## 3 Transient Response Simulation
### Research on Incentive and Damping Parameters  
#### 3.3.1 Transient Excitation Calculation  
In the dynamic response calculation of this ship's propulsion shaft system, the excitation mainly considers the transient excitation of the diesel engine and the propeller excitation (which corresponds to \"Q\" in Equation (1)). The transient excitation of the diesel engine primarily consists of the gas excitation torque and the reciprocating inertia torque. Unlike the selection of the excitation torque of the diesel engine at each harmonic during steady-state torsional vibration calculations in the frequency domain, the excitation torque is calculated for each harmonic at every rotational speed across the full speed range. \begin{table}
\begin{tabular}{c c c c c c c c} \hline \hline \multirow{2}{*}{**System Component**} & **Inertia** & \multirow{2}{*}{
\begin{tabular}{c} **System** \\ **Component** \\ **(kgm\({}^{2}\))** \\ \end{tabular} } & **System** & **Outer** & **Inner** & **Internal** & **Damping** & **Damping** \\  & **Component** & **Diameter** & **Diameter** & **Diameter** & **Damping** & **Coefficient** & **Factor** \\ \hline damper & 9890 & 71.0 & - & 0 & 155,000 & 0 & 0 \\ TVD & 5572 & 7812.5 & 1220 & 0 & 0 & 0.005 & 0 \\ cylinder & 91,505 & 6172.8 & 1220 & 0 & 0 & 0.005 & 0.0085 \\ cylinder & 91,505 & 6211.2 & 1220 & 0 & 0 & 0.005 & 0.0085 \\ cylinder & 91,505 & 6060.6 & 1220 & 0 & 0 & 0.005 & 0.0085 \\ camshaft & 18,833 & 9090.9 & 1220 & 0 & 0 & 0.005 & 0.0085 \\ drive & 91,505 & 6060.6 & 1220 & 0 & 0 & 0.005 & 0.0085 \\ cylinder & 91,505 & 6172.8 & 1220 & 0 & 0 & 0.005 & 0.0085 \\ cylinder & 91,505 & 6410.3 & 1220 & 0 & 0 & 0.005 & 0.0085 \\ cylinder & 91,505 & 9708.7 & 1220 & 0 & 0 & 0.005 & 0.0085 \\ thrust bearing & 13,385 & 14,925.4 & 1240 & 200 & 0 & 0 & 0.0085 \\ turn + rotor1 & 37,843 & 1280.4 & 840 & 0 & 0 & 0 & 0.005 \\ rotor1 + rotor2 & 28,551 & 703.2 & 840 & 0 & 0 & 0 & 0 \\ rotor2 + I/S & 5748 & 279.3 & 840 & 0 & 0 & 0 & 0 \\ I/S + P/S & 7917 & 407.8 & 920 & 0 & 0 & 0 & 0 \\ P/S + prop & 502,256 & - & - & 0 & 0 & 0 & 0 \\ \hline \hline \end{tabular}
\end{table} 
Table 2: Equivalent parameter model data. Figure 2: Lumped-parameter model diagram. The calculation of the gas excitation torque for a two-stroke diesel engine is shown in Equation (9) [31]. \[T_{n}=T_{n0}+\frac{\pi}{4}D^{2}\mathrm{R}\sum_{v}^{\infty}a_{v}\cos\
u\omega t+b_ {v}\sin\
u\omega t=T_{n0}+\frac{\pi}{4}D^{2}\mathrm{R}\sum_{v}^{\infty}C_{v} \sin(\
u\omega t+\Psi_{v}) \tag{9}\]  
where \(T_{n}\) is the average torque of a single cylinder, Nm; \(D\) is the diameter of the diesel engine cylinder, mm; \(R\) is the crank radius of the diesel engine, mm; \(a_{v}\) is the sine component of the \(v\)th harmonic tangential force, \(\mathrm{N/mm^{2}}\); \(b_{v}\) is the cosine component of the \(v\)th harmonic tangential force, \(\mathrm{N/mm^{2}}\); \(C_{v}\) is the harmonic coefficient, \(\mathrm{N/mm^{2}}\); \(\Psi_{v}\) is the initial phase angle of the \(v\)th harmonic tangential force, rad; and \(\omega t\) is the crankshaft angle, rad. The amplitude of the excitation torque of a certain harmonic order produced by the gas pressure in the cylinder is shown in Equation (10) [31]. \[T_{v}=\frac{\pi}{4}D^{2}RC_{V}\ast 10^{-3} \tag{10}\]  
The excitation torque produced by the reciprocating inertia force of a diesel engine cylinder is shown in Equation (11) [31]. \[T_{I}=mR^{2}\omega^{2}\left[ \begin{array}{c}\frac{\pi}{4}sin2(\omega t-\delta )-\left(\frac{3\lambda}{4}+\frac{9\lambda^{2}}{32}\right)sin3(\omega t-\delta )-\\ \frac{\lambda^{2}}{4}sin4(\omega t-\delta)+\frac{5\lambda^{2}}{32}sin5(\omega t -\delta)\end{array} \right]\bullet 10^{-6} \tag{11}\]  
where \(T_{I}\) is the reciprocating inertia force excitation torque, Nm; \(m\) is the total mass, kg; \(\delta\) is the firing interval angle between any cylinder and the 1 st cylinder, rad; \(\lambda\) is the ratio of the crank radius to the connecting rod length, \(\lambda=R/L\); and \(\omega t\) is the crankshaft angle, rad.