Chapter 4 left a ship hanging: push KG through the metacentre and the wall sided formula promises she will not capsize but settle at a definite angle, holding a small positive GM there as if nothing were wrong. This chapter is about that ship. She is the most dangerous vessel in this volume precisely because she looks so ordinary: a few degrees of heel, a lazy roll, and a master who reads it as a list and corrects it as a list will roll her over. Everything here runs on MV Ninja in her post discharge condition: 26120 t, KM 10.400 m, KB 4.384 m, so BM = 6.016 m.
No single mistake produces a loll; a season of small ones does. On a ship whose fuel is carried in the double bottom the bunkers burn out from low down and G creeps up (MV Ninja carries her heavy fuel oil high, in wing tanks at Kg 12.65 m and 10.27 m, so on this ship burning fuel lowers G a little, though the emptying tanks still go slack). The emptying tanks go slack and free surface lifts the effective G further. A timber deck cargo absorbs rain and spray, and the usual allowance is ten per cent of its own weight; seas ship aboard and lie on deck; ice grows aloft where nobody weighs it. Each rung is routine. Together they can lift G through M, and the first anyone hears of it is a ship that will not sit upright.
Run the wall sided formula through a family of GM values and the whole taxonomy of initial stability appears in one picture. Positive GM: the curve leaves the origin climbing, stiff or tender by slope. Zero GM: it leaves flat, with only the wedge term to build on. A small negative GM: it dips below the axis, negative levers pushing her away from upright, until the climbing metacentre repays the deficit and the curve crosses back to zero. That crossing is the angle of loll: the angle where she is once again in equilibrium, and where she will sit, on either side indifferently, because the geometry is symmetrical. With a large negative GM the crossing lies beyond the deck edge, where the wall sided formula no longer applies and the ship’s real curve must be used; if that curve never becomes positive she capsizes.
After discharge and a hard week, MV Ninja floats at 26120 t with a solid KG of 10.41 m. The after peak tank (503.0 m³ of sea water when full) is slack. The booklet gives KM 10.400 m and KB 4.384 m at this displacement, and for the after peak i = 3976 m⁴. Show she is lolled and find the angle.
FSM = i × density of the contents = 3976 × 1.025 = 4075 t m. FSC = FSM ÷ Δ = 4075 ÷ 26120 = 0.156 m, so fluid KG = 10.41 + 0.156 = 10.566 m.
GM (fluid) = KM − KG = 10.400 − 10.566 = −0.166 m: negative upright, so she cannot sit upright.
BM = KM − KB = 10.400 − 4.384 = 6.016 m.
tan θloll = √(−2 × (−0.166) ÷ 6.016) = √0.0552 = 0.235, so θloll = 13.2°, port or starboard as the sea last left her. The booklet’s maximum KG at this displacement is 10.250 m; her fluid KG is 0.316 m over it.
An angle of list is caused by weight distributed unsymmetrically about the centreline: G is off centre, GM is positive, and the ship heels to the side of the offending weight and no other. An angle of loll is caused by a negative upright GM with G on the centreline: she may settle to either side, and a passing swell can flop her from one to the other. The clinometer cannot tell them apart. The officer must, because the cure for a list, moving weight towards the high side, is the one action guaranteed to capsize a lolled ship.
MV Ninja lies at her berth heeled 13° to port in the condition of Worked example 5.1. The mate proposes pumping ballast from the port double bottom tanks to starboard to bring her up. Set out the diagnosis that must precede any such action.
Interrogate the weights. The deadweight ledger shows nothing loaded, discharged or shifted off centre: no cause for a list exists on paper.
Interrogate the behaviour. The lashing gang report she lay to starboard at dawn and flopped to port when the tug passed: a listed ship holds her side; only a lolled ship changes it. Her rolling near the angle is sluggish and reluctant.
Interrogate the arithmetic. The sums of Worked example 5.1: fluid GM −0.166 m, predicted loll 13.2°, agreeing with the clinometer.
Verdict: loll, not list. The proposed transfer to starboard is vetoed: it is the forbidden move of Section 5.6, and Worked example 5.5 prices what it would have done.
By Chapter 4’s formula she holds GM = (−2 × initial GM) ÷ cos θ at the loll, a genuinely positive figure, and this is why she sits there rather than rolling over. It is a thin comfort. Everything else about her stability is mutilated, and the examiner expects the catalogue.
Find the metacentric height MV Ninja holds at her 13.2° loll, and state why the figure must not reassure anybody.
GM at loll = (−2 × (−0.166)) ÷ cos 13.2° = 0.331 ÷ 0.974 = +0.340 m.
