Last-Minute Cargo Loading: Draft and Freeboard
A small coastal cargo vessel is moored in calm fresh water at a river port. Its hull cross-section is box-shaped. Length at waterline 40.0 m, beam 8.0 m, and the vertical distance from keel to main deck edge is 4.0 m. The empty vessel displaces 320 tonnes and floats with a draft of 1.0 m. Crew, fuel and stores add 80 tonnes, already on board and included in that draft. The captain must load steel coils whose total mass is 240 tonnes. The port authority requires a minimum freeboard (deck edge above waterline) of 1.2 m at all times. Seawater density is 1.025 t/m³, fresh water 1.000 t/m³, and the vessel will later sail to sea without changing its load.
1. Determine whether the vessel can take the full 240 t of coils in fresh water while keeping at least 1.2 m freeboard.
Read the worked answer
Step 1 - Find the waterplane area. Box hull: A = L × B = 40.0 × 8.0 = 320 m².
Step 2 - Draft after loading. In fresh water, added mass 240 t needs extra displaced volume 240/1.000 = 240 m³. Extra draft = 240 / 320 = 0.75 m. New draft = 1.0 + 0.75 = 1.75 m.
Step 3 - Freeboard. Freeboard = 4.0 − 1.75 = 2.25 m, which exceeds the 1.2 m minimum.
Conclusion: Yes, the full 240 t can be loaded in fresh water; freeboard is 2.25 m, a margin of 1.05 m.
2. The ship then enters seawater without changing load. Explain what happens to draft and freeboard, and check the 1.2 m rule again.
Read the worked answer
Step 1 - Total displacement. Empty-plus-stores displacement is 320 t at draft 1.0 m, consistent with 320 m² × 1.0 m × 1.000 t/m³. After loading: 320 + 240 = 560 t.
Step 2 - Draft in seawater. Volume displaced = 560 / 1.025 = 546.3 m³. Draft = 546.3 / 320 = 1.707 m.
Step 3 - Freeboard. Freeboard = 4.0 − 1.707 = 2.29 m. The ship rises slightly because seawater is denser; freeboard increases by about 0.04 m.
Judgement: The rule is satisfied in both waters, and the critical case is fresh water (deepest draft, 1.75 m). The captain should verify the fresh-water condition, not the sea condition.
3. Suppose the captain wants a safety margin so that freeboard never falls below 1.5 m even in fresh water. What maximum coil mass may be carried?
Read the worked answer
Step 1 - Maximum draft allowed. Draft ≤ 4.0 − 1.5 = 2.5 m.
Step 2 - Available extra draft. From the current 1.0 m: 2.5 − 1.0 = 1.5 m.
Step 3 - Convert to mass. Extra volume = 320 × 1.5 = 480 m³; in fresh water mass = 480 t.
Answer: Up to 480 t of coils could be carried under the stricter rule. Since only 240 t is planned, the margin is large; the binding constraint is the 1.5 m freeboard target, not the cargo itself.