A draught survey weighs a ship with no scales. The surveyor reads the draught marks, enters the hydrostatic tables, applies a short list of corrections, and subtracts everything on board that is not cargo. Two surveys, one before loading and one after, give the cargo weight as a simple difference. The result is a commercial document: bills of lading and freight payments rest on it, so every line of the standard form exists to remove one known source of error. This chapter works the full form once, using the tools built in Chapters 6 to 9.
Six draughts are read: forward, midships and aft, on both sides. The port and starboard pairs are averaged to remove any small list. After that, the form runs five stages: correct the readings to the perpendiculars, combine them into the quarter mean, enter the tables and apply the two trim corrections, correct for the dock water density, and finally subtract the deductibles. Each stage is short. The discipline is doing all of them, every time, in order.
The hydrostatic tables are built for draughts at the perpendiculars. The painted marks are almost never exactly there: on MV Ninja the forward marks are 2.0 m abaft the FP and the aft marks are 3.0 m forward of the AP, so the distance between the marks is 143.0 m rather than 148.0 m. When the ship is trimmed, the waterline slopes, and a mark that is not at its perpendicular reads a slightly different draught. The fix is a small slide along the waterline.
Final survey. Readings (port and starboard already averaged): forward 8.99 m, midships 9.43 m, aft 9.77 m. Forward marks 2.0 m abaft the FP; aft marks 3.0 m forward of the AP; distance between marks 143.0 m. Find the draughts at the perpendiculars and the true trim.
Observed trim at the marks = 9.77 − 8.99 = 0.78 m by the stern.
FP correction = 2.0 × 0.78 ÷ 143 = 0.011 m. The ship is deeper aft, and the forward marks sit abaft the FP, so the marks read deeper than the FP itself: draught at FP = 8.99 − 0.011 = 8.979 m.
AP correction = 3.0 × 0.78 ÷ 143 = 0.016 m. The aft marks sit forward of the AP, so they read shallower than the AP: draught at AP = 9.77 + 0.016 = 9.786 m.
True trim = 9.786 − 8.979 = 0.807 m by the stern. The corrections are small, but the survey is a business document, and they go in every time.
A loaded hull bends. If the middle sits deeper than the two ends suggest, the ship is sagged; if it sits shallower, she is hogged. The underwater volume is concentrated around the middle of the hull, so the midships reading carries more information about the displacement than the end readings do. The form handles this by weighting: midships gets six votes out of eight.
Continue the survey. The midships marks are 0.8 m abaft midships. Find the corrected midships draught and the quarter mean.
Midships correction = 0.8 × true trim ÷ LBP = 0.8 × 0.807 ÷ 148 = 0.004 m. The marks sit abaft midships on a stern trim, so they read deep: corrected midships draught = 9.43 − 0.004 = 9.426 m.
Check the bend first: the mean of the end draughts is (8.979 + 9.786) ÷ 2 = 9.3825 m, carried as 9.383 m, but midships reads 9.426 m. The ship is sagged by 9.426 − 9.383 = 0.043 m, or 4.3 cm, which is normal for a loaded bulk carrier.
Quarter mean = (8.979 + 6 × 9.426 + 9.786) ÷ 8 = 75.321 ÷ 8 = 9.415 m (9.4151 unrounded).
Using the simple mean of the ends instead would understate the draught by 9.4151 − 9.3825 = 0.0326 m, or 3.3 cm, which at TPC 35.22 is about 115 t of cargo. That is why the form weights midships and why the midships marks are read as carefully as the ends.
The quarter mean now enters the hydrostatic tables, exactly as in Chapter 9. But the tables assume an even keel, and this ship is trimmed 0.807 m. Two printed corrections put that right. The first is the layer correction from Chapter 9, converted to tonnes: it moves the draught from midships to the centre of flotation, where the true mean lives. The second, called the form correction, allows for the hull changing shape over the length of the trim; it uses the change in MCTC half a metre above and below the quarter mean, and it is always added.
Enter the tables at 9.415 m and apply both trim corrections (rows 9.40 m: 29751 t, TPC 35.22, MCTC 403.6, LCF 71.91; 9.60 m: 30456 t, TPC 35.28, MCTC 405.7, LCF 71.84; for the second correction, MCTC at 9.915 m is 408.9 and at 8.915 m is 398.3).
Fraction = (9.415 − 9.40) ÷ 0.20 = 0.075. Displacement = 29751 + 0.075 × 705 = 29804 t; TPC 35.22; LCF 71.90 m foap.
