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A ship of 19200 M.T. displacement, 235m long has KM 10.2m, KG 9.4m, TPC 32 MCTC 254tm is floating at draft forward of 9.42m and aft 10.88m in water of RD 1.005. The centre of floatation is 5m abaft amidships, centre of buo yancy is 3m forward of amidships. Vessel is drydocked in water of density 1.020. Calculate her virtual GM on taking blocks all over.
Solution –
Given W = 19200, LBP = 235m, KM = 10.2m, KG = 9.4m, TPC = 32, MCTC = 254tm, draft fwd = 9.42m, draft aft = 10.88m in RD 1.005
In this Question first change of trim has to be considered due to change of density.
We have BB1 =
(change in v/w vol.) ⨯ (Distance between LCB + LCF) /Final volume
We can calculate, v/w vol. in 1.005
= 19200/1.005
= 19104.477m2
v/w vol. in 1.020
= 19200/1.020
= 18823.529m3
BB1 = 280.95 ⨯ 8/18823.529
= 0.119m
Since COF is abaft midship vessel will him by stern
Change in trim = (W⨯BB1/ MCTC)
= 19200 ⨯ 0.119/252.76 ⨯ 100
MCTC in 1.025 = 254
MCTC in 1.020 = 252.76
Trim caused ( TC) = 0.090m
Change in trim aft = (0.090 ⨯11.5)/235 = 0.043m
Change in trim fwd = (0.090 – 0.043) = 0.047m
Also,
FWA = (W40 ⨯ TPC)
= 19200 /40 ⨯ 32
= 15cm,
DWA = 0.015 ⨯ 15/0.025
= 9cm
