An Effective Approach of Collision Avoidance for UAV

Abstract

In the last decade, the collision avoidance of Unmanned Aerial Vehicles (UAVs) has become increasingly important for the safe operation of UAVs. In this article, an effective conflict detection and alerting principle is firstly proposed based on mixed collision cone and alerting criterion for the collision avoidance of UAV. Second, a reactive collision avoidance and trajectory recovery strategy (RCATRS) is presented based on the model of relative kinematics. In this strategy, by acting an acceleration vector with different magnitude and direction on UAV, a horizontal collision avoidance maneuver is realized in situations of different relative position and velocity vector. When UAV has bypassed the obstacle, the horizontal trajectory recovery maneuver is initiated to make UAV to return to original flight paths. Thus, different recovery trajectories corresponding with different relative velocity vector are planned. Finally, the safety controller is designed to apply RCATRS to autonomous quadrotor. Since a few of computation is off-line, RCATRS is simple in implement and can satisfy the request of running in real time. The results of simulation show the validity of the proposed RCATRS.

Appendices

Appendix A

By inserting the Eq. (17) into the Eq. (7), we have

$$\begin{array}{c}{x}_{a}\left({t}_{e}\right)={v}_{ab}\left({t}_{0}\right)\mathrm{sin}\left(\gamma \right)\frac{{v}_{ab}\left({t}_{0}\right)\mathrm{sin }\left({\theta }_{h}-\gamma \right)}{{a}_{h}}+\frac{1}{2}{a}_{h}\mathrm{cos}\left({\theta }_{h}\right){\left(\frac{{v}_{ab}\left({t}_{0}\right)\mathrm{sin }\left({\theta }_{h}-\gamma \right)}{{a}_{h}}\right)}^{2}\\ =\frac{{({v}_{ab}\left({t}_{0}\right))}^{2}}{2{a}_{h}}(2\mathrm{sin }\left(\gamma \right)\mathrm{sin }\left({\theta }_{h}-\gamma \right)+\mathrm{cos}\left({\theta }_{h}\right){\left(\mathrm{sin}\left({\theta }_{h}-\gamma \right)\right)}^{2})\end{array}$$

(54)

From the Eqs. (19) and (14), the Eq. (54) is written as

$${x}_{a}\left({t}_{e}\right)=\frac{{({v}_{ab}\left({t}_{0}\right))}^{2}}{2{a}_{h}}{\mathrm{cos}\left({\theta }_{h}\right)(\mathrm{sin }\left({\theta }_{c}\right))}^{2}=\frac{r}{{(\mathrm{sin }\left({\theta }_{c}\right))}^{2}}{\mathrm{cos}\left({\theta }_{h}\right)(\mathrm{sin }\left({\theta }_{c}\right))}^{2}=r\mathrm{cos}\left({\theta }_{h}\right)$$

(55)

By inserting the Eq. (17) into the Eq. (8), we have

$${y}_{a}\left({t}_{e}\right)={y}_{a}\left({t}_{0}\right)-{v}_{ab}\left({t}_{0}\right)\mathrm{cos }\left(\gamma \right)\frac{{v}_{ab}\left({t}_{0}\right)\mathrm{sin }\left({\theta }_{h}-\gamma \right)}{{a}_{h}}+\frac{1}{2}{a}_{h}\mathrm{sin}({\theta }_{h}){(\frac{{v}_{ab}\left({t}_{0}\right)\mathrm{sin }\left({\theta }_{h}-\gamma \right)}{{a}_{h}})}^{2}$$

