Question

A ship of 3000 tonnes displacement is 100m long, has KM6m, KG5.5m.The centre of flotation is 2m aft of amidships and MCTC40 tonnes m. Find the maximum trim for the ship to enter a dry dock if the metacentric height at the critical instant before the ship takes the blocks forward and aft is to be not less than 0.3m.

KM = 6.0m

KG = 5.5m

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Original GM = 0.5m

Virtual GM = 0.3m

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Virtual loss = 0.2m

Method (a)

Virtual loss of GM (MM1) = P X KM / W

P = virtual loss X W / KW

P = 0.2 X 3600 / 6.

Maximum P = 100 tonnes.

But

P = MCTC X t / l

Or

Maximum t = P x l / MCTC

= 100 x 48 / 40.

Answer maximum trim = 120cm by stern.

Method (b)

Virtual loss GM (GG1) = P x KG / W - P.

0.2 = P x 5.5 / 3000 – P

600 – 0.2P = 5.5P, 5.7P = 600.

Maximum P = 600/ 5.7 = 105.26 tonnes.

But

P = MCTC x t / l = 105.26 x 48 / 40

Answer. Maximum trim = 126.3 cm by stern.

There are, therefore, two possible answers to this question, depending on the method of solution used. The reason for this is that although the effective metacentric height at the critical instant in each case will be the same, the righting moments at equal angles of heel will not be the same.