Maximum list during derrick operation.

Initial condition: displacement = 9900 t; KM = 7.3 m; KG = 6.4 m, so initial GM = KM - KG = 7.3 - 6.4 = 0.90 m.

The first 50 t lift is picked up off the quay (inshore side). While it is suspended at the derrick head, its position is above and outboard of the crane; as it swings, it acts as a weight at the head which is 15 m above the keel and 12 m out from the centreline. The maximum list occurs when the weight is suspended fully outboard (12 m out) before it is placed on deck at 6 m out, because the heeling moment is greatest then.

Effective heeling of the first lift. Treated the 50 t as suspended at (y=12 m outboard, z=15 m above keel):

• Transverse shift of the centre of gravity, GG' = w x y / Delta = 50 x 12 / 9900 = 0.0606 m outboard (about 6.06 cm).
• Vertical rise of the centre of gravity, since the load hangs at the derrick head above deck: compare to its final on-deck position at 6 m out, 9 m above keel. The movement from quay to head raises the effective load centre; the vertical rise of G = w x (rise)/(Delta). For the list the relevant GM is reduced by the raised KG.

A cleaner approach is to compute the moment. At maximum list the weight is fully outboard at 12 m; its KG contribution raises the centre of gravity by GGv = 50 x (15 - 9)/9900 = 50 x 6/9900 = 0.0303 m. New KG = 6.4 + 0.0303 = 6.4303 m; the effective GM falls to 7.3 - 6.4303 = 0.8697 m.

List, GGh = 50 x 12 / 9900 = 0.0606 m outboard.

tan(list) = GGh / GM = 0.0606 / 0.8697 = 0.0697.

So list = atan(0.0697) = 3.99 deg, about 4 deg.

Answer: the maximum list during the operation is about 4 degrees towards the outboard (derrick) side. (Note: after the first lift is placed on deck at 6 m out and the second lift is begun, the situation changes slightly; the critical maximum effect is when the suspended load is furthest outboard.)