Free surface, list and the press-up decision
MV Tista is completing loading with displacement 4,500 t, KM 7.10 m and solid KG 6.35 m. A rectangular double-bottom tank 14 m long and 9 m wide holds 2.0 m of seawater (relative density 1.025) and is slack. Sixty tonnes of deck cargo, already on deck at KG 5.5 m, must be shifted 7 m to starboard to clear the hatch. The berth is upriver and the harbour authority requires a list of not more than 5° while the vessel manoeuvres. The stability booklet gives a minimum required GM of 0.15 m and notes that pressing the tank up raises the ballast's centre of gravity by 1.0 m. The master intends to press the tank up before the shift if it improves matters. Beam is 18 m.
1. Calculate the solid GM, the free surface correction and the fluid GM, and say whether the vessel meets the booklet minimum.
Read the worked answer
Solid GM = KM − KG = 7.10 − 6.35 = 0.75 m.
Free surface correction. For a rectangular free surface, i = l·b³/12 = 14 × 9³/12 = 850.5 m⁴. Free surface moment = i × density of the liquid = 850.5 × 1.025 = 871.8 t·m. FSC = 871.8 / 4,500 = 0.194 m.
Fluid GM = 0.75 − 0.194 = 0.556 m.
This is well above the booklet minimum of 0.15 m, so the vessel is stable. The point of the calculation is not the pass mark but the size of the loss: roughly a quarter of the solid GM has been given away by one slack tank, which is why slack tanks must be counted properly rather than assumed to be harmless.
2. Calculate the list after the 60 t shift with the tank slack, and state what additional correction is needed to get within the 5° limit.
Read the worked answer
Listing moment = 60 × 7 = 420 t·m. tan θ = 420 / (4,500 × 0.556) = 0.1678, so θ = 9.5° - outside the 5° limit and enough to make the shift unsafe at the berth.
Allowable moment at 5° = 4,500 × 0.556 × tan 5° = 4,500 × 0.556 × 0.0875 = 219 t·m. The excess is 420 − 219 = 201 t·m, which must be countered by weight on the port side: for example 40 t transferred 5.0 m to port (200 t·m), or 25 t over 8.0 m. Pressing up the tank alone (GM about 0.693 m) only reduces the list to about 7.7° (tan θ = 420/(4,500 × 0.693) = 0.135), so a counter-moment is still required. Do the counter-ballast first, then the shift, and check the list at each stage.
3. Decide whether the tank should be pressed up before the transverse shift, supporting the decision with the effect on GM, KG and roll period.
Read the worked answer
Pressing up raises KG but restores GM. Ballast weight = 14 × 9 × 2.0 × 1.025 = 258.3 t. Raising its centre of gravity by 1.0 m gives ΔKG = 258.3 × 1.0 / 4,500 = 0.057 m, so solid GM becomes 0.75 − 0.057 = 0.693 m, but the free surface correction disappears. Fluid GM after pressing up = 0.693 m, a net gain of about 0.137 m over 0.556 m.
Roll period. k ≈ 0.38 × 18 = 6.84 m. With GM 0.556 m: T = 2π × 6.84 / √(9.81 × 0.556) = 43.0 / 2.335 ≈ 18.4 s. With GM 0.693 m: T ≈ 43.0 / 2.608 ≈ 16.5 s.
Decision. Press up. The KG penalty is small and is outweighed by the removal of free surface; GM rises, the list for the same transverse moment falls, and the roll shortens towards a stiffer but acceptable range. Press up before the shift, then verify with an inclining check or the stability computer, not by eye.