题目
说明积分曲线、重积分曲线与原曲线的关系.并以水线面面积曲线为例说明积分曲线、重积分曲线的应用。⏺Exercise 1-1某海洋客船L=155m,B=18m,d=7.1m,V=10900m3,Am=115m2,Aw=1980m2。试求Cb, Cp, Cw, Cm, Cvp。已知: L=155m,B=18m,d=7.1m,V=10900m3,Am=115m2,Aw=1980m2求:Cb=V/LBd=10.00/(155*18*7.1)=0.550 Cp=V/Lam=10900/(155*115)=0.62Cw=Aw/BL=1980./(18*155)=0.710Cm=Am/Bd=115/(18*7.1)=0.900Exercise 2-4已知:海船中横剖面矩形,Lw=128m.Bm=15.2m.d(m)CwAwTpcAwZAwAwZ6.10.80156..4815.9595.5.53----------4.88.781517.5715.567405.741517.577405.743.66.72140..8314.365127.04140..835127.042.44.621206.2712.362943.301206.272943.301.22.2446..944.79269.6746..94269.670.00.047..820.807..82Σ6205.9125241.284639.4315.85.75ε817.1547.7.777.7.703702.87Σ’538..762046.523841.7411982.88Exercise 2-4d(m)CwAwTpcAwZAwAwZΣ’538..762046.523841.7411982.88最高水线:V=Σ’δd=538..76*1.22=6574.29 (m3)Δ=ωV=1.025*6574.29=6738.64 (t)Zb=20463.52/538..76=3.797 (m)Cb=V/LBd=6.br>5.4.29/(128*5.2*6.1)=0.553Cp=V/AmL= 6.br>5.4.29/(15.2*6.1*128)=0.554Cvp=V/Awd= 6.74.29/(1566.48*6.1)=0.692Exercise 2-4d(m)CwAwTpcAwZAwAwZΣ’538..762046.523841.7411982.88次高水线:V=Σ’δd=3841.47*1.22=46.6.59 (m3)Δ=ωV=1.025*46..59=4803.76 (t)Zb=1.982.88/ 3.41.47 =3.119 (m)⏺Exercise 2-5某船L=60m,其水线以下横剖面为半圆形(其形心距水面高为4./3.),从尾向首其半宽为0.3, 1.6, 4.3, 5.0, 4.6, 3.3m,试求水线面漂心xF,排水体积▽,浮心xb, zb,方形系数Cb。用梯形法解:δL=60/5=12 my0.300 1.600 4.300 5.000 4.600 3.300 19.100x-30.000 -18.000 -6.000 6.000 18.000 30.000 yx-9.000 -28.800 -25.800 30.000 82.800 9..000 148.200A0.141 4.021 29.044 39.270 3..238 17.106 12..820z0.127 0.679 1.825 2.122 1.952 1.401 Az0.018 2.731 53.005 83.333 64.891 23.958 227.935Ax-4.241 -72.382 -174.264 235.619 598.284 513.179 1096.195Aw=2.1.*(∑-ε)=2*12*(19.1-(0.3.3.3)/2)=4.5.2 m2 xF=12*(148.2-(-9+99)/2)/207.6.5.965 m ▽=12*(122.82-(0.141+17.106)/2)=1270.362 m3 Zb’=12*(227.935-(0.018+23.958)/2)/1270.362=1.891 m xb=12*(1096.195-(-4.241+513.179)/2)/1270.362=7.371 mExercise 2-6某船L=60.,其水线以下横剖面为等边三角形,从尾向首其半宽为0.3. 1.6, 4.3, 5.0, 4.6, 3.3m,试求水线面漂心xF,排水体积▽,浮心xb, zb,方形系数Cb。