多波束测线问题
- 问题 1
- 建立模型
- 覆盖宽度
- 海水深度
- 重叠长度
- 重叠率
- 问题二
- 问题三
- 问题四
问题 1
与测线方向垂直的平面和海底坡面的交线构成一条与水平面夹角为
α
\alpha
α的斜线（如下图），称
α
\alpha
α为坡度。请建立多波束测深的覆盖宽度及相邻条带之间重叠率的数学模型。
若多波束换能器的开角为
120
120
120度，坡度为 1.5度，海域中心点处的海水深度为 70 m，利用上述模型计算表 1 中所列位置的指标值，将结果以表 1 的格式放在正文中，同时保存到result1.xlsx文件中。
建立模型
覆盖宽度
根据题目意思画出了另一个图：
根据上述示意图
其中
D
,
E
D,E
D,E表示船的不同位置，线段
D
E
DE
DE距离表示记为
d
d
d
假设
D
D
D点为中心点，由题可知线段
D
F
DF
DF长度
D
D
D即海域中心点处的海水深度为70m
设
A
F
AF
AF为
W
L
W_L
WL ；
F
H
FH
FH为
W
R
W_R
WR ； 覆盖宽度
A
H
AH
AH 为
W
W
W
那么
A
H
=
A
F
+
F
H
AH = AF + FH
AH=AF+FH 即
W
=
W
L
+
W
R
W = W_L + W_R
W=WL+WR
由正弦定理 a sin a = b sin b = c sin c \dfrac{a}{\sin a} =\dfrac{b}{\sin b}=\dfrac{c}{\sin c} sinaa=sinbb=sincc
在三角形AFD中有:
W L sin θ 2 = D sin x 1 \dfrac{W_L}{\sin \frac{\theta}{2}} =\dfrac{D}{\sin x_1} sin2θWL=sinx1D
π
=
θ
2
+
x
1
+
(
π
−
(
π
2
−
α
)
)
\pi = \dfrac{\theta}{2} + x_1 + (\pi - (\frac{\pi}{2}-\alpha))
π=2θ+x1+(π−(2π−α))
x
1
=
π
−
θ
2
−
π
+
π
2
−
α
x_1 = \pi - \dfrac{\theta}{2} - \pi + \frac{\pi}{2}-\alpha
x1=π−2θ−π+2π−α
解得
x
1
=
π
2
−
α
−
θ
2
x_1 = \dfrac{\pi}{2}-\alpha- \dfrac{\theta}{2}
x1=2π−α−2θ
W L sin θ 2 = D sin ( π 2 − α − θ 2 ) \dfrac{W_L}{\sin \frac{\theta}{2}} =\dfrac{D}{\sin (\frac{\pi}{2}-\alpha- \frac{\theta}{2})} sin2θWL=sin(2π−α−2θ)D
在三角形FDH中用正弦定理:
W R sin θ 2 = D sin x 2 \dfrac{W_R}{\sin \frac{\theta}{2}} =\dfrac{D}{\sin x_2} sin2θWR=sinx2D
π = θ 2 + x 2 + ( π 2 − α ) \pi = \dfrac{\theta}{2} + x_2 + (\frac{\pi}{2}-\alpha) π=2θ+x2+(2π−α)
解得
x 2 = π 2 + α − θ 2 x_2 = \dfrac{\pi}{2}+\alpha- \dfrac{\theta}{2} x2=2π+α−2θ
W R sin θ 2 = D sin ( π 2 + α − θ 2 ) \dfrac{W_R}{\sin \frac{\theta}{2}} =\dfrac{D}{\sin (\frac{\pi}{2}+\alpha- \frac{\theta}{2})} sin2θWR=sin(2π+α−2θ)D
总结:
W R sin θ 2 = D sin ( π 2 + α − θ 2 ) \dfrac{W_R}{\sin \frac{\theta}{2}} =\dfrac{D}{\sin (\frac{\pi}{2}+\alpha- \frac{\theta}{2})} sin2θWR=sin(2π+α−2θ)D
W L sin θ 2 = D sin ( π 2 − α − θ 2 ) \dfrac{W_L}{\sin \frac{\theta}{2}} =\dfrac{D}{\sin (\frac{\pi}{2}-\alpha- \frac{\theta}{2})} sin2θWL=sin(2π−α−2θ)D
覆盖宽度 W = W L + W R W = W_L + W_R W=WL+WR
海水深度
过点
U
U
U做
V
U
VU
VU平行于
F
G
FG
FG
由题已知角
G
F
C
GFC
GFC为
α
\alpha
α度
又内错角知识可得角
G
F
C
GFC
GFC 等于 角
C
U
V
CUV
CUV
线段
U
V
=
d
UV = d
UV=d，在三角形
C
U
V
CUV
CUV中由三角函数和图形可得
D
′
=
D
−
d
∗
t
a
n
α
D' = D - d*tan\alpha
D′=D−d∗tanα
所以已知测线距中心点处的距离
d
d
d就可以直接解出改点的海水深度
D
′
D'
D′
重叠长度
根据上图
在三角形
