Worked example
From the following information calculate the corrections to perpendiculars and the draughts at the perpendiculars. Also calculate the true trim.
Vessel LBP 181.8 metres Density at the time of draught reading 1.0185 t/m3 port side stbd side distance marks from perp.
Forward draughts; 4.61m Midships draughts 4.93m Aft draughts
5.58m
4.65m Fd = 2.94 m aft 5.10m Md = 1.44 m aft 5.60m Ad = 7.30 m forward
Forward mean = (4.61 + 4.65) / 2 = 4.63m Midships mean = (4.93 + 5.10) / 2 = 5.015m Aft mean
= (5.58 + 5.6) / 2 = 5.59m
So apparent trim is: 5.59 - 4.63 = 0.96m And LBM is: 181.8 - 2.94 - 7.30 = 171.56m
Forward corr’n= Apparent trim x Fd = 0.96 x -2.94 = LBM
171.56 Midships corr’n = LBM
Apparent trim x Md = 0.96 x -1.44 = 171.56
Aft corr’n = Apparent trim x Ad = 0.96 x 7.3 = LBM
171.56 Now:
Forward draught is 4.63 – 0.0165 m = 4.6135 m Midships draught is 5.015 – 0.0081 m = 5.0069 m Aft draught is 5.59 + 0.0408 m = 5.6308 m True trim is
5.6308 – 4.6135 m = +0.0408m -0.0081m -0.0165m
1.0173 metres = 101.73 cm
¾ mean draught = (6 x 5.0069) + 4.6135 + 5.6308 8
= 5.0357 m
From the original survey the following data was given in the vessels hydrostatic particulars: Scale density of hydrostatic particulars 1.025 t/m3
Draught Displacement TPC LCF Draught MCTC Draught MCTC 5.00 5.10
19743 20167
42.32 -4.354 42.37 -4.289
5.50 5.60
445.5 446.6
4.50 4.60
434.9 435.9
The stability book stated that a negative (-) sign for Lcf indicated forward of midships. Interpolating the data from the table (it is easier to use centimetres in the interpolation rather than metres) The difference in the tabulated draughts is 10 cm and the draught we are looking for is 3.57 cm more than 5 metres
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