﻿ 大地线航法在智能船舶上的应用
 舰船科学技术  2023, Vol. 45 Issue (1): 180-185    DOI: 10.3404/j.issn.1672-7649.2023.01.033 PDF

Application of geodesic sailing on intelligent ship
WU Zu-xin, ZHENG Zhong-yi
Navigation College, Dalian Maritime University, Dalian 116026, China
Abstract: Sailing along the shortest path can help build an efficient and clean shipping. Geodesic is the shortest path between two points on the surface of space. Due to geodesic's complicated calculation, varying course, and only shorten 5% voyage by average compared with rhumb line, the geodesic sailing (GS) is almost impossible in traditional nautical practice. The intelligent ship can maximize the advantage of the shortest geodetic path by using its calculation, data acquisition, course control, etc, In this paper, a fast and high-precision algorithm is presented to solve GS (direct/inverse) problems firstly. Then, apply the GS to the unmanned ship by calculating the course dynamically based on the current position and destination, and the new order is issued to the autopilot. Finally, alter course restriction mechanism (ACRM) with different strategies are proposed to reduce the autopilot’s work load by avoiding frequent course alternation. The simulation results show that ACRM can help to obtain a balance between the shortest path and the minimum number of alter course.
Key words: geodesic sailing (GS)     rhumb line sailing (RLS)     intelligent ship     alter course restriction mechanism (ACRM)
0 引　言

1 大地线航法

 图 1 大地线航法及其辅助球体示意图 Fig. 1 Geodesic sailing diagram and its auxiliary sphere
 $\frac{s_{0i}}{b} = \int_0^{{\sigma _{0i}}} {\sqrt {1 + {k^2}{{\sin }^2}\sigma}{\rm{ d}}\sigma } ，$ (1)
 ${\lambda _{0i}} = {\omega _{oi}} - f\sin {\alpha _0}\int_0^{{\sigma _{0i}}} {\frac{{2 - f}}{{1 + (1 - f)\sqrt {1 + {k^2}{{\sin }^2}\sigma } }}}{\rm{ d}}\sigma 。$ (2)

1.1 正解 φ1, λ1, α1, s12 ${\boldsymbol{\Rightarrow}}$ φ2, λ2, α2

 ${\beta _1} = \arctan ((1 - f)\tan {\varphi _1}) ，$ (3)

 ${\alpha _0} = \arctan (\frac{{\sin {\alpha _1}\cos {\beta _1}}}{{\sqrt {{{\cos }^2}{\alpha _1} + {{\sin }^2}{\alpha _1}{{\sin }^2}{\beta _1}} }}) ，$ (4)
 ${\sigma _{01}} = \arctan (\tan {\beta _1}/\cos {\alpha _1})，$ (5)
 ${\omega _{01}} = \arctan (\sin {\alpha _0}\tan {\sigma _{01}})，$ (6)

 ${s_{01}} = b{A_1}({\sigma _{01}} + \sum\limits_{l = 1}^\infty {{C_{1l}}} \sin 2l{\sigma _{01}}) ，$ (7)
 ${s_{02}} = {s_{01}} + {s_{12}}，$ (8)
 ${\tau _2} = {s_{02}}/(b{A_1})，$ (9)
 ${\sigma _{02}} = {\tau _2} + \sum\limits_{l = 1}^\infty {C_{1l}'} \sin 2l{\tau _2} 。$ (10)

 $\varepsilon = \dfrac{{\sqrt {1 + {k^2}} - 1}}{{\sqrt {1 + {k^2}} + 1}}，\quad {A_1} = \dfrac{{\left(1 + \dfrac{1}{4}{\varepsilon ^2} + \dfrac{1}{{64}}{\varepsilon ^4} + \dfrac{1}{{256}}{\varepsilon ^6} + \cdots \right)}}{{\left( {1 - \varepsilon } \right)}}，$
 $\begin{split}& {C_{11}} = - \frac{1}{2}\varepsilon + \frac{3}{{16}}{\varepsilon ^3} - \frac{1}{{32}}{\varepsilon ^5} + \cdots ，\\ & {C_{12}} = - \frac{1}{{16}}{\varepsilon ^2} + \frac{1}{{32}}{\varepsilon ^4} -\frac{9}{{2048}}{\varepsilon ^6} + \cdots ，\end{split}$
 ${C_{13}} = - \frac{1}{{48}}{\varepsilon ^3} + \frac{3}{{256}}{\varepsilon ^5} - \cdots ，\quad {C_{14}} = - \frac{5}{{512}}{\varepsilon ^4} + \frac{3}{{512}}{\varepsilon ^6} - \cdots，$
 ${C_{15}} = - \frac{7}{{1280}}{\varepsilon ^5} + \cdots ，\quad \quad {C_{16}} = - \frac{7}{{2048}}{\varepsilon ^6} + \cdots，$
 $\begin{split}& C_{11}' = \frac{1}{2}\varepsilon - \frac{9}{{32}} {\varepsilon ^3} + \frac{{205}}{{1536}}{\varepsilon ^5} + \cdots，\\ & C_{12}' = \frac{5}{{16}}{\varepsilon ^2} - \frac{{37}}{{96}}{\varepsilon ^4} +\frac{{1335}}{{4096}}{\varepsilon ^6} + \cdots，\end{split}$
 $C_{13}' = \frac{{29}}{{96}}{\varepsilon ^3} - \frac{{75}}{{128}}{\varepsilon ^5} + \cdots ，\quad C_{14}' = \frac{{539}}{{1536}}{\varepsilon ^4} - \frac{{2391}}{{2560}}{\varepsilon ^6} + \cdots，$
 $C_{15}' = \frac{{3760}}{{7680}}{\varepsilon ^5} + \cdots ，\quad \quad C_{16}' = \frac{{38081}}{{61440}}{\varepsilon ^6} + \cdots。$

