# Numerical Analysis of the Forces on the Components of a Direct Diesel Engine
## 3 Kinetics and Dynamics of the Test Engine
### Kinetics and Dynamics of the Piston Mechanism  
According to [18], assuming the displacement of the piston in the horizontal direction is trivial, applying the sine theorem in the OAB triangle (Figure 2), we have:  
\[\frac{l_{2}}{\sin\theta}=\frac{r}{\sin\phi}\Rightarrow\sin\phi=\frac{r}{l}\sin\theta. \tag{6}\]  
If we derive the above expression, we have:  
\[-\dot{\phi}\cos\phi=-\frac{r}{l_{2}}\theta\cos\theta\Rightarrow\dot{\phi}= \frac{r\dot{\theta}\cos\theta}{l_{2}\cos\phi}. \tag{7}\]  
The \(z\) displacement of the piston in the vertical direction is defined by:  
\[z=r\cos\theta+l_{2}\cos\phi. \tag{8}\]  
We can infer that the velocity of the piston in the vertical direction is determined by Equation (8) and the acceleration of the piston in the vertical direction can be calculated with Equation (9).  
\[\dot{z}=-r\dot{\theta}\sin\theta-l_{2}\dot{\phi}\sin\phi, \tag{9}\]  
\[\ddot{z}=-\bigg{(}r\ddot{\theta}\sin\theta+r\dot{\theta}^{2}\cos\theta\bigg{)} -\bigg{(}l_{2}\ddot{\phi}\sin\phi+l_{2}\dot{\phi}^{2}\cos\phi\bigg{)}, \tag{10}\]  
The piston moves only along the cylinder, and thus the acceleration in the horizontal and vertical directions across the piston are determined in Equations (10) and (11), respectively.  
\[a_{PY}=0, \tag{11}\]  
\[a_{PY}=\ddot{z}=\dot{\theta}^{2}\bigg{(}\frac{(r\cos\theta)^{2}}{l\cos^{3} \phi}-r\cos\theta-r\sin\theta\tan\phi\bigg{)}+\ddot{\theta}(r\cos\theta\tan \phi-r\sin\theta), \tag{12}\]  
The cylinder has actually become elliptical due to the rotation of the piston about the wrist pin when the piston moves toward the bottom dead center. In addition, the high temperature during the operation process also affects the deformation of the piston and cylinder. Consequently, the forces exerted on the piston are not symmetric in the x direction. However, in this research, we assumed that the forces exerted on the piston were symmetric in the x direction. Therefore, the forces acting on the piston include \(Q(t)\), the total external force acting on the piston, \(m_{P,g}\), the force of gravity, and \(F_{BX}\), \(F_{BY}\), the link reaction forces in the \(x\) and \(y\) directions, respectively [17; 19], as described in Figure 4.  
Figure 4: The modelled forces on the piston.  
Since the piston does not move in the \(x\) direction, the total force acting on \(x\) is zero, inferring that:  
\[\sum F_{X}= F_{BX}-S=0\Rightarrow F_{BX}=S. \tag{13}\]  
The total force acting in the \(y\) direction is:  
\[\sum F_{Y}=F_{BY}-m_{P}g-Q(t)=m_{P}a_{PY}, \tag{14}\]  
where \(Q(t)\) is the total external force acting on the piston, including the force acting on the piston (\(P_{T}\)), the friction caused by the impacts on the cylinder wall (\(F_{r}\)), friction from the piston surface and cylinder (\(\mu_{P}S\)), the force caused by pressure acting on the piston bottom (\(P_{B}\) is crankcase pressure), and the inertia force \(m_{P}a_{PY}\).  
From Equation (13), we infer that the reaction force on the piston in the y direction is:  
\[F_{BY}=m_{P}g+Q(t)+m_{P}a_{PY}, \tag{15}\]  
in which:  
\[Q(t)=P_{T}+F_{r}\pm\mu_{P}S+P_{B}. \tag{16}\]  
Note that the sign of \(\mu_{P}S\) depends on the direction of piston movements. Thus, we obtained a graph illustrating the total force on the pistons of the test engine, as shown in Figure 5.