Before entering the dock the ship must have adequate and known stability. The requirements are:

• The ship should be stable with an adequate GM, so that the small upward force (reaction) at the keel, building during docking, cannot produce a list. Dock support is provided on the keel blocks and any excessive list during docking would subject the structure and blocks to unequal loading.
• As the water is pumped out, the reaction of the keel blocks reduces the effective displacement of the ship, lowering the KB and BM and hence GM; if all buoyancy were removed the ship would rest wholly on the blocks. The ship must retain sufficient GM throughout the docking operation (for example by keeping the GM such that the maximum inclination, where the righting moment at the critical point, does not become negative).
• Ballast should be arranged to give even keel (or slight trim) and list-free condition; tanks should be pressed up or emptied to avoid free-surface effects, and draft read before docking.
• Weight and trim must be such that the keel blocks contact evenly over the full keel length (avoid excessive trim that overloads the block crown or aft blocks).
• Cargo/weights should not be changed and the dock must be level; adequate dock pumping should keep the vessel central over the keel line.

If a beam vessel develops instability while docking (transverse GM small or negative), it can heel and capsize on the blocks; hence a suitable stability margin (e.g. GM not less than a minimum) is insisted on and the docking weight, draft and trim are checked by stability data (e.g. from hydrostatics and the Docking Plan).

Box-shaped vessel 50 m long, 10 m wide, floats on even keel in salt water at a draft of 4 m. A centreline longitudinal watertight bulkhead runs full depth/size. A midship compartment on the starboard side is 15 m long and has permeability 30 per cent. KG = 3 m. Calculate the list when bilged.

Waterplane area intact Aw = 50 x 10 = 500 m2; the flooded wing compartment has waterplane area = 15 x 5 = 75 m2 (only starboard half up to the centreline bulkhead).

Lost buoyancy volume of the compartment below the original waterline, with permeability 30%: Vlost = length x breadth x draft x permeability = 15 x 5 x 4 x 0.30 = 90 m3. Corresponding lost weight W = 90 x 1.025 = 92.25 t.

Sinkage: the region once flooded provides no increase of buoyancy; effective sinking waterplane = 500 - 75 = 425 m2. Mean sinkage = 90/425 = 0.212 m.

List: the lost buoyancy acts at the centroid of the lost volume, which is at 5/4 = 2.5 m out from the centreline (half-way between centreline bulkhead and side). Heeling moment = W x y = 92.25 x 2.5 = 230.6 t-m.

New GM. Using the lost-buoyancy method, displacement remains 2050 t. The second moment of area of the intact waterplane about the centreline loses the flooded starboard compartment: I(intact)=50x10^3/12 = 4166.7 m4; I(flooded)=15x5^3/3 = 625 m4; so I(new)=3541.7 m4. BM=new = I(new)/V = 3541.7/2000 = 1.771 m. KB = d/2 = 2 m (approximately, ignoring sinkage and list); GM = KB + BM - KG = 2 + 1.77 - 3 = 0.77 m.

List: tan(list) = heeling moment/(Delta x GM) = 230.6/(2050 x 0.77) = 230.6/1579 = 0.146. List = atan(0.146) = 8.3 deg.

Answer: the vessel lists about 8.3 deg to starboard.