# Equivalent Dynamic Analysis of a Cable-Driven Snake Arm Maintainer
## 3 Methods
### Equivalent Joint Dynamics
In Figure 7b, the cable traction force (\(\mathbf{F_{1}^{j}}\)) of the \(j\)-th joint is drawn as a spatial parallelogram (\(\mathbf{O},\mathbf{L},\mathbf{N},\mathbf{M}\)), and its diagonal (\(\mathbf{O}\mathbf{N}\)) is \(\mathbf{F_{1}^{i}}\), which is the force acting on the \(i\)-th joint. Using coordinate transformation, the \(\mathbf{O},\mathbf{L},\mathbf{N},\mathbf{M}\) points are respectively transformed into the coordinate system \((x_{2},y_{2},z_{2})\) and \((x_{3},y_{3},z_{3})\), with \(\mathbf{A_{2}}\) as the coordinate origin. In the \(\mathbf{A_{2}}(x_{2},y_{2},z_{2})\) coordinate system, the coordinates of point \(\mathbf{O}(x_{O2},y_{O2},z_{O2})\), as well as the size and direction of the vector (\(\mathbf{F_{1}^{i}}\)) are known. The coordinates of \(\mathbf{L}(x_{12},y_{L2},z_{L2})\) and \(\mathbf{M}(x_{M2},y_{M2},z_{M2})\) can be obtained using matrix transformation and the distance formula  
\[\mathbf{F_{x2F1}^{10}}=\frac{(x_{B2}-x_{E2})}{\sqrt{\left(x_{B2}-x_{E2}\right)^{2} +\left(y_{B2}-y_{E2}\right)^{2}+\left(z_{B2}-z_{E2}\right)^{2}}}\mathbf{F_{1}^{10}} \tag{12}\]0 Force analysis of any joint in any posture (the \(i\)-th joint was taken as an example). (**a**) \(i\)-joint coordinate system and cable forces. (**b**) Local enlargement of \(\mathbf{F_{1}^{j}}\). between two points. From the geometric properties of the parallelogram in space, the coordinates of \(\mathbf{N}(x_{N2},y_{N2},z_{N2})\) can be obtained as follows:  
\[ \begin{cases}x_{N2}=(x_{M2}-x_{O2})+x_{L2}\\ y_{N2}=(y_{M2}-y_{O2})+y_{L2}\\ z_{N2}=(z_{M2}-z_{O2})+z_{L2}\end{cases}  \tag{20}\]  
From this, the unit direction vector (\(\overset{\rightarrow}{\mathbf{O}}\)N of \(\mathbf{F^{i}_{j1}}\)) can be obtained as follows:  
\[\overset{\rightarrow}{\mathbf{O}}\mathbf{N}=\frac{(x_{N2}-x_{O2},\ y_{N2}-y_{O2},\ z _{N2}-z_{O2})}{\sqrt{(x_{N2}-x_{O2})^{2}+(y_{N2}-y_{O2})^{2}+(z_{N2}-z_{O2})^{2 }}} \tag{21}\]  
Meanwhile, from the calculation formula of the angle between two non-zero vectors:  
\[\cos(\varphi)=\frac{\overset{\rightarrow}{\mathbf{O}}\mathbf{M}\overset{\rightarrow }{\mathbf{O}}\mathbf{L}}{|OM||OL|} \tag{22}\]  
It is clear that the size of \(\mathbf{F^{i}_{j1}}\) is as follows:  
\[F^{i}_{j1}=2F^{i}_{1}\cos(\frac{\varphi}{2}) \tag{23}\]  
According to the force vector decomposition method, \(\mathbf{F^{i}_{j1}}\) is decomposed to \(\mathbf{A_{2}}(x_{2},y_{2},z_{2})\) on the parallel lines of each axis:  
\[\mathbf{F^{i}_{x2 Fj1}}=\frac{(x_{N2}-x_{O2})}{\sqrt{(x_{N2}-x_{O2})^{2}+(y_{N2}-y_{ O2})^{2}+(z_{N2}-z_{O2})^{2}}}\mathbf{F^{i}_{j1}} \tag{24}\]  
\[\mathbf{F^{i}_{y2 Fj1}}=\frac{(y_{N2}-y_{O2})}{\sqrt{(x_{N2}-x_{O2})^{2}+(y_{N2}-y_{ O2})^{2}+(z_{N2}-z_{O2})^{2}}}\mathbf{F^{i}_{j1}} \tag{25}\]  
\[\mathbf{F^{i}_{z2 Fj1}}=\frac{(z_{N2}-z_{O2})}{\sqrt{(x_{N2}-x_{O2})^{2}+(y_{N2}-y_{ O2})^{2}+(z_{N2}-z_{O2})^{2}}}\mathbf{F^{i}_{j1}} \tag{26}\]  
The torque produced by \(\mathbf{F^{i}_{j1}}\) on the \(x_{2}\) axis at \(\mathbf{A_{2}}(x_{2},y_{2},z_{2})\) can be obtained as follows:  
\[\mathbf{T^{i}_{x2 Fj1}}=\mathbf{F^{i}_{y2 Fj1}}z_{O}+\mathbf{F^{i}_{z2 Fj1}}y_{O} \tag{27}\]  
In the same way, close to \(\mathbf{F^{i}_{2}}\), \(\mathbf{F^{i}_{3}}\), there are also cable forces of \(\mathbf{F^{i}_{j2}}\) and \(\mathbf{F^{i}_{j3}}\). Accordingly, the torques generated on the \(x_{2}\) axis of the \(\mathbf{A_{2}}(x_{2},y_{2},z_{2})\) coordinate system are \(\mathbf{T^{i}_{x2 Fj2}}\) and \(\mathbf{T^{i}_{x2 Fj3}}\).