用梯形法解:δL=60/5=12 my0.300 1.600 4.300 5.000 4.600 3.300 19.100x-30.000 -18.000 -6.000 6.000 18.000 30.000 yx-9.000 -28.800 -25.800 30.000 82.800 9..000 148.200A0.156 4.429 31.988 43.250 36.607 18.840 135.269z0.173 0.925 2.486 2.890 2.659 1.908 Az0.027 4.096 79.507 125.000 97.336 35.937 341.903Ax-4.671 -79.718 -1.1.926 259.500 658.922 5.5.191 1207.298Aw=2*12*(∑-ε)=2*12*(19.1-(0.3.3.3)/2)=415.2 m2xF=1.*(4.4.8.(18+0./2)/207.6=24.03.br>5.m ▽=12*(135.269-(0.156+18.84)/2)=810.678 m3 Zb’=12*(341.903-(0.027+35.937)/2)/810.678 =4.795 m xb=12*(1207.298-(-4.671+565.191)/2)/810.678 =13.722 m⏺Exercise 2-7某船的一个煤舱长为24m,自尾至首各横剖面面积为5.7,8.7,1..3.10.1,8.8(单位m2)这些剖面的形心在基线以上的高度分别为3.7,3.5,3.3,3.5,3.6(单位 m)。剖面之间的间距为6m。设煤舱的积载因数(每吨煤所占体积m3)为1.56m3/t。试列表计算:(1)该船载煤吨数,(2)该舱的重心位置(基线以上距离及距煤舱尾舱壁的距离)。煤舱L=24m,Asi方法1方法2方法3计算ZXiXiAsiXiXiAsiXiXiAsiZiZiAsi5.7-12-68.4-2-1..43.721.0918.7-6-52.2652.2-1-8.73.530.4521..312135.63.337.29310.1660.6181.1.8110.13.535.3548.812105.62421..2217.63.631.68Σ4..645.6580.87.615..86ε7.2518.6105.63.126.385Σ’37.3527.0475.24.5129.475AsAsXAsX AsXAsZΣ’37.3527.0475.24.5129.475V=δx*Σ’ = 6*37.35 = 224.1 m3P=V/1.56 = 224.1/1.56 =143.65 (t)X’g= Σ’AsX*δx / Σ’As*δx =27.37.35 =0.723 (m)Xg=12.0.723=12.723 (m)Zg= Σ’AsZ*δx / Σ’As*δx =129.47./3..35 =3.466 (m)位置AsXiAsXiZiAsZi辛氏数5.73.721.09118.7652.23.530.45421..312135.63.337.242310.1181.1.83.535.35448.82421..23.631.681Σ加权112.31418.3390.55V=2./12*112.3=224.6 m3Xg=(2.br>4.12*1418.3)/224.6=12.630 mP=224.6/1.56=143.97 tZg=(24.12*3.0.55)/224.6=3.478 m⏺某海船各水线的排水量为10804, 8612, 6511, 4550, 2810,1.31, 263 t。各水线间距为1.2m。求吃水7.8m时的浮心垂某海船各水线面面积为20., 185, 160, 125, 30m2,试求该水线面的每厘米吃水吨数TPC,并绘制TPC曲线。如水线间距为0.5m,求排水体积▽,并绘制V=f(d)曲线(列表计算)。某海船水线面面积如表,水线间距为δd=1.1m,用梯形法列表计算排水量曲线,每厘米吃水吨数曲线。并绘图。某货船在A港吃水d=5.35m,进入B港其吃水不能超过d1=4.6m,船在d2=5.5m时TPC2=18.6t/cm,在d3=4.5m时TPC3=14.8t,假定TPC对于吃水的变化是一直线,求进入B港前必须卸下的货物重量。已知:d=4.6m:Tpc=1.4.br>8.(18.6-14.8)/1*(4.6-4.5.=15.18 t/cm d=5.35m:Tpc=1.4.+3.8.(5.35-4.5)=18.03 t/cm p=(15.18+18.03)/2*100*(5.35-4.6)=1245.4 (t)某船船长L=164m,船宽B=19.7m,方形系数Cb=0.50,水线面系数Cw=<b.>0.73,在海水中平均吃水d=8.2m,求船进入淡水中的平均吃水。已知:L=1.4m,B=19.7m,Cb=0.50,Cw=0.73,海水d=8.20m δd =(1.025-1.0)Cb.LBd/CwLB= (<b.>1.025-1.0)Cb.d/Cw=(1.0.5-1.0)*0.50*8.2/0.73=0.140 (m) d(淡)=8.20+0.14=8.34 m δd/d=-Cb/Cw*dω/ω δd=-0.5/0.73*(1.025-1.0)/1.025*8.2=0.137 m d=8.2+0.137=8.337 m⏺Exercise 2-13某船由淡水进入海水,必须增加载荷p=175 t,才能使其在海水中的吃水与淡水中的吃水相等。求增加载重后的排水量。
