Q8 (10 Marks) Ship Stability 🔥 Repeated 11x in exams
SC&S • Written Exam

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

(b) A ship 120m long floats at draughts of 5.50m forward and 5.80m aft; MCT 1cm 80 tonne m, TPC 13, LCF 2.5m forward of midships. Calculate the new draughts when a mass of 110 tonne is added 24m aft of midships. (10)

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✓ 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.

LCF in fwd and trim by stern

$$Bodily \space sinkage \space = \space {{mass \space added} \over TPC} \space = \space {{110} \over 13} \space$$

$$=\:8.5\operatorname{\mathrm{cm}}\:=0.085m$$

$$Trim \space = \space {{m \times d} \over MCT_{1cm}}$$

$$ = \space {{110 \times (24 + 2.5)} \over 80}$$

$$Trim=36.43\operatorname{cm}=0.364m\:$$

Change in fwd draught:

$$d_{F}=\frac{-t}{L}\left\lbrack\frac{L}{2}-LCF\right\rbrack$$

$$=\frac{-36.43}{120}\left\lbrack\frac{120}{2}-2.5\right\rbrack$$

$$=-17.45\operatorname{cm}=-0.1745m$$

Change in Aft draught:

$$d_{A}=\frac{+t}{L}\left\lbrack\frac{L}{2}+LCF\right\rbrack$$

$$=\frac{+36.43}{120}\left\lbrack\frac{120}{2}+2.5\right\rbrack$$

$$=+18.97\operatorname{cm}=0.189m$$

New draughts:

$$D_{F}=5.5+0.085-0.175=5.41m$$

$$D_{A}=5.8+0.085+0.18=6.065m$$

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