# Dynamic Analysis and Seat Selection of Bus Driving Comfort under Different Road Conditions
## 3 Mechanical Analysis of a Bus Running on an Uneven Road
### Road Condition I  
The bus moves forward. The front wheel drives into the uneven road section, but the rear wheel is still in the flat road section; thus, the bus travel distance is \(0\leq x\leq(b+c)\). The vertical displacement increment of two points \(A\) and \(B\) of the bus is denoted as \(\delta_{A}\) and \(\delta_{B}\). Since the front wheel has driven into the pit, there is a geometric relationship:  
\[\delta_{A}=u_{A}-y_{A}\qquad\delta_{B}=u_{B}-y_{B} \tag{6}\]  
where \(y_{A}\) and \(y_{B}\) are the vertical coordinates of the front and rear wheels of the bus.  
\[y_{A}=h\cos\left(\frac{2\pi}{s}x\right)\quad\ y_{B}=h \tag{7}\]  
All bodies of the bus are regarded as rigid bodies. The vertical displacement coordinate \(u_{G}\) at the center of gravity can be solved by using a similar triangle method:  
\[u_{G}=\frac{cu_{A}+bu_{B}}{b+c} \tag{8}\]  
The upper part of the bus is analyzed by the free-body method. The force analysis is shown in Figure 4:  
The equation of motion is  
\[k\delta_{A}+k\delta_{B}=-m\frac{d^{2}u_{G}}{dt^{2}} \tag{9}\]  
The equilibrium equation for calculating the moment of force at point \(G\) is  
\[k\delta_{A}b-k\delta_{B}c=0 \tag{10}\]  
Figure 4: Sketch of the force analysis of the upper part of the bus.  
Figure 3: A simplified mechanical analysis of uneven pavement.  
Combining Equations (8)-(10) obtains the differential equation of motion:  
\[k(b+c)(u_{A}-y_{A})=-m\frac{b^{2}+c^{2}}{b+c}\frac{d^{2}u_{A}}{dt^{2}} \tag{11}\]  
Its general solution is obtained with  
\[u_{A}=y_{A}+C_{1}\cos\left[\frac{(b+c)}{\sqrt{b^{2}+c^{2}}}\sqrt{\frac{k}{m}}t \right]+C_{2}\sin\left[\frac{(b+c)}{\sqrt{b^{2}+c^{2}}}\sqrt{\frac{k}{m}}t\right] \tag{12}\]  
where \(C_{1}\) and \(C_{2}\) are the coefficients to be determined by boundary conditions.  
Combining Equations (6) and (10) obtains \(u_{B}\):  
\[u_{B}=\frac{b}{c}(u_{A}-y_{A})+y_{B} \tag{13}\]  
When \(x=0\) and the front wheel moves to point \(O\), the boundary condition is  
\[u_{A}=h\ \ \ \ \ \frac{du_{A}}{dt}=0 \tag{14}\]  
Therefore, the coefficients in Equation (12) can be determined as  
\[C_{11}=h\left[1-\cos\left(\frac{2\pi}{s}x\right)\right]\ \ \ \ \ C_{21}=0 \tag{15}\]  
The vertical displacement of bus points \(A\) and \(B\) is then expressed as  
\[\left\{\begin{array}{l}u_{A1}=h\cos\left(\frac{2\pi}{s}x\right)+h\left[1- \cos\left(\frac{2\pi}{s}x\right)\right]\cos\left[\frac{(b+c)}{\sqrt{b^{2}+c^{2 }}}\sqrt{\frac{k}{m}}t\right]\\ u_{B1}=h+\frac{bh}{c}\left[1-\cos\left(\frac{2\pi}{s}x\right)\right]\cos\left[ \frac{(b+c)}{\sqrt{b^{2}+c^{2}}}\sqrt{\frac{k}{m}}t\right]\end{array}\right. \tag{16}\]  
The last subscript 1 in Equations (15) and (16) denotes _rad condition I_.  
This paper uses MATLAB to solve Equation (16). Figure 5 shows the result of only net displacement, which means the change in displacement and acceleration of the front and rear wheels in the first stage. The maximum net displacement difference is 5.93 cm at the front wheel and 11.86 cm at the rear wheel. When the front wheel drives on the uneven road and the rear wheel is still on the flat road, the displacement near the rear wheel position is greater. The net vertical acceleration of the front and rear wheels is more than 3 m/s\({}^{2}\), the peak net acceleration of the rear wheels is 5.31 m/s\({}^{2}\), and the peak acceleration of the front wheels is 3.32 m/s\({}^{2}\). The rear wheel acceleration is generally higher than the front wheel acceleration. Thus, the rear wheels are bumpier in the _road condition I_.