说明积分曲线、重积分曲线与原曲线的关系.并以水线面面积曲线为例说明积分曲线、重积分曲线的应用。⏺Exercise 1-1某海洋客船L=155m,B=18m,d=
7.1m,V=10900m3,Am=115m2,Aw=1980m2。试求Cb, Cp, Cw, Cm, Cvp。已知: L=155m,B=18m,d=
7.1m,V=10900m3,Am=115m2,Aw=1980m2求:Cb=V/LBd=1
0.00/(155*18*
7.1)=0.550 Cp=V/Lam=10900/(155*115)=0.62Cw=Aw/BL=198
0./(18*155)=0.710Cm=Am/Bd=115/(18*
7.1)=
0.900Exercise 2-4已知:海船中横剖面矩形,Lw=128
m.Bm=1
5.2m.d(m)CwAwTpcAwZAwAwZ
6.10.8015
6..481
5.959
5.5.53----------
4.88.78151
7.571
5.56740
5.74151
7.57740
5.74
3.66.7214
0..831
4.36512
7.0414
0..83512
7.04
2.44.62120
6.271
2.36294
3.30120
6.27294
3.30
1.22.244
6..94
4.7926
9.674
6..9426
9.67
0.00.04
7..82
0.80
7..82Σ620
5.912524
1.28463
9.431
5.85.75ε81
7.154
7.7.77
7.7.70370
2.87Σ’53
8..76204
6.52384
1.741198
2.88Exercise 2-4d(m)CwAwTpcAwZAwAwZΣ’53
8..76204
6.52384
1.741198
2.88最高水线:V=Σ’δd=53
8..76*
1.22=657
4.29 (m3)Δ=ωV=
1.025*657
4.29=673
8.64 (t)Zb=2046
3.52/53
8..76=3.797 (m)Cb=V/LBd=
6.br>5.
4.29/(128*5.2*6.1)=
0.553Cp=V/AmL=
6.br>5.
4.29/(15.2*6.1*128)=
0.554Cvp=V/Awd=
6.7
4.29/(1566.48*6.1)=
0.692Exercise 2-4d(m)CwAwTpcAwZAwAwZΣ’53
8..76204
6.52384
1.741198
2.88次高水线:V=Σ’δd=384
1.47*1.22=4
6.6.59 (m3)Δ=ωV=
1.025*4
6..59=480
3.76 (t)Zb=
1.98
2.88/
3.41.47 =3.119 (m)⏺Exercise 2-5某船L=60m,其水线以下横剖面为半圆形(其形心距水面高为
4./
3.),从尾向首其半宽为
0.3,
1.6, 4.3,
5.0, 4.6, 3.3m,试求水线面漂心xF,排水体积▽,浮心xb, zb,方形系数Cb。用梯形法解:δL=60/5=12 my
0.300
1.600
4.300
5.000
4.600
3.300 1
9.100x-3
0.000 -1
8.000 -
6.000
6.000 1
8.000 3
0.000 yx-
9.000 -2
8.800 -2
5.800 3
0.000 8
2.800
9..000 14
8.200A
0.141
4.021 2
9.044 3
9.270
3..238 1
7.106 1
2..820z
0.127
0.679
1.825
2.122
1.952
1.401 Az
0.018
2.731 5
3.005 8
3.333 6
4.891 2
3.958 22
7.935Ax-
4.241 -7
2.382 -17
4.264 23
5.619 59
8.284 51
3.179 109
6.195Aw=
2.
1.*(∑-ε)=2*12*(1
9.1-(
0.
3.3.3)/2)=
4.
5.2 m2 xF=12*(14
8.2-(-9+99)/2)/20
7.
6.5.965 m ▽=12*(122.82-(0.141+17.106)/2)=1270.362 m3 Zb’=12*(227.935-(0.018+23.958)/2)/1270.362=1.891 m xb=12*(1096.195-(-4.241+513.179)/2)/1270.362=7.371 mExercise 2-6某船L=6
0.,其水线以下横剖面为等边三角形,从尾向首其半宽为0.