C
U
V
CUV
CUV中可得线段
C
U
CU
CU长度为
d
c
o
s
α
\dfrac{d}{cos \alpha}
cosαd
所以重叠长度 C = W R 1 + W L 2 − d c o s α C = WR1 + WL2 - \dfrac{d}{cos \alpha} C=WR1+WL2−cosαd
总结:
重叠长度
C
=
W
R
i
+
W
L
i
+
1
−
d
c
o
s
α
C = W_{R_i} + W_{L_{i+1}} - \dfrac{d}{cos \alpha}
C=WRi+WLi+1−cosαd
重叠率
重叠率为 η = C W 1 + W 2 − C \eta = \dfrac{C}{W_1+W_2-C} η=W1+W2−CC
问题二
问题三
重叠率为
η
=
C
W
1
+
W
2
−
C
\eta = \dfrac{C}{W_1+W_2-C}
η=W1+W2−CC
因为
η
=
0.1
\eta = 0.1
η=0.1
已知
W
1
,
W
R
1
W_1,W_{R_1}
W1,WR1 , 求其他未知量
因为
C
=
W
R
i
+
W
L
i
+
1
−
d
c
o
s
α
C = W_{R_i} + W_{L_{i+1}} - \dfrac{d}{cos \alpha}
C=WRi+WLi+1−cosαd
0.1
=
W
R
1
+
W
L
2
−
d
c
o
s
α
W
1
+
W
2
−
(
W
R
1
+
W
L
2
−
d
c
o
s
α
)
0.1 = \dfrac{W_{R_1} + W_{L_{2}} - \dfrac{d}{cos \alpha}}{W_1+W_2-(W_{R_1} + W_{L_{2}} - \dfrac{d}{cos \alpha})}
0.1=W1+W2−(WR1+WL2−cosαd)WR1+WL2−cosαd
又因为
W R = s i n θ 2 D s i n ( π 2 + α − θ 2 ) W_R =sin \frac{\theta}{2}\dfrac{D}{sin (\frac{\pi}{2}+\alpha- \frac{\theta}{2})} WR=sin2θsin(2π+α−2θ)D
W L = s i n θ 2 D s i n ( π 2 − α − θ 2 ) W_L =sin \frac{\theta}{2} \dfrac{D}{sin (\frac{\pi}{2}-\alpha- \frac{\theta}{2})} WL=sin2θsin(2π−α−2θ)D
W = W L + W R W = W_L + W_R W=WL+WR
D i = D − d ∗ t a n α D_i = D - d*tan\alpha Di=D−d∗tanα
d = D 中心 − D i t a n α d = \dfrac{D_{中心} - D_i}{tan\alpha} d=tanαD中心−Di
已知$D_{中心} = 110 m $
可得
0.1
=
W
R
1
+
s
i
n
θ
2
D
2
s
i
n
(
π
2
−
α
−
θ
2
)
−
∣
D
1
−
D
2
∣
t
a
n
α
c
o
s
α
W
1
+
s
i
n
θ
2
D
2
s
i
n
(
π
2
−
α
−
θ
2
)
+
s
i
n
θ
2
D
2
s
i
n
(
π
2
+
α
−
θ
2
)
−
(
W
R
1
+
s
i
n
θ
2
D
2
s
i
n
(
π
2
−
α
−
θ
2
)
−
∣
D
1
−
D
2
∣
t
a
n
α
c
o
s
α
)
0.1 = \dfrac{W_{R_1} + sin \frac{\theta}{2} \dfrac{D_2}{sin (\frac{\pi}{2}-\alpha- \frac{\theta}{2})} - \dfrac{\frac{|D_1-D_2|}{tan\alpha}}{cos \alpha}}{W_1+sin \frac{\theta}{2} \dfrac{D_2}{sin (\frac{\pi}{2}-\alpha- \frac{\theta}{2})} + sin \frac{\theta}{2}\dfrac{D_2}{sin (\frac{\pi}{2}+\alpha- \frac{\theta}{2})} -(W_{R_1} + sin \frac{\theta}{2} \dfrac{D_2}{sin (\frac{\pi}{2}-\alpha- \frac{\theta}{2})} - \dfrac{\frac{|D_1-D_2|}{tan\alpha}}{cos \alpha})}
0.1=W1+sin2θsin(2π−α−2θ)D2+sin2θsin(2π+α−2θ)D2−(WR1+sin2θsin(2π−α−2θ)D2−cosαtanα∣D1−D2∣)WR1+sin2θsin(2π−α−2θ)D2−cosαtanα∣D1−D2∣
