## 2 Materials and Methods
### Single-Cable-Pulley Transmission
\begin{array}{ll}\frac{\sqrt{T_{pk}T_{2}R\varphi(\rho, \mu,\alpha)}}{E\Lambda}+\frac{T_{1}R\varphi(\rho,\mu,\left(\theta_{slip}- \alpha\right))}{E\Lambda(\rho,\mu,\left(\theta_{slip}-\alpha\right))}-\frac{ T_{pr}R}{E\Lambda}\theta_{slip},&if\ \text{sgn}(\dot{\theta}_{in})=1\\ \frac{\sqrt{T_{pk}T_{2}R}}{E\Lambda}\frac{\varphi(\rho,\mu,\alpha)}{\lambda( \rho,\mu,\alpha)}+\frac{T_{1}R}{E\Lambda}\sigma\left(\rho,\mu,\left(\theta_ {slip}-\alpha\right)\right)-\frac{T_{pr}R}{E\Lambda}\theta_{slip},&\ other\end{array} \right., \tag{21}\]  
As shown in Figure 4, there are three types of supports for the pulley: (a) fixed to the base; (b) rotating around the pulley shaft; and (c) rotating around the rolling bearing. Therefore, the tension loss \(F_{loss}\) on both sides of the pulley is mainly caused by the friction \(F_{c}\) between the cable and the pulley or the friction \(F_{s}\) between the pulley and the pulley shaft, which can be calculated as:  
\[F_{loss}=\min\{F_{c},F_{s}\}=T_{2}-T_{1}, \tag{22}\]  
Figure 3: Tension distribution of the cable: (**a**) Tension distribution of the cable under loading; (**b**) Tension distribution of the cable under unloading. For the pulley of type (a), the tension loss can be calculated from Equation (12) as:  
\[F_{loss}=F_{c}=T_{2}-T_{1}=T_{1}(\lambda(\rho,\mu,\theta_{w})-1), \tag{23}\]  
where \(\theta_{w}\) is the wrap angle of the cable wrapped on the pulley. For the pulley of type (b), if \(F_{s}>F_{c}\), then \(F_{loss}=F_{c}=T_{1}(\lambda(\rho,\mu,\theta_{w})-1)\); if \(F_{s}<F_{c}\), then \(F_{loss}=F_{s}\), and the friction \(F_{s}\) between the pulley and pulley shaft can be expressed as:  
\[\left\{ \begin{array}{l}F_{s}=\mu_{s}\sqrt{N_{sx}^{2}+N_{sy}^{2}}=\mu_{s} \sqrt{T_{1}^{2}+T_{2}^{2}-2T_{1}T_{2}\cos(\theta_{w})}\\ N_{sx}=-T_{1}\sin(\theta_{w})\\ N_{sy}=-T_{1}\cos(\theta_{w})+T\end{array} \right., \tag{24}\]  
where \(N_{s}\) is the normal force applied by the shaft to the pulley, whose components in the \(x\) and \(y\) directions are represented by \(N_{sx}\) and \(N_{sy},r_{s}\) is the shaft radius, \(T_{1}\) and \(T_{2}\) are the tension of the cable on both sides of the pulley, and \(\mu_{s}\) is the friction coefficient between the pulley and shaft. As shown in Figure 4b, the applied external moment causes the driven cable to move clockwise along the pulley, ignoring the moment of inertia of the cable-pulley system, the following formula that describes the change of tension along the driven pulley can be formulated as:  
\[(M_{load}-r_{s}F_{s})=R(T_{2}-T_{1}), \tag{25}\]  
Substituting Equation (24) into Equation (25) can yield:  
\[\frac{M_{load}}{R}+T_{1}-T_{2}=\frac{r_{s}}{R}\widetilde{\mu}_{s}\sqrt{T_{1}^{ 2}+T_{2}^{2}-2T_{1}T_{2}\cos(\theta_{w})}, \tag{26}\]  
where \(\widetilde{\mu}_{s}=\mu_{s}\text{sgn}(\hat{\theta})\) is the signed friction coefficient used instead of \(\mu_{s}\), as the direction of the frictional force \(F_{s}\) is related to the direction of the rotation of the driven pulley, if the cable moves clockwise along the driven pulley, then \(\text{sgn}(\hat{\theta})=1\), otherwise \(\text{sgn}(\hat{\theta})=-1\). Let \(\lambda=T_{1}/T_{2}\), according to Equation (26), then the following can be formulated as:  
\[\left(\frac{M_{load}}{T_{1}R}+1\right)\lambda-1=\frac{r_{s}\widetilde{\mu}_{s }}{R}\sqrt{\lambda^{2}+1-2\lambda\cos(\theta_{w})}, \tag{27}\]  
Let \(k=\left(M_{load}/\left(T_{1}R\right)+1\right)\), \(\gamma=\left(r_{s}\widetilde{\mu}_{s}/R\right)^{2}\), \(w=\cos(\theta_{w})\), then Equation (25) can be expressed as:  
\[(k^{2}-\gamma)\lambda^{2}+2(\gamma w-k)\lambda+(1-\gamma)=0, \tag{28}\]  
Since \(0<\lambda<1\), Equation (28) can be solved as:  
\[\lambda=\frac{(k-\gamma w)-\sqrt{(k-\gamma w)^{2}-(k^{2}-\gamma)(1-\gamma)}}{ (k^{2}-\gamma)}, \tag{29}\]  
Figure 4: Free-body diagram of pulleys with three supporting forms: (**a**) Pulley fixed to the base; (**b**) Pulley supported on the shaft; (**c**) Pulley supported on the rolling bearing.