Therefore, GML = 149.93 = 150
W = (80 ⨯ 16 ⨯ 3.5 ⨯ 1.025)
= 4592 tonnes
MCTC = 4592 ⨯ 150/80 ⨯ 100
= 86.1 tm
P = 100 ⨯ 86.1/40
= 215.25 tonnes
GG1 = P ⨯ KM/W
= 0.367m
GM old = (KM – KG)
= (7.845 – 4.2)m
= 3.645m
GG1 = 0.367m
Virtual GM = (3.645 – 0.367)m
= 3.278m
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A vessel of length 140m, W = 19200t, KM = 6.7m, KG = 6.2m, CF = 4m fwd of midship, MCTC = 145 Tm is to be drydocked. Calculate the max. permissible tm. By the stern to ensure that she would have a positive GM of 30 cm on taking blocks F & A.
Solution –
GM now = 0.5m
GM required = 0.3m
GG1 = 0.2m
We know that –
GG1 = P ⨯ KM/W
0.2 = P ⨯ 6.7/19200
P = 573.13 tonnes
Therefore, permissible P = 573.13 tonnes
We know P ⨯ d = trim ⨯ MCTC
= 573.13 ⨯ 74/145 ⨯ 100 = trim in metres
