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.
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Acknowledgements
This paper was supported by the Natural Science Foundation of Jilin Province, China under Grants 20210101171JC; the National Natural Science Foundation of China under Grants 61571462.
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This research was supported by the Natural Science Foundation of Jilin Province, China under Grants 20210101171JC; the National Natural Science Foundation of China under Grants 61571462. But, it did not receive any funding from the commercial or not-for-profit sectors.
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A) Jianwu Tao contributes the main ideas in this research, and proposes the effective conflict detection and alerting principle and the reactive collision avoidance and trajectory recovery strategy (RCATRS). B) Wei-chao Ji participates partly the design of safety controller in this research. C) Qiong-jian Fan participates partly the Numerical simulation in this research.
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Appendices
Appendix A
By inserting the Eq. (17) into the Eq. (7), we have
From the Eqs. (19) and (14), the Eq. (54) is written as
By inserting the Eq. (17) into the Eq. (8), we have
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
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
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
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
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
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
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
Thus, we obtain
where \(t\in [0,{\Delta t}_{r1}]\). When \(t={\Delta t}_{r1}\), we have
-
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,
From (70), we have
where \(t\in [0,{\Delta t}_{r2}]\). Since \({\Delta t}_{r2}={\Delta t}_{r1}\), we have
-
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,
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,
From (74), (76) and (77), we have
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
-
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
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
-
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
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
When \({a}_{1}>{a}_{max}\), we have
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
From (82), (85) and (90), \({{(x}_{a}\left(t\right))}^{2}+{{(y}_{a}\left(t\right))}^{2}\ge {r}^{2}\) is obtained where \(t\in [{t}_{e},{t}_{r3}]\).
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Tao, Jw., Ji, Wc. & Fan, Qj. An Effective Approach of Collision Avoidance for UAV. J Intell Robot Syst 108, 18 (2023). https://doi.org/10.1007/s10846-023-01869-4
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DOI: https://doi.org/10.1007/s10846-023-01869-4