As a double-bottom tank is filled, two effects act on GM:

• The added weight of ballast is low in the ship (deep in the double bottom), so it lowers the centre of gravity, which increases GM (stiffening the ship). This is the dominant effect once the tank is full.
• While the tank is partially full, a free surface exists and produces a free-surface effect that reduces the effective GM. The free-surface correction is rho x i / Delta, where i is the second moment of area of the free surface about its longitudinal axis (i = L B^3/12). For a wide double-bottom tank this can be significant.

Hence during filling, the effective GM first falls (as the free surface develops) and then rises as the tank approaches full and the free surface disappears; when the tank is completely full (pressed up) the free-surface effect is zero and the GM is increased by the low weight. The net effect of a full double-bottom tank is an increase in GM (stiffer ship), but the transient free-surface loss during filling must be watched, especially in a tender ship.

Oil tanker 160 m long, 22 m beam, floats at a draught of 9 m in sea water. Cw = 0.865. The midship section is a rectangle with 1.2 m radius at the bilges. A midship tank 10.5 m long has twin longitudinal bulkheads and contains oil of 1.4 m3/t to a depth of 11.5 m. The tank is holed to the sea for the whole of its transverse section. Find the new draught.

Waterplane area Aw = Cw x L x B = 0.865 x 160 x 22 = 3044.8 m2.

Midship section area (rectangle with bilge radius r=1.2 m): Ams = B x d - (4 - pi) r^2 = 22 x 9 - 0.858 x 1.44 = 198 - 1.236 = 196.76 m2.

Volume of the tank below the original waterline = Ams x 10.5 = 196.76 x 10.5 = 2066 m3.

The tank contains oil of density rho_o = 1/1.4 = 0.714 t/m3. When holed, sea water (1.025 t/m3) replaces the oil, so the net loss of buoyancy is the volume times the relative density difference:

Vlost = 2066 x (1 - 0.714/1.025) = 2066 x 0.303 = 626 m3.

The flooded tank provides no increase of buoyancy, so the effective sinking waterplane = Aw - (10.5 x 22) = 3044.8 - 231 = 2813.8 m2.

Sinkage = Vlost/effective waterplane = 626/2813.8 = 0.222 m.

New draught = 9 + 0.22 = 9.22 m.

Answer: the new draught is about 9.2 m.