Loading a Deck Weight with a Slack Tank
Your 120-ft inspected vessel has displacement 250 long tons, KM 8.0 ft, and KG 7.2 ft. You plan to load a 20-long-ton deck package at a vertical center of gravity 12 ft above the keel. At the same time, a rectangular double-bottom tank 20 ft long and 10 ft wide is partially filled with seawater. The tank is open to atmosphere and has no longitudinal subdivision. The vessel currently has no list. The owner wants the load carried on this trip. Minimum acceptable GM for the voyage is 0.50 ft by company policy.
1. Compute GM before and after loading, ignoring free surface.
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Before loading: GM = KM − KG = 8.0 − 7.2 = 0.80 ft.
After loading, new KG = (250 × 7.2 + 20 × 12) ÷ 270 = (1,800 + 240) ÷ 270 = 7.56 ft. Assuming KM remains about 8.0 ft, GM = 8.0 − 7.56 = 0.44 ft. The load has already cut GM below the company minimum before free surface is considered.
2. Compute the free-surface correction and the final GM. Is the load acceptable?
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Volume displacement = 250 long tons × 35 ft³/long ton = 8,750 ft³. Tank moment of inertia i = (20 × 10³) ÷ 12 = 1,667 ft⁴. Free-surface correction = i ÷ V = 1,667 ÷ 8,750 = 0.19 ft.
Final GM = 0.44 − 0.19 = 0.25 ft, which is below the 0.50-ft minimum. The load is not acceptable as proposed.
3. At what loading height would GM be zero without free surface, and what do you recommend?
Read the worked answer
Set new KG equal to KM: (1,800 + 20h) ÷ 270 = 8.0 → 1,800 + 20h = 2,160 → h = 18 ft.
At an 18-ft loading height, solid GM would be zero, and with free surface it would be negative. Recommendation: lower the package, fill or press the slack tank to eliminate free surface, or add low ballast before sailing. Do not sail with 0.25 ft GM.