(56)

where ${a}_{h}$ is given by Eq. (15) or (16). From Eq. (14), we have $\frac{{{(v}_{ab}\left({t}_{0}\right))}^{2}}{{a}_{h}}=\frac{2r}{{(\mathrm{sin }\left({\theta }_{c}\right))}^{2}}$. Let ${y}_{a}\left({t}_{0}\right)={v}_{ab}\left({t}_{0}\right)\mathrm{cos }\left(\gamma \right){T}_{l}$, where ${T}_{l}=\sqrt{\frac{2r}{{a}_{max}}}$ if ${v}_{ab}\left({t}_{0}\right)>{v}_{T}$, and ${T}_{l}=\frac{2r}{{c}_{f}{V}_{ab}\left({t}_{0}\right)}$ if ${v}_{ab}\left({t}_{0}\right)\le {v}_{T}$. Thus, we have

$$\begin{array}{c}{y}_{a}\left({t}_{e}\right)=2r\mathrm{cos }\left(\gamma \right)\frac{(\mathrm{sin }\left({\theta }_{c}\right)-\mathrm{sin }\left({\theta }_{h}-\gamma \right))}{{(\mathrm{sin }\left({\theta }_{c}\right))}^{2}}+r\mathrm{sin}({\theta }_{h}){(\frac{\mathrm{sin }\left({\theta }_{h}-\gamma \right)}{\mathrm{sin }\left({\theta }_{c}\right)})}^{2}\\ =2r\mathrm{cos }\left(\gamma \right)\frac{(\mathrm{sin }\left({\theta }_{c}\right)-\mathrm{sin }\left({\theta }_{h}-\gamma \right))}{{(\mathrm{sin }\left({\theta }_{c}\right))}^{2}}-r\mathrm{sin}\left({\theta }_{h}\right)\frac{\left({\left(\mathrm{sin}\left({\theta }_{c}\right)\right)}^{2}-{\left(\mathrm{sin}\left({\theta }_{h}-\gamma \right)\right)}^{2}\right)}{{\left(\mathrm{sin}\left({\theta }_{c}\right)\right)}^{2}}+r\mathrm{sin}\left({\theta }_{h}\right)\\ =r\frac{\left(\mathrm{sin}\left({\theta }_{c}\right)-\mathrm{sin}\left({\theta }_{h}-\gamma \right)\right)}{{\left(\mathrm{sin}\left({\theta }_{c}\right)\right)}^{2}}(2\mathrm{cos }\left(\gamma \right)-\mathrm{sin}\left({\theta }_{h}\right)\left(\mathrm{sin}\left({\theta }_{c}\right)+\mathrm{sin}\left({\theta }_{h}-\gamma \right)\right))+r\mathrm{sin}\left({\theta }_{h}\right)\end{array}$$

(57)

Since $\mathrm{sin}\left({\theta }_{c}\right)-\mathrm{sin}\left({\theta }_{h}-\gamma \right)\ge 0$, we have ${y}_{a}\left({t}_{e}\right)\ge r\mathrm{sin}({\theta }_{h})$ if the following equations are satisfied

$$2\mathrm{cos }\left(\gamma \right)-\mathrm{sin}\left({\theta }_{h}\right)\left(\mathrm{sin}\left({\theta }_{c}\right)+\mathrm{sin}\left({\theta }_{h}-\gamma \right)\right)\ge 0$$

(58)

Since $\mathrm{cos }\left(\gamma \right)-\mathrm{sin}\left({\theta }_{h}\right)\mathrm{sin}\left({\theta }_{h}-\gamma \right)=\mathrm{cos}\left({\theta }_{h}\right)\mathrm{cos}\left({\theta }_{h}-\gamma \right)$, the Eq. (58) can be written as

$$\mathrm{cos }\left(\gamma \right)-\mathrm{sin}\left({\theta }_{h}\right)\mathrm{sin}\left({\theta }_{c}\right)+\mathrm{cos}\left({\theta }_{h}\right)\mathrm{cos}\left({\theta }_{h}-\gamma \right)\ge 0$$

(59)

Since $\mathrm{cos }\left({\theta }_{c}+{\theta }_{h}\right)-\mathrm{cos}\left({\theta }_{h}\right)\mathrm{cos}\left({\theta }_{c}\right)=-\mathrm{sin}\left({\theta }_{h}\right)\mathrm{sin}\left({\theta }_{c}\right)$, the Eq. (59) can be written as

$$\mathrm{cos }\left(\gamma \right)+\mathrm{cos }\left({\theta }_{c}+{\theta }_{h}\right)-\mathrm{cos}\left({\theta }_{h}\right)\mathrm{cos}\left({\theta }_{c}\right)+\mathrm{cos}\left({\theta }_{h}\right)\mathrm{cos}\left({\theta }_{h}-\gamma \right)\ge 0$$

