# Equivalent Dynamic Analysis of a Cable-Driven Snake Arm Maintainer
## 3 Methods
### Equivalent Joint Dynamics
The unit direction vector of \(\mathbf{F_{1}^{10}}\) in the coordinate system \((x_{2},y_{2},z_{2})\) can be obtained as follows:  
\[\overset{\rightarrow}{E}\mathbf{B}=\frac{(x_{B2}-x_{E2},\ y_{B2}-y_{E2},\ z_{B2}- z_{E2})}{\sqrt{\left(x_{B2}-x_{E2}\right)^{2}+\left(y_{B2}-y_{E2}\right)^{2}+ \left(z_{B2}-z_{E2}\right)^{2}}} \tag{11}\]  
Then, decompose \(\mathbf{F_{1}^{10}}\) into the parallel lines of each axis of the coordinate system \(\mathbf{A_{2}}(x_{2},y_{2},z_{2})\) according to the force vector decomposition method:  
\[\mathbf{F_{x2F1}^{10}}=\frac{(x_{B2}-x_{E2})}{\sqrt{\left(x_{B2}-x_{E2}\right)^{2} +\left(y_{B2}-y_{E2}\right)^{2}+\left(z_{B2}-z_{E2}\right)^{2}}}\mathbf{F_{1}^{10}} \tag{12}\]  
\[\mathbf{F_{y2F1}^{10}}=\frac{(y_{B2}-y_{E2})}{\sqrt{\left(x_{B2}-x_{E2}\right)^{2} +\left(y_{B2}-y_{E2}\right)^{2}+\left(z_{B2}-z_{E2}\right)^{2}}}\mathbf{F_{1}^{10}} \tag{13}\]  
\[\mathbf{F_{z2F1}^{10}}=\frac{(z_{B2}-z_{E2})}{\sqrt{\left(x_{B2}-x_{E2}\right)^{2} +\left(y_{B2}-y_{E2}\right)^{2}+\left(z_{B2}-z_{E2}\right)^{2}}}\mathbf{F_{1}^{10}} \tag{14}\]  
Similarly, \(\mathbf{F_{2}^{10}},\mathbf{F_{3}^{10}}\) can be decomposed to the parallel lines of each axis of the coordinate system \(\mathbf{A_{2}}(x_{2},y_{2},z_{2})\) according to the vector decomposition method of force, with the following results: \(\mathbf{F_{x2F2}^{10}},\mathbf{F_{y2F2}^{10}},\mathbf{F_{z2F2}^{10}}\) and \(\mathbf{F_{x2F3}^{10}},\mathbf{F_{y2F3}^{10}},\mathbf{F_{z2F3}^{10}}\). Figure 6: Force analysis of end joints in any posture. Therefore, the torques produced by \(\mathbf{F_{1}^{10}},\mathbf{F_{2}^{10}},\mathbf{F_{3}^{10}}\) on the \(x_{2}\) axis of the coordinate system of point \(\mathbf{A_{2}}(x_{2},y_{2},z_{2})\) can be obtained as follows:  
\[T_{x2\mathbf{F1}}^{10}=F_{y2\mathbf{F1}}^{10}z_{E2}+F_{22\mathbf{F1}}^{10}y_{E2} \tag{15}\]  
\[T_{x2\mathbf{F2}}^{10}=F_{y2\mathbf{F2}}^{10}z_{E2}+F_{22\mathbf{F2}}^{10}y_{F2} \tag{16}\]  
\[T_{x2\mathbf{F3}}^{10}=F_{y2\mathbf{F3}}^{10}z_{G2}+F_{22\mathbf{F3}}^{10}y_{G2} \tag{17}\]  
From the forcibly equivalent transformation, the following can be obtained:  
\[\mathbf{T_{1}^{10}}=T_{x2\mathbf{F1}}^{10}+T_{x2\mathbf{F2}}^{10}+T_{x2\mathbf{F3}}^{10} \tag{18}\]  
Similarly, the forcibly equivalent transformation of \(\mathbf{T_{2}^{10}}\) on the \(z_{3}\) axis in the coordinate system \(\mathbf{A_{2}}(x_{3},y_{3},z_{3})\) can be used to obtain:  
\[\mathbf{T_{2}^{10}}=T_{2\mathbf{F1}}^{10}+T_{2\mathbf{F2}}^{10}+T_{2\mathbf{F3}}^{10} \tag{19}\]  
Upon setting the minimum traction force to \(\mathbf{F_{1}^{10}}=10\) N, Equation (18) can be combined with Equation (19) to solve real-time \(\mathbf{F_{1}^{10}}\) and \(\mathbf{F_{2}^{10}}\). The force characteristics of any joint (except the end joint) of the SAM are more complicated than those of the end joint. In addition to the cable traction force of the joint itself, all cables after the joint have forces acting on them. As shown in Figure 7, we took the \(i\)-th joint (\(i<10\)) as an example for the force analysis and decoupling calculations. Compared with the end joint shown in Figure 6, the \(i\)-th joint has its own cable traction forces (\(\mathbf{F_{1}^{i}},\mathbf{F_{2}^{i}},\mathbf{F_{3}^{i}}\)), in addition to the cable traction forces from the \((i+1)\)-th,..., the 10 th joints (end joints). As shown in Figure 7a, \(\mathbf{F_{1}^{i}},\mathbf{F_{2}^{i}},\mathbf{F_{3}^{i}}\) are the forces generated by the cables of the \(j\)-th joint on the midline of the cable clamping angle (\(i<j\leq 10\)). For the cable traction force (\(\mathbf{F_{1}^{i}},\mathbf{F_{2}^{i}},\mathbf{F_{3}^{i}}\)) of the \(i\)-th joint shown in Figure 7a, the solution algorithm is similar to that of the end joint and will not be repeated here; rather, we focus on analyzing the influence of the traction force of the \(j\)-th joints (\(i<j\leq 10\)) on the \(i\)-th joint.