First correction = 80.7 × (74.00 − 71.90) × 35.22 ÷ 148 = 80.7 × 2.10 × 35.22 ÷ 148 = +40 t. The trim is by the stern and the LCF is abaft midships, so the true mean draught is deeper than the quarter mean and the correction is added. Had the LCF been forward of midships, the same formula would have come out negative.
Second correction = 50 × 0.807² × (408.9 − 398.3) ÷ 148 = 50 × 0.651 × 10.6 ÷ 148 = +2 t. Small here, but it grows with the square of the trim.
Corrected displacement = 29804 + 40 + 2 = 29846 t.
The tables are built for salt water of relative density 1.025. Dock water is usually lighter. At the same draught the ship displaces the same volume, but a lighter volume weighs less, so the table figure is scaled by the measured dock density. The density sample is taken from the dock at survey time with a hydrometer, and it matters: on this ship the third decimal place is worth about 29 t.
The dock water sample reads RD 1.018. Deductibles on board at the final survey, from the tank soundings (each volume within its MV Ninja tank capacity, times the relative density of the contents: salt water ballast 1.025, heavy fuel 0.950, diesel 0.850, fresh water 1.000): ballast 1253.6 t (the after peak pressed, 503.0 m³, and 720.0 m³ in the two No.5 double bottoms), heavy fuel 877.8 t, diesel 32.3 t, fresh water 182.0 t, lubricating oil 28 t (the chief engineer's figure). Find the net figure.
Dock water displacement = 29846 × 1.018 ÷ 1.025 = 29642 t.
Deductibles = 1253.6 + 877.8 + 32.3 + 182.0 + 28.0 = 2373.7 t, taken as 2374 t. These are everything on board that is not cargo and not the ship herself, and every tank is sounded at survey time to prove the numbers; a sounding that gives more than the tank can hold (the Diesel Oil Tank, 41.8 m³, holds at most 35.5 t of diesel) is a sign of an error.
Net figure = 29642 − 2374 = 27268 t. This is the ship plus her constant plus the cargo. On its own it names no cargo weight: for that we need the same figure from before loading.
The same form was run at arrival, before loading. Its results: draughts at the perpendiculars 5.070 m forward and 5.170 m aft, corrected midships draught 5.115 m (hogged 0.5 cm), so quarter mean (5.070 + 6 × 5.115 + 5.170) ÷ 8 = 5.116 m, true trim 0.10 m by the stern, dock water RD 1.018. Deductibles then, from the soundings: ballast 8690.9 t (every ballast tank pressed except the after peak at 250.0 m³ and the empty No.5 double bottoms, with 2280.0 m³ still in No.3 hold), heavy fuel 891.1 t, diesel 34.0 t, fresh water 194.0 t, lubricating oil 28 t (rows 5.00 m: 14798 t, TPC 32.28, LCF 77.51; 5.20 m: 15445 t, TPC 32.44, LCF 77.24). Find the cargo loaded.
Tables at 5.116 m: fraction 0.58, displacement = 14798 + 0.58 × 647 = 15173 t; TPC 32.37; LCF 77.35 m foap.
First correction = 10 × (74.00 − 77.35) × 32.37 ÷ 148 = −7 t. In ballast the LCF sits forward of midships, so with a stern trim the true mean is shallower than the quarter mean: this time the formula comes out negative, and the sign takes care of itself. The second correction at 0.10 m of trim is under a tenth of a tonne: recorded as nil.
Corrected displacement = 15173 − 7 = 15166 t; dock water displacement = 15166 × 1.018 ÷ 1.025 = 15062 t; deductibles 8690.9 + 891.1 + 34.0 + 194.0 + 28.0 = 9838.0 t, taken as 9838 t; net figure = 15062 − 9838 = 5224 t. The light displacement is 4950 t, so the ship's constant is 5224 − 4950 = 274 t.
Cargo loaded = 27268 − 5224 = 22044 t. Check the balance: the ship got heavier by 29642 − 15062 = 14580 t while 9838 − 2374 = 7464 t of ballast, fuel and water left her; 14580 + 7464 = 22044 t. The two routes agree.
Six readings, averaged port and starboard; then correct each station to its perpendicular along the trimmed waterline.
Quarter mean = (dFP + 6 × dM + dAP) ÷ 8: midships gets six votes because that is where the underwater volume is.
Two trim corrections: the layer (Chapter 9, in tonnes, sign from the formula) and the form correction (always added, grows with trim squared).
Scale by the measured dock density: W × RD ÷ 1.025. Then subtract the sounded deductibles.
Cargo = final net figure minus arrival net figure. Check it against the weight change plus the deductibles that left the ship.