(60)

Since $\mathrm{cos }\left(\gamma \right)+\mathrm{cos }\left({\theta }_{c}+{\theta }_{h}\right)=2\mathrm{cos }\left(\frac{{\theta }_{c}+{\theta }_{h}+\gamma }{2}\right)\mathrm{cos }\left(\frac{{\theta }_{c}+{\theta }_{h}-\gamma }{2}\right)$, the Eq. (60) can be written as

$$2\mathrm{cos }\left(\frac{{\theta }_{c}+{\theta }_{h}+\gamma }{2}\right)\mathrm{cos }\left(\frac{{\theta }_{c}+{\theta }_{h}-\gamma }{2}\right)-\mathrm{cos}\left({\theta }_{h}\right)(\mathrm{cos}\left({\theta }_{c}\right)-\mathrm{cos}\left({\theta }_{h}-\gamma \right))\ge 0$$

(61)

Since $\mathrm{cos}\left({\theta }_{c}\right)-\mathrm{cos}\left({\theta }_{h}-\gamma \right)=-2\mathrm{sin }\left(\frac{{\theta }_{c}+{\theta }_{h}-\gamma }{2}\right)\mathrm{sin }\left(\frac{{\theta }_{c}-{\theta }_{h}+\gamma }{2}\right)$, the Eq. (61) can be written as

$$2\mathrm{cos }\left(\frac{{\theta }_{c}+{\theta }_{h}+\gamma }{2}\right)\mathrm{cos }\left(\frac{{\theta }_{c}+{\theta }_{h}-\gamma }{2}\right)+2\mathrm{cos}\left({\theta }_{h}\right)\mathrm{sin }\left(\frac{{\theta }_{c}+{\theta }_{h}-\gamma }{2}\right)\mathrm{sin }\left(\frac{{\theta }_{c}-{\theta }_{h}+\gamma }{2}\right)\ge 0$$

(62)

Since $0\le {\theta }_{h}\le {\theta }_{c}+\gamma$, we have ${\theta }_{c}-{\theta }_{h}+\gamma \ge 0$. If ${\theta }_{c}+{\theta }_{h}+\gamma \le 180^\circ$, the Eq. (62) is satisfied. Thus, we have ${y}_{a}\left({t}_{e}\right)\ge r\mathrm{sin}({\theta }_{h})$.

Appendix B

B.1 the case of ${a}_{1}\le {a}_{max}$ and ${a}_{h}={a}_{1}$

• 1) When $t\in [0,{\Delta t}_{r1}]$, we have $(x_\alpha{(t_e\;+\;t))}^2\;+\;(y_\alpha{(t_e\;+\;t))}^2\;=\;(x_\alpha{(t_e))}^2\;+\;(y_\alpha{(t_e))}^2\;+\;\Delta r_1\;=\;r^2\;+\;\Delta r_1$ from Eqs. (23) and (24), where

  $$\begin{array}{c}{{\Delta {r}_{1}={(v}_{ab,y}\left({t}_{e}\right)t)}^{2}-{x}_{a}\left({t}_{e}\right){a}_{h}{t}^{2}{-2{y}_{a}\left({t}_{e}\right)v}_{ab,y}\left({t}_{e}\right)t+{2{x}_{a}\left({t}_{e}\right)v}_{ab,x}\left({t}_{e}\right)t+(v}_{ab,x}\left({t}_{e}\right){t-\frac{1}{2}{a}_{h}{t}^{2})}^{2}\\ ={t}^{2}\left({{v}_{ab,y}\left({t}_{e}\right)}^{2}-{x}_{a}\left({t}_{e}\right){a}_{h}\right)+2t\left({-{y}_{a}\left({t}_{e}\right)v}_{ab,y}\left({t}_{e}\right)+{{x}_{a}\left({t}_{e}\right)v}_{ab,x}\left({t}_{e}\right)\right)+{(v}_{ab,x}\left({t}_{e}\right){t-\frac{1}{2}{a}_{h}{t}^{2})}^{2}\end{array}$$

