-effector is disturbed by a heavy load. First, the relationship between the rotation angles and the end-probe coordinates is established through transformation matrices:  
\[{}^{O_{1}T_{O_{1}}}{}^{o_{1}}T_{C}{}^{C}T_{O_{2}}{}^{O_{2}}T_{O_{1}} \begin{bmatrix} 0\\ 0\\ H\\ 1\end{bmatrix} = \begin{bmatrix}x_{pr}\\ y_{pr}\\ z_{pr}\\ 1\end{bmatrix}  \tag{16}\]  
By taking the derivative of both sides, a Jacobian matrix is established, yielding  
\[ \begin{bmatrix}\dot{x}_{pr}\\ \dot{y}_{pr}\\ \dot{z}_{pr}\end{bmatrix} = \begin{bmatrix}J_{11}&J_{12}&\ldots&J_{17}\\ J_{21}&J_{22}&\ldots&J_{27}\\ J_{31}&J_{32}&\ldots&J_{37}\end{bmatrix}\begin{bmatrix}\dot{\theta}_{m_{1}} \\ \dot{\theta}_{r_{1}}\\ \dot{\theta}_{b_{1}}\\ \dot{\alpha}\\ \dot{\theta}_{m_{2}}\\ \dot{\theta}_{b_{2}}\end{bmatrix}  \tag{17}\]  
The Jacobian matrix is a three-row-seven-column matrix with rank \((J)=3\). Its pseudo-inverse matrix is given by  
\[J^{+}=J^{\mathrm{T}}(JJ^{\mathrm{T}})^{-1} \tag{18}\]  
Thus, the transformation relationship between the torques corresponding to each rotation angle and the end-effector output force is obtained, where \([F_{x}\ F_{y}\ F_{z}]^{\mathrm{T}}\) is the external force on the end, and \([\tau_{m1}\ \tau_{r1}\ \tau_{rb1}\ \tau_{\alpha}\ \tau_{m2}\ \tau_{r2}\ \tau_{b2}]^{\mathrm{T}}\) is the equivalent moment of each joint angle:  
\[ \begin{bmatrix}F_{x}\\ F_{y}\\ F_{z}\end{bmatrix} =(J^{+})^{T} \begin{bmatrix}\tau_{m_{1}}\\ \tau_{r_{1}}\\ \tau_{b_{1}}\\ \tau_{\alpha}\\ \tau_{m_{2}}\\ \tau_{r_{2}}\\ \tau_{b_{2}}\end{bmatrix}  \tag{19}\]Furthermore, based on the established kinematic relationships, the relationship between the tension provided by the mechanical arm's driving cables \(F_{II}(i=1,2,\ldots,10)\) and the torque at each corresponding rotation angle is as follows:  
\[ \begin{split}\tau_{m_{1}}&=\sqrt{h^{2}+0.25d^{2}} \sin\Bigl{(}\theta_{b_{1}}+\arctan\frac{d}{2h}\Bigr{)}\cdot\frac{\sqrt{0.5d^{2} (1-\sin\theta_{b_{1}})}}{l_{1}}F_{I_{1}}\\ &-\sqrt{h^{2}+0.25d^{2}}\sin\Bigl{(}\theta_{b_{1}}+\arctan\frac{ d}{2h}\Bigr{)}\cdot\frac{\sqrt{0.5d^{2}(1+\sin\theta_{b_{1}})}}{l_{2}}F_{I_{2}}\\ &+\sqrt{h^{2}+0.25d^{2}}\sin\Bigl{(}\arctan\frac{d}{2h}-\theta_{ b_{1}}\Bigr{)}\cdot\frac{\sqrt{0.5d^{2}(1-\sin\theta_{b_{1}})}}{l_{3}}F_{I_{3}}\\ &-\sqrt{h^{2}+0.25d^{2}}\sin\Bigl{(}\arctan\frac{d}{2h}-\theta_{ b_{1}}\Bigr{)}\cdot\frac{\sqrt{0.5d^{2}(1+\sin\theta_{b_{1}})}}{l_{4}}F_{I_{4}} \end{split}  \tag{20}\]  
\[\tau_{r_{1}}=0.707 dF_{I_{1}}-0.707 dF_{I_{2}}+0.707 dF_{I_{3}}-0.707 dF_{I_{4}} \tag{21}\]  
\[ \begin{split}\tau_{b_{1}}&=\Bigl{[}\sqrt{h^{2}+0.25 d^{2}}\cos\Bigl{(}\theta_{b_{1}}+\arctan\frac{d}{2h}\Bigr{)}+h\Bigr{]}\cdot \frac{\sqrt{l_{1}^{2}-0.5d^{2}(1-\sin\theta_{b_{1}})}}{l_{1}}F_{I_{1}}\\ &+\Bigl{[}\sqrt{h^{2}+0.25d^{2}}\cos\Bigl{(}\theta_{b_{1}}+ \arctan\frac{d}{2h}\Bigr{)}+h\Bigr{]}\cdot\frac{\sqrt{l_{2}^{2}-0.5d^{2}(1- \sin\theta_{b_{1}})}}{l_{2}}F_{I_{1}}\\ &-\Bigl{[}\sqrt{h^{2}+0.25d^{2}}\cos\Bigl{(}\arctan\frac{d}{2h}- \theta_{b_{1}}\Bigr{)}+h\Bigr{]}\cdot\frac{\sqrt{l_{3}^{2}-0.5d^{2}(1-\sin \theta_{b_{1}})}}{l_{3}}F_{I_{3}}\\ &-\Bigl{[}\sqrt{h^{2}+0.25d^{2}}\cos\Bigl{(}\arctan\frac{d}{2h}- \theta_{b_{1}}\Bigr{)}+h\Bigr{]}\cdot\frac{\sqrt{l_{4}^{2}-0.5d^{2}(1-\sin \theta_{b_{1}})}}{l_{4}}F_{I_{4}}\\ \end{split}  \tag{22}\]  
\[ \begin{split}\tau_{a}&=\frac{k_{1}k_{2}}{l_{5}} \cos\pi F_{l_{5}}-\frac{k_{1}k_{1}}{l_{6}}\cos\alpha F_{l_{6}}\end{split}  \tag{23}\]  
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