3.
1.6,
4.3,
5.0, 4.6, 3.3m,试求水线面漂心xF,排水体积▽,浮心xb, zb,方形系数Cb。用梯形法解:δL=60/5=12 my
0.300
1.600
4.300
5.000
4.600
3.300 1
9.100x-3
0.000 -1
8.000 -
6.000
6.000 1
8.000 3
0.000 yx-
9.000 -2
8.800 -2
5.800 3
0.000 8
2.800
9..000 14
8.200A
0.156
4.429 3
1.988 4
3.250 3
6.607 1
8.840 13
5.269z
0.173
0.925
2.486
2.890
2.659
1.908 Az
0.027
4.096 7
9.507 12
5.000 9
7.336 3
5.937 34
1.903Ax-
4.671 -7
9.718 -
1.1.926 25
9.500 65
8.922
5.5.191 120
7.298Aw=2*12*(∑-ε)=2*12*(1
9.1-(
0.
3.3.3)/2)=41
5.2 m2xF=
1.*(
4.4.
8.(18+
0./2)/20
7.6=24.0
3.br>
5.m ▽=12*(135.269-(0.156+18.84)/2)=810.678 m3 Zb’=12*(341.903-(0.027+35.937)/2)/810.678 =4.795 m xb=12*(1207.298-(-4.671+565.191)/2)/810.678 =13.722 m⏺Exercise 2-7某船的一个煤舱长为24m,自尾至首各横剖面面积为
5.7,
8.7,
1..
3.1
0.1,8.8(单位m2)这些剖面的形心在基线以上的高度分别为3.7,3.5,3.3,3.5,3.6(单位 m)。剖面之间的间距为6m。设煤舱的积载因数(每吨煤所占体积m3)为1.56m3/t。试列表计算:(1)该船载煤吨数,(2)该舱的重心位置(基线以上距离及距煤舱尾舱壁的距离)。煤舱L=24m,Asi方法1方法2方法3计算ZXiXiAsiXiXiAsiXiXiAsiZiZiAsi
5.7-12-6
8.4-2-
1..4
3.72
1.091
8.7-6-5
2.265
2.2-1-
8.7
3.53
0.452
1..31213
5.6
3.33
7.2931
0.166
0.618
1.1.811
0.1
3.53
5.354
8.81210
5.6242
1..221
7.6
3.63
1.68Σ
4..64
5.658
0.8
7.61
5..86ε
7.251
8.610
5.6
3.12
6.385Σ’3
7.352
7.047
5.2
4.512
9.475AsAsXAsX AsXAsZΣ’3
7.352
7.047
5.2
4.512
9.475V=δx*Σ’ = 6*3
7.35 = 22
4.1 m3P=V/
1.56 = 22
4.1/1.56 =14
3.65 (t)X’g= Σ’AsX*δx / Σ’As*δx =2
7.37.35 =
0.723 (m)Xg=1
2.
0.723=12.723 (m)Zg= Σ’AsZ*δx / Σ’As*δx =12
9.4
7./
3..35 =3.466 (m)位置AsXiAsXiZiAsZi辛氏数
5.7
3.72
1.0911
8.765
2.2
3.53
0.4542
1..31213
5.6
3.33
7.24231
0.118
1.1.8
3.53
5.3544
8.8242
1..2
3.63
1.681Σ加权11
2.3141
8.339
0.55V=
2./12*112.3=22
4.6 m3Xg=(
2.br>4.12*141
8.3)/224.6=12.630 mP=22
4.6/
1.56=14
3.97 tZg=(2
4.12*
3.
