Learning objectives
By the end of this chapter, and of Volume One, you will be able to:
- draw and read the metacentric diagram: KB and KM plotted against draught;
- explain, through BM = I ÷ V, why the KM curve falls, flattens through a minimum, and rises again;
- interpolate any quantity in the hydrostatic table, entering by draught or by displacement;
- assemble a complete departure condition from the light ship to the summer marks;
- apply the fresh and dock water allowances to the sailing draught;
- follow a condition through a voyage: consumption, free surfaces, and the arrival figures;
- diagnose and correct an arrival list from asymmetric consumption;
- run the six step method that solves every condition problem in this volume.
Eleven chapters ago this book began with a hull pushed into the water and the water pushing back. Chapter by chapter the machinery accumulated: displacement from the draughts, moments about the keel, the metacentre, the righting lever, the free surface, the measured light ship, the suspended weight. Every one of those chapters borrowed numbers from the same place, the booklet's hydrostatic table, and mostly took them on trust. This closing chapter pays the trust back. First it draws the table as a picture, the metacentric diagram, and shows why its central column behaves as it does. Then it puts the whole volume to work at once: MV Ninja loaded from her measured light ship to her summer marks, sailed, and brought in, every step from the booklet, nothing on trust any more.
12.1 The hydrostatic table, drawn as a picture
Take the booklet's forty rows, plot KB and KM against draught, and the columns of figures become two curves. The result is the metacentric diagram, and one glance at it tells you things the table whispers only to the patient.
Three features deserve a long look. KB climbs steadily and almost straight: the centre of the underwater volume rises at a little more than half the rate of the draught (KB is 0.52 to 0.53 of the draught at every row). KM starts enormously high, above 19 m with the ship at the lightest tabulated draught of 2.60 m, then falls steeply, flattens, and passes through a shallow minimum of 10.328 m at a draught of 9.40 m, her winter draught, before beginning to rise again. And the two curves never meet: the gap between them is BM, shrinking from nearly 18 m at 2.60 m to less than 5 m at 10.40 m as she loads. Plot the ship's G at her working draught and the diagram becomes an instrument: the vertical gap from G up to the KM curve is the metacentric height itself.
12.2 Why the curve bends, and reading between the rows
The shape of the KM curve is not an accident of MV Ninja: every conventional ship's diagram shares it, and one line of Chapter 7 explains why. KM = KB + BM, and BM = I ÷ V. The waterplane's second moment I depends principally on the beam, which changes little once the bilge is under water: for MV Ninja, I = BM × V grows by about a quarter from the lightest row of the table to the deepest, while the underwater volume V grows more than four times over the same range. So BM collapses as the draught grows, fastest when V is smallest, while KB climbs steadily underneath it. Near the light draughts the collapse of BM dominates and KM plunges; near the working draughts the two effects come into balance and the curve flattens through its minimum; beyond it, KB's patient climb wins and KM rises again.
That flatness is a quiet professional mercy: exactly where the ship spends her loaded life, KM is least sensitive to draught, and a small error in the draught costs almost nothing in KM. But the ship rarely floats exactly on a printed row, so the officer must read between them, and the tool is linear interpolation: over a 0.20 m step every hydrostatic curve is close enough to straight. The straight line is exact to the last printed digit at the working draughts; at the lightest draughts, where the KM curve bends hardest, it is within about two centimetres of KM across a single step, and the displacement column is straight to within 3 t everywhere.
MV Ninja floats at a mean draught of 9.30 m. From the booklet rows (9.20 m: KM 10.332, KB 4.829; 9.40 m: KM 10.328, KB 4.935), find her KM and KB.
fraction of step = (9.30 − 9.20) ÷ (9.40 − 9.20) = 0.10 ÷ 0.20 = 0.5
KM = 10.332 + 0.5 × (10.328 − 10.332) = 10.332 − 0.002 = 10.330 m
KB = 4.829 + 0.5 × (4.935 − 4.829) = 4.829 + 0.053 = 4.882 m
Note the sign care in the KM line: near the minimum the change across the step is negative, and the interpolation must be allowed to subtract. The formula never changes; only the sign of the step's change does.
