Q9 (16 Marks) Ship Stability 🔥 Repeated 8x in exams
SC&S • Written Exam

(a) Define longitudinal centre of gravity (LCG) and longitudinal centre of buoyancy (LCB). (6)

(b) The immersed cross-sectional areas of a ship 120m long, commencing from aft are 2, 40, 79, 100, 103, 104, 104, 103, 97, 58 and 0 m2. Calculate:

(i) Displacement

(ii) Longitudinal position of the centre of buoyancy. (10)

Appeared In: Sep 2025Apr 2025Apr 2023Feb 2021Dec 2019Sep 2019Apr 2019Aug 2018

Verified Model Answer (Text Solution)

Structured for DG Shipping MEO Class II examination scoring criteria.

Exam Ready
Part (a)

Longitudinal Centre of Gravity (LCG):

  • The Longitudinal Centre of Gravity (LCG) is the point along the length of the vessel where the total weight of the ship is considered to act vertically downward.
  • It represents the balance point of the ship's weight distribution and is measured as a distance forward or aft of the midship.

Longitudinal Centre of Buoyancy (LCB):

  • The Longitudinal Centre of Buoyancy (LCB) is the point along the length of the vessel through which the total buoyant force, acting vertically upward, is considered to act.
  • It represents the balance point of the underwater volume of the ship and is also given as a distance forward or aft of the midship.
Part (b)

Given:

$$Common \space interval \space (h) \space = \space {{L} \over h} \space = \space {{120} \over 10} \space = \space 12 $$

Cross-sectional area

SM

Product of volume

Lever

Product of 1st moment

2

1

2

+5

+10

40

4

160

+4

+640

79

2

158

+3

+474

100

4

400

+2

+800

103

2

206

+1

+206

ΣMA = +2130

104

4

416

0

0

104

2

208

-1

-208

103

4

412

-2

-824

97

2

194

-3

-582

58

4

232

-4

-928

0

1

0

-5

0

Σ∇ = 2388

ΣMF = -2542

$$Displacement \space = \space \rho \times {{h} \over 3} \times \sum ∇ \space tonne $$

$$=1.025\times{{12}\over3}\times2388$$

$$Displacement \space = \space 9790.8 tonne$$

Centre of buoyancy from midship (LCB)

$$LCB\:=\:h\times({{\sum M_{A}+\sum M_{F}}\over\sum\nabla})$$

$$=12\times({{2130-2542}\over2388})$$

$$LCB \space = \space -2.07m fwd$$

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