# 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 Crankshaft  
The crankshaft converts the up and down movements of the pistons into rotary motion and drives the external drive. The crankshaft is connected to the pistons via the transmission rods. Similar to the determination of the rod acceleration, we can determine the crankshaft acceleration at the center using the acceleration at the common point between the transmission rod and the crankshaft \(\dddot{a}_{C}=\dddot{a}_{A}+\dddot{a}_{C/A}\) and project it in the horizontal and vertical directions:  
\[a_{CX}=-\dot{\theta}^{2}hr\sin\theta+\ddot{\theta}hr\cos\theta, \tag{23}\] \[a_{CY}=\dot{\theta}^{2}hr\cos\theta+\ddot{\theta}hr\sin\theta. \tag{24}\]  
The total force acting on the crankshaft is illustrated in Figure 7.  
Figure 6: The model of forces on the rod. Figure 7: The forces acting on the crankshaft. The total force in \(x\) direction is:  
\[\sum F_{X}=F_{OX}-F_{AX}=m_{C}a_{CX}, \tag{25}\]  
The total force acting in the \(y\) direction is:  
\[\sum F_{Y}=F_{OY}-F_{AY}-m_{C}g=m_{C}a_{CY}, \tag{26}\]  
The total torque around crankshaft is:  
\[\sum M_{C}=I_{C}\ddot{\theta}_{C}, \tag{27}\]  
\[T+F_{AX}(1+h)r\cos\theta+F_{AY}(1+h)r\sin\theta-F_{OX}hr\cos\theta-F_{OY}hr \sin\theta=I_{C}\ddot{\theta}_{C}. \tag{28}\]  
Combining these above equations, we have:  
\[ \begin{array}{l}S=-r\ddot{\theta}\left[\frac{I_{B}\cos\theta}{(I\cos\phi)^{ 2}}+m_{P}\tan\phi(\cos\theta\tan\phi-\sin\theta)+jm_{R}\left(\frac{j\cos\theta }{\cos^{2}\phi}-\sin\theta\tan\phi-\cos\theta\right)\right]-\\ -r\ddot{\theta}^{2}\left[\frac{I_{B}}{(I\cos\phi)^{2}}\left(\frac{r\cos^{2} \theta\tan\phi}{I\cos\phi}-\sin\theta\right)+m_{P}\tan\phi\left(\frac{r\cos^ {2}\theta}{I\cos^{3}\phi}-\sin\theta\tan\phi-\cos\theta\right)+\right.\\ \left.+jm_{R}\left(\frac{ir\cos^{2}\theta\tan\phi}{I\cos^{3}\phi}-\cos\theta \tan\phi+\sin\theta-\frac{j\sin\theta}{\cos^{2}\phi}\right)\right]-g\tan\phi (m_{P}+jm_{R})-Q(t)\tan\phi\end{array} . \tag{29}\]  
Substituting the known forces in Equation (8), we can obtain the torque acting on the crankshaft:  
\[ \begin{array}{l}T=I_{C}\ddot{\theta}+m_{C}a_{CX}hr\cos\theta-m_{R}a_{RX}r \cos\theta+m_{C}a_{CY}hr\sin\theta-m_{R}a_{RY}r\sin\theta-m_{P}a_{PY}r\sin \theta+\\ +m_{C}hgr\sin\theta-m_{R}gr\sin\theta-m_{P}gr\sin\theta-Q(t)r\sin\theta-Sr\cos \theta\end{array} . \tag{30}\]  
Substituting the known accelerations into the above equation, we have:  
\[T=\ddot{\theta}I(\theta)+\frac{1}{2}\dot{\theta}^{2}I^{\prime}(\theta)+g( \theta)+Q(t,\theta), \tag{31}\]  
where \(I(\theta)\) is the inertia function, defined by:  
\[ \begin{array}{l}I(\theta)=I_{C}+m_{C}h^{2}r^{2}+I_{R}\left(\frac{r\cos \theta}{I\cos\phi}\right)^{2}+m_{P}r^{2}(\cos\theta\tan\phi-\sin\theta)^{2}+ \\ +m_{R}r^{2}\left[(1-j)^{2}\cos^{2}\theta+(j\cos\theta\tan\phi-\sin\theta)^{2} \right]\end{array} , \tag{32}\]  
where \(I^{\prime}(\theta)\) is the rate of the inertial variable based on the rotational angle \(\theta\) of the crankshaft determined by:  
\[ \begin{array}{l}I^{\prime}(\theta)=2I_{R}\left(\frac{r\cos\theta}{I\cos\phi }\right)^{2}\left(\frac{r\cos\theta}{I\cos\phi}\tan\phi-\tan\theta\right)+\\ +2m_{P}r^{2}(\cos\theta\tan\phi-\sin\theta)\left(\frac{r\cos^{2}\theta}{I\cos^{ 3}\phi}-\cos\theta-\sin\theta\tan\phi\right)-\\ -2m_{R}r^{2}(1-j)^{2}\sin\theta\cos\theta+2m_{R}r^{2}(j\cos\theta\tan\phi-\sin \theta)\left(\frac{ir\cos^{2}\theta}{I\cos^{3}\phi}-\cos\theta-j\sin\theta \tan\phi\right)\end{array}