A ship of 14000 tonne displacement is 145m long and floats at draughts of 7.9 m forward and 8.5 m aft. The TPC is 19, GML 120 m and LCF 3 m forward of midships. It is required to bring the vessel to an even keel draught of 8.5m. Calculate the mass which should be added and the distance of the centre of the mass from midships. (16)
Structured for DG Shipping MEO Class II examination scoring criteria.
Mass to be added to bring the vessel to an even keel of 8.5 m.
Ship 14,000 t displacement, 145 m long, floats at draughts 7.9 m forward and 8.5 m aft. TPC = 19, GML = 120 m, LCF 3 m forward of midships. It is required to bring the vessel to an even keel draught of 8.5 m. Calculate the mass to be added and the distance of its centre from midships.
Current mean draught = (7.9 + 8.5)/2 = 8.2 m. Required even-keel draught = 8.5 m, so the mean draught must increase by 0.3 m.
MCT1cm = Delta x GML/(100 x L) = 14000 x 120/(100 x 145) = 1,680,000/14,500 = 115.9 t-m per cm.
Mass to add for the mean sinkage of 0.3 m (30 cm): w = 30 x TPC = 30 x 19 = 570 t.
The current trim is 0.6 m by the stern (8.5 - 7.9). To bring the vessel to an even keel, the trim must be removed, i.e. a change of trim of 0.6 m (60 cm) is required. The change of trim moment = 60 x MCT1cm = 60 x 115.9 = 6954 t-m.
The added mass must be placed forward of the centre of flotation so that it both sinks the vessel and removes the stern trim. Distance of the mass from the LCF: x = change of trim moment/w = 6954/570 = 12.2 m forward of the LCF.
Since the LCF is 3 m forward of midships, the mass is at 12.2 + 3 = 15.2 m forward of midships.
Answer: add 570 t with its centre 15.2 m forward of midships; the final even-keel draught is 8.5 m.