  (63)

Further, we have

$${t}^{2}\left({{v}_{ab,y}\left({t}_{e}\right)}^{2}-{x}_{a}\left({t}_{e}\right){a}_{h}\right)={t}^{2}\left({\left({v}_{ab}\left({t}_{0}\right){(\mathrm{cos }\left({\theta }_{c}\right))}^{2}\right)}^{2}-r\mathrm{cos}({\theta }_{c}){({V}_{ab}\left({t}_{0}\right)\mathrm{cos }\left({\theta }_{c}\right))}^{2}\mathrm{cos }\left({\theta }_{c}\right)/r\right)=0$$

(64)

$$2t({-{y}_{a}\left({t}_{e}\right)v}_{ab,y}\left({t}_{e}\right)+{{x}_{a}\left({t}_{e}\right)v}_{ab,x}\left({t}_{e}\right))=2t{v}_{ab}\left({t}_{0}\right)\mathrm{cos }\left({\theta }_{c}\right)r\left(\mathrm{cos}\left({\theta }_{c}\right)\mathrm{sin}\left({\theta }_{c}\right)-\mathrm{sin}\left({\theta }_{c}\right)\mathrm{cos}\left({\theta }_{c}\right)\right)=0$$

(65)

$${(v}_{ab,x}\left({t}_{e}\right){t-\frac{1}{2}{a}_{h}{t}^{2})}^{2}={v}_{ab,x}\left({t}_{e}\right){t}^{2}\left({v}_{ab,x}\left({t}_{e}\right)-{a}_{h}t\right)+{(\frac{1}{2}{a}_{h}{t}^{2})}^{2}={(\frac{1}{2}{a}_{h}{t}^{2})}^{2}$$

(66)

Thus, we obtain

$${{(x}_{a}\left({t}_{e}+t\right))}^{2}+{{(y}_{a}\left({t}_{e}+t\right))}^{2}{=r}^{2}+\Delta {r}_{1}{=r}^{2}+{(\frac{1}{2}{a}_{h}{t}^{2})}^{2}\ge {r}^{2}$$

(67)

where $t\in [0,{\Delta t}_{r1}]$. When $t={\Delta t}_{r1}$, we have

$${{(x}_{a}\left({t}_{r1}\right))}^{2}+{{(y}_{a}\left({t}_{r1}\right))}^{2}{=r}^{2}+{(\frac{1}{2}{a}_{h}{{\Delta t}_{r1}}^{2})}^{2}$$

(68)

• 2) When $t\in [0,{\Delta t}_{r2}]$, we have ${{(x}_{a}\left({t}_{r1}+t\right))}^{2}+{{(y}_{a}\left({t}_{r1}+t\right))}^{2}={{(x}_{a}\left({t}_{r1}\right))}^{2}+{{(y}_{a}\left({t}_{r1}\right))}^{2}+\Delta {r}_{2}{=r}^{2}+{(\frac{1}{2}{a}_{h}{{\Delta t}_{r1}}^{2})}^{2}+\Delta {r}_{2}$ from Eqs. (27) and (28), where

  $$\begin{array}{c}{\Delta {r}_{2}={(v}_{ab,y}\left({t}_{e}\right)t)}^{2}-{x}_{a}\left({t}_{r1}\right){a}_{h}{t}^{2}{-2{y}_{a}\left({t}_{r1}\right)v}_{ab,y}\left({t}_{e}\right)t+({\frac{1}{2}{a}_{h}{t}^{2})}^{2}\\ ={t}^{2}\left({{v}_{ab,y}\left({t}_{e}\right)}^{2}-{x}_{a}\left({t}_{e}\right){a}_{h}\right)-{a}_{h}{t}^{2}\left({v}_{ab,x}\left({t}_{e}\right){\Delta t}_{r1}-\frac{1}{2}{a}_{h}{{\Delta t}_{r1}}^{2}\right)+{(\frac{1}{2}{a}_{h}{t}^{2})}^{2}\end{array}$$

