Form stability is the stability that depends on the shape (form) of the hull, i.e. the position of the metacentre and the righting levers that arise from the geometry of the waterplane and the hull. It is measured by the metacentric height and the GZ curve, and depends on the beam, waterplane area, draught and the shape of the hull. A ship with a wide beam and a large waterplane area has high form stability (large BM and GM).

Weight stability is the stability that depends on the position of the centre of gravity, i.e. the distribution of the weights in the ship. It is the effect of the KG on the stability: a low centre of gravity gives high weight stability (large GM), while a high centre of gravity reduces it. The total stability is the combination of the form stability (position of the metacentre) and the weight stability (position of the centre of gravity), since GM = KM - KG, where KM is the metacentric height above the keel (form) and KG is the height of the centre of gravity (weight).

The end bulkhead of the wing tank of an oil tanker has widths at 3 m intervals, commencing at the deck: 6.0, 6.0, 5.3, 3.6 and 0.6 m. Calculate the load on the bulkhead and the position of the centre of pressure if the tank is full of oil of relative density 0.8.

The bulkhead is a vertical plane of width varying with depth. The depth below the deck (and below the liquid surface, since the tank is full) at the five stations is 0, 3, 6, 9, 12 m (spacing 3 m).

For a vertical plane surface, the load (total force) = rho g x (first moment of area about the liquid surface), and the centre of pressure is at depth = (second moment)/(first moment).

First-moment integrand (per unit width) about the liquid surface: M1 = integral of z dz from 0 to H = H^2/2.

Second-moment integrand: M2 = integral of z^2 dz from 0 to H = H^3/3.

Widths b: 6.0, 6.0, 5.3, 3.6, 0.6. Depths H: 0, 3, 6, 9, 12.

M1 values (b x H^2/2): 0, 6x4.5=27, 5.3x18=95.4, 3.6x40.5=145.8, 0.6x72=43.2.

M2 values (b x H^3/3): 0, 6x9=54, 5.3x72=381.6, 3.6x243=874.8, 0.6x576=345.6.

Using Simpson's rule with spacing 3 m and 5 ordinates:

M1 = (3/3)[0 + 0.6 + 4(27 + 145.8) + 2(95.4)] = 1[0.6 + 4x172.8 + 190.8] = 0.6 + 691.2 + 190.8 = 882.6.

M2 = (3/3)[0 + 345.6 + 4(54 + 874.8) + 2(381.6)] = 345.6 + 4x928.8 + 763.2 = 345.6 + 3715.2 + 763.2 = 4824.

Load (force) = rho g x M1 = 0.8 x 1.025 x 9.81 x 882.6 (using sea-water density 1.025 t/m3 for the oil of RD 0.8, i.e. rho = 0.8 x 1.025 = 0.82 t/m3 = 820 kg/m3).

Load = 820 x 9.81 x 882.6 = 7,099,000 N = 7099 kN, or in tonnes-force = 0.82 x 882.6 = 723.7 t.

Centre of pressure depth below the liquid surface = M2/M1 = 4824/882.6 = 5.47 m.

Answer: the load on the bulkhead is about 7100 kN (about 724 t), and the centre of pressure is about 5.5 m below the liquid surface (i.e. about 5.5 m below the deck).