Ship Stability Flashcards

What information does the GZ curve (curve of statical stability) provide to the Master?

Answer: The righting lever at different angles of heel, showing range of stability, maximum GZ, and angle of vanishing stability

The GZ curve plots the righting lever (GZ) against the angle of heel. It shows: the initial GM (slope at origin), the maximum righting lever and the angle at which it occurs, the range of stability (from 0° to the angle of vanishing stability where GZ returns to zero), and the dynamical stability (area under the curve). This is essential for assessing the vessel's ability to resist capsizing.

Under the MCA's stability criteria (based on the IMO Intact Stability Code), what is the minimum required GM for a cargo vessel in the departure condition?

Answer: 0.15 metres

The IMO Intact Stability Code (incorporated by the MCA) requires a minimum initial GM of 0.15 metres in all loading conditions. Additionally, the area under the GZ curve up to 30° must be at least 0.055 metre-radians, and the maximum GZ must occur at an angle of heel preferably exceeding 25° but not less than 25° for most vessel types.

What is 'angle of loll' and how does it differ from a list caused by asymmetric loading?

Answer: Angle of loll occurs when GM is negative and the vessel heels to find a new equilibrium; list is caused by G being off-centreline due to weight distribution

A list is caused by the centre of gravity being off the vessel's centreline (asymmetric loading), and can be corrected by redistributing weight. An angle of loll occurs when GM is negative — the vessel is initially unstable and heels until the waterplane area increases sufficiently for the metacentre to rise above G, creating a new, precarious equilibrium. Loll is corrected by lowering G (adding low weight, not shifting weight).

When a vessel grounds at low water on an even keel, what is the primary stability concern?

Answer: The upward ground reaction force reduces displacement, effectively raising G above M and potentially causing the vessel to heel dangerously

When a vessel grounds, the seabed exerts an upward force (ground reaction) that reduces the effective displacement. This is equivalent to removing buoyancy, which raises the effective KG relative to the metacentre. If the ground reaction is large enough, the effective GM can become negative, causing the vessel to become unstable and heel to one side (loll), potentially with catastrophic consequences.

What is the 'moment to change trim by one centimetre' (MCTC or MCT1cm) and how is it used?

Answer: The moment required to change the vessel's total trim by one centimetre — used to calculate the effect of loading, discharging, or shifting weights on trim

MCTC (moment to change trim by 1 cm) is the longitudinal moment (in tonne-metres) required to change the vessel's total trim by one centimetre. It is used to calculate the change in trim when weights are loaded, discharged, or shifted longitudinally. The formula is: Change in trim (cm) = (w × d) / MCTC, where w is the weight and d is the distance from the centre of flotation.

A vessel displacing 10,000 tonnes has KM = 8.5 m and KG = 7.2 m. A weight of 500 tonnes is shifted vertically upward by 4 metres. What is the new GM?

Answer: 1.10 metres

Original GM = KM - KG = 8.5 - 7.2 = 1.30 m. Rise of G due to shifting weight upward: GG1 = (w × d) / W = (500 × 4) / 10,000 = 0.20 m. New KG = 7.2 + 0.20 = 7.40 m. New GM = 8.5 - 7.40 = 1.10 m. The metacentric height has decreased because the centre of gravity has risen, reducing initial stability.