0.1 ∗ ( W 1 + s i n θ 2 D 2 s i n ( π 2 − α − θ 2 ) + s i n θ 2 D 2 s i n ( π 2 + α − θ 2 ) − ( W R 1 + s i n θ 2 D 2 s i n ( π 2 − α − θ 2 ) − ∣ D 1 − D 2 ∣ t a n α c o s α ) ) = W R 1 + s i n θ 2 D 2 s i n ( π 2 − α − θ 2 ) − ∣ D 1 − D 2 ∣ t a n α c o s α 0.1*(W_1+sin \frac{\theta}{2} \dfrac{D_2}{sin (\frac{\pi}{2}-\alpha- \frac{\theta}{2})} + sin \frac{\theta}{2}\dfrac{D_2}{sin (\frac{\pi}{2}+\alpha- \frac{\theta}{2})} -(W_{R_1} + sin \frac{\theta}{2} \dfrac{D_2}{sin (\frac{\pi}{2}-\alpha- \frac{\theta}{2})} - \dfrac{\frac{|D_1-D_2|}{tan\alpha}}{cos \alpha})) = W_{R_1} + sin \frac{\theta}{2} \dfrac{D_2}{sin (\frac{\pi}{2}-\alpha- \frac{\theta}{2})} - \dfrac{\frac{|D_1-D_2|}{tan\alpha}}{cos \alpha} 0.1∗(W1+sin2θsin(2π−α−2θ)D2+sin2θsin(2π+α−2θ)D2−(WR1+sin2θsin(2π−α−2θ)D2−cosαtanα∣D1−D2∣))=WR1+sin2θsin(2π−α−2θ)D2−cosαtanα∣D1−D2∣
因为 D 1 ≥ D 2 D_1 \ge D_2 D1≥D2
0.1 s i n θ 2 D 2 s i n ( π 2 − α − θ 2 ) + 0.1 s i n θ 2 D 2 s i n ( π 2 + α − θ 2 ) − 0.1 s i n θ 2 D 2 s i n ( π 2 − α − θ 2 ) + 0.1 D 1 t a n α c o s α − 0.1 D 2 t a n α c o s α − s i n θ 2 D 2 s i n ( π 2 − α − θ 2 ) + D 1 t a n α c o s α − D 2 t a n α c o s α = W R 1 − 0.1 W 1 + 0.1 W R 1 0.1sin \frac{\theta}{2} \dfrac{D_2}{sin (\frac{\pi}{2}-\alpha- \frac{\theta}{2})} + 0.1sin \frac{\theta}{2}\dfrac{D_2}{sin (\frac{\pi}{2}+\alpha- \frac{\theta}{2})} -0.1sin \frac{\theta}{2} \dfrac{D_2}{sin (\frac{\pi}{2}-\alpha- \frac{\theta}{2})} +0.1\dfrac{D_{1}}{tan\alpha cos \alpha}-0.1\dfrac{D_{2}}{tan\alpha cos \alpha}- sin \frac{\theta}{2} \dfrac{D_2}{sin (\frac{\pi}{2}-\alpha- \frac{\theta}{2})} + \dfrac{D_{1}}{tan\alpha cos \alpha}-\dfrac{D_{2}}{tan\alpha cos \alpha} = W_{R_1} - 0.1W_1 + 0.1W_{R_1} 0.1sin2θsin(2π−α−2θ)D2+0.1sin2θsin(2π+α−2θ)D2−0.1sin2θsin(2π−α−2θ)D2+0.1tanαcosαD1−0.1tanαcosαD2−sin2θsin(2π−α−2θ)D2+tanαcosαD1−tanαcosαD2=WR1−0.1W1+0.1WR1
0.1 s i n θ 2 D 2 s i n ( π 2 − α − θ 2 ) + 0.1 s i n θ 2 D 2 s i n ( π 2 + α − θ 2 ) − 0.1 s i n θ 2 D 2 s i n ( π 2 − α − θ 2 ) − 0.1 D 2 t a n α c o s α − s i n θ 2 D 2 s i n ( π 2 − α − θ 2 ) − D 2 t a n α c o s α = W R 1 − 0.1 W 1 + 0.1 W R 1 − 0.1 D 1 t a n α c o s α − D 1 t a n α c o s α 0.1sin \frac{\theta}{2} \dfrac{D_2}{sin (\frac{\pi}{2}-\alpha- \frac{\theta}{2})} + 0.1sin \frac{\theta}{2}\dfrac{D_2}{sin (\frac{\pi}{2}+\alpha- \frac{\theta}{2})} -0.1sin \frac{\theta}{2} \dfrac{D_2}{sin (\frac{\pi}{2}-\alpha- \frac{\theta}{2})} -0.1\dfrac{D_{2}}{tan\alpha cos \alpha}- sin \frac{\theta}{2} \dfrac{D_2}{sin (\frac{\pi}{2}-\alpha- \frac{\theta}{2})} -\dfrac{D_{2}}{tan\alpha cos \alpha} = W_{R_1} - 0.1W_1 + 0.1W_{R_1}-0.1\dfrac{D_{1}}{tan\alpha cos \alpha}- \dfrac{D_{1}}{tan\alpha cos \alpha} 0.1sin2θsin(2π−α−2θ)D2+0.1sin2θsin(2π+α−2θ)D2−0.1sin2θsin(2π−α−2θ)D2−0.1tanαcosαD2−sin2θsin(2π−α−2θ)D2−tanαcosαD2=WR1−0.1W1+0.1WR1−0.1tanαcosαD1−tanαcosαD1
