= 5.415m
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A vessel about to dry dockn is in the following condition :
draft : For’d 6.10m, Aft 6.70m
KMₒ = 7.20m, KGₒ = 6.8m
MCTC = 155 tm, TPC = 22t, LCF = 80m for’d of AP
Length = 180m, Disp = 11000 tonnes
Find : (a) GM of the vessel at critical instant
(B)The righting moment of 1° heel
(c)The draft For’d and at the critical instant.
Solution –
Question same as Q. No. – 8
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A vessel with length of 185m, displacement of 29000 is floting in a graving dock with a depth of 10m at drafts F 8.00 and A 9.38m. KM = 11.50m, KG = 10.95m, MCTC = 410 t/m, TPC = 29.6, LCF = 89.5m. find the effective GM and draft fore and aft of the vessel after the water level has dropped by 1.25m.
Solution –
After level dropped by 1.25m,
New water Level = (10 – 1.25) m
= 8.75m
Correction to aft draft = (1.38 ⨯ 89.5)/ 185
= 0.668m
(hydrostatic draft ) Hd = 8.712m
Clearly since level of water 8.75m ˃ Hd 8.712m
It means stern has taken to the block and fwd end is still afloat.
Reduction in aft draft = 0.63m
Or, 63 = P/29.6 + P ⨯ 89.5/4100 ⨯ 89.5/185
Or, P = 451.96 tonnes
We know that –
GG1 = P ⨯ KM/ W
= 451.96 ⨯ 11.50/29000
= 0.179m
Therefore, GM = 0.371m
Trim caused = Tc = 451.96 ⨯ 89.5/410 ⨯ 100
= 0.986m
