## 3 Results and Discussion
### Perforation Diameter Analysis
#### 3.2.3 Quantitative Analysis Method for Perforation Diameter  
Based on the principle of quantitative analysis, the size of the perforation is related to  
\[d=f(d_{p},l_{p},c_{p},\rho_{p},E_{k},Q_{s},Y_{p},V,T_{t},c_{t},\rho_{t},Y_{t}) \tag{15}\]  
According to the basic principle of the analysis of the quantities and \(\pi\) theorem, (as shown in Table 8) the basic physical quantities are \(L,M,\) and \(T.\) The basic physical quantities \(d_{p},\rho_{t},Y_{t}\) are taken as the basic physical quantities. \[\frac{d}{d_{p}}=f\left(\frac{l_{p}}{d_{p}},\frac{c_{p}}{\rho_{t}^{-1}Y_{t}^{ \frac{1}{2}}},\frac{\rho_{p}}{\rho_{t}},\frac{E_{k}}{d_{p}^{3}Y_{t}},\frac{Q_{s} }{d_{t}^{3}Y_{t}},\frac{Y_{p}}{\rho_{t}^{-\frac{1}{2}}Y_{t}^{\frac{1}{2}}}, \frac{V}{d_{p}^{-\frac{1}{2}}Y_{t}^{\frac{1}{2}}}\right) \tag{16}\]  
A hypothesis was proposed based on the impact process of the reactive fragments while assuming that 80% of the chemical reaction energy is used to increase the hydrostatic pressure generated by the impact process and that the degree of chemical reaction is exponentially correlated with the impact velocity. Then, the chemical reaction energy released could be considered as  
\[Q_{s}=0.8\times Q\times(1-\exp(-V)) \tag{17}\]  
In the case of the same reactive fragments impacting the same steel plate at different velocities:  
\[(d_{p},l_{p},c_{p},\rho_{p},Y_{p},T_{t},c_{t},\rho_{t},Y_{t})=const \tag{18}\]  
So, the functional relationship equation is simplified as  
\[\frac{d}{d_{p}}=g\left(\frac{E_{k}}{d_{p}^{3}Y_{t}},\frac{V}{\rho_{t}^{-\frac{ 1}{2}}Y_{t}^{\frac{1}{2}}},\frac{Q_{s}}{d_{p}^{3}Y_{t}}\right) \tag{19}\]  
It is thought that the chemical energy \(Q_{s}\), kinetic energy \(E_{k}\), and impact velocity \(V\) all work together to determine the steel plate's perforation diameter. Based on the results of the simulation, the following assumptions were made regarding the above function expressions:  
\[d=d_{p}\times\left(A\left(\frac{Q_{s}}{d_{p}^{3}Y_{t}}\right)^{B}+C\left(\frac {V}{\rho_{t}^{-\frac{1}{2}}Y_{t}^{\frac{1}{2}}}\right)^{D}+E\left(\frac{E_{K} }{d_{p}^{3}Y_{t}}\right)^{F}\right) \tag{20}\] \begin{table}
\begin{tabular}{c c c c c} \hline \hline
**Object** & **Physical Quantities** & **Symbols** & **Units** & **Dimensions** \\ \hline \multirow{8}{*}{Reactive fragment} & Diameter & \(d_{\text{p}}\) & mm & \(L\) \\ \cline{2-5}  & Cylindrical length & \(l_{\text{p}}\) & mm & \(L\) \\ \cline{2-5}  & Sound velocity & \(c_{\text{p}}\) & km/s & \(L\)T\({}^{-1}\) \\ \cline{2-5}  & Density & \(\rho_{\text{p}}\) & g/cm\({}^{3}\) & \(M\)L\({}^{-3}\) \\ \cline{2-5}  & Kinetic energy & \(E_{\text{k}}\) & kJ & \(L^{2}M\)T\({}^{-2}\) \\ \cline{2-5}  & Chemical energy released & \(Q_{\text{s}}\) & kJ & \(L^{2}M\)T\({}^{-2}\) \\ \cline{2-5}  & Yield strength & \(Y_{\text{p}}\) & GPa & \(L^{-1}M\)T\({}^{-2}\) \\ \cline{2-5}  & Impact velocity & \(V\) & km/s & \(L\)T\({}^{-1}\) \\ \hline \multirow{4}{*}{Steel plate} & Thickness & \(T_{\text{t}}\) & mm & \(L\) \\ \cline{2-5}  & Sound velocity & \(c_{\text{t}}\) & km/s & \(L\)T\({}^{-1}\) \\ \cline{2-5}  & Density & \(\rho\) & g/cm\({}^{3}\) & \(M\)L\({}^{-3}\) \\ \cline{2-5}  & Yield strength & \(\gamma_{\text{t}}\) & GPa & \(L^{-1}M\)T\({}^{-2}\) \\ \hline \hline \end{tabular}
\end{table} 
Table 8: Physical units and dimensions. The coefficients \(A\), \(B\), \(C\), \(D\), \(E\), and \(F\) were produced as shown in Table 9 when the perforation diameters in the table data were fitted by using regression analysis in MATLAB for various reactive fragments. The equation for the steel plate's perforated diameter based on the fit is displayed below.