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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