  (69)

From (64), ${y}_{a}\left({t}_{r1}\right)=0$ and ${v}_{ab,x}\left({t}_{e}\right)={a}_{h}{\Delta t}_{r1}$, we have $\Delta {r}_{2}=-\frac{1}{2}{a}_{h}{{\Delta t}_{r1}}^{2}{a}_{h}{t}^{2}+({\frac{1}{2}{a}_{h}{t}^{2})}^{2}$. Thus,

$${(\frac{1}{2}{a}_{h}{{\Delta t}_{r1}}^{2})}^{2}+\Delta {r}_{2}={(\frac{1}{2}{a}_{h}{{\Delta t}_{r1}}^{2})}^{2}-\frac{1}{2}{a}_{h}{{\Delta t}_{r1}}^{2}{a}_{h}{t}^{2}+({\frac{1}{2}{a}_{h}{t}^{2})}^{2}=\frac{1}{4}{a}_{h}{{{(\Delta t}_{r1}}^{2}-{t}^{2})}^{2}\ge 0$$

(70)

From (70), we have

$${{(x}_{a}\left({t}_{r1}+t\right))}^{2}+{{(y}_{a}\left({t}_{r1}+t\right))}^{2}{=r}^{2}+\frac{1}{4}{a}_{h}{{{(\Delta t}_{r1}}^{2}-{t}^{2})}^{2}\ge {r}^{2}$$

(71)

where $t\in [0,{\Delta t}_{r2}]$. Since ${\Delta t}_{r2}={\Delta t}_{r1}$, we have

$${{(x}_{a}\left({t}_{r2}\right))}^{2}+{{(y}_{a}\left({t}_{r2}\right))}^{2}{=r}^{2}$$

(72)

• 3) When $t\in [0,{\Delta t}_{r3}]$, we have ${{(x}_{a}\left({t}_{r2}+t\right))}^{2}+{{(y}_{a}\left({t}_{r2}+t\right))}^{2}={g}_{1}+{g}_{2}+{g}_{3}$ from Eqs. (31) and (32), where

  $${g}_{1}={\left({x}_{a}\left({t}_{r2}\right)-{v}_{ab,x}\left({t}_{r2}\right)t\right)}^{2}+{({y}_{a}\left({t}_{r2}\right)-{v}_{ab,y}\left({t}_{e}\right)t)}^{2}$$

  (73)
  $${g}_{2}={(\frac{1}{2}{a}_{h}{\mathrm{cos}\left({\theta }_{c}\right)t}^{2})}^{2}+{(\frac{1}{2}{a}_{h}{\mathrm{sin}\left({\theta }_{c}\right)t}^{2})}^{2}={(\frac{1}{2}{a}_{h}{t}^{2})}^{2}$$

  (74)
  $${g}_{3}={({x}_{a}\left({t}_{r2}\right)-{v}_{ab,x}\left({t}_{r2}\right)t){a}_{h}{\mathrm{cos}\left({\theta }_{c}\right)t}^{2}-(y}_{a}\left({t}_{r2}\right)-{v}_{ab,y}\left({t}_{e}\right)t){a}_{h}{\mathrm{sin}\left({\theta }_{c}\right)t}^{2}$$

  (75)

Since ${x}_{a}\left({t}_{r2}\right)={x}_{a}\left({t}_{e}\right)$, ${y}_{a}\left({t}_{r2}\right)={-y}_{a}\left({t}_{e}\right)$ and ${v}_{ab,x}\left({t}_{r2}\right)={v}_{ab,x}\left({t}_{e}\right)$, we have ${\left({x}_{a}\left({t}_{r2}\right)\right)}^{2}+{({y}_{a}\left({t}_{r2}\right))}^{2}={r}^{2}$; ${{x}_{a}\left({t}_{r2}\right){v}_{ab,x}\left({t}_{r2}\right)+y}_{a}\left({t}_{r2}\right){v}_{ab,y}\left({t}_{e}\right)=0$; ${\left({v}_{ab,x}\left({t}_{r2}\right)t\right)}^{2}+{({v}_{ab,y}\left({t}_{e}\right)t)}^{2}={({v}_{ab}\left({t}_{0}\right)t)}^{2}$. Thus,

$${g}_{1}={r}^{2}+{({v}_{ab}\left({t}_{0}\right)t)}^{2}$$

(76)

