• Design newship dimension base on mothership:
Item Unit Content& Value
Boundary Condition
Ship Type - Container 1,650 TEU
LBP M 162 [1team] Fixed Length, Vs = 19.5 knots
B M 27 [2 team] Fixed Breadth (Crane reach), Vs = 21 knots
Tor d M 8.7 [3 team] Fixed Draft (Habor), Vs= 21 knots
Dispt. M3 25.115 Displacement is constant.
vs knots 20.5
L = 168.75
T = 9.85
B = 27
The lenght and draft were both increased weight of ship is also increasing.
Item Unit Content& Value
Boundary Condition
Ship Type
- Container 1,650 TEU
LBP M 162 [1team] Fixed Length, Vs = 19.5 knots
B M 27 [2 team] Fixed Breadth (Crane reach), Vs =21 knots
Tor d M 8.7 [3 team] Fixed Draft (Habor), Vs= 21 knots
Dispt. M3 25.115 Displacement is constant.
vs knots 20.5
Fix Lenght from mother ship L= 162
Disp' 25.115
B = 27.0
Cb = H = L/B B/H
Cb of Mother ship 0.660 8.7
6.0
3.10
Case 1 0.652 8.8 3.07
Case 2 0.645 8.9 3.03
Case 3 0.638 9.0 3.00
Case 4 0.631 9.1 2.97
Case 5 0.624 9.2 2.93
Block coefficient
Cb 0.573H 8.7 9.1 9.6 9.8 10 10.2
Disp 22888.9 23941.3 25256.7 25782.9 26309.1 26835.3
There’re 4 suitable case above, but what is the best choice?
Cb 0.573H 8.7 9.1 9.6 9.8 10 10.2
Disp 22889.0 23941.3 25256.8 25783.0 26309.1 26835.3LWT*1.3 9006.5 9159.2 9344.3 9416.6 9488.1 9558.6
Disp - LWT 13882.4 14782.1 15912.5 16366.3 16821.1 17276.7DWT 15069.0
How can we choice the bestest one?
Method 1 Method 2New ship 1 New ship 2 New ship 3
L 168.75 162 170B 27T 9.85 10.2 9.6
Cb 0.589 0.563 0.573Disp 26431.5 25.115 25256.8
Method 1 Method 2New ship 1 New ship 2 New ship 3
L 168.75 162 170B 27T 9.85 10.2 9.6
Cb 0.589 0.563 0.573Disp 26431.5 25.115 25256.8
Disp - LWT 17119.6 16333.8 15912.4
DWT 15069.0
Ship 1 and ship 3 has the lenght bigger than ship 2 higher building cost - due with longitudinal streght.Ship 2 is the best choice!