A ship upright with GM 0.34 m would be lawful. A ship lolled at 13.2° with the same figure is wounded: her righting levers only exist beyond the loll. On the booklet KN curve at 26120 t her GZ is negative to 12.8°, her maximum is 0.205 m at 30° and she vanishes at 41.2°, against 0.370 m and 47.5° for a stable twin with GM +0.166 m; the area under her positive curve is 0.051 m rad against 0.145 m rad; the deck edge (23.0° at this draught of 8.365 m) is only 9.8 degrees further on; and every further rise of G deepens the angle. Far less wave energy is now enough to put her over.
The disease is G too high, so the cure is G brought down, and nothing else. The sequence is fixed: verify the diagnosis; strike or lower topweight where possible and press up slack tanks one at a time to kill free surface, starting with centreline tanks or the tank on the low side (never a slack tank on the high side while GM is negative: pressing it up adds a high side moment); then, if double bottom ballast is needed, choose a divided tank to keep the new free surface small, fill one tank at a time, and fill the tank on the low side first. Expect her to lie deeper while it fills: the off centre weight and the fresh free surface both act before the lowering of G does. Hold your nerve; when the tank presses full the free surface vanishes, G is lower, and she rides listed but stable. Only then fill the high side tank to bring her upright.
Bring the ship of Worked example 5.1 upright. She lies lolled 13.2° to port at 26120 t, solid KG 10.41 m, with the after peak slack (FSM 4075 t m). Available: the after peak is 90% full and can be pressed up with 52 t of sea water entering at its centroid, Kg 9.52 m on the centreline; and No. 1 double bottom, port and starboard, is empty. The booklet gives each No. 1 tank 386.5 m³ (396.2 t of sea water) at Kg 1.43 m, centroid 6.21 m off the centreline, i = 1949 m⁴, so 1949 × 1.025 = 1998 t m of free surface while slack. Follow the drill, stage by stage, taking KM and KB from the booklet at each displacement and using the wall sided balance (GM + ½ BM tan² θ) tan θ = GGh for any heel with weight off the centreline.
| Stage | Δ (t) | KG solid (m) | FSM (t m) | KG fluid (m) | KM (m) | GM (m) | Attitude |
|---|---|---|---|---|---|---|---|
| As found | 26120 | 10.410 | 4075 | 10.566 | 10.400 | −0.166 | lolled 13.2° port |
| 1 After peak pressed up (52 t at Kg 9.52) | 26172 | 10.408 | 0 | 10.408 | 10.398 | −0.010 | lolled 3.3° |
| 2a No. 1 DB port half full (198 t) | 26370 | 10.341 | 1998 | 10.417 | 10.391 | −0.026 | heeled 14.7° port |
| 2b No. 1 DB port full (396.2 t at Kg 1.43) | 26568 | 10.274 | 0 | 10.274 | 10.384 | +0.109 | listed 15.4° port |
| 3 No. 1 DB starboard full | 26964 | 10.144 | 0 | 10.144 | 10.370 | +0.226 | upright |
Stage 1, press up the after peak. New KG = (26120 × 10.41 + 52 × 9.52) ÷ 26172 = 10.408 m with no free surface left; KM at 26172 t is 10.398 m, so GM = 10.398 − 10.408 = −0.010 m. Free surface was almost the whole of the problem, but nearly whole is not whole: with BM 6.006 m she still lolls at tan θ = √(0.020 ÷ 6.006), 3.3°, and holds only 0.020 m of GM there. The tank was on the centreline, so nothing has been added to one side.
Stage 2, fill No. 1 port, the low side. Mid fill is the worst moment: 198 t aboard at 6.21 m off centre and 1998 t m of live free surface. Δ 26370 t, solid KG 10.341 m, FSC 1998 ÷ 26370 = 0.076 m, fluid KG 10.417 m, KM 10.391 m, fluid GM −0.026 m: worse than when the stage began. The listing moment is 198 × 6.21 = 1230 t m, GGh = 0.047 m, and the wall sided balance (−0.026 + ½ × 5.97 tan² θ) tan θ = 0.047 heels her to about 14.7° to port: deeper than the 3.3° she showed at the end of stage 1. This is the moment the drill exists for: the plan is working, hold your nerve. Tank pressed full: Δ 26568 t, KG 10.274 m, KM 10.384 m, GM +0.109 m, a stable ship. She rides listed to port on the full tank’s 396.2 × 6.21 = 2460 t m, GGh = 0.093 m; because GM is still small the list must come from the wall sided balance, not from tan θ = GGh ÷ GM (which would give an absurd 40°): (0.109 + ½ × 5.931 tan² θ) tan θ = 0.093 gives 15.4° to port. A large list, but an ordinary one, on a ship with positive GM.
Stage 3, fill No. 1 starboard. The listing moments cancel, Δ 26964 t, KG 10.144 m, KM 10.370 m, GM +0.226 m, upright, and above the 0.15 m the criteria demand. She has taken 844 t of ballast in all. The pressing up continues on the next pair until the margin is respectable.