Similarly, we have ${{x}_{a}\left({t}_{r2}\right)\mathrm{cos}\left({\theta }_{c}\right)-y}_{a}\left({t}_{r2}\right)\mathrm{sin}\left({\theta }_{c}\right)=r$ and $-{v}_{ab,x}\left({t}_{r2}\right)\mathrm{cos}\left({\theta }_{c}\right)+{v}_{ab,y}\left({t}_{e}\right)\mathrm{sin}\left({\theta }_{c}\right)=0$. Thus,

$${g}_{3}=r{a}_{h}{t}^{2}$$

(77)

From (74), (76) and (77), we have

$${{(x}_{a}\left({t}_{r2}+t\right))}^{2}+{{(y}_{a}\left({t}_{r2}+t\right))}^{2}{=r}^{2}+{({v}_{ab}\left({t}_{0}\right)t)}^{2}+{(\frac{1}{2}{a}_{h}{t}^{2})}^{2}+r{a}_{h}{t}^{2}\ge {r}^{2}$$

(78)

where $t\in [0,{\Delta t}_{r3}]$. From (67), (71) and (78), ${{(x}_{a}\left(t\right))}^{2}+{{(y}_{a}\left(t\right))}^{2}\ge {r}^{2}$ is obtained where $t\in [{t}_{e},{t}_{r3}]$.

B.2 the case of ${a}_{1}>{a}_{max}$ and ${a}_{h}={a}_{max}$

• 1) When $t\in [0,{\Delta t}_{r1}]$, we have ${{(x}_{a}\left({t}_{e}+t\right))}^{2}+{{(y}_{a}\left({t}_{e}+t\right))}^{2}={{(x}_{a}\left({t}_{e}\right))}^{2}+{{(y}_{a}\left({t}_{e}\right))}^{2}+\Delta {r}_{1}{=r}^{2}+\Delta {r}_{1}$, where $\Delta {r}_{1}$ is given by (B1). From (B2), we have ${{v}_{ab,y}\left({t}_{e}\right)}^{2}={x}_{a}\left({t}_{e}\right){a}_{h}$. Since ${a}_{1}>{a}_{max}$, we have ${{v}_{ab,y}\left({t}_{e}\right)}^{2}={x}_{a}\left({t}_{e}\right){a}_{1}>{x}_{a}\left({t}_{e}\right){a}_{max}$. Further, we have ${t}^{2}\left({{v}_{ab,y}\left({t}_{e}\right)}^{2}-{x}_{a}\left({t}_{e}\right){a}_{max}\right)\ge 0$. From (B3) and (B4), we have

  $$\Delta {r}_{1}={t}^{2}\left({{v}_{ab,y}\left({t}_{e}\right)}^{2}-{x}_{a}\left({t}_{e}\right){a}_{max}\right)+{(\frac{1}{2}{a}_{max}{t}^{2})}^{2}\ge 0$$

  (81)

Thus, we obtain

$${{(x}_{a}\left({t}_{e}+t\right))}^{2}+{{(y}_{a}\left({t}_{e}+t\right))}^{2}{=r}^{2}+\Delta {r}_{1}\ge {r}^{2}$$

(82)

