Before docking, the ship must be stable with an adequate GM, and the docking operation must not cause a list or loss of stability. As the water is pumped out, the keel-block reaction reduces the effective buoyancy, lowering KB and BM and hence GM; the ship must retain sufficient GM throughout so that it does not heel over on the blocks. Requirements:

• Adequate initial GM and even keel (or slight trim), with no list; ballast arranged to give even keel and to press up or empty tanks to avoid free surface.
• The docking weight, draft and trim must be such that the keel blocks contact evenly over the full keel length; excessive trim overloads the aft blocks.
• The ship must be central over the keel line and the dock level; the reaction builds gradually and the GM must remain positive at all stages.
• No weights moved during docking; the stability data (hydrostatics, docking plan) used to check the condition.

If the GM becomes too small or negative during docking, the vessel can heel and capsize on the blocks, so a stability margin is insisted on.

Ship 8,000 t displacement, 110 m long, floats in sea water (1.024 t/m3) at draughts 6.0 m forward and 6.3 m aft. TPC = 16, LCB 0.6 m aft of midships, LCF 3 m aft of midships, MCT1cm = 65 t-m. The vessel moves into fresh water (1.000 t/m3). Calculate the distance a 50 t mass must be moved to bring the vessel to an even keel, and the final draught.

Step 1 - change of mean draught due to density.

Waterplane area A = TPC x 100/rho = 16 x 100/1.024 = 1562.5 m2.

Volume in sea water = 8000/1.024 = 7812.5 m3; volume in fresh water = 8000/1.000 = 8000 m3. Increase in volume = 187.5 m3.

Increase in mean draught = 187.5/1562.5 = 0.12 m. New mean draught = 6.15 + 0.12 = 6.27 m.

Step 2 - change of trim due to density.

The added buoyancy (187.5 t) acts at the centre of flotation (3 m aft of midships), while the original centre of buoyancy is at LCB 0.6 m aft of midships. The moment about the centre of flotation = 187.5 x (3 - 0.6) = 187.5 x 2.4 = 450 t-m.

Change of trim = moment/MCT1cm = 450/65 = 6.92 cm. Since the added buoyancy is aft of the original CB, it lifts the stern, reducing the stern-down trim. Original trim = 0.3 m (6.3 - 6.0) by the stern. New trim = 0.3 - 0.069 = 0.231 m by the stern.

Step 3 - move the 50 t mass to bring the vessel to an even keel.

To remove the 0.231 m (23.1 cm) stern trim, the required change of trim moment = 23.1 x 65 = 1502 t-m. For a 50 t mass, the distance moved = 1502/50 = 30.0 m. The mass must be moved forward (towards the bow) by 30 m.

Final draught: since the displacement is unchanged, the even-keel draught equals the new mean draught = 6.27 m.

Answer: move the 50 t mass 30 m forward; final even-keel draught = 6.27 m.