MV Ninja's cargo plan gives a final displacement of 29000 t. Entering the table by displacement (9.00 m: 28343 t, KM 10.339; 9.20 m: 29046 t, KM 10.332), find the corresponding mean draught and KM.
fraction = (29000 − 28343) ÷ (29046 − 28343) = 657 ÷ 703 = 0.935
draught = 9.00 + 0.935 × 0.20 = 9.187 m
KM = 10.339 + 0.935 × (10.332 − 10.339) = 10.339 − 0.007 = 10.332 m
Entering by displacement is the everyday direction: the moments table finishes with a displacement, and the booklet must be opened at that displacement, not at a guessed draught. The fraction comes from whichever column you enter by; it then serves every other column in the row.
Using the booklet, find KB, KM and hence BM at draughts of 3.00 m, 6.00 m and 9.60 m, and comment on the pattern.
| draught (m) | KB (m) | KM (m) | BM = KM − KB (m) |
|---|---|---|---|
| 3.00 | 1.558 | 17.248 | 15.690 |
| 6.00 | 3.125 | 11.355 | 8.230 |
| 9.60 | 5.041 | 10.330 | 5.289 |
KB is a little over half the draught each time, about 0.52 of it, as it should be for a full formed hull. Between 3.00 m and 9.60 m the displacement grows from 8501 t to 30456 t, a factor of 3.6, and BM falls to a third of its light value: I grows only slowly, being set mainly by the beam, while V has more than tripled. The sum, KM, falls by almost 7 m from 3.00 m to 9.60 m and then changes by only 44 mm from 9.60 m to 10.40 m: the flattening of Figure 12.1, in three rows of arithmetic.
12.3 From keel to load line: the departure
Now the volume goes to work as one machine. The starting entry is the light ship that Chapter 10 measured: 4950 t at KG 8.86 m. The cargo is grain in all five holds, the centres straight from the booklet's capacity table; bunkers and fresh water ride high in the topside tanks. The moments table of Chapter 6 does what it has always done, and the total lands, by design of the loading plan, exactly on the summer displacement.
Assemble MV Ninja's departure condition from the table below, and find her sailing KG and GM at the summer draught (booklet KM 10.330 m). The fuel, diesel and fresh water tanks are all slack.
| Item | w (t) | Kg (m) | moment (t m) |
|---|---|---|---|
| Light ship | 4950 | 8.86 | 43857 |
| No.1 hold grain | 4503 | 7.94 | 35754 |
| No.2 hold grain | 5308 | 7.70 | 40872 |
| No.3 hold grain | 5038 | 7.72 | 38893 |
| No.4 hold grain | 5308 | 7.70 | 40872 |
| No.5 hold grain | 4640 | 8.02 | 37213 |
| Heavy fuel oil (topside) | 509 | 12.65 | 6439 |
| Diesel oil | 35 | 11.45 | 401 |
| Fresh water | 165 | 11.86 | 1957 |
| Totals | 30456 | 246258 |
KG = Σ Vertical Moments ÷ ∆ = 246258 ÷ 30456 = 8.086 m, carried as 8.09 m
GM = KM − KG = 10.330 − 8.086 = 2.244 m, carried as 2.24 m
These are the figures of Worked example 6.4, and meeting them again is the point: then, the light ship line was a number on trust; now it is Chapter 10's measurement, and every entry in the table has been earned somewhere in this volume. The 2.24 m is the solid GM. The 509 t of oil fills the four heavy fuel oil tanks only three quarters full, and the diesel oil and fresh water tanks are slack as well, so, as Worked example 9.4 showed, the seven free surfaces (each i × relative density of the liquid) total 815.2 t m and cost 815.2 ÷ 30456 = 0.027 m: she sails with a fluid KG of 8.086 + 0.027 = 8.113 m and a fluid GM of 2.244 − 0.027 = 2.217 m.
One matter remains before she sails: the berth is dock water. Chapter 5's allowances decide how far past the summer mark she may lawfully sink, knowing the sea will give it back.
MV Ninja completes loading at a berth where the dock water density is 1.012 t/m³. Using the booklet's summer figures (W 30456 t, TPC 35.28), find her fresh water allowance, her dock water allowance, and the maximum draught to which she may load at the berth.
FWA = W ÷ (4 × TPC) = 30456 ÷ (4 × 35.28) = 30456 ÷ 141.12 = 215.8 ≈ 216 mm
DWA = FWA × (1.025 − ρdock) ÷ 0.025 = 216 × (1.025 − 1.012) ÷ 0.025 = 216 × 0.52 = 112 mm
maximum berth draught = 9.600 + 0.112 = 9.712 m
Both formulas are the MCA sheet's (September 2020). The 112 mm is not a concession but a prediction: it is precisely the rise the denser sea will supply, so that she crosses into salt water with her summer mark just awash, and not a millimetre under.
