Half of this volume has been spent reading three columns out of the hydrostatic booklet: TPC, MCTC and LCF. All three are made from the same thing, the waterplane, the flat shape the ship cuts in the sea surface. This chapter looks at that shape directly: how to measure it, why it changes as the ship sinks, why the three columns drift at different speeds, and how to handle the special case where the area changes size in one jump. Nothing here is new machinery; it is the reason the machinery of Chapters 6 to 12 behaves the way it does.
The waterplane does three jobs at once. Its area sets the TPC, because loading a tonne means pressing that area one small step deeper. Its centroid is F, the point the ship trims about. And its spread about F, the second moment of area, builds BML and with it the MCTC. (Its spread about the centreline, the transverse second moment, also fixes BM = I ÷ V and with it KM, the fourth job, which Volume One, Chapter 7 introduced.) Change the waterplane and all three change together, but not by the same amount, and the difference is the story of this chapter.
A ship shaped waterplane is always smaller than the rectangle drawn round it, because the bow is pointed and the stern is rounded. The fraction of the rectangle the ship actually uses is the waterplane coefficient, Cw. The booklet never prints the area, but the TPC column contains it: turn the TPC formula around and the area falls out.
From MV Ninja’s booklet, find her waterplane area and Cw at 4.80 m and at 9.60 m (TPC 32.10 and 35.28; length 148 m, breadth 24.2 m).
Area = TPC × 100 ÷ 1.025. At 4.80 m: 32.10 × 100 ÷ 1.025 = 3131.7 m². At 9.60 m: 35.28 × 100 ÷ 1.025 = 3442.0 m².
The rectangle is 148 × 24.2 = 3581.6 m² at both draughts. So Cw = 3131.7 ÷ 3581.6 = 0.874 at 4.80 m, and 3442.0 ÷ 3581.6 = 0.961 at 9.60 m.
Between the two draughts the waterplane gained 310.2 m², about 9.9%. The ship fills out her own rectangle as she sinks, which is ordinary hull form: fine ends near the keel, full ends near the load line.
Chapter 9 noticed that LCF falls, walks aft, as the draught rises. The waterplane areas explain it. If the centroid of the whole moved, the new area must have arrived off centre, and a one line balance of moments says exactly where.
Using the areas of Worked example 13.1 and the booklet’s LCF values (77.79 m foap at 4.80 m; 71.84 m foap at 9.60 m), find where the added 310.2 m² arrived.
Moments about the after perpendicular: new area × new LCF = old area × old LCF + added area × x.
Carrying the areas to two decimals (3131.71 and 3441.95 m², a difference of 310.24 m²): 3441.95 × 71.84 = 3131.71 × 77.79 + 310.24 × x, so x = (247270 − 243616) ÷ 310.24 = 3654 ÷ 310.24 = 11.8 m foap. The result is sensitive to the areas: one square metre more in the new area moves x by about 0.19 m, so the TPC values must be used as printed and not rounded on the way.
Nearly all the new waterplane arrived at the stern counter, 71.84 − 11.8 = 60.0 m abaft of F. That is why F walks aft as she sinks, and it is about to explain something bigger in Worked example 13.5.
Chapter 12 worked bilging on a box with MV Ninja’s length and beam. How wrong is that box as a model of the real ship? The waterplane numbers give an honest answer, and the answer is also the standard way to estimate TPC when only the dimensions are known.
Find the TPC and MCTC of the box (148 × 24.2, draught 4.00 m, salt water) and compare with the real ship’s booklet row at the same draught (4.00 m: TPC 31.48, MCTC 293.0).
Box TPC = 3581.6 × 1.025 ÷ 100 = 36.71. The real ship reads 31.48: the ratio 31.48 ÷ 36.71 = 0.8575, which is simply her Cw at this draught (3071.2 ÷ 3581.6 = 0.8575, printed 0.857). As a check, 148 × 24.2 × 0.857 × 1.025 ÷ 100 = 31.46, the last figure lost in the rounding of the coefficient.
Box MCTC: BML = 24.2 × 148³ ÷ 12 ÷ 14326.4 = 456.3 m, so MCTC = 14684.6 × 456.3 ÷ (100 × 148) = 452.8 t m. The real ship reads 293.0: the box overshoots by more than half as much again.
The lesson is the shape of the error: the box overstates TPC by 17% but MCTC by 55%, because the area the real ship is missing sits at her fine ends, exactly where the second moment counts it hardest. (At the same displacement rather than the same draught, 14684.6 t floats the ship at 4.965 m, TPC 32.25 and MCTC 312.9, and the box is still 14% high on TPC and 45% high on MCTC.) Box answers are for boxes; the booklet is for the ship.
Usually the waterplane grows smoothly. A watertight flat can make it change size suddenly instead. Chapter 12 met the easy cases; the interesting one is a flat sitting just above the waterline, with the space below it bilged. While the waterline is below the flat, the flooded slice of waterplane is out of action. The moment the ship sinks past the flat, the watertight top starts working and the full waterplane comes back. One problem, two waterplane areas, and the sum runs in two stages.
The Chapter 12 box (148 × 24.2 m) floats at 3.90 m. A full beam compartment 11 m long has a watertight flat 4.10 m above the keel, and is bilged below the flat. Find the final draught.
Volume of lost buoyancy = 11 × 24.2 × 3.90 = 1038.18 m³.
Stage 1, below the flat: the intact waterplane is (148 − 11) × 24.2 = 3315.4 m². Sinking from 3.90 to the flat at 4.10 recovers 3315.4 × 0.20 = 663.08 m³: not enough, so she sinks past the step.
Stage 2, above the flat: still to recover = 1038.18 − 663.08 = 375.10 m³, and the watertight top now works, so the full 3581.6 m² carries it: extra sinkage = 375.10 ÷ 3581.6 = 0.1047 m, 0.105 m.
Final draught = 4.10 + 0.105 = 4.205 m. Always test whether the sinkage crosses the step before finishing the sum: had the first stage recovered everything, the second waterplane would never have entered the problem.
Between 4.80 m and 9.60 m, MV Ninja’s TPC rises 9.9% but her MCTC rises 31.2% (309.3 to 405.7). Explain the difference using Worked example 13.2.
TPC counts area: 310.2 m² more area on 3131.7 is 9.9%, and that is the whole TPC story.
MCTC counts area times distance from F, squared. Worked example 13.2 put the added area at 11.8 m foap, which is 60 m from F at 71.84 m foap. Each of those square metres therefore adds about 60² = 3600 m⁴ of inertia, while a square metre added at F itself would add nothing.
So the same 310 m² that nudged the TPC by a tenth pushed the MCTC by nearly a third. The general rule: area added at the ends of the ship feeds MCTC far harder than TPC, and area added near F feeds TPC alone. The booklet’s columns drift at different speeds because they measure different moments of the same shape.
The waterplane does three jobs: its area is the TPC, its centroid is F, its spread about F is the MCTC.
The booklet hides the area inside the TPC column: area = TPC × 100 ÷ density, and Cw = area ÷ (L × B).
As the draught rises the hull fills its rectangle: MV Ninja’s Cw grows from 0.874 to 0.961, and the new area arrives right aft, walking F with it.
A watertight flat can change the working waterplane in one jump: test whether the sinkage crosses the step, and run the sum in two stages if it does.
Area added far from F counts by distance squared: that is why MCTC outruns TPC, and why box estimates overshoot MCTC worst of all.