| Fwd | Aft | |
| Draft | 5.23m | 5.74m |
| Rise | -0.200m | -.200m |
| -0.730m | +0.796m | |
| 4.300m | 6.336m |
Q3. A vessel of displacement 7200 tonnes. Length 120m, MCTC 110tm, KG 6.0m, TPC 16 tonnes, center of floatation 2 mts forward of midship ground on a rock 10mts. Abaft her forward perpendicular. Given initial draft F : 4.2m, A : 5.1m.
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Find the fall in tide in cm, that will make the vessel unstable, given KM at that moment equal to 6.4 mts.
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What will be the drafts forward and aft at that moment.
Solution –
- For vessel to unstable GM should be zero.
GG1 = 0.4m
So,
0.4 = P ⨯ 6.4/ 7200
P = 450 tonnes
We can calculate fall in tide –
Fall in tide = P/ TPC + ( P x AL / MCTC) + AL/ LBP
= 450/16 + 450 ⨯ 48/110 ⨯ 48/120
= 1.067m