• 2) When$t\in [0,{\Delta t}_{r2}]$, we have ${{(x}_{a}\left({t}_{r1}+t\right))}^{2}+{{(y}_{a}\left({t}_{r1}+t\right))}^{2}={{(x}_{a}\left({t}_{r1}\right))}^{2}+{{(y}_{a}\left({t}_{r1}\right))}^{2}+\Delta {r}_{2}$, where $\Delta {r}_{2}$ is given by (B7). From ${y}_{a}\left({t}_{r1}\right)<0$ and${v}_{ab,x}\left({t}_{e}\right)={a}_{max}{\Delta t}_{r1}$, we have

  $$\Delta {r}_{2}={t}^{2}\left({{v}_{ab,y}\left({t}_{e}\right)}^{2}-{x}_{a}\left({t}_{e}\right){a}_{max}\right)+{2\left|{y}_{a}\left({t}_{r1}\right)\right|v}_{ab,y}\left({t}_{e}\right)t-\frac{1}{2}{a}_{max}{{\Delta t}_{r1}}^{2}{a}_{max}{t}^{2}+({\frac{1}{2}{a}_{max}{t}^{2})}^{2}$$

  (83)

Since ${{(x}_{a}\left({t}_{r1}\right))}^{2}+{{(y}_{a}\left({t}_{r1}\right))}^{2}{=r}^{2}+\Delta {r}_{1}{|}_{{\Delta t}_{r1}}$, where $\Delta {r}_{1}{|}_{{\Delta t}_{r1}}$ is given by (B17) at $t={\Delta t}_{r1}$, we have

$$\Delta {r}_{1}{|}_{{\Delta t}_{r1}}+\Delta {r}_{2}={(t}^{2}+{{\Delta t}_{r1}}^{2})\left({{v}_{ab,y}\left({t}_{e}\right)}^{2}-{x}_{a}\left({t}_{e}\right){a}_{max}\right)+{2\left|{y}_{a}\left({t}_{r1}\right)\right|v}_{ab,y}\left({t}_{e}\right)t+\frac{1}{4}{a}_{max}{{{(\Delta t}_{r1}}^{2}-{t}^{2})}^{2}$$

(84)

Since ${{(v}_{ab,y}\left({t}_{e}\right)}^{2}-{x}_{a}\left({t}_{e}\right){a}_{max})>0$, we have $\Delta {r}_{1}{|}_{{\Delta t}_{r1}}+\Delta {r}_{2}>0$. Thus, we obtain

$${{(x}_{a}\left({t}_{r1}+t\right))}^{2}+{{(y}_{a}\left({t}_{r1}+t\right))}^{2}{=r}^{2}+\Delta {r}_{1}{|}_{{\Delta t}_{r1}}+\Delta {r}_{2}>{r}^{2}$$

(85)

• 3) When $t\in [0,{\Delta t}_{r3}]$, we have ${{(x}_{a}\left({t}_{r2}+t\right))}^{2}+{{(y}_{a}\left({t}_{r2}+t\right))}^{2}={{(x}_{a}\left({t}_{r2}\right))}^{2}+{{(y}_{a}\left({t}_{r2}\right))}^{2}+\Delta {r}_{3}$ from Eqs. (37) and (38), where

  $${\Delta {r}_{3}={(v}_{ab,y}\left({t}_{e}\right)t)}^{2}{-2{y}_{a}\left({t}_{r2}\right)v}_{ab,y}\left({t}_{e}\right)t+{2x}_{a}\left({t}_{r2}\right){v}_{ab,x}\left({t}_{r2}\right)t+{x}_{a}\left({t}_{r2}\right){a}_{max}{t}^{2}+({v}_{ab,x}\left({t}_{r2}\right)t+{\frac{1}{2}{a}_{max}{t}^{2})}^{2}$$

  (86)

