Volume One ended with the metacentric diagram and the hydrostatic tables: the small angle world, where one number, GM, told the whole story. This volume begins where that story runs out. A ship in a gale does not heel by three degrees; she heels by thirty, and at thirty degrees the metacentre has wandered off its perch and GM alone can mislead. What never misleads is the lever itself: the righting lever GZ, computed angle by angle and drawn as the curve of statical stability. This chapter builds that curve for MV Ninja, teaches the examination protocol for drawing and reading it, and then puts it to work: what a rising G does to it, what the hull form gives it, and what a shifted weight steals from it.
Heel a ship with an external force, wind or wave or tug, and hold her there for an instant. Her weight still acts vertically downwards through G, which has not moved. Her buoyancy acts vertically upwards through B, which has: the underwater volume now bulges into the low side, and B follows it. Two equal forces, opposite in direction, separated by a horizontal distance: a couple. The perpendicular distance between the two lines of action is the righting lever GZ, and the couple it makes is the moment of statical stability, the ship’s answer to the sea.
Everything in this chapter is bookkeeping on that one picture. While the lever points the right way the ship fights back; the moment she heels past the angle where GZ passes through zero, the same couple changes sides and works to capsize her. The officer’s job is to know, before sailing, exactly how much lever the ship holds at every angle.
At small angles of heel the metacentre M stands effectively still on the centreline, and the triangle GZM is right angled at Z. One line of trigonometry gives the whole of initial stability:
MV Ninja lies in her summer departure condition, Δ 30456 t, KG 8.09 m against the booklet KM of 10.330 m, so GM = 2.24 m. A beam squall holds her at a steady 6° of heel. Find the righting lever and the moment of statical stability opposing the squall.
GZ = GM × sin θ = 2.24 × sin 6° = 2.24 × 0.10453 = 0.23414 = 0.234 m
MSS = Δ × GZ = 30456 × 0.23414 = 7131 t m
Six degrees is comfortably inside the small angle world, so the GM route is sound. The moment is carried through with the five figure lever; the three figure lever 0.234 m would give 7127 t m, four tonne metres short, which is the usual price of rounding too early. The question this chapter answers is what happens when the squall becomes a storm and 6° becomes 36°.
Beyond the deck edge no tidy formula survives: the shape of the emerging and immersing hull takes over, and the naval architect computes the lever numerically for a family of angles and displacements, publishing the results as the cross curves of stability. Because KG changes from voyage to voyage, the curves assume G rests on the keel: the tabulated lever is called KN, and correcting it to the real G costs one term. Chapter 2 dismantles the KN machinery in full; this chapter simply uses it, the way the working officer does.
The KG in the formula is the fluid KG of Volume One, Chapter 9, with every free surface allowed for. The 8.09 m used throughout this chapter is the solid KG of the departure condition, and the curves are drawn on that stated solid condition so that the KN arithmetic can be followed on round figures; with the seven slack tanks of Volume One the fluid figures are KG 8.113 m and GM 2.217 m, and every lever below would be 0.023 sin θ smaller. The MV Ninja booklet tabulates KN at ten angles, 5°, 10°, 12°, 20°, 30°, 40°, 50°, 60°, 70° and 80°, against displacement from 8000 t to 30500 t in steps of 1500 t. A displacement between two printed rows is dealt with by linear interpolation, exactly as for the hydrostatic table, and the summer displacement of 30456 t lies 44 t short of the last row.
At MV Ninja’s summer displacement of 30456 t, with the departure KG of 8.09 m, tabulate the righting lever at every angle of the booklet’s KN table, ready for plotting.
