# An Improved Analytical Algorithm on Main Cable System of Suspension Bridge
## 2 Segmental Catenary Theory of Main Cable
### No Solution Cases for Cable Segment Equation  
The solution to the governing equation requires the Newton-Raphson type iteration while using initial trials of the force vector of the left node in the first cable element. However, the convergence of the gradient-based Newton-Raphson approach strongly depends on the initial value, and the estimation of initial value remains a challenge.  
Generally, there are two states for numerical analysis of main cable system: one is the main cable system at finished state for the whole bridge; the other is at construction state, only the main cable installation is finished [37]. The tension force at one end need to be assumed (or determined) for the main cable system calculation, the coordinates are iterated with convergence conditions. At the finished state, the tension force at one end and the horizontal distance between two ends are given, the unstressed cable length and the elevation between two end points can be solved, that is, \(l\), \(H_{i}\), \(V_{i}\) are known, to solve \(S_{0}\), \(h\). If the end tension force is assumed unreasonably, then there will be no solution for Equations (1) and (2).  
To solve unstressed length \(S_{0}\), Equation (1) is rewritten as:  
\[f(S_{0})=-\frac{H_{i}\cdot S_{0}}{EA}-\frac{H_{i}}{q}\left\{\ln(V_{i}+\sqrt{H_ {i}^{2}+V_{i}^{2}})-\ln(V_{i}-S_{0}\cdot q+\sqrt{H_{i}^{2}+(V_{i}-S_{0}\cdot q )^{2}}\right\}-l \tag{11}\]  
Suppose that \(l\), \(H_{i}\), \(EA\) are constants, and \(EA\) > 0, \(q\) > 0, 0 < \(S_{0}\) < 5000 m (the length of main cable for single-span suspension bridge is currently less than 5000 m), there will be no solution for Equation (1) in the following three conditions:  
**Condition 1**.: _When \(V_{i}\) is positive and the absolute value of \(V_{i}\) is large enough, \(l\) and \(H_{i}\) have the same sign, there will be no solution for Equation (1). It can be proved, as follows:_  
\[f(S_{0})=-\frac{H_{i}\cdot S_{0}}{EA}-\frac{H_{i}}{q}\ln\frac{1+\sqrt{\frac{H _{i}^{2}}{V_{i}^{2}}+1}}{1-\frac{S_{0}\cdot q}{V_{i}^{2}}+\sqrt{\frac{H_{i}^{ 2}}{V_{i}^{2}}+\frac{(V_{i}-S_{0})^{2}}{V_{i}^{2}}}}-l\approx-\frac{H_{i} \cdot S_{0}}{EA}-\frac{H_{i}}{q}\ln 1-l=-\frac{H_{i}\cdot S_{0}}{EA}-l\
eq 0 \tag{12}\]  
**Condition 2**.: _When \(V_{i}\) is negative and the absolute value of \(V_{i}\) is large enough, \(l\) and \(H_{i}\) have the same sign, there will be no solution for Equation (1). It can be proved, as follows:_  
\[\begin{array}{l}f(S_{0})=-\frac{H_{i}\cdot S_{0}}{EA}-\frac{H_{i}}{q}\ln \frac{1-\sqrt{\frac{H_{i}^{2}}{V_{i}^{2}}+1}}{1-\frac{S_{0}q}{V_{i}}-\frac{V_{ 0}q}{V_{i}}}\sqrt{1+\frac{H_{i}^{2}}{(V_{i}-S_{0}q)^{2}}}-l\\ \approx-\frac{H_{i}\cdot S_{0}}{EA}-\frac{H_{i}}{q}\ln\frac{1-\left(1+\frac{1} {2}\frac{H_{i}^{2}}{V_{i}^{2}}\right)}{1-\frac{S_{0}q}{V_{i}^{2}}-\left(1- \frac{S_{0}q}{V_{i}}\right)\left[1+\frac{1}{2}\frac{H_{i}^{2}}{(V_{i}-S_{0}q)^ {2}}\right]}-l\approx-\frac{H_{i}\cdot S_{0}}{EA}-l\
eq 0\end{array} \tag{13}\]  
**Condition 3**.: _When the absolute value of \(V_{i}\) is small enough, \(l\) and \(H_{i}\) have the same sign, Equation (14) is obtained from Equation (11), there will be no solution for Equation (1), It can be proved, as follows:_  
\[f(S_{0})=-H_{i}\left(\frac{S_{0}}{EA}+\frac{1}{q}\ln\frac{|H_{i}|}{\sqrt{H_{i} ^{2}+\left(S_{0}\cdot q\right)^{2}-S_{0}\cdot q}}\right)-l<-H_{i}\frac{S_{0}} {EA}-l\
eq 0 \tag{14}\]