## 5 Relationship between Cable Unstrained Length and Cable Tension  
### Simplify the Problem  
In order to formulate the relationship between \(L\) and \(\mathbf{T}\), \(\Psi\) that is the set of all feasible solutions \(\mathbf{S}\) of (19) with given desired position \(\mathbf{B}^{d}\) and \(\varsigma\), must be established. However, the construction of \(\Psi\)  
Figure 7: The relationship between \(\beta\) and \(h_{b}\) with given \(\varsigma\). directly is not simple. Meanwhile, in Statement 2, we assumed that \(\Psi\subset\Pi\), as long as \(\rho gL<w_{\max}\) and \(T_{x}>T_{x_{\min}}\). By successfully constructing the projection of \(\Pi\) onto the \(\Upsilon\), \(\pi\), the relationship between \(T_{x}\) and \(T_{z}\) was obtained. Hence, the relationship between \(L\) and \(\mathbf{T}\) can be formed by constructing the relationship between \(L\) and any component of tension such as \(T_{x}\), \(T_{z}\), \(T\), \(\alpha\) instead of directly constructing \(\Psi\). Since the relationship between \(T_{z}\) and \(T_{x}\) is described as (21), if \(b_{\pi}\
eq 0\) then the minimum magnitude of tension \(T_{\min}\) that satisfies a certain point \(\mathbf{B}^{d}\) with \(0<\beta<90^{\circ}\) and \(D>0\), is always greater than 0, i.e., the function that expresses the relationship between \(L\) and \(T\) has its vertical asymptote that can be estimated as (31)  
\[T_{\min}=\frac{|b_{\pi}|}{\sqrt{{a_{\pi}}^{2}+{b_{\pi}}^{2}}} \tag{31}\]  
where \(a_{\pi}\) and \(b_{\pi}\) are the parameters that define \(\pi\). On the other hand, when \(T_{x}\) or \(T_{z}\) is considered instead of \(T\), the vertical asymptotes of the function that expresses the relationship between \(T_{x}\) or \(T_{z}\), and \(d\) are \(T_{x}=0\) or \(T_{z}=b_{\pi}\) respectively. Hence, to ease the calculation and effortlessly identify the model, \(T_{x}\) is chosen. Moreover, we have \(L=D+d\) and \(D\) is determined so the relationship between \(L\) and \(T_{x}\) can be replaced by \(d\) and \(T_{x}\). Let \(\Psi\) be the new solution set that can be expressed as (32). \[\Psi=\{(T_{x},d):d=d(T_{x})\} \tag{32}\]  
Theoretically, if the cable is unstretchable then the adjusting length \(d\) cannot be lesser than 0 and vice versa. \(\Psi\) of unstretchable cable and stretch cable with given certain \(\mathbf{B}^{d}\) and \(\varsigma\) are illustrated in the Figure 8.  
The negative part of \(d\) of the stretchable cable illustrated in Figure 8 is caused by the elasticity of the cable and it has the boundary that obeys Hooke's law. In general, modelling \(\Psi\) in the case of unstretchable cable is easier, since it has \(d=0\) as its horizontal asymptote. Meanwhile, \(\Psi\) in the case of stretchable cable has an inclined asymptote. To avoid misunderstanding, from now on, \(L\) denotes the unstrained length of stretchable cable. Let \(s\) and \(\Delta l(s)\) be the Lagrangian coordinates of an unstrained stretchable cable and elongation of the cable under external force at s, respectively. \(\Delta l(s)\) can be formulated as Equation (33)  
\[\Delta l(s)=\sqrt{\Delta l_{x}(s)+\Delta l_{z}(s)} \tag{33}\]  
Figure 8: The relationship between d and \(T_{x}\) of both of unstretchable and stretchable cable in the instance. where \(\Delta l_{x}\), \(\Delta l_{z}\) are the strained length of cable at s along the \(X\) and \(Z\) axis, respectively. They can be formed by Equations (34) and (35), derived from Hooke's law. \[\Delta l_{x}(s)=\frac{T_{x}s}{EA} \tag{34}\]  
\[\Delta l_{z}(s)=\frac{T_{z}s}{EA}-\frac{\rho s(L-s)s}{EA} \tag{35}\]  
Substituting (34), (35) to (33), the relationship between the elongation of overall cable and tension can be formulated by (36)  
\[\Delta l=\frac{TL}{EA} \tag{36}\]  
where \(\Delta l=\Delta l(L)\). In terms of \(T_{x}\), (36) can be written as (37).