The displacement lies between the 29000 t and 30500 t rows, a fraction (30456 − 29000) ÷ 1500 = 1456 ÷ 1500 = 0.971 of the way from the first to the second. At 30°, for instance, the two rows read 5.261 m and 5.146 m, so KN30 = 5.261 − 0.971 × (5.261 − 5.146) = 5.261 − 0.971 × 0.115 = 5.149 m, and GZ30 = KN − KG sin θ = 5.149 − 8.09 × 0.5000 = 5.149 − 4.045 = 1.104 m. The table repeats the arithmetic at all ten angles (KN and GZ to 3 dp; GZ from the unrounded chain).
| θ | KN at 29000 t (m) | KN at 30500 t (m) | KN at 30456 t (m) | KG × sin θ (m) | GZ (m) |
|---|---|---|---|---|---|
| 5° | 0.902 | 0.901 | 0.901 | 0.705 | 0.196 |
| 10° | 1.808 | 1.807 | 1.807 | 1.405 | 0.402 |
| 12° | 2.172 | 2.171 | 2.171 | 1.682 | 0.489 |
| 20° | 3.650 | 3.632 | 3.633 | 2.767 | 0.866 |
| 30° | 5.261 | 5.146 | 5.149 | 4.045 | 1.104 |
| 40° | 6.576 | 6.327 | 6.334 | 5.200 | 1.134 |
| 50° | 7.525 | 7.383 | 7.387 | 6.197 | 1.190 |
| 60° | 8.033 | 7.917 | 7.920 | 7.006 | 0.914 |
| 70° | 8.193 | 8.113 | 8.115 | 7.602 | 0.513 |
| 80° | 8.052 | 8.010 | 8.011 | 7.967 | 0.044 |
The column of GZ values is the raw material of the whole chapter. At 5° the table gives 0.196 m where the small angle rule gives 2.24 × sin 5° = 0.195 m: the two routes agree, as they must, because KN at 5° is nothing more than KM sin 5°. And at 80° the lever is still positive, though only just: the curve has not yet crossed zero at the last angle the booklet prints, so the angle of vanishing stability will have to be found by carrying the last trend a little beyond the table. The plot will show where.
Plotting is not decoration; in the examination room and the ship’s office alike it is evidence, and it is marked. The protocol below turns the table of Worked example 1.2 into a defensible curve.
1. Scale the axes generously: heel to 90° along the base, GZ in metres up the side, and plot every tabulated point including the origin.
2. Erect the value of GM at 57.3° (one radian) and rule a light line from the origin to it. The first part of the curve is tangential to this line: it fixes the launch slope before a single point is faired.
3. If the angle of deck edge immersion is known, drop a light vertical there: the curve changes from curving upwards to curving downwards as it crosses it, the point of contraflexure.
4. Fair a smooth curve through the points, leaving the tangent gently, peaking where the points say and running on through zero: never join the dots with straight lines.
From the curve of Figure 1.4, read off the principal features of the summer departure condition, verify the initial slope against the known GM, and check the deck edge angle against the formula sheet.
Maximum GZ: the largest tabulated lever is 1.190 m at 50° of heel, the hardest she can ever fight back in this condition. A fair curve through the 40°, 50° and 60° points (1.134, 1.190, 0.914 m) peaks a little earlier, at about 47° and about 0.02 m higher, but the tabulated value is what the examiner and the criteria use, and it is what to quote.
Angle of vanishing stability: the lever is 0.513 m at 70° and 0.044 m at 80°, so, carrying the same straight line on beyond the table, θv = 80 + 10 × 0.044 ÷ (0.513 − 0.044) = 80.9°, about 81° (an extrapolation, since the crossing lies beyond the last tabulated angle); range of stability 0° to about 81°.
Tangent check: the initial slope should be GM per radian, 2.24 ÷ 57.3 = 0.0391 m per degree, and the first chord of the table, 0.196 m in 5°, is 0.0392 m per degree. The line from the origin through the early curve, extended to the 57.3° ordinate, reaches 2.24 m there, exactly the GM of Worked example 1.1. A curve that fails this check is wrong before the first mark is lost.
