# Research on Material Viscoelasticity and Its Influence on Indentation Rolling Resistance
## 5 Research on Calculation of Indentation Rolling Resistance Based on Relaxation Model Parameters
### Theoretical Basis of Calculation of Indentation Rolling Resistance  
The calculation of indentation rolling resistance, viewed from a temporal perspective, mainly addresses the operational resistance encountered in belt conveyors. As the conveyor \begin{table}
\begin{tabular}{c c c c c} \hline \hline
**Type of Spectral** & **Number of** & **Logt[\(-\)16,\(-\)14]** & **Logt[\(-\)12,\(-\)10]** & **Logt[\(-\)8,\(-\)6]** \\ \hline \multirow{3}{*}{Slope-type} & 61 & 7.461 \(\times\) 10\({}^{-15}\) & 6.872 \(\times\) 10\({}^{-15}\) & 2.881 \(\times\) 10\({}^{-16}\) \\ \cline{2-5}  & 121 & 2.122 \(\times\) 10\({}^{-14}\) & 1.972 \(\times\) 10\({}^{-14}\) & 1.010 \(\times\) 10\({}^{-15}\) \\ \cline{2-5}  & 241 & 3.424 \(\times\) 10\({}^{-14}\) & 3.196 \(\times\) 10\({}^{-14}\) & 1.811 \(\times\) 10\({}^{-15}\) \\ \hline \multirow{3}{*}{Trapezoid-type} & 61 & 7.330 \(\times\) 10\({}^{-18}\) & 1.441 \(\times\) 10\({}^{-12}\) & 1.669 \(\times\) 10\({}^{-12}\) \\ \cline{2-5}  & 121 & 7.079 \(\times\) 10\({}^{-18}\) & 2.819 \(\times\) 10\({}^{-12}\) & 3.247 \(\times\) 10\({}^{-12}\) \\ \cline{2-5}  & 241 & 7.032 \(\times\) 10\({}^{-18}\) & 3.852 \(\times\) 10\({}^{-12}\) & 4.430 \(\times\) 10\({}^{-12}\) \\ \hline \multirow{3}{*}{Lorentz-type} & 61 & 1.042 \(\times\) 10\({}^{-26}\) & 3.625 \(\times\) 10\({}^{-19}\) & 3.358 \(\times\) 10\({}^{-12}\) \\ \cline{2-5}  & 121 & 1.088 \(\times\) 10\({}^{-26}\) & 3.790 \(\times\) 10\({}^{-19}\) & 3.860 \(\times\) 10\({}^{-12}\) \\ \cline{2-5}  & 241 & 1.088 \(\times\) 10\({}^{-26}\) & 3.871 \(\times\) 10\({}^{-19}\) & 4.122 \(\times\) 10\({}^{-12}\) \\ \hline \multirow{3}{*}{Gaussian-type} & 61 & 1.116 \(\times\) 10\({}^{-17}\) & 8.749 \(\times\) 10\({}^{-14}\) & 5.403 \(\times\) 10\({}^{-12}\) \\ \cline{2-5}  & 121 & 1.404 \(\times\) 10\({}^{-17}\) & 1.304 \(\times\) 10\({}^{-13}\) & 9.720 \(\times\) 10\({}^{-12}\) \\ \cline{2-5}  & 241 & 1.568 \(\times\) 10\({}^{-17}\) & 1.575 \(\times\) 10\({}^{-13}\) & 1.280 \(\times\) 10\({}^{-11}\) \\ \hline \hline \end{tabular}
\end{table} 
Table 1: Residual sum of squares of relaxation modulus data before and after simplifying the model. belt moves on the rollers, it undergoes compression, leading to deformation. Subsequently, as the conveyor belt moves, the rollers gradually disengage from contact with the conveyor belt, resulting in a residual deformation that persists. This phenomenon of hysteresis creates indentation rolling resistance between the conveyor belt and the rollers. To conceptualize the scenario of contact between the conveyor belt and the roller, we established a model: the interface between the conveyor belt and the roller was likened to the surface of a viscoelastic material, whereas the roller was treated as a rigid cylinder. A schematic representation of the model established in this study is shown in Figure 7.  
In the illustration of the model, \(h\) refers to the thickness of the covering layer; \(v\) denotes the speed of the conveyor belt movement; \(a\) represents the initial contact point between the conveyor belt and the roller; \(-b\) denotes the position where the conveyor belt and the roller disengaged; \(R\) represents the radius of the roller; \(h_{0}(t)\) denotes the depth of indentation during stable motion; and \(\delta(t)\) represents the depth of indentation at different positions. Previous findings indicate that, assuming that \(\delta(t)\) is significantly smaller than \(R\), after sequential integration and other numerical operations [23], the following equations can be obtained:  
\[\frac{E_{0}}{2}(1+\xi)(1-\xi)+\sum\limits_{i=1}^{N}E_{i}k_{i}[k_{i}-\xi-(1+k_ {i})e^{-(1+\xi)/k_{i}}]=0 \tag{15}\]  
\[ \begin{array}{l}\frac{Rh}{a^{3}}W=\frac{E_{0}}{6}(2+3\xi-\xi^{3})+\sum \limits_{i=1}^{N}E_{i}k_{i}\Big{\{}\frac{1}{2}(1-\xi^{2})-k_{i}[(1+k_{i})\\ (1-e^{-(1+\xi)/k_{i}})-(1+\xi)]\Big{\}}\\ F=\frac{E_{0}a^{4}}{8R^{2}i}(1-2\xi^{2}+\xi^{4})+\sum\limits_{i=1}^{N}\frac{E_ {0}a^{4}}{R^{2}i}[k_{i}^{3}-\frac{k_{i}}{2}(1+\xi^{2})+\frac{1}{3}\\ (1+\xi^{3})-k_{i}(1+k_{i})(k_{i}+\xi)e^{-(1+\xi)/k_{i}}]\end{array}