# Multi-Objective Motion Control Optimization for the Bridge Crane System
## 3 Control Design
### Trajectory Planning Control  
Based on the geometrical analysis, an efficient trajectory imposed by Constraints 1 to 3 can be acquired. The second-order ordinary differential equation and its derivative equation are solved based on Equation (5):  
\[\theta(t)=\theta(t_{0})\cos\omega_{n}t+\frac{\dot{\theta}(t_{0})}{\omega_{n}} \sin\omega_{n}t-\frac{a}{g}(1-\cos\omega_{n}t) \tag{10}\]  
\[\dot{\theta}(t)=-\theta(t_{0})\omega_{n}\sin\omega_{n}t+\dot{\theta}(t_{0}) \cos\omega_{n}t-\frac{\omega_{n}a}{g}\sin\omega_{n}t \tag{11}\]  
where \(\omega_{n}=\sqrt{g/L}\), \(a=\dot{x}_{T}\), \(\theta(t_{0})\) and \(\dot{\theta}(t_{0})\) denote the initial swing and the angle velocity of the payload, respectively. Since the payload does not oscillate at the initial motion point, \(\theta(t_{0})\) and \(\dot{\theta}(t_{0})\) equal zero. Based on Equations (10) and (11), we can get:  
\[\left[\theta(t)+\frac{a}{g}\right]^{2}+\dot{\Phi}^{2}(t)=\left[\frac{a}{g} \right]^{2} \tag{12}\]  
where \(\dot{\Phi}(t)=\dot{\theta}(t)/\omega_{n}\). Now, the phase plane is employed in terms of Equation (12) for further analysis, of which \(\dot{\Phi}(t)\) and \(\theta(t)\) refer to the longitudinal coordinate and the horizontal coordinate, respectively, as shown in Figure 3. To implement the intuitive illustration, the following three cases are discussed according to different values of the acceleration. **Case 1. **_When \(a>0\), the vector in the phase plane moves in the clockwise direction at a constant angular velocity \(\omega_{n}\);_  
**Case 2. **_When \(a<0\), the vector moves in the counterclockwise direction at the same velocity;_  
Figure 3: Phase plane of the payload. **Case 3**. : _When \(a=0\), the vector stays in the original point; namely, there does not exist any relative motion between the crane and the payload._  
Here, an appropriate trajectory is determined considering the geometric properties. Based on the above analysis, the primary expression can be furnished as:  
\[\ddot{x}=\left\{ \begin{array}{ll}a_{\max},&0\leq t\leq t_{acc}\\ -a_{\max},&t_{acc}+t_{cust}\leq t\leq 2t_{acc}+t_{cust}\\ 0,&else\end{array} \right. \tag{13}\]  
where \(t_{acc}\) is the duration of the acceleration, and \(t_{cust}\) is the duration for the constant speed stage. As shown in Figure 3, \(t_{acc}\), \(t_{cust}\) and \(a_{\max}\) can be determined according to the geometric properties. At the beginning of acceleration, the vector starts to move from the original point towards \(O_{2}\) with the angular velocity \(\omega_{n}\). After reaching \(A_{2}\), the maximum payload swing is achieved, and the relationship with respect to \(\theta_{\max}\) and \(a_{\max}\) can be determined, i.e.,:  
\[|a_{\max}|=\frac{g|\theta_{\max}|}{2} \tag{14}\]  
and they should be satisfied with the following constraints:  
\[|\theta_{\max}|\leq\theta_{lim},|a_{\max}|\leq a_{\lim} \tag{15}\]  
The payload swing reaches zero after one complete acceleration cycle. At \(T\), the crane stops accelerating and keeps its speed unchanged, while the load's swing angle maintains zero, as stated in Case 2. In addition, during the deceleration process presented in Case 3, the similar phenomenon can also be observed: that one cycle is required to be undergone in order to reach the target position through point \(A_{2}\), during which the residual angle can be eliminated. Based on the previous analysis, the duration in terms of acceleration/deceleration stages requires the same time cost, i.e., \(T\). Given the chosen target position, the motion states can be determined after an integral calculation:  
\[\dot{x}(t)=\left\{ \begin{array}{ll}a_{\max}t,&0\leq t\leq t_{acc}\\ a_{\max}t_{acc},&t_{acc}\leq t\leq t_{acc}+t_{cust}\\ -a_{\max}(t-T_{total}),&t_{acc}+t_{cust}\leq t\leq t_{total}\\ 0,&else\end{array} \right. \tag{16}\]  
\[x(t)=\left\{ \begin{array}{ll}\frac{1}{2}a_{\max}t^{2},&0\leq t\leq t_{acc} \\ a_{\max}t_{acc}t,&t_{acc}\leq t\leq t_{acc}+t_{cust}\\ -a_{\max}(\frac{1}{2}t^{2}-T_{total}t),&t_{acc}+t_{cust}\leq t\leq t_{total}\\ 0,&else\end{array} \right.