## Appendix A  
To achieve flexible impact, it is required that the force change over time in a differentiable manner. Therefore, the displacement-time curve must be at least a fourth-order function. The relationships of the physical quantities on the S-shaped curve are as follows:  
\[s^{(4)}(t)=v^{\prime\prime\prime\prime}(t)=a^{\prime\prime\prime}(t)=J^{\prime \prime}(t)=d(t) \tag{1}\]  
In the formula, \(t\) represents time, \(s\) represents displacement, \(v\) represents velocity, a represents acceleration, \(J\) represents acceleration, and \(d\)\(t_{1}\) represents acceleration (spasm). The schematic diagram of the fourth-order curve for time displacement optimization is shown in Figure 1.  
In Figure A1, \(t_{1}\) represents the time for the variable acceleration phase, \(t_{4}\) represents the time for the constant acceleration phase, and \(t_{8}\) represents the time for the constant speed phase. Let \(d_{1}=d\), \(d_{2}=-d\), \(d_{3}=0\), \(d_{4}=-d\), and \(d_{5}=d\). Based on the equations and continuity conditions, we have the following:  
\[ \begin{cases}J_{1}=\int_{0}^{t}d_{1}\,dt=dt\\ J_{2}=\int_{t_{1}}^{t}d_{2}\,dt=-dt+2 dt_{1}\\ J_{3}=\int_{2t_{1}}^{t}d_{3}\,dt=0\\ J_{4}=\int_{2t_{1}+t_{4}}^{t}d_{4}\,dt=-dt+2 dt_{1}+dt_{4}\\ J_{5}=\int_{3t_{1}+t_{4}}^{t}d_{5}\,dt=dt-4 dt_{1}-dt_{4}\\ \end{cases}  \tag{12}\]  
For acceleration, we have the following:  
\[ \begin{cases}a_{1}=\int_{0}^{t}I_{1}\,dt=\frac{1}{2}dt^{2}\\ a_{2}=\int_{t_{1}}^{t}J_{2}\,dt=-\frac{1}{2}dt^{2}+2 dt_{1}t-dt_{1}^{2}\\ a_{3}=\int_{2t_{1}}^{t}J_{3}\,dt=dt_{1}^{2}\\ a_{4}=\int_{2t_{1}+t_{4}}^{t}J_{4}\,dt=-\frac{1}{2}dt^{2}+(2 dt_{1}+dt_{4})t-dt _{1}^{2}-2 dt_{1}t_{4}-\frac{1}{2}dt_{4}^{2}\\ a_{5}=\int_{3t_{1}+t_{4}}^{t}J_{5}\,dt=\frac{1}{2}dt^{2}-(4 dt_{1}+dt_{4})t+8 dt _{1}^{2}+4 dt_{1}t_{4}+\frac{1}{2}dt_{4}^{2}\\ \end{cases}  \tag{13}\]  
For velocity, we have the following:\[ \begin{cases}v_{1}=\int_{0}^{t}v_{1}\,dt=24 dt^{4}\\ v_{2}=\int_{t_{1}}^{t}v_{2}\,dt=-24 dt^{4}+3 dt_{1}t^{3}-2 dt_{1}^{2}t^{2}+3 dt_{1}^{ 3}t-12 dt_{1}^{4}\\ v_{3}=\int_{2t_{1}}^{t}v_{3}\,dt=2 dt_{1}^{2}t^{2}-dt_{1}^{3}t+\frac{12}{7}dt_{1 }^{4}\\ v_{4}=\int_{2t_{1}+t_{4}}^{t}v_{4}\,dt=-24 dt^{4}+6(2 dt_{1}+dt_{4})t^{3}-2(dt_{1 }^{2}+2 dt_{1}t_{4}-2 dt_{4}^{2})t^{2}\\ \qquad+\Big{(}3 dt_{1}^{3}+2 dt_{1}^{2}t_{4}+dt_{1}t_{4}^{2}+6 dt_{4}^{3}\Big{)}t- 12 dt_{1}^{4}-3 dt_{1}^{3}t_{4}-dt_{1}^{2}t_{4}^{2}-3 dt_{1}t_{4}^{3}-24 dt_{4}^{4} \\ v_{5}=\int_{3t_{1}+t_{4}}^{t}v_{5}\,dt=24 dt^{4}-6(4 dt_{1}+dt_{4})t^{3}+2\Big{(} 8 dt_{1}^{2}+4 dt_{1}t_{4}+2 dt_{4}^{2}\Big{)}t^{2}\\ \qquad-\Big{(}\frac{3}{26}dt_{1}^{3}+7 dt_{1}^{2}t_{4}+2 dt_{1}t_{4}^{2}+6 dt_{4} ^{3}\Big{)}t+\frac{3}{20}dt_{1}^{4}\\ \qquad+\frac{3}{23}dt_{1}^{3}t_{4}+\frac{2}{7}dt_{1}^{2}t_{4}^{2}+\frac{3}{2} dt_{1}t_{4}^{3}+6 dt_{4}^{4}\end{cases}  \tag{10}\]  
For displacement, we have the following:  
\[ \begin{cases}s_{1}=\int_{0}^{t}v_{1}\,dt=\frac{1}{24}dt^{4}\\ s_{2}=\int_{t_{1}}^{t}v_{2}\,dt=-\frac{1}{24}dt^{4}+\frac{1}{3}dt_{1}t^{3}- \frac{1}{2}dt_{1}^{2}t^{2}+\frac{1}{3}dt_{1}^{3}t-\frac{1}{12}dt_{1}^{4}\\ s_{3}=\int_{2t_{1}}^{t}v_{3}\,dt=\frac{1}{2}dt_{1}^{2}t^{2}-dt_{1}^{3}t+\frac{ 7}{12}dt_{1}^{4}\\ s_{4}=\int_{2t_{1}+t_{4}}^{t}v_{4}\,dt=-\frac{1}{24}dt^{4}+\frac{1}{6}(2 dt_{1 }+dt_{4})t^{3}-\frac{1}{2}(dt_{1}^{2}+2 dt_{1}t_{4}-\frac{1}{2}dt_{4}^{2})t^{2} \\ \qquad+\Big{(}\frac{1}{3}dt_{1}^{3}+2 dt_{1}^{2}t_{4}+dt_{1}t_{4}^{2}+\frac{ 1}{6}dt_{4}^{3}\Big{)}t-\frac{1}{2}dt_{1}^{4}-\frac{4}{3}dt_{1}^{3}t_{4}-dt_{1 }^{2}t_{4}^{2}-\frac{1}{3}dt_{1}t_{4}^{3}-\frac{1}{24}dt_{4}^{4}\\ s_{5}=\int_{3t_{1}+t_{4}}^{t}v_{5}\,dt=\frac{1}{24}dt^{4}-\frac{1}{6}(dt_{1 }+dt_{4})t^{3}+\frac{1}{2}(8 dt_{1}^{2}+4 dt_{1}t_{4}+\frac{1}{2}dt_{4}^{2})t^{ 2}\\ \qquad-\Big{(}\frac{26}{3}dt_{1}^{3}+7 dt_{1}^{2}t_{4}+2 dt_{1}t_{4}^{2}+\frac{ 1}{6}dt_{4}^{3}\Big{)}t\\ \qquad+\frac{20}{3}dt_{1}^{4}+\frac{23}{3}dt_{1}^{3}t_{4}+\frac{7}{2}dt_{1}^{2 }t_{4}^{2}+\frac{2}{3}dt_{1}t_{4}^{3}+\frac{1}{6}dt_{4}^{4}\end{cases}