D 2 ( 0.1 s i n θ 2 1 s i n ( π 2 + α − θ 2 ) − s i n θ 2 1 s i n ( π 2 − α − θ 2 ) − 0.1 1 t a n α c o s α − 1 t a n α c o s α ) = W R 1 − 0.1 W 1 + 0.1 W R 1 − 0.1 D 1 t a n α c o s α − D 1 t a n α c o s α D_2( 0.1sin \frac{\theta}{2}\dfrac{1}{sin (\frac{\pi}{2}+\alpha- \frac{\theta}{2})} -sin \frac{\theta}{2} \dfrac{1}{sin (\frac{\pi}{2}-\alpha- \frac{\theta}{2})} -0.1\dfrac{1}{tan\alpha cos \alpha}-\dfrac{1}{tan\alpha cos \alpha}) = W_{R_1} - 0.1W_1 + 0.1W_{R_1}-0.1\dfrac{D_{1}}{tan\alpha cos \alpha}- \dfrac{D_{1}}{tan\alpha cos \alpha} D2(0.1sin2θsin(2π+α−2θ)1−sin2θsin(2π−α−2θ)1−0.1tanαcosα1−tanαcosα1)=WR1−0.1W1+0.1WR1−0.1tanαcosαD1−tanαcosαD1
D 2 = W R 1 − 0.1 W 1 + 0.1 W R 1 − 0.1 D 1 t a n α c o s α − D 1 t a n α c o s α ( 0.1 s i n θ 2 1 s i n ( π 2 + α − θ 2 ) − s i n θ 2 1 s i n ( π 2 − α − θ 2 ) − 0.1 1 t a n α c o s α − 1 t a n α c o s α ) D_2 =\dfrac{ W_{R_1} - 0.1W_1 + 0.1W_{R_1}-0.1\dfrac{D_{1}}{tan\alpha cos \alpha}- \dfrac{D_{1}}{tan\alpha cos \alpha}}{( 0.1sin \frac{\theta}{2}\dfrac{1}{sin (\frac{\pi}{2}+\alpha- \frac{\theta}{2})} -sin \frac{\theta}{2} \dfrac{1}{sin (\frac{\pi}{2}-\alpha- \frac{\theta}{2})} -0.1\dfrac{1}{tan\alpha cos \alpha}-\dfrac{1}{tan\alpha cos \alpha})} D2=(0.1sin2θsin(2π+α−2θ)1−sin2θsin(2π−α−2θ)1−0.1tanαcosα1−tanαcosα1)WR1−0.1W1+0.1WR1−0.1tanαcosαD1−tanαcosαD1
问题四
W R sin θ 2 = D sin ( π 2 + α − θ 2 ) \dfrac{W_R}{\sin \frac{\theta}{2}} =\dfrac{D}{\sin (\frac{\pi}{2}+\alpha- \frac{\theta}{2})} sin2θWR=sin(2π+α−2θ)D
W L sin θ 2 = D sin ( π 2 − α − θ 2 ) \dfrac{W_L}{\sin \frac{\theta}{2}} =\dfrac{D}{\sin (\frac{\pi}{2}-\alpha- \frac{\theta}{2})} sin2θWL=sin(2π−α−2θ)D
覆盖宽度 W = W L + W R W = W_L + W_R W=WL+WR
D
′
=
D
−
d
∗
t
a
n
α
D' = D - d*tan\alpha
D′=D−d∗tanα
所以
α
=
arctan
D
−
D
′
d
\alpha = \arctan{\frac{D - D'}{d}}
α=arctandD−D′
W L = D sin ( π 2 − α − θ 2 ) sin θ 2 W_L =\dfrac{D}{\sin (\frac{\pi}{2}-\alpha- \frac{\theta}{2})}\sin \frac{\theta}{2} WL=sin(2π−α−2θ)Dsin2θ
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