12.4 The voyage: the condition is a moving target
A condition is not a fact about a ship: it is a fact about a moment. The instant she sails, consumption begins rewriting it, and the officer's job is to know the new figures before the sea asks for them.
At departure (Worked example 12.4) the No.2 heavy fuel oil tanks hold 150 t each and the No.1 tanks 104.5 t each, the fresh water tanks 90 t (port) and 75 t (starboard), and the diesel oil tank 35 t: all seven slack. On a ten day passage MV Ninja burns 400 t of heavy fuel oil (Kg 12.65 m), the No.2 pair first until it is dry and then from the No.1 pair, and consumes 100 t of fresh water (Kg 11.86 m), the starboard tank first until it is dry; no diesel oil is used. The booklet tabulates the free surface inertia i as 147 m&sup4; for each No.1 fuel tank, 145 m&sup4; for the port fresh water tank and 28 m&sup4; for the diesel oil tank, and the relative densities are 0.950, 1.000 and 0.850. Find the arrival displacement, mean draught, KG, and the solid and fluid GM. (Interpolate the booklet at the arrival displacement.)
arrival ∆ = 30456 − 400 − 100 = 29956 t
arrival moments = 246258 − 400 × 12.65 − 100 × 11.86 = 246258 − 5060 − 1186 = 240012 t m
KG = 240012 ÷ 29956 = 8.012 m, say 8.01 m
Entering the table at 29956 t (between 29751 and 30456): fraction = 205 ÷ 705 = 0.291; draught = 9.400 + 0.291 × 0.20 = 9.458 m; KM = 10.328 + 0.291 × 0.002 = 10.329 m
solid GM = 10.329 − 8.012 = 2.317 m, say 2.32 m
Which tanks are slack on arrival? The No.2 pair (300 t at departure) is dry; the No.1 pair has given up the other 100 t and holds 109 t, slack. The starboard fresh water tank (75 t) is dry and the port tank holds 90 − 25 = 65 t, slack. The diesel oil tank is slack as it was. The free surface moment of each is i × relative density (Chapter 9), never i alone:
ΣFSM = 2 × 147 × 0.950 + 145 × 1.000 + 28 × 0.850 = 279.3 + 145.0 + 23.8 = 448.1 t m
FSC = ΣFSM ÷ ∆ = 448.1 ÷ 29956 = 0.015 m, so fluid GM = 2.317 − 0.015 = 2.302 m, say 2.30 m
She arrives stiffer than she sailed on both counts: the consumption came from high tanks, so G moved down and the solid GM rose from 2.24 m to 2.32 m; and burning the No.2 pair dry and draining one fresh water tank closed three of the seven free surfaces, so the correction fell from 0.027 m to 0.015 m and the fluid GM rose from 2.217 m to 2.30 m. Had the same 400 t been burned from double bottom tanks at Kg 1.12 m the arrival KG would have been (246258 − 448 − 1186) ÷ 29956 = 8.166 m, a rise; both results depend on which tanks are used, which is why the consumption plan is a stability document, not just an engineering one.
Approaching the pilot station, MV Ninja is found to be listing steadily. Investigation shows the last 100 t of fuel were drawn from the No.1 H.F.O. tank (port) only, whose centre the booklet puts 9.34 m from the centreline. Using the arrival condition of Worked example 12.6 (∆ 29956 t, fluid GM 2.30 m), find the list, and the mass of fuel that must be transferred from the starboard tank of the pair to bring her upright for the pilot.
Weight removed from the port side moves G away from it, to starboard:
GGH = (100 × 9.34) ÷ 29956 = 934 ÷ 29956 = 0.0312 m to starboard
tan(List) = GGH ÷ GM = 0.0312 ÷ 2.30 = 0.0136, so List = 0.8° to starboard
The starboard tank of the pair is the mirror of the port one, so a transfer from starboard to port moves each tonne through 2 × 9.34 = 18.68 m:
correcting transfer: w = (GGH × ∆) ÷ d = 934 ÷ 18.68 = 50 t from starboard to port
A textbook forensic: the list formula run forwards to explain the symptom, then backwards to prescribe the cure, with the fluid GM of the actual arrival condition doing the honest work in both directions. Chapter 7 diagnosed; Chapter 11 sized the transfer; Chapter 9 insisted the GM be the fluid one. Check the tanks: the port tank held 104.5 t at departure, so it could give up 100 t; afterwards it holds 4.5 t and the starboard tank 104.5 t, and the 50 t transfer leaves 54.5 t in each, both still slack, so the free surface correction does not change.