Deck edge: tan θdei = freeboard ÷ (½ × B) = 3.92 ÷ 12.10 = 0.324, so θdei = 17.95°, call it 18°, exactly where the contraflexure sits on the plotted curve: the chords of the table steepen up to the 12° to 20° interval and flatten after it. Two independent routes, one answer.
One caution: the booklet gives the angle of flooding at the summer draught as 52.8°. Beyond it water enters through openings that cannot be closed, so the useful curve in this condition ends there, not at 81°.
Every term in GZ = KN − (KG × sin θ) belongs to the hull except KG. Raise G and the subtraction bites harder at every angle: the whole curve settles, the peak drops and walks inboard, and the vanishing angle marches towards the upright. Control of KG, the discipline of Volume One, is control of this curve.
On passage MV Ninja burns fuel from her double bottoms and takes heavy spares on deck, and her effective KG creeps from 8.09 m to 8.60 m at effectively constant displacement. Retabulate the levers, and state what has happened to the curve.
At each angle the loss is GGV × sin θ with GGV = 0.51 m: at 30° it is 0.51 × 0.5000 = 0.255 m, so GZ = 1.104 − 0.255 = 0.849 m; at 60° it is 0.51 × 0.8660 = 0.442 m, so GZ = 0.914 − 0.442 = 0.472 m (0.473 m from the unrounded chain). At the ten booklet angles the levers become 0.151, 0.314, 0.383, 0.691, 0.849, 0.806, 0.799, 0.473, 0.034 and −0.458 m at 80°.
The new curve peaks at 0.849 m at 30° and, with 0.034 m left at 70°, vanishes at 70 + 10 × 0.034 ÷ (0.034 + 0.458) = 70.7°: the range has surrendered about ten degrees without a single wave touching her. A further half metre, KG 9.10 m, gives a peak of 0.599 m at 30° and a vanishing angle of 60.8°, another ten degrees gone.
GM falls to 10.330 − 8.60 = 1.73 m (1.23 m at KG 9.10 m), and the tangent check confirms the flatter launch. Both raised conditions are still inside the booklet’s maximum permissible KG of 9.64 m at this displacement, and the deck edge angle has not moved. The lesson travels: a curve is drawn for one loading condition and dies with it. Change the condition, redraw the curve.
The hull form sets the rest. More freeboard leaves the small angle curve alone, then keeps the deck dry longer: the contraflexure arrives later, the peak grows and the range stretches. More beam feeds BM and therefore GM, so the curve launches more steeply, but the wider deck edge dips sooner, so the improvement lives mainly at small angles. Both effects reverse exactly when freeboard or beam is reduced.
A weight that moves aboard the ship moves G with it, and the curve keeps the account. A vertical rise GGV subtracts GGV sin θ at every angle, nothing at the upright and most at large heel: the curve sinks. A transverse shift GGH subtracts GGH cos θ, most at the upright and nothing at 90°: the curve tilts, and where the falling heeling arm crosses the rising righting lever, the ship comes to rest. That crossing is the list.
In the summer departure condition a 500 t parcel of cargo shifts 7.0 m to starboard and 2.0 m upwards. Find the list by two independent routes, and the surviving range of stability.
GGH = (w × s) ÷ Δ = (500 × 7.0) ÷ 30456 = 0.115 m; GGV = (500 × 2.0) ÷ 30456 = 0.033 m.
Route one, the small angle rule: KG rises to 8.123 m so GM = 10.330 − 8.123 = 2.207 m, and tan (List) = GGH ÷ GM = 0.115 ÷ 2.207 = 0.0521, List = 3.0° to starboard.
Route two, the curve: tabulate the sunken lever KN − 8.123 sin θ and the heeling arm 0.115 cos θ; their difference is the residual lever. At the upright the residual is −0.115 m (the heeling arm alone) and at 5° it is +0.079 m (0.193 m less the arm of 0.115 m, by the unrounded chain), so the residual curve crosses zero at 5 × 0.115 