0.55)/224.6=3.478 m⏺某海船各水线的排水量为10804, 8612, 6511, 4550, 2810,
1.31, 263 t。各水线间距为1.2m。求吃水
7.8m时的浮心垂某海船各水线面面积为2
0., 185, 160, 125, 30m2,试求该水线面的每厘米吃水吨数TPC,并绘制TPC曲线。如水线间距为0.5m,求排水体积▽,并绘制V=f(d)曲线(列表计算)。某海船水线面面积如表,水线间距为δd=
1.1m,用梯形法列表计算排水量曲线,每厘米吃水吨数曲线。并绘图。某货船在A港吃水d=
5.35m,进入B港其吃水不能超过d1=
4.6m,船在d2=
5.5m时TPC2=1
8.6t/cm,在d3=4.5m时TPC3=1
4.8t,假定TPC对于吃水的变化是一直线,求进入B港前必须卸下的货物重量。已知:d=4.6m:Tpc=
1.
4.br>8.(18.6-14.8)/1*(4.6-4.
5.=15.18 t/cm d=5.35m:Tpc=
1.
4.+
3.
8.(
5.35-4.5)=18.03 t/cm p=(15.18+18.03)/2*100*(5.35-4.6)=1245.4 (t)某船船长L=164m,船宽B=1
9.7m,方形系数Cb=
0.50,水线面系数Cw=<
b.>
0.73,在海水中平均吃水d=
8.2m,求船进入淡水中的平均吃水。已知:L=
1.4m,B=1
9.7m,Cb=0.50,Cw=0.73,海水d=8.20m δd =(1.025-1.0)Cb.LBd/CwLB= (<
b.>
1.025-1.0)Cb.d/Cw=(
1.
0.5-1.0)*0.50*
8.2/0.73=0.140 (m) d(淡)=8.20+0.14=8.34 m δd/d=-Cb/Cw*dω/ω δd=-0.5/0.73*(1.025-1.0)/1.025*8.2=0.137 m d=8.2+0.137=8.337 m⏺Exercise 2-13某船由淡水进入海水,必须增加载荷p=175 t,才能使其在海水中的吃水与淡水中的吃水相等。求增加载重后的排水量。
7.1m,V=10900m3,Am=115m2,Aw=1980m2。试求Cb, Cp, Cw, Cm, Cvp。已知: L=155m,B=18m,d=
7.1m,V=10900m3,Am=115m2,Aw=1980m2求:Cb=V/LBd=1
0.00/(155*18*
7.1)=0.550 Cp=V/Lam=10900/(155*115)=0.62Cw=Aw/BL=198
0./(18*155)=0.710Cm=Am/Bd=115/(18*
7.1)=
0.900Exercise 2-4已知:海船中横剖面矩形,Lw=128
m.Bm=1
5.2m.d(m)CwAwTpcAwZAwAwZ
6.10.8015
6..481
5.959
5.5.53----------
4.88.78151
7.571
5.56740
5.74151
7.57740
5.74
3.66.7214
0..831
4.36512
7.0414
0..83512
7.04
2.44.62120
6.271
2.36294
3.30120
6.27294
3.30
1.22.244
6..94
4.7926
9.674
6..9426
9.67
0.00.04
7..82
0.80
7..82Σ620
5.912524
1.28463
9.431
5.85.75ε81
7.154
7.7.77
7.7.70370
2.87Σ’53
8..76204
6.52384
1.741198
2.88Exercise 2-4d(m)CwAwTpcAwZAwAwZΣ’53
8..76204
6.52384
1.741198
2.88最高水线:V=Σ’δd=53
8..76*
1.22=657
4.29 (m3)Δ=ωV=
1.025*657
4.29=673
8.64 (t)Zb=2046
3.52/53
8..76=3.797 (m)Cb=V/LBd=
6.br>5.
4.29/(128*5.2*6.1)=
0.553Cp=V/AmL=
6.br>5.