 ${\beta _2} = \arctan \left(\frac{{\cos {\alpha _0}\sin {\sigma _{02}}}}{{\sqrt {{{\cos }^2}{\sigma _{02}} + {{\sin }^2}{\alpha _0}{{\sin }^2}{\sigma _{02}}} }}\right)，$ (11)
 ${\alpha _2} = \arctan (\tan {\alpha _0}/\cos {\sigma _2})，$ (12)
 ${\omega _{02}} = \arctan (\sin {\alpha _0}\tan {\sigma _{02}})，$ (13)

 ${\lambda }_{01}={\omega }_{01}-f\mathrm{sin}{\alpha }_{0}{A}_{3}\left({\sigma }_{01}+{\displaystyle \sum _{l=1}^{\infty }{C}_{3l}\mathrm{sin}2l{\sigma }_{01}}\right)，$ (14)
 ${\lambda }_{02}={\omega }_{02}-f\mathrm{sin}{\alpha }_{0}{A}_{3}\left({\sigma }_{02}+{\displaystyle \sum _{l=1}^{\infty }{C}_{3l}\mathrm{sin}2l{\sigma }_{02}}\right)，$ (15)
 ${\lambda _{12}} = {\lambda _{02}} - {\lambda _{01}}，$ (16)
 ${\lambda _2} = {\lambda _1} + {\lambda _{12}}。$ (17)

 \begin{aligned}{A}_{3}= &1-\Biggr(\frac{1}{2}-\frac{n}{2}\Biggr)\varepsilon -\Biggr(\frac{1}{4}+\frac{n}{8}-\frac{3{n}^{2}}{8}\Biggr){\varepsilon }^{2}-\Biggr(\frac{1}{16}+\frac{3n}{16}+\frac{{n}^{2}}{16}\Biggr){\varepsilon }^{3}-\\ &\Biggr(\frac{3}{64}+\frac{n}{32}\Biggr){\varepsilon }^{4}-\frac{3}{128}{\varepsilon }^{5}+\cdots ，\end{aligned}
 \begin{aligned}{C}_{31}= &\Biggr(\frac{1}{4}-\frac{n}{4}\Biggr)\varepsilon +\Biggr(\frac{1}{8}-\frac{{n}^{2}}{8}\Biggr){\varepsilon }^{2}+\Biggr(\frac{3}{64}+\frac{3n}{64}-\frac{{n}^{2}}{64}\Biggr){\varepsilon }^{3}+\\ &\Biggr(\frac{5}{128}+\frac{n}{64}\Biggr){\varepsilon }^{4}+\frac{3}{128}{\varepsilon }^{5}+\cdots，\end{aligned}
 \begin{aligned}{C}_{32}= &\Biggr(\frac{1}{16}-\frac{3n}{32}+\frac{{n}^{2}}{32}\Biggr){\varepsilon }^{2}+\Biggr(\frac{3}{64}-\frac{n}{32}-\frac{3{n}^{2}}{64}\Biggr){\varepsilon }^{3}+\\ &\Biggr(\frac{3}{128}+\frac{n}{128}\Biggr){\varepsilon }^{4}+ \frac{5}{256}{\varepsilon }^{5}+\cdots，\end{aligned}
 ${C}_{33}=\Biggr(\frac{5}{192}-\frac{3n}{64}+\frac{5{n}^{2}}{192}\Biggr){\varepsilon }^{3}+\Biggr(\frac{3}{128}-\frac{5n}{192}\Biggr){\varepsilon }^{4}+\frac{7}{512}{\varepsilon }^{5}+\cdots ，$
 ${C}_{34}=\Biggr(\frac{7}{52}-\frac{7n}{256}\Biggr){\varepsilon }^{4}+\frac{7}{512}{\varepsilon }^{5}+\cdots，$
 ${C_{35}} = \frac{{21}}{{2560}}{\varepsilon ^5} + \cdots。$
1.2 反解 φ1, λ1, φ2, λ2, ${\boldsymbol{\Rightarrow}}$ α1, s12, α2

1.2.1 初始航向预估

 ${\alpha _1} = \arctan \Biggr( - \frac{{x/(1 + v)}}{{y/v}}\Biggr) 。$ (18)

