A ship of 22000 tonne displacement is 160 m long and MCTI cm 280tonne m, waterplane area 3060 m2 centre of buoyancy 1 m aft of midships and centre of flotation 4 m aft of midships. It floats in water of 1.007 t/ m3 at draughts of 8.15 m forward and 8.75 m aft. Calculate the new draughts if the vessel moves into sea water of 1.026 t/ m2 Calculate the metacentric height of the vessel.
Given
Δ = 22000 t
L = 160 m
MCT1 = 280 tm
Aw = 3060 m2
COB = 1 m aft
COF = 4 m aft
ƍ = 1.007 t/m3
df = 8.15 m
da = 8.75 m
ƍs = 1.026 t/m3
To find
1.df new
2.da new
3. GM
Solution
Change in mean draught = ((100 x Δ) / Aw) x (୧s - ୧) / (୧s x ୧)
= (100 x 22000)/3060) x (1.026 - 1.007) / (1.026 x 1.007)
= 13.22 cm reduction (As moves to SW)
Distance from COF to COB = COF - COB (As both are aft of midship)
FB = 4 - 1
= 3 m
Change in trim = (Δ x FB (୧s - ୧)) / (MCT1 cm x ୧s)
= (22000 x 3 (1.026 - 1.007)) / (280 x 1.026)
t = 4.5 cm by stern (Ship moves from fw to sw trim by stern)
Draught change FWD = - t/L (L/2 + COF)
= - 4.5 / 160 x (80 + 4)
= - 2.36 cm
Draught change AFT = + t/L (L/2 - COF)
= + 4.5 / 160 x (80 - 4)
= + 2.14 cm
1. df new = Initial draught FWD + change in mean draught + draught change FWD
= 8.15 - 0.132 - 0.024
= 7.994 m
2. da new = Initial draught AFT + change in mean draught + draught change AFT
= 8.75 - 0.132 + 0,021
= 8.639 m
3.MCT1 cm = Δ x GML / 100 x L
3060 =(22000 x GM) / (100 x 160)
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