12.5 The method, and the judgement
Strip away the stories and every condition problem in this volume, and in the examination hall, is the same six steps in the same order.
Run the judgement, step six, on MV Ninja's departure condition of Worked example 12.4: ∆ 30456 t, solid KG 8.09 m and GM 2.24 m, fluid KG 8.113 m and GM 2.217 m (Worked example 9.4), upright. The booklet's maximum KG table gives 9.753 m at 30000 t and 9.631 m at 30500 t, so 9.642 m at 30456 t.
The marks: ∆ = 30456 t is exactly the summer displacement; with the DWA of Worked example 12.5 in hand she may load to 9.712 m at the berth and lift to her marks in salt water. Compliant.
Maximum KG: it is the fluid KG that is checked, as Chapter 9 requires: 8.113 m against a permitted 9.642 m, a margin of 1.53 m. Compliant, by a street.
The GM itself: 2.217 m fluid, against the 0.15 m minimum of the intact stability criteria and the 0.69 m that the maximum KG implies at this displacement (10.330 − 9.642): comfortable rather than punishingly stiff, so she will be seakindly as well as safe.
The list: the horizontal moments of the plan cancel; she sails upright.
Four verdicts, all clear, and each one traceable through this volume to a measurement or a booklet page. That traceability, more than any single number, is what the signature on the loading plan certifies.
12.6 The end of Volume One
The metacentric diagram was the last piece of trust to repay, and with it repaid, the foundation is complete. You can now take a ship from her measured light ship to her marks, prove her stable, sail her, and bring her in, using nothing but her own approved booklet and the discipline of moments. Volume Two turns the same machinery fore and aft: trim and longitudinal stability, dry docking, bilging and the flooded compartment, and the full statutory judgement of the IS Code, all aboard the same MV Ninja, whose booklet you now know from its first measured page to its last interpolated row.
Interactive: the metacentric diagram explorer
The booklet's forty rows, live. Slide the draught and the cursor rides both curves; set a KG and the diagram reads your GM by eye, exactly as Figure 12.1 promised.
Interactive: the interpolation trainer
The trainer deals a draught or a displacement between the rows and shows you the two rows either side, just as the booklet would. Interpolate with pencil and paper, answer, and build a streak. Tolerance ±0.002 m on KM, ±0.005 m on draught.
Interactive: the grand loading laboratory
Worked example 12.4 as a machine. Edit any weight or Kg; the laboratory recomputes the totals, interpolates KM from the booklet at your displacement, and judges the condition against the summer displacement and the maximum KG table, live. The GM shown is the solid GM; the free surface correction of the slack tanks (0.027 m for the preset condition) is Worked example 9.4's business.
Interactive: the voyage player
Press Sail and watch ten days of Worked example 12.6 unfold: 40 t of fuel and 10 t of water a day, KG and GM recomputed continuously, the draught easing up the marks, and the free surface correction of the seven slack tanks (815.2 t m) narrowing to that of four (448.1 t m) on day 7.5, when the No.2 fuel pair and the starboard fresh water tank run dry. Tick the box to hide the free surface correction and see the solid GM alone.
Interactive: the dock water checker
Worked example 12.5 for any berth. Slide the dock density; the checker computes the FWA from the booklet's summer figures and hands you the lawful berth draught.
Chapter summary, and the volume's
- The metacentric diagram is the hydrostatic table drawn: KB and KM against draught, with G plotted at the working draught and GM read as the gap.
- KM = KB + BM with BM = I ÷ V: BM collapses as V grows while KB climbs, so KM falls, flattens through a minimum near the working draughts, and rises again.
- Linear interpolation reads between the rows: fraction of the step times the change across it, entering by draught or by displacement alike.
- The departure condition is one moments table from the measured light ship to the marks; the dock water allowance governs the berth draught.
- Consumption rewrites the condition continuously: rerun the machinery at arrival, with fresh interpolation and the free surface moment, i × relative density of the liquid, of every tank left slack.
- The six step method, displacement, moments, interpolation, GM and surfaces, the horizontal story, judgement, solves every condition problem in this volume.
Self test questions
Work each question with pencil and paper first. Your score appears in the bar below.