 $x = \frac{{{\lambda _{12}} - \text{π} }}{{f \text{π} \cos {\beta _1}}},y = \frac{{{\beta _1} + {\beta _2}}}{{f \text{π} {{\cos }^2}{\beta _1}}},\frac{{{x^2}}}{{1 + {v^2}}} + \frac{{{y^2}}}{{{v^2}}} = 1。$ (19)
1.2.2 混合解

 ${\varphi _1} \leqslant 0,{\varphi _1} \leqslant {\varphi _2} \leqslant - {\varphi _1},0 \leqslant {\lambda _{12}} \leqslant \text{π}。$ (20)

 图 2 混合问题中起始纬度交换示意图 Fig. 2 Demonstration of swapping points in hybrid geodesic problem

 $\cos {\alpha _2} = \frac{{\sqrt {{{\cos }^2}{\alpha _1}{{\cos }^2}{\beta _1} + {{\cos }^2}{\beta _2} - {{\cos }^2}{\beta _1}} }}{{\cos {\beta _2}}}，$ (21)

 ${\sigma _{02}} = \arctan (\tan {\beta _2}/\cos {\alpha _2}) ，$ (22)
 ${\omega _{02}} = \arctan (\sin {\alpha _0}\tan {\sigma _{02}})。$ (23)

 $\delta {\lambda _{12}} = {\lambda _{12}} - \lambda _{12}^{(0)}，$ (24)

 $\delta {\alpha _1} = \frac{{ - \delta {\lambda _{12}}a\cos {\alpha _2}\cos {\beta _2}}}{{{\eta _{12}}b}} ，$ (25)
 \begin{aligned} {\eta _{12}} = &\cos {\sigma _{01}}\sin {\sigma _{02}}\sqrt {1 + {k^2}{{\sin }^2}{\sigma _{02}}} - \\ &\cos {\sigma _{02}}\sin {\sigma _{01}}\sqrt {1 + {k^2}{{\sin }^2}{\sigma _{01}}} - \\ & \cos {\sigma _{01}} \cos {\sigma _{02}}((J({\sigma _{02}}) - J({\sigma _{01}}))\theta ，\end{aligned} (26)
 $\begin{split}J({\sigma _{0i}}) = &{A_1}\left({\sigma _{0i}} + \sum\limits_{l = 1}^\infty {{C_{1l}}} \sin 1l{\sigma _{0i}}\right) -\\ &{A_2}\left({\sigma _{0i}} + \sum\limits_{l = 1}^\infty {{C_{2l}}} \sin 2l{\sigma _{0i}}\right)。\end{split}$ (27)

 ${A_2} = \dfrac{{\left(1 - \dfrac{3}{4}{\varepsilon ^2} - \dfrac{7}{{64}}{\varepsilon ^4} - \dfrac{{11}}{{256}}{\varepsilon ^6} + \cdots \right)}}{{(1 + \varepsilon )}} ，$
 $\begin{split}& {C_{21}} = \frac{1}{2}\varepsilon + \frac{1}{{16}}{\varepsilon ^3} - \frac{1}{{32}}{\varepsilon ^5} + \cdots，\\ &{C_{22}} = \frac{3}{{16}}{\varepsilon ^2} + \frac{1}{{32}}{\varepsilon ^4} + \frac{{35}}{{2048}}{\varepsilon ^6} + \cdots，\end{split}$
 ${C_{23}} = \frac{5}{{48}}{\varepsilon ^3} + \frac{5}{{256}}{\varepsilon ^5} + \cdots ，\qquad {C_{24}} = \frac{{35}}{{512}}{\varepsilon ^4} + \frac{7}{{512}}{\varepsilon ^6} + \cdots，$
 ${C_{25}} = \frac{{63}}{{1280}}{\varepsilon ^5} + \cdots ，\qquad {C_{26}} = \frac{{77}}{{2048}}{\varepsilon ^6} + \cdots。$

α1 = α1 + δα1作为新的初始航向，并重复上述迭代过程，直到δλ12趋于0。通常经过几次迭代后，经差误差便小于10−15

 ${s_{12}} = {s_{02}} - {s_{01}}。$ (28)

2 航向变更限制机制

ACRM旨在限制频繁的转向，新的舵令只有在满足一定条件后才会被触发，否则保持原航向。ACRM最常见的策略有但不限于：按一定角度/时间/距离转向。在模拟仿真中，GPS位置不是直接获得的，而是通过恒向线算法计算而得，且航向交替过程中没有时间损失和延迟。因此，仿真结果会与实际情况存在些许差异。仿真中，假定GPS刷新率为10 Hz，即船舶每0.1 s步长为1 m。

2.1 按一定角度转向

 图 3 ACRM执行流程图：按一定角度转向 Fig. 3 Diagram of ACRM process: alter course at a certain angle

2.2 按一定时间转向

2.3 按一定距离转向

 图 4 不同ACRM策略下的航线示意图 Fig. 4 Diagram of voyage under different ACRM strategies

3 结　语

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