**Drydocking and Stability Maths (Derrett Stability)**

**A ship of 3000 tonnes displacement is 100 m long, has KM = 6 m, and KG = 5.5 m. The center of flotation is 2 m aft of amidships and MCTC 40 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 0.3 m.**

KM = 6.0 m

KG = 5.5 m

Original GM = 0.5 m

Virtual GM = 0.3 m

Virtual loss = 0.2 m

**Method (a)**

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

or, P = (Virtual loss x W) / KM

= (0.2 x 3000) / 6

Maximum P = 100 tonnes

But, P = (MCTC X t) / l

Maximum t = (P x l) / MCTC

= (100 x 48) / 40

*Maximum trim 120 cm by the stern. (Ans.)*

**Method (b)**

Virtual loss of 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

Maximum t = (P x l) / MCTC

(105.26 x 48) / 40

**Maximum trim= 126.3 cm by the stern. (Ans.)**

**Drydocking and Stability Maths **

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.