Q7 (16 Marks) Ship Stability 🔥 Repeated 5x in exams
SC&S • Written Exam

(a) Describe briefly the significance of the factor of subdivision.

(b) A ship 120m long has a light displacement of 4000 tonne and LCG in this condition 2.5m aft of midships. The following items are then added:

Cargo 10000 tonne LCG 3.0m forward of midships

Fuel 1500 tonne LCG 2.0 m aft of midships

Water 400 tonne LCG 8.0m aft of midships

Stores 100 tonnes LCG 10.0m forward of midships

Using the following hydrostatic data, calculate the final draughts:

Draught (m)

Displacement (t)

MCT1cm (tm)

LCB from midships

LCF from midships

8.50

16650

183

1.94F

1.29A

8.00

15350

175

2.10F

0.60F

Appeared In: Apr 2026Mar 2026Dec 2025Dec 2024Apr 2024

Verified Model Answer (Text Solution)

Structured for DG Shipping MEO Class II examination scoring criteria.

Exam Ready
Part (a)

Factor of Subdivision

The factor of subdivision introduces a safety measure by reducing the size of the compartments to limit the effects of flooding. It ensures that the ship's draft or trim has less chance of touching the margin line during flooding or heeling.

Permissible Length Formula:

$$Permissible\:length=\frac{Floodable\:length}{Factor\:of\:Subdivision}$$

A smaller factor of subdivision leads to a smaller permissible length, requiring more numerous and smaller compartments. This reduces the potential for catastrophic flooding, as a smaller flooded area is less likely to exceed the ship's reserve buoyancy and cause it to sink. The factor of subdivision is determined by the ship's length and its intended service. The nature of service is quantified by a "criterion of service" (Cs) number, which considers the proportion of passenger and machinery spaces to the total volume of the ship. A higher Cs number (indicating more passenger space) typically results in a lower factor of subdivision and therefore smaller compartments.

Part (b)

Mass added

LCG from midship

Mass moments

F

A

10000

3.0 Fwd

30000

1500

2.0 Aft

3000

400

8.0 Aft

3200

100

10.0 Fwd

1000

4000

2.5 Aft

10000

16000

31000

16200

$$Excess\:moment=31000-16200=14800$$

$$LCG=\frac{\sum M}{\sum m}=\frac{14800}{16000}$$

$$LCG=0.925m\:fwd\:of\:midship$$

Draught

Displacement

MTC 1cm

LCB from midship

LCF from midship

8.5

16650

183

1.94 Fwd

1.20 Aft

8.25

16000

179

2.03 Fwd

0.57 Aft

8.0

15350

175

2.10 Fwd

0.06 Fwd

$$LCG=0.925m\:Fwd\:of\:midship$$

$$LCB=2.02m\:Fwd\:of\:midship$$

$$Trimming\:lever=LCB-LCG$$

$$2.02-0.925$$

$$=1.09m$$

$$Trimming\:moment=m\times d$$

$$=16000\times1.09$$

$$=17440tm$$

$$Change\:of\:trim=\frac{Trimming\:moment}{MCT_{1\operatorname{\mathrm{cm}}}}$$

$$=\frac{17440}{179}$$

$$97.43\operatorname{cm}$$

$$Draft\:fwd=8.25-\frac{97.43}{100\times120}\left(\frac{120}{2}+0.57\right)$$

$$=7.756m$$

$$Draft\:aft=8.25+\frac{97.43}{100\times120}\left(\frac{120}{2}-0.57\right)$$

$$=8.735m$$

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