4.29/(15.2*6.1*128)=
0.554Cvp=V/Awd=
6.7
4.29/(1566.48*6.1)=
0.692Exercise 2-4d(m)CwAwTpcAwZAwAwZΣ’53
8..76204
6.52384
1.741198
2.88次高水线:V=Σ’δd=384
1.47*1.22=4
6.6.59 (m3)Δ=ωV=
1.025*4
6..59=480
3.76 (t)Zb=
1.98
2.88/
3.41.47 =3.119 (m)⏺Exercise 2-5某船L=60m,其水线以下横剖面为半圆形(其形心距水面高为
4./
3.),从尾向首其半宽为
0.3,
1.6, 4.3,
5.0, 4.6, 3.3m,试求水线面漂心xF,排水体积▽,浮心xb, zb,方形系数Cb。用梯形法解:δL=60/5=12 my
0.300
1.600
4.300
5.000
4.600
3.300 1
9.100x-3
0.000 -1
8.000 -
6.000
6.000 1
8.000 3
0.000 yx-
9.000 -2
8.800 -2
5.800 3
0.000 8
2.800
9..000 14
8.200A
0.141
4.021 2
9.044 3
9.270
3..238 1
7.106 1
2..820z
0.127
0.679
1.825
2.122
1.952
1.401 Az
0.018
2.731 5
3.005 8
3.333 6
4.891 2
3.958 22
7.935Ax-
4.241 -7
2.382 -17
4.264 23
5.619 59
8.284 51
3.179 109
6.195Aw=
2.
1.*(∑-ε)=2*12*(1
9.1-(
0.
3.3.3)/2)=
4.
5.2 m2 xF=12*(14
8.2-(-9+99)/2)/20
7.
6.5.965 m ▽=12*(122.82-(0.141+17.106)/2)=1270.362 m3 Zb’=12*(227.935-(0.018+23.958)/2)/1270.362=1.891 m xb=12*(1096.195-(-4.241+513.179)/2)/1270.362=7.371 mExercise 2-6某船L=6
0.,其水线以下横剖面为等边三角形,从尾向首其半宽为0.
3.
1.6,
4.3,
5.0, 4.6, 3.3m,试求水线面漂心xF,排水体积▽,浮心xb, zb,方形系数Cb。用梯形法解:δL=60/5=12 my
0.300
1.600
4.300
5.000
4.600
3.300 1
9.100x-3
0.000 -1
8.000 -
6.000
6.000 1
8.000 3
0.000 yx-
9.000 -2
8.800 -2
5.800 3
0.000 8
2.800
9..000 14
8.200A
0.156
4.429 3
1.988 4
3.250 3
6.607 1
8.840 13
5.269z
0.173
0.925
2.486
2.890
2.659
1.908 Az
0.027
4.096 7
9.507 12
5.000 9
7.336 3
5.937 34
1.903Ax-
4.671 -7
9.718 -
1.1.926 25
9.500 65
8.922
5.5.191 120
7.298Aw=2*12*(∑-ε)=2*12*(1
9.1-(
0.
3.3.3)/2)=41
5.2 m2xF=
1.*(
4.4.
8.(18+
0./2)/20
7.6=24.0
3.br>
5.m ▽=12*(135.269-(0.156+18.84)/2)=810.678 m3 Zb’=12*(341.903-(0.027+35.937)/2)/810.678 =4.795 m xb=12*(1207.298-(-4.671+565.191)/2)/810.678 =13.722 m⏺Exercise 2-7某船的一个煤舱长为24m,自尾至首各横剖面面积为
5.7,
8.7,
1..
3.1
0.1,8.8(单位m2)这些剖面的形心在基线以上的高度分别为3.7,3.5,3.3,3.5,3.6(单位 m)。剖面之间的间距为6m。设煤舱的积载因数(每吨煤所占体积m3)为1.56m3/t。试列表计算:(1)该船载煤吨数,(2)该舱的重心位置(基线以上距离及距煤舱尾舱壁的距离)。煤舱L=24m,Asi方法1方法2方法3计算ZXiXiAsiXiXiAsiXiXiAsiZiZiAsi
5.7-12-6
8.4-2-
1..4
3.72
1.091
8.7-6-5
2.265
2.2-1-
8.7
3.53
0.452
1..31213
5.6
3.33
7.2931
0.166
0.618
1.1.811
0.1
3.53
5.354
8.81210
5.6242
1..221
7.6
3.63
1.68Σ
4..64
5.658
0.8
7.61
5..86ε
7.251
8.610
5.6
3.12
6.385Σ’3
7.352
7.047
5.2
4.512
9.475AsAsXAsX AsXAsZΣ’3
7.352
7.047
5.2
4.512
9.475V=δx*Σ’ = 6*3
7.35 = 22
4.1 m3P=V/
1.56 = 22
4.1/1.56 =14
3.65 (t)X’g= Σ’AsX*δx / Σ’As*δx =2
7.37.35 =
0.723 (m)Xg=1
2.
