The breadths at the load water-plane of a ship 90 metres long, measured at equal intervals from forward, are as follows: 0, 3.96, 8.53, 11.58, 12.19, 12.5, 11.58, 5.18, 3.44 and 0.30m, respectively. If the load draft is 5 metres, and the block coefficient is 0.6, find the FWA and the position of the centre of flotation, from amidships.
The breadths at the load water-plane of a ship 90 metres long, measured at equal intervals from forward, are as follows: 0, 3.96, 8.53, 11.58, 12.19, 12.5, 11.58, 5.18, 3.44 and 0.30m, respectively. If the load draft is 5 metres, and the block coefficient is 0.6, find the FWA and the position of the centre of flotation, from amidships.
Reference: Ship
Stability for Masters & Mates (6th edition), Exercise – 10.3
Solution:
|
|
Breadth |
SM |
f(A) |
Levers |
f(M) |
|
FP |
0 |
1 |
0 |
-5 |
0 |
|
|
3.96 |
3 |
11.88 |
-4 |
-47.52 |
|
|
8.53 |
3 |
25.59 |
-3 |
-76.77 |
|
|
11.58 |
2 |
23.16 |
-2 |
-46.32 |
|
|
12.19 |
3 |
36.57 |
-1 |
-36.57 |
|
|
12.5 |
3 |
37.5 |
0 |
0 |
|
|
11.58 |
2 |
23.16 |
+1 |
+23.16 |
|
|
5.18 |
3 |
15.54 |
+2 |
+31.08 |
|
|
3.44 |
3 |
10.32 |
+3 |
+30.96 |
|
AP |
0.3 |
1 |
0.3 |
+4 |
+1.2 |
|
|
|
|
∑ = 184.02 |
+5 |
∑ = -120.78 |
Common interval, h = 90/9 = 10 m
Area = 3h/8 * ∑f(A)
= 3*10/8 *
184.02
= 690.075 m2
Center of flotation = ∑f(m) / ∑f(A) * h
= -120.78/184.02 * 10
= -6.56 m FOR from midship
Center of flotation from AFT to FOR = 40 + 6.65 = 46.56 m
TPC = AW*1.025/100
=
690.075*1.025/100
= 7.0733 ton/cm
Cb = ∇ / L*B*d
=> ∇ = 0.6 * 90 * 12.5 * 5
=> ∇ = 3375 m3
∆ = 3375*1.025 = 3459.375 tons
FWA = ∆/4*TPC
=
3459.375/4*7.0733
= 122.268 mm
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