﻿ 全回转拖轮操纵运动建模与仿真
 舰船科学技术  2017, Vol. 39 Issue (2): 114-120 PDF

1. 大连海事大学 航海动态仿真与控制交通行业重点实验室, 辽宁 大连 116026;
2. 交通运输部北海航海保障中心天津航标处, 天津 300456

Modeling and simulation on maneuvering motion of azimuth stern drive tug
YANG Jing1, REN Jun-sheng1, HUO Hu-wei2
1. Key Laboratory of Marine Simulation & Control, Dalian Maritime University, Dalian 116026, China;
2. Tianjin Aid to Navigation Station, NGCN, MOT, Tianjin 300456, China
Abstract: In order to improve the behavioural realism of tug ship-handling simulator, a ship motion mathematical model of azimuth tug is established by using the theory of maneuverability and separate modeling framework and this model is suitable for full-mission maneuvering range in four quadrants. By processing the ducted propeller JD75 character curves and four-quadrant propeller Nordstrom test curves, a ducted propeller thrust curve across four quadrants is put forward, then a hull hydrodynamic model for tug applicable to all speed range is proposed. The established simulation platform carries out a series of typical maneuverability tests, including speed trial test, spiral test, zigzag test, turning test and stopping test. In comparison with simulation results and test data provided by Force Technology company, the model is reasonable and satisfactory, and its simulation precision can meet the need of navigation.
Key words: azimuth stern drive tug     mathematical model     ducted propeller     maneuverability test

1 ASD 拖轮操纵运动数学模型

 \left\{ {\begin{aligned} & {(m+{m_x})\dot u - (m+{m_y})vr = {X_H}{\rm{+}}{X_P}}\text{，}\\ & {(m+{m_y})\dot v+(m+{m_x})ur = {Y_H}+{Y_P}}\text{，}\\ & {({I_{zz}}+{J_{zz}})\dot r = {N_H}+{N_P}}\text{。} \end{aligned}} \right. (1)

 图 1 ASD 拖轮水平面运动坐标系 Fig. 1 ASD tug horizontal plane coordinate system

ASD 拖轮操纵运动的轨迹和姿态方程如下，运用四阶龙格-库塔法解算：

 \left\{ {\begin{aligned} & {{{\dot x}_0} = u\cos \psi - v\sin \psi }\text{，}\\ & {{{\dot y}_0} = u\sin \psi+v\cos \psi }\text{，}\\ & {\dot \psi = r\cos \psi }\text{。} \end{aligned}} \right. (2)

2 拖船阻力模型 2.1 惯性力和粘性力

 \left\{ {\begin{aligned} & {{X_H} = X(u)+{X_{vv}}{v^2}+{X_{vr}}vr+{X_{rr}}{r^2}}\text{，}\\ & {{Y_H} = {Y_v}v+{Y_r}r+{Y_{|v|v}}|v|v+{Y_{|v|r}}|v|r+{Y_{|r|r}}|r|r}\text{，}\\ & {{N_H} = {N_v}v+{N_r}r+{N_{{\rm{|}}r|r}}|r|r+{N_{vvr}}{v^2}r+{N_{vrr}}v{r^2}}\text{。} \end{aligned}} \right. (3)

 ${X'_{{H}}} = X'(u')+{X'_{vv}}{v'^2}+{X'_{vr}}v'r'+{X'_{rr}}{r'^2}\text{。}$ (4)

2.2 直航阻力的计算

ASD 拖轮直航阻力的计算采用二因次法：

 $X'(u') \!=\! {X'_{uu}}{u'^2} \!=\! - \frac{S}{{Ld}}{C_{\rm{t}}}{u'^2} \!=\! - \frac{S}{{Ld}}\left( {{C_{\rm{f}}} \!+\! {C_{\rm{r}}} \!+\! \Delta {C_{\rm{f}}}} \right){u'^2}\text{。}$ (5)

 图 2 特格尔特拖轮剩余阻力曲线 Fig. 2 Target’s tug residual resistance curves

 图 3 ASD 拖轮塔格尔特剩余阻力曲线拟合 Fig. 3 ASD tug residual resistance curve fitting
3 导管桨推力模型 3.1 联合推力模型

ASD 拖轮操纵运动时，2 个导管桨操作方向相同或相反，横向力及转矩大小相同，方向相同或相反。其联合推力模型如下：

 \left\{ {\begin{aligned} & {{X_P} = {T_{(p)}}cos({\delta _{(p)}})+{T_{(s)}}cos({\delta _{(s)}})}\text{，}\\ & {{Y_P} = - {T_{(p)}}sin({\delta _{(p)}}) - {T_{(s)}}sin({\delta _{(s)}})}\text{，}\\ & {{N_P} \!=\! ({T_{(p)}}cos({\delta _{(p)}}) \!-\! {T_{(s)}}sin({\delta _{(s)}})) \cdot {L_{(ps)}}/2\! - \!\!{Y_P} \cdot {x_p}}\text{。} \end{aligned}} \right. (6)