0.723=12.723 (m)Zg= Σ’AsZ*δx / Σ’As*δx =12
9.4
7./
3..35 =3.466 (m)位置AsXiAsXiZiAsZi辛氏数
5.7
3.72
1.0911
8.765
2.2
3.53
0.4542
1..31213
5.6
3.33
7.24231
0.118
1.1.8
3.53
5.3544
8.8242
1..2
3.63
1.681Σ加权11
2.3141
8.339
0.55V=
2./12*112.3=22
4.6 m3Xg=(
2.br>4.12*141
8.3)/224.6=12.630 mP=22
4.6/
1.56=14
3.97 tZg=(2
4.12*
3.
0.55)/224.6=3.478 m⏺某海船各水线的排水量为10804, 8612, 6511, 4550, 2810,
1.31, 263 t。各水线间距为1.2m。求吃水
7.8m时的浮心垂某海船各水线面面积为2
0., 185, 160, 125, 30m2,试求该水线面的每厘米吃水吨数TPC,并绘制TPC曲线。如水线间距为0.5m,求排水体积▽,并绘制V=f(d)曲线(列表计算)。某海船水线面面积如表,水线间距为δd=
1.1m,用梯形法列表计算排水量曲线,每厘米吃水吨数曲线。并绘图。某货船在A港吃水d=
5.35m,进入B港其吃水不能超过d1=
4.6m,船在d2=
5.5m时TPC2=1
8.6t/cm,在d3=4.5m时TPC3=1
4.8t,假定TPC对于吃水的变化是一直线,求进入B港前必须卸下的货物重量。已知:d=4.6m:Tpc=
1.
4.br>8.(18.6-14.8)/1*(4.6-4.
5.=15.18 t/cm d=5.35m:Tpc=
1.
4.+
3.
8.(
5.35-4.5)=18.03 t/cm p=(15.18+18.03)/2*100*(5.35-4.6)=1245.4 (t)某船船长L=164m,船宽B=1
9.7m,方形系数Cb=
0.50,水线面系数Cw=<
b.>
0.73,在海水中平均吃水d=
8.2m,求船进入淡水中的平均吃水。已知:L=
1.4m,B=1
9.7m,Cb=0.50,Cw=0.73,海水d=8.20m δd =(1.025-1.0)Cb.LBd/CwLB= (<
b.>
1.025-1.0)Cb.d/Cw=(
1.
0.5-1.0)*0.50*
8.2/0.73=0.140 (m) d(淡)=8.20+0.14=8.34 m δd/d=-Cb/Cw*dω/ω δd=-0.5/0.73*(1.025-1.0)/1.025*8.2=0.137 m d=8.2+0.137=8.337 m⏺Exercise 2-13某船由淡水进入海水,必须增加载荷p=175 t,才能使其在海水中的吃水与淡水中的吃水相等。求增加载重后的排水量。
题目解答
答案
解:前后排水体积相同
Δ/ω淡=(Δ+175)/ω海
(1.025-1.0)Δ=175
Δ=175/0.025=7000 (t)
Exercise 2-14
某船在海水中的正常吃水d=2.2m,排水量△=930t,水线面面积Aw=606m2,型深D=3.35m,在甲板处的水线面面积Aw=658m2假定船的水上部分舷侧是直线形状,求储备浮力占排水量的百分数。
海船:储备排水量:
Δ=ω(D-d)*(Aw+A甲)/2
=1.025*(3.35-2.2)*(658+606)/2
=944.97 (t)
百分数:744.97/930=80%
Exercise 2-15
某内河船吃水d=2.4m,方形系数Cb=0.654,水线面系数Cw=0.785,卸下货物p=80%排水量,求船舶的平均吃水(设在吃水变化范围内船体为直舷)。
解:p=0.08△=0.08ωCbLBd δd=p/ωCwLB=0.08ωCbLBd/ωCwLB
=0.08dCb/Cw=0.08*2.4*0.654/0.785=0.16 m d1=d-δd=2.4-0.16= 2.24 m