Q10 (10 Marks) Ship Stability 🔥 Repeated 2x in exams
SC&S • Written Exam

The following data are available from the hydrostatic curves ofa vessel.

Draught (m) 4.9 5.2

KB (m) 2.49 2.61

KM (m) 10.73 10.79

I(m4) 65250 68860

Calculate the TPC at a draught of 5.05m.

Appeared In: Sep 2023Jul 2022

Verified Model Answer (Text Solution)

Structured for DG Shipping MEO Class II examination scoring criteria.

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TPC at a draught of 5.05 m from hydrostatic data.

Data: Draught 4.9 m: KB 2.49, KM 10.73, I 65250 m4. Draught 5.2 m: KB 2.61, KM 10.79, I 68860 m4.

BM = KM - KB. At 4.9 m: BM = 10.73 - 2.49 = 8.24 m. At 5.2 m: BM = 10.79 - 2.61 = 8.18 m.

Volume of displacement V = I/BM. At 4.9 m: V = 65250/8.24 = 7919 m3. At 5.2 m: V = 68860/8.18 = 8418 m3.

Displacement = V x 1.025. At 4.9 m: 8117 t. At 5.2 m: 8629 t.

The waterplane area between these draughts is the rate of change of volume with draught: A = (V2 - V1)/(draught change) = (8418 - 7919)/(5.2 - 4.9) = 499/0.3 = 1663 m2.

TPC = A x 1.025/100 = 1663 x 1.025/100 = 17.05 t/cm.

Answer: the TPC at a draught of 5.05 m is about 17.0 t/cm.

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