3.2 导管桨四象限推力模型 3.2.1 导管桨第一象限推力系数计算模型

 \left\{ {\begin{aligned} & {T = (1 - {t_{\rm{p}}})\rho {n^2}{D^4}{K_{\rm{T}}}(J)}\text{，}\\ & {J = (1 - {w_{\rm{p}}})u/(nD)}\text{，}\\ & {{K_T} = \sum {\sum {{A_{ij}}{{(H/D)}^i}{{(J)}^j}} } }\text{，}\\ & {{K_P} = \sum {\sum {{B_{ij}}{{(H/D)}^i}{{(J)}^j}} } }\text{，}\\ & {{K_{TN}} = \sum {\sum {{C_{ij}}{{(H/D)}^i}{{(J)}^j}} } }\text{。} \end{aligned}} \right. (7)

 图 4 JD75 导管+Ka4-70 螺旋桨系列正车敞水试验 Fig. 4 JD75 Ducted+Ka4-70 propeller’s open character curves
3.2.2 四象限螺旋桨推力系数计算模型

 $\left\{ \begin{array}{l} P = (1 - {t_{\rm{p}}})\rho {D^2}({v_{\rm{P}}}^2+{n^2}{D^2}){K_{\rm{p}}}^\prime (J')\text{，}\\[5pt] {v_{{P}}} = (1 - {w_{\rm{P}}})u\text{，}\\[5pt] J' = {v_{\rm{P}}}/\sqrt {{v_{\rm{P}}}^2+{n^2}{D^2}} = \left\{ \begin{array}{l} J\sqrt {1+{J^2}} ;\;\;\;\;n > 0{\rm{ }}\text{，}\\ - J/\sqrt {1+{J^2}} ;\;n < 0{\rm{ }}\text{，} \end{array} \right.\\[5pt] {K_{{p}}}^\prime (J') = P/\rho {D^2}({v_{{P}}}^2+{n^2}{D^2}) = {K_{{p}}}/1+{J^2}\text{。} \end{array} \right.$ (8)

 图 5 四象限螺旋桨推力 ${K{'}_{{P}}} - J{'}$ 曲线图谱 Fig. 5 Four-quadrant propeller thrust ${K{'}_{{P}}} - J{'}$ curves
3.2.3 导管桨四象限推力系数计算模型

 ${x^\prime } = x/\sqrt {1+{x^2}} ;{y^\prime } = y/1+{x^2}\text{，}$ (9)

 图 6 导管桨第一象限 ${K'_{{T}}} - J'$ 曲线 Fig. 6 Ducted propeller first quadrant ${K'_{{T}}} - J'$ curve

 $\left\{ \begin{array}{l} \!\!\!\! a = {{y'}_0}/{y_1}\text{，}\\ \!\!\!\! {b_1} = {{x'}_0}/{x_1};\;\;\;\;\;\;\;\;\;\;\;{b_2} = \left( {1 - {{x'}_0}} \right)/\left( {1 - {x_1}} \right)\text{，}\\[5pt] \!\!\!\! n > 0,{\rm{ }}0 \leqslant J' < {x_1}{\rm{;}}\;\;x' = {b_1}x,{\rm{ }}y' = ay\text{，}\\[5pt] \!\!\!\! n \!\!> \!\!0,{\rm{ }}{x_1} \leqslant J'\!\! \leqslant \!\!1;\;\;\;x' \!\!=\!\! {{x'}_0}+{b_2}(x - {x_1}),{\rm{ }}y' \!\!=\!\! ay\text{，}\\[5pt] \!\!\!\! n > 0, - 1 \leqslant J' < 0;\;\;x' = x,{\rm{ }}y' = ay\text{，}\\[5pt] \!\!\!\! n < 0, - 1 \leqslant J' \leqslant 1;\;\;x' = x,{\rm{ }}y' = ay\text{。} \end{array} \right.$ (10)

 图 7 导管桨四象限推力 ${K'_{{T}}} - J'$ 曲线 Fig. 7 Ducted propeller four-quadrant thrust ${K'_{{T}}} - J'$ curve

 $T = (1 - {t_{{p}}})\rho {D^2}({v_{{P}}}^2+{n^2}{D^2}){K_{{T}}}^\prime (J')\text{。}$ (11)
4 操纵性试验及分析

 图 8 Modeler 仿真测试平台 Fig. 8 Modeler simulation platform
4.1 速度试航试验

4.2 螺线试验

 图 9 螺线试验 Fig. 9 Spiral test

4.3 Z 形试验

 图 10 Z 形试验 Fig. 10 Zigzag test
4.4 回转试验

 图 11 35°回转试验 Fig. 11 35° turning test

4.5 停船试验

 图 12 停船试验 Fig. 12 Stopping test

5 结 语

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