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Class 2 (2nd Engineer) SC&S 📅 Jan 2019

Exam Question

(a) Explain why the GM must remain positive until the critical instant at which the ship takes the blocks overall. (6)

(b) A ship of displacement 10,010 tones has a container of 10t at Kg = 7.5m. The container is shifted transversely. A pendulum of length 7.5m deflects through 13.5cm. GM of the ship = 0.76m, KM = 6.7m. Find the distance through which the container is shifted. Also find the new KG if the

container is removed. (10)

Exam question diagram

Reference Answer

(a) For safe dry-docking, a ship must meet two key stability requirements:
Positive GM (Metacentric Height):  The ship needs a positive GM. GM is the distance between the centre of gravity (G) and the metacentre (M). A positive GM indicates inherent stability; the ship will right itself if tilted. During dry-docking, the loss of buoyancy as the ship rests on the blocks reduces GM. Insufficient positive GM increases the risk of the ship heeling over or capsizing.
Trim by the stern:  The vessel should be trimmed slightly by the stern (aft end lower than the bow) to ensure the aft end sits on the keel blocks first. This controlled settling minimises the risk of instability during the docking process. An even keel is generally preferred for the initial floating condition before the dry-docking procedure begins.
At the critical moment , the centre of gravity G rises faster than the metacentre M , so G catches up with M, and the vessel stability is compromised. So, GM needs to be positive until the critical moment.
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