# Research on the Transportation of Low-Strength Composites Sheets
## 3 Problem Solution  
When the thin-sheet material is transported from one roll to another, its projection curve along the rolls' axes between two contact points with rolls can be described with Equation (5). As shown in Figure 4, the upstream roll is located at coordinate \(O_{1}(0,D_{h})\), the downstream roll is located at coordinate \(O_{2}(2D_{l},0)\). The radius of the upstream roll and the downstream roll is \(R_{1}\) and \(R_{2}\) respectively. The touchpoints between the sheet and the two rolls are \(B\) and \(C\). The upper external co-tangents of \(O_{1}\) and \(O_{2}\) is \(AD\). The angles between the vertical direction and \(O_{1}B\) and \(O_{2}C\) are \(\theta_{1}\) and \(\theta_{2}\), respectively. The first derivative of the track is continuous at point \(B\) and point \(C\). The boundary conditions of this problem are:  
\[\left\{ \begin{aligned} y_{B}&=D_{h}+R_{1}\cos\theta_{1 }\\ y_{C}&=R_{2}\cos\theta_{2}\\ y_{B}&=-\tan\theta_{1}\\ \dot{y}_{C}&=\tan\theta_{2}\end{aligned} \right. \Rightarrow\left\{ \begin{aligned} y(R_{1}\sin\theta_{1})=D_{h}+R_{ 1}\cos\theta_{1}\\ y(2D_{l}-R_{2}\sin\theta_{2})=R_{2}\cos\theta_{2}\\ \dot{y}(R_{1}\sin\theta_{1})=-\tan\theta_{1}\\ \dot{y}(2D_{l}-R_{2}\sin\theta_{2})=\tan\theta_{2}\end{aligned} \right. \tag{12}\]  
Substitute Equation (5) in Equation (12):  
\[\left\{ \begin{aligned} -D_{h}-R_{1}\cos\theta_{1}+\frac{C_{1}}{ \rho btg}\cosh\left(\frac{\rho btg}{C_{1}}(R_{1}\sin\theta_{1}-C_{2})\right) +\frac{\lambda-\rho btv^{2}}{\rho btg}=0\\ -R_{2}\cos\theta_{2}+\frac{C_{1}}{\rho btg}\cosh\left(\frac{\rho btg}{C_{ 1}}(2D_{l}-R_{2}\sin\theta_{2}-C_{2})\right)+\frac{\lambda-\rho btv^{2}}{\rho b tg }=0\\ \tan\theta_{1}+\sinh\left(\frac{\rho btg}{C_{1}}(R_{1}\sin\theta_ {1}-C_{2})\right)=0\\ -\tan\theta_{2}+\sinh\left(\frac{\rho btg}{C_{1}}(2D_{l}-R_{2} \sin\theta_{2}-C_{2})\right)=0\end{aligned} \right. \tag{13}\]  
Define dimensionless variables as follows:  
\[\tilde{D_{l}}\equiv\frac{D_{l}}{R_{1}},\tilde{r}\equiv\frac{R_{2}}{R_{1}}, \tilde{D_{h}}\equiv\frac{D_{h}}{R_{1}},\tilde{C_{1}}\equiv\frac{C_{1}}{R_{1} \rho btg},\tilde{C_{2}}\equiv\frac{C_{2}}{R_{1}},\tilde{C_{3}}\equiv\frac{ \lambda-\rho btv^{2}}{R_{1}\rho btg} \tag{14}\]  
Figure 4: Flexible thin-sheet material between rolls. So, the dimensionless form of Equation (13) is:  
\[\left\{ \begin{aligned} -\vec{D_{h}}-\cos\theta_{1}+\vec{C_{1}}\cosh \frac{\sin\theta_{1}-\vec{C_{2}}}{\vec{C_{1}}}+\vec{C_{3}}=0\\ -\hat{r}\cos\theta_{2}+\vec{C_{1}}\cosh\frac{2\vec{D_{1}}-r\sin \theta_{2}-\vec{C_{2}}}{\vec{C_{1}}}+\vec{C_{3}}=0\\ \tan\theta_{1}+\sinh\frac{\sin\theta_{1}-\vec{C_{2}}}{\vec{C_{1}}}=0\\ -\tan\theta_{2}+\sinh\frac{2\vec{D_{1}}-r\sin\theta_{2}-\vec{C_{2 }}}{\vec{C_{1}}}=0\end{aligned} \right. \tag{15}\]  
These four equations contains four unknown varibles. For any given \(\theta_{1}\in(\theta_{s},\frac{\pi}{2})\), the unkown \(\theta_{2},\vec{C_{1}},\vec{C_{2}},\vec{C_{3}}\) can be solved with Equation (15) where \(\theta_{s}\) is the angle between \(AD\) and the horizontal direction and given by  
\[\theta_{s}=\arctan\frac{\vec{D_{h}}}{2\vec{D_{l}}}-\arcsin\frac{\tilde{r}-1}{ \sqrt{\vec{D_{h}}^{2}+4\vec{D_{l}}^{2}}} \tag{16}\]  
It is hard to obtain analytical solution to Equation (15). Numerical methods are used to solve it. In fact, Equation (15) has multiple solutions, as shown in Figure 5, and only the one with \(\theta_{2}\in(-\theta_{s},\frac{\pi}{2})\) accords with the actual circumstance. In many cases, the numerical methods cannot converge because of the high nonlinearity of Equation (15). It is found that the convergence of many numerical methods and which solution of the multiple solutions will be given are highly sensitive to the initial value \([\theta_{2}^{(0)},\vec{C_{1}}^{(0)},\vec{C_{2}}^{(0)},\vec{C_{3}}^{(0)}]^{T}\). To solve this problem, an approximate value of the solution is given in Equation (17).