Next, we demonstrate $\Delta {r}_{3}\ge 0$. First, we demonstrate${-2{y}_{a}\left({t}_{r2}\right)v}_{ab,y}\left({t}_{e}\right)t+{2x}_{a}\left({t}_{r2}\right){v}_{ab,x}\left({t}_{r2}\right)t\ge 0$. From (25), (28), ${y}_{a}\left({t}_{r1}\right)<0$ and${x}_{a}\left({t}_{r2}\right)={x}_{a}\left({t}_{r1}\right)/2$, we have

$$\begin{array}{c}{{k}_{1}=-{2y}_{a}\left({t}_{r2}\right)v}_{ab,y}\left({t}_{e}\right)+{2x}_{a}\left({t}_{r2}\right){v}_{ab,x}\left({t}_{r2}\right)=-{x}_{a}\left({t}_{r1}\right){a}_{max}\Delta {t}_{r2}{-2{y}_{a}\left({t}_{r1}\right)v}_{ab,y}\left({t}_{e}\right)+2{{(v}_{ab,y}\left({t}_{e}\right))}^{2}{\Delta t}_{r2}\\ =\left(2{{(v}_{ab,y}\left({t}_{e}\right))}^{2}-{x}_{a}\left({t}_{r1}\right){a}_{max}\right)\Delta {t}_{r2}+{2\left|{y}_{a}\left({t}_{r1}\right)\right|v}_{ab,y}\left({t}_{e}\right)\end{array}$$

(87)

From (21) and (23), we have ${k}_{2}=2{{(v}_{ab,y}\left({t}_{e}\right))}^{2}-{x}_{a}\left({t}_{r1}\right){a}_{max}={{(v}_{ab,y}\left({t}_{e}\right))}^{2}-{x}_{a}\left({t}_{e}\right){a}_{max}+{{(v}_{ab,y}\left({t}_{e}\right))}^{2}-\frac{1}{2}{{(v}_{ab,x}\left({t}_{e}\right))}^{2}$. Let ${{{k}_{3}=(v}_{ab,y}\left({t}_{e}\right))}^{2}-\frac{1}{2}{{(v}_{ab,x}\left({t}_{e}\right))}^{2}={{(v}_{ab}\left({t}_{0}\right)\mathrm{cos }\left({\theta }_{c}\right))}^{2}({\left(\mathrm{cos}\left({\theta }_{c}\right)\right)}^{2}-\frac{1}{2}{\left(\mathrm{sin}\left({\theta }_{c}\right)\right)}^{2})$. Further, let

$${k}_{4}={\left(\mathrm{cos}\left({\theta }_{c}\right)\right)}^{2}-\frac{1}{2}{\left(\mathrm{sin}\left({\theta }_{c}\right)\right)}^{2}=1-\frac{3}{2}{\left(\mathrm{sin}\left({\theta }_{c}\right)\right)}^{2}$$

(88)

When ${a}_{1}>{a}_{max}$, we have

$${a}_{1}=\frac{{({v}_{ab}\left({t}_{0}\right)\mathrm{cos }\left({\theta }_{c}\right))}^{2}\mathrm{cos}\left({\theta }_{c}\right)}{r}>{a}_{max}=\frac{{({v}_{ab}\left({t}_{0}\right)\mathrm{sin }\left({\theta }_{c}\right))}^{2}}{2r}$$

(89)

By solving (89), we obtain ${\theta }_{c}\le {48}^{^\circ }$ if ${a}_{1}>{a}_{max}$. When ${\theta }_{c}\le {48}^{^\circ }$, thus, we have ${k}_{4}>0\Rightarrow {k}_{3}>0$. Since ${{(v}_{ab,y}\left({t}_{e}\right))}^{2}-{x}_{a}\left({t}_{e}\right){a}_{max}>0$, we have ${k}_{2}>0\Rightarrow {k}_{1}>0$. Since all components in (B22) are not less than zero, $\Delta {r}_{3}\ge 0$. From (85), we have

$${{(x}_{a}\left({t}_{r2}+t\right))}^{2}+{{(y}_{a}\left({t}_{r2}+t\right))}^{2}={{(x}_{a}\left({t}_{r2}\right))}^{2}+{{(y}_{a}\left({t}_{r2}\right))}^{2}+\Delta {r}_{3}>{r}^{2}$$

(90)

From (82), (85) and (90), ${{(x}_{a}\left(t\right))}^{2}+{{(y}_{a}\left(t\right))}^{2}\ge {r}^{2}$ is obtained where $t\in [{t}_{e},{t}_{r3}]$.