## 2 Materials and Methods
### Closed-Circuit Cable-Pulley System Transmission Backlash and Grasping Force Estimation
#### 2.3.1. Cable Elongation and Grasping Force of the Cable-Pulley System during the Working Phase  
Assuming that the system is in the working stage at time \(t_{1}\), the tension \(T_{0,1}^{t}\) of cable \(T\) increases (decreases) while the tension \(T_{0,1}^{b}\) of cable \(B\) decreases (increases), when the external torque \(M_{in}\) drives the driving pulley to rotate counterclockwise (clockwise). From Equation (15), it can be obtained as follows:  
\[\left\{ \begin{array}{l}\Delta L_{total}^{t}=\frac{T_{0,1}^{t}\kappa_{11}}{EA} -\frac{T_{0,1}^{t}L_{total}^{t}}{EA},\;if\text{sgn}(\dot{\theta}_{in})=1\\ \Delta L_{total}^{t}=\frac{T_{0,1}^{t}\kappa_{21}}{EA}-\frac{T_{0,1}^{t}L_{total }^{t}}{EA},\;\text{other}\end{array} \right., \tag{39}\]  
\[\Delta L_{total}^{b}=\left\{ \begin{array}{l}\frac{T_{0,1}^{b}\kappa_{12}}{EA} -\frac{T_{0,1}^{b}L_{total}^{b}}{EA},\;if\text{sgn}(\dot{\theta}_{in})=1\\ \frac{T_{0,1}^{b}\kappa_{22}}{EA}-\frac{T_{0,1}^{b}L_{total}^{b}}{EA},\;\text{ other}\end{array} \right., \tag{40}\]  
where \(L_{total}^{t}=L_{0,1}^{t}+\sum_{i=1}^{4}\left(L_{i,i+1}^{t}+\theta_{s,i}^{t}\right)\), \(L_{total}^{b}=L_{0,1}^{b}+\sum_{i=1}^{4}\left(L_{i,i+1}^{b}+\theta_{s,i}^{b}\right)\), \(\kappa_{11}=L_{0,1}^{t}+\sum_{i=1}^{4}\left(L_{i,i+1}^{b}+\theta_{s,i}^{b} \right)\), \(\kappa_{12}=L_{0,1}^{b}+\sum_{i=1}^{4}\left(L_{i,i+1}^{b}+R_{1}^{b}K_{1,i-1}^{ b}\varphi(\rho_{i}^{b},\mu_{c},\theta_{si}^{b})\right)\), \(\kappa_{21}=L_{0,1}^{t}+\sum_{i=1}^{4}\left(L_{i,i+1}^{b}+R_{1}^{b}K_{1,i-1}^{ b}\varphi(\rho_{i}^{b},\mu_{c},\theta_{si}^{b})\right)\), \(\kappa_{22}=L_{0,1}^{b}+\sum_{i=1}^{4}\left(\frac{L_{i-1,i}^{b}}{K_{1,i}^{b}}+ \frac{R_{1}^{b}\varphi(\rho_{i}^{b},\mu_{c},\theta_{si}^{b})}{K_{1,i}^{b}}\right)\). Substituting Equations (39) and (40) into Equation (36), it can be obtained as follows:  
\[\left\{ \begin{array}{l}T_{0,1}^{t}\kappa_{11}+T_{0,1}^{b}\kappa_{12}-\kappa _{3}=0,\;if\text{sgn}(\dot{\theta}_{in})=1\\ T_{0,1}^{t}\kappa_{21}+T_{0,1}^{b}\kappa_{22}-\kappa_{3}=0,\;other\end{array} \right., \tag{41}\]  
where \(\kappa_{3}=T_{pre}^{t}L_{total}^{t}+T_{pre}^{b}L_{total}^{b}\). Substituting Equation (32) into Equation (41) yields:  
\[\left\{ \begin{array}{l}T_{0,1}^{t}=\frac{\kappa_{3}}{(\kappa_{11}+\kappa_{12 })}+\frac{M_{in}}{R_{0}}\frac{\kappa_{12}}{(\kappa_{11}+\kappa_{12})},\;if \text{sgn}(\dot{\theta}_{in})=1\\ T_{0,1}^{t}=\frac{\kappa_{3}}{(\kappa_{12}+\kappa_{22})}+\frac{M_{in}}{R_{0}} \frac{(\kappa_{22})}{(\kappa_{21}+\kappa_{22})},\;other\end{array} \right., \tag{42}\]  
\[\left\{ \begin{array}{l}T_{0,1}^{b}=\frac{\kappa_{3}}{(\kappa_{11}+\kappa_{12 })}-\frac{M_{in}}{R_{0}}\frac{\kappa_{11}}{(\kappa_{11}+\kappa_{12})},\;if \text{sgn}(\dot{\theta}_{in})=1\\ T_{0,1}^{b}=\frac{\kappa_{3}}{(\kappa_{21}+\kappa_{22})}-\frac{M_{in}}{R_{0}} \frac{(\kappa_{21})}{(\kappa_{21}+\kappa_{22})},\;other\end{array} \right., \tag{43}\]  
Substituting Equations (42) and (43) into Equation (35), the following can be obtained:  
\[\left\{ \begin{array}{l}M_{out}=\frac{M_{in}R_{5}}{R_{0}}\frac{(\kappa_{11}^{t }\kappa_{21}^{t}\kappa_{21}+\kappa_{12})}{K_{1,4}^{t}(\kappa_{11}+\kappa_{12 })}+\frac{(1-\kappa_{11}^{t}\kappa_{11}^{t}\kappa_{21}^{t})R_{5}\kappa_{3}}{K_{ 1,4}^{t}(\kappa_{11}+\kappa_{12})},\;if\text{sgn}(\dot{\theta}_{in})=1\\ M_{out}=\frac{M_{in}R_{5}}{R_{0}}\frac{(\kappa_{21}+\kappa_{11}^{t}\kappa_{21}^{t} \kappa_{22})}{K_{1,4}^{t}(\kappa_{21}+\kappa_{22})}-\frac{(1-\kappa_{11}^{t} \kappa_{11}^{t}\kappa_{21}^{t})R_{5}\kappa_{3}}{K_{1,4}^{t}(\kappa_{21}+\kappa_ {22})},\;other\end{array} \right., \tag{44}\]Combining Equations (37) and (41)-(43), the transmission backlash of the cable-pulley system can be calculated as:  
\[\Delta\theta_{out}=\left\{