# Phase Plane Trajectory Planning for Double Pendulum Crane Anti-Sway Control
## 3 Trajectory Planning
At this time, the acceleration remains \(a_{c}=0\), and Equation (13) becomes:  
\[\ddot{\theta}_{11}(t)+{w_{n}}^{2}\theta_{11}(t)=\frac{{w_{n}}^{2}}{g}a_{1} \tag{18}\]1Solving the differential Equation (27), we obtain:  
\[\ddot{\theta}_{11}(t)+{w_{n}}^{2}\theta_{11}(t)=\frac{{w_{n}}^{2}}{g}a_{1} \tag{18}\]2  
From the initial conditions:  
\[\ddot{\theta}_{11}(t)+{w_{n}}^{2}\theta_{11}(t)=\frac{{w_{n}}^{2}}{g}a_{1} \tag{18}\]3  
\[\ddot{\theta}_{11}(t)+{w_{n}}^{2}\theta_{11}(t)=\frac{{w_{n}}^{2}}{g}a_{1} \tag{18}\]4  
Substituting the initial conditions into Equation (28):  
\[\ddot{\theta}_{11}(t)+{w_{n}}^{2}\theta_{11}(t)=\frac{{w_{n}}^{2}}{g}a_{1} \tag{18}\]5  
Therefore, Equation (28) is obtained as:  
\[\ddot{\theta}_{11}(t)+{w_{n}}^{2}\theta_{11}(t)=\frac{{w_{n}}^{2}}{g}a_{1} \tag{18}\]6  
Taking the first-order derivative of Equation (29) with respect to time, achieves:  
\[\ddot{\theta}_{11}(t)+{w_{n}}^{2}\theta_{11}(t)=\frac{{w_{n}}^{2}}{g}a_{1} \tag{18}\]7  
Divide \(w_{n}\) on both sides of Equation (30) to obtain:  
\[\ddot{\theta}_{11}(t)+{w_{n}}^{2}\theta_{11}(t)=\frac{{w_{n}}^{2}}{g}a_{1} \tag{18}\]8  
Equations (29) and (31) obtain:  
\[\ddot{\theta}_{11}(t)+{w_{n}}^{2}\theta_{11}(t)=\frac{{w_{n}}^{2}}{g}a_{1} \tag{18}\]9  
The angle \(\alpha_{2}\) corresponding to the \(\widehat{CO}\) segment is calculated as follows:  
\[\theta_{11}(t)=C_{1}\cos(w_{n}t)+C_{2}\sin(w_{n}t)-C_{3}\frac{{w_{n}}^{2}}{g} \tag{19}\]0  
Arriving at point C, the second section of the uniform velocity trajectory ends. The load trajectory is \(\widehat{ABC}\), which is a part of a circle with (0, 0) as the center and \(R\) as the radius. The third section is the second acceleration process of the load. The trajectory is symmetrical with the first stage. The acceleration amplitude of this process is the same as that of the first section. This stage is the same as the first stage, and the deceleration stage is similar to the acceleration process. \[\theta_{11}(t)=C_{1}\cos(w_{n}t)+C_{2}\sin(w_{n}t)-C_{3}\frac{{w_{n}}^{2}}{g} \tag{19}\]1  
The acceleration amplitude and switching time are calculated by the system condition constraints. According to the kinematic equation, the speed and displacement of each section of the trolley are calculated as follows:\[\dot{x}(t)=\left\{ \begin{array}{l}a_{1}t,0\leq t<t_{1}\\ a_{1}t_{1},t_{1}\leq t<t_{2}\\ a_{1}t_{1}+a_{1}(t-t_{2}),t_{2}\leq t<t_{3}\\ a_{1}t_{1}+a_{1}(t_{3}-t_{2}),t_{3}\leq t<t_{4}\\ a_{1}t_{1}+a_{1}(t_{3}-t_{2})-a_{1}(t-t_{4}),t_{4}\leq t<t_{5}\\ a_{1}t_{1}+a_{1}(t_{3}-t_{2})-a_{1}(t_{5}-t_{4}),t_{5}\leq t<t_{6}\\ a_{1}t_{1}+a_{1}(t_{3}-t_{2})-a_{1}(t_{5}-t_{4})-a_{1}(t-t_{6}),t_{6}\leq t<t_{7 }\end{array} \right. \tag{34}\]  
After integrating Equation (34) once over time, the target \(x_{d}\) is obtained as:  
\[x_{d}={4a_{1}{t_{1}}^{2}}+{2a_{1}t_{1}}(t_{2}-t_{1})+{2a_{1}t_{1}}(t_{4}-t_{3}) \tag{35}\]  
The maximum speed of the trolley during the whole process is:  
\[v_{\max}={a_{1}t_{1}}+{a_{1}}(t_{3}-t_{2}) \tag{36}\]  
From the phase plane trajectory, it can be seen that the maximum swing angle corresponding to the load in the phase plane is:  
\[\theta_{\max}=\frac{\sqrt{2a_{1}(1-\cos(w_{n}t_{1}))}}{g} \tag{37}\]  
From the phase plane trajectory, we can see that at point A, the trolley has the same speed before and after switching acceleration, and the loads at points A and C have the same swing angle, obtaining:  
\[\alpha_{1}=\alpha_{2} \tag{38}\]  
From Equations (24), (25) and (33)-(38), the acceleration amplitude \(a_{1}\) and acceleration switching time \(t_{1},t_{2},t_{3},t_{4}\) of the trolley can be calculated.