## 3 Trajectory Planning  
### Path Planning  
Delta robots are often applied to implement PPO in manufacture line. The PPO path is shown in Figure 2. From Figure 2, we can see PPO path includes vertical segments, which are \(AH\) and \(IG\), and a horizontal segment, which is \(HI\). In order to smooth the PPO path, curve transitions are adopted, that is, curve \(BC\) is used to replace the right angle formed by \(BH\) and \(HC\), and curve \(EF\) is used to replace the right angle formed by \(EI\) and \(IF\). Figure 1: Schematic of a Delta robot. Figure 2: Geometric path for pick-and-place operation (PPO). Quintic PH curves, namely segments \(BC\) and \(EF\), are used for the PPO path. In Figure 2, \(|AG|=w\), \(|HC|=|EI|=k\), \(|BH|=|FI|=m\), \(|AB|=|GF|=j\), \(|AH|=|GI|=h\). The quintic PH curves in the PPO path illustrated in Figure 3 are given [35]  
\[P(\gamma)=(x(\gamma),y(\gamma))=\sum_{i=0}^{n}\left( \begin{array}{c}n\\ i\end{array} \right)P_{i}(1-\gamma)^{n-i}\gamma^{i},\gamma\in[0,1], \tag{2}\]  
where  
\[\left\{ \begin{aligned} x(\gamma)&=\frac{2}{5}(u_{0} \upsilon_{2}+u_{2}\upsilon_{2})\gamma^{5}-u_{0}\upsilon_{2}\gamma^{4}+\frac{2} {3}u_{0}\upsilon_{2}\gamma^{3}\\ y(\gamma)&=(\frac{1}{5}u_{0}^{2}+\frac{1}{15}u_{0}u_{2}) \gamma^{5}+(u_{0}^{2}+\frac{1}{3}u_{0}u_{2})\gamma^{4}(1-\gamma)\\ &\quad+(2u_{0}^{2}+\frac{2}{3}u_{0}u_{2})\gamma^{3}(1-\gamma)^{2}+2u_ {0}^{2}\gamma^{2}(1-\gamma)^{3}\\ &\quad+u_{0}^{2}\gamma(1-\gamma)^{4},\end{aligned} \right. \tag{3}\]  
\[\left\{ \begin{aligned} u_{0}&=\sqrt{\frac{5(35m+k-\sqrt{m^ {2}+k^{2}+70 mk})}{34}}\\ u_{2}&=\sqrt{\frac{5(m+35k-\sqrt{m^{2}+k^{2}+70 mk})}{68 }}\\ v_{2}&=\sqrt{\frac{5(m+35k-\sqrt{m^{2}+k^{2}+70 mk})}{68 }}.\end{aligned} \right. \tag{4}\]  
In the PPO path, for the purpose of simplifying the calculation and meeting the real-time requirements, assuming \(m\) equals \(k\), thus the maximum deviation between the right angle and the PH curve can be obtained when \(\gamma=0.5\), and see Proposition 1 for details. The parameter \(d\) in Figure 3 represents the maximum deviation between the right angle and the PH curve in the PPO path, which is called prescribed geometrical constraint. Therefore, we can get the quintic PH curve if \(d\), the prescribed geometrical constraint, is given, and it is proved in the following proposition. **Proposition 1**. : _If \(m=k\), the maximum deviation between the right angle and the PH curve can be obtained when \(\gamma=0.5\), on this basis, if \(d\), the prescribed geometrical constraint, is given, the PH curve in the PPO path can be solved._  
Figure 3: The Pythagorean–Hodograph (PH) curve in the PPO path. Proof. : \(\kappa(\gamma)\), the curvature of PH curves, can be obtained by [24; 30]  
\[\kappa(\gamma)=\frac{P^{\prime}(\gamma)\times P^{\prime\prime}(\gamma)}{\|P^{ \prime}(\gamma)\|^{3}}=\frac{-2\big{[}u^{\prime}(\gamma)v(\gamma)-u(\gamma)v^{ \prime}(\gamma)\big{]}}{\big{(}u^{2}(\gamma)+v^{2}(\gamma)\big{)}^{2}}. \tag{5}\]  
To ensure a \(G^{2}\) connection between the horizontal and vertical segments, curvature \(\kappa(\gamma)\) must satisfy \(\kappa(0)=0\) and \(\kappa(1)=0\). Then we can get the following equation  
\[\left\{ \begin{array}{l}u(\tau)=u_{0}(1-\tau)^{2}+u_{2}\tau^{2}\\ v(\tau)=v_{2}\tau^{2}\end{array} \right. \tag{6}\]  
By substituting (6) into (5), the following equation can be obtained  
\[\kappa(\gamma)=\frac{4u_{0}v_{2}(\gamma-\gamma^{2})}{\big{(}u^{2}(\gamma)+v^{ 2}(\gamma)\big{)}^{2}}. \tag{7}\]  
The derivative of curvature can be calculated as  
\[\kappa^{\prime}(\gamma)=\left(\frac{4u_{0}v_{2}(\gamma-\gamma^{2})}{\big{(}u ^{2}(\gamma)+v^{2}(\gamma)\big{)}^{2}}\right)^{\prime}=4u_{0}v_{2}\varphi(u^{ 2}(\gamma)+v^{2}(\gamma)), \tag{8}\]  
where \(\varphi=\frac{(1-2\gamma)(u^{2}(\gamma)+v^{2}(\gamma))-4(\gamma-\gamma^{2})(v (\gamma)v^{\prime}(\gamma)+u(\gamma)u^{\prime}(\gamma))}{(u^{2}(\gamma)+v^{2} (\gamma))^{4}}\).