The draught survey method
1The lesson in plain words
A draught survey finds the cargo weight from the ship’s displacement before and after loading. Every correction matters: one centimetre can be 50 tonnes.
The idea
Read the draughts and find displacement in the hydrostatic tables. Deduct ballast, fresh water, bunkers and other measured variable weights. The remainder, called net displacement, is lightship plus the constant and any cargo. The constant covers unaccounted stores, spares and sediment. Cargo loaded is final minus initial net displacement, assuming lightship and constant are unchanged. The UN/ECE Code of 1992 for coal cargoes sets out the uniform procedure.
Six draughts and the perpendiculars
Read forward, midships and aft on both sides in calm conditions. Average corresponding readings. Correct forward and aft marks to their perpendiculars, and offset midships marks to geometric midships. Each trim correction has magnitude observed trim × longitudinal offset ÷ length between forward and aft marks. Determine its sign from whether the reference point is deeper or shallower than the mark.
Mean of means
Use the corrected forward (F), midships (M) and aft (A) draughts. The mean of means is (F + A + 2M) ÷ 4. The quarter mean, or mean of mean of means, is (F + A + 6M) ÷ 8. It allows for hog or sag, assuming the hull bends in a smooth curve. Enter the hydrostatic table with the quarter mean.
Trim corrections
Use true trim: corrected aft minus forward draught, in metres. Use its magnitude here. First correction (t) = trim × LCF (centre of flotation) from midships × TPC × 100 ÷ LBP. Add when LCF is towards the deeper end; subtract otherwise. Second correction (t) = 50 × trim² × ΔMCTC ÷ LBP. Here ΔMCTC = MCTC at QM + 0.5 m minus MCTC at QM − 0.5 m. It is normally added.
Density and deductibles
Tables are usually for salt water of 1.025 t/m³. Sample the dock water at several places and depths and use a draught survey hydrometer. Density correction = Δ × (ρ − 1.025) ÷ 1.025. Here Δ is displacement after both trim corrections; ρ is dock-water density in t/m³. Then sound every ballast and fresh water tank, take bunker figures from the engine room, and deduct them all.
2Key data
- Perpendicular correction
- Observed trim × mark offset ÷ length between marks
- Quarter mean
- (F + A + 6M) ÷ 8
- First trim correction (t)
- Trim × LCF × TPC × 100 ÷ LBP
- Second trim correction (t)
- 50 × trim² × ΔMCTC ÷ LBP
- ΔMCTC
- MCTC at QM + 0.5 m minus MCTC at QM − 0.5 m
- Density correction (t)
- Δ × (ρ − 1.025) ÷ 1.025
3Key points
- 1
Correct draughts to the forward perpendicular, amidships and after perpendicular before calculating the quarter mean.
- 2
Use the quarter mean (F + A + 6M) ÷ 8 to allow for hog or sag.
- 3
First trim correction: its sign follows the LCF and the trim; the second is normally added.
- 4
Correct for water density; deduct ballast, fresh water, bunkers and other measured variable weights.
4Frequent mistakes
Wrong: LCF in the table is always measured from midships.
Right: Some tables give LCF from the after perpendicular. Convert it to a distance from midships before using the formula.
Wrong: Reading the draughts in a swell from the quay is good enough.
Right: Read close to the water, from a boat if possible, and average several readings; use a tube or manometer in waves.
Wrong: Density only matters in fresh water.
Right: At the same measured draught, water of 1.018 t/m³ gives about 287 t less displacement than a 42,000 t table value for 1.025 t/m³. The ship’s mass does not change merely because water density changes.
5Practice case & exam keys
Final survey: trim 1.20 m by the stern, LCF 4.0 m aft of midships, TPC 45 t/cm, LBP 160 m, ΔMCTC 36 t·m/cm. Find both trim corrections. A colleague subtracts both corrections. Do you accept the result?
Show the answer
No. Check each sign rather than treating corrections as deductions. First: 1.20 × 4.0 × 45 × 100 ÷ 160 = +135 t. Trim by the stern and LCF aft point towards the same end, so add. Second: 50 × 1.20² × 36 ÷ 160 = +16.2 t, added. The total, +151.2 t, goes onto the table displacement.
Exam keys
- Perpendicular correction = observed trim × mark offset ÷ length between marks.
- QM = (F + A + 6M) ÷ 8.
- 1st trim correction = t × LCF × TPC × 100 ÷ LBP; 2nd = 50 t² ΔMCTC ÷ LBP.
- Density correction = Δ (ρ − 1.025) ÷ 1.025; deduct ballast, fresh water and bunkers.
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