Q1. The TPC of a ship are as follows:
Draft (m) | 6.0 | 5.0 | 4.0 | 3.0 | 2.0 |
TPC | 22.45 | 22.04 | 21.53 | 20.91 | 19.68 |
The displacement and KB at 2m draft are 3280t and 1.2m respectively. Find the displacement and KB at 6m draft in SW.
Solution –
Given,
Draft | TPC | Area (TPC= A/100 ⨯ density) | SM | Product | Lever | Moment |
6.0 | 22.45 | (100)⨯22.45/1.025 | 1 | 22.45x | 4h | 89.8xh |
5.0 | 22.04 | 22.04x | 4 | 88.16x | 3h | 264.48xh |
4.0 | 21.53 | 21.53x | 2 | 43.06x | 2h | 86.12xh |
3.0 | 20.91 | 20.91x | 4 | 83.64x | 1h | 83.64xh |
2.0 | 19.68 | 19.68x | 1 | 19.68x | 0h | 0 |
Sum of product of volume = 256.99x
Volume = 1/3 ⨯ 256.99 ⨯ 100/1.025
= 8357.398m3
Sum of product of moment = 524.04 ⨯ 1/3 ⨯ 100/1.025 ⨯ 1
= 17041.95m4
Total volume including appendage =?
Given displacement of appendage = 3280 tonnes at draft 2m
We know displacement = ( v/w volume X density of water displaced)
Volume of appendage = 3280/1.025
= 3200m3
Total volume including appendage = (8357.398 + 3200) m3
= 11557.398 m3
COG of the structure without appendage
= volume/Moment
= 2.039m
COG from 2.0 mtr draft = 2.039 m
Now total moment = moment of appendage + moment of structure
11846.33 ⨯ X = (3280 ⨯ 2) + (8566.33 ⨯ 4.039)
X = 3.253m
Q2. The breadths of a ship’s water-plane 120m long, measured at equal intervals from aft, are: 1.2, 9.6, 13.2, 15.0, 15.3, 15.6, 15.6, 14.7, 12.9, 9.0 and 0.0m respectively. Find:
The water plane area;
TPC in sea water;
FWA if displacement is 6690t.
Solution –
Ordinate | SM | Product | |
1.2 | 1 | 1.2 | |
9.6 | 4 | 38.4 | |
13.2 | 2 | 26.4 | |
15.0 | 4 | 60.0 | |
15.3 | 2 | 30.6 | |
15.6 | 4 | 62.4 | |
15.6 | 2 | 31.2 | |
14.7 | 4 | 58.8 | |
12.9 | 2 | 25.8 | |
9.0 | 4 | 36.0 | |
0.0 | 1 | 0.0 | |
370.8 |
Water plane Area = ( H x SOP )/ 3
= 1483.2m2
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