Suppose the mate’s original proposal had been carried out: from the as found condition (lolled 13.2° to port, after peak still slack with its 4075 t m), No. 1 double bottom starboard, the high side, is filled first. Trace what happens as the tank fills, and where she ends up.
A quarter of the tank in (99 t). Δ 26219 t; solid KG 10.376 m; free surface 4075 + 1998 = 6073 t m, a correction of 0.232 m; fluid KG 10.608 m; KM 10.396 m; fluid GM −0.212 m, worse than when she started, because the filling tank’s free surface has been added to the after peak’s. The starboard moment is 99 × 6.21 = 615 t m, GGh = 0.023 m.
With GM negative the wall sided balance (GM + ½ BM tan² θ) tan θ = GGh is a cubic with, for a small moment, three solutions: a stable one on each side and an unstable one between. The port solution exists only while the moment is smaller than the deepest negative lever the curve reaches on the port side, which is ⅔ |GM| √(|GM| ÷ 1.5 BM). With GM −0.212 m and BM 5.997 m that is 0.141 × 0.153 = 0.022 m, a moment of 568 t m at this displacement. The quarter tank supplies 615 t m: the port equilibrium ceased to exist at about 95 t in, and the only equilibrium left is on the starboard side, at 17.2° by the wall sided cubic and 17.3° on the booklet KN curve.
The swing. Released from rest at 13.2° to port with nothing to hold her, she rolls through upright gathering momentum, and the work done by the heeling lever on the way down is only recovered by the righting lever on the way up. Integrating the net lever on the booklet curve, KN − 10.608 sin θ − 0.023 cos θ, from 13.2° port until the net work returns to zero puts the end of the swing at 28.8° to starboard, 5.8° past the deck edge (23.0°), a swing of 42° in one uncommanded roll, without damping and without the sea.
Half the tank in (198 t): fluid GM −0.181 m, GGh 0.047 m, starboard equilibrium 18.2°, and the same integration never returns to zero before the booklet curve vanishes at about 39°: on paper she capsizes. Tank full (396 t, Δ 26516 t, solid KG 10.276 m, fluid KG 10.430 m, KM 10.386 m): GM still −0.044 m, static equilibrium 18.3° to starboard, 31.5° from where she began, and again the swing from the port loll does not stop. The momentum of the roll may carry the ship past her angle of vanishing stability and capsize her; short of that, it can shift the cargo and injure everyone standing. Contrast stage 2 of Worked example 5.4, where the same tank on the low side took her deeper on the same side, never through upright, and delivered a stable ship. This is why the low side is filled first, always.
MV Ninja sails at 26120 t carrying 600 t of timber on deck at Kg 15.6 m: solid KG 10.28 m, FSM 2600 t m. On passage the timber is assumed to absorb 15% of its weight in water (a heavier soaking than the usual ten per cent allowance), 180 t of heavy fuel oil is burned from the No. 1 tanks at Kg 12.65 m (this ship carries her fuel high), and the slack tanks’ FSM grows to 3400 t m. Examine her stability on departure and on arrival (booklet at 26030 t: KM 10.404 m, KB 4.371 m).
Departure: FSC = 2600 ÷ 26120 = 0.100 m; fluid KG 10.380 m; GM = 10.400 − 10.380 = +0.021 m. Lawful for a timber deck cargo she is not: the minimum on departure is 0.10 m, and this departure should never have happened.
Arrival: Δ = 26120 + 90 − 180 = 26030 t. Solid KG = (26120 × 10.28 + 90 × 15.6 − 180 × 12.65) ÷ 26030 = (268514 + 1404 − 2277) ÷ 26030 = 10.282 m. FSC = 3400 ÷ 26030 = 0.131 m; fluid KG 10.413 m; GM = 10.404 − 10.413 = −0.008 m.
BM = 10.404 − 4.371 = 6.034 m; tan θloll = √(0.016 ÷ 6.034), so she arrives lolled at about 3.0°. Where the 0.029 m of GM went: the fuel, carried high, lowered KG by 0.016 m and helped; the absorbed water raised it by 0.018 m; the growth of free surface cost 0.031 m; KM rose 0.004 m. Had the fuel come from a double bottom at Kg 0.60 m, as it does on many ships, the arrival GM would have been −0.092 m and the loll 9.9°. The lesson is the syllabus’s own: examine the stability of the arrival condition before departure, because the changes at sea are almost always adverse.
Loll: G on the centreline but above M; she settles either side at tan θ = √(−2GM ÷ BM) and holds (−2GM ÷ cos θ) of metacentric height there.
List: G off the centreline, GM positive, one side only. Diagnose before touching a valve.
The cure is G downwards, nothing else: topweight off, slack tanks pressed, divided double bottoms, one tank at a time.
Low side first, and expect her to lie deeper mid fill; the high side first is how lolled ships capsize.
Check the arrival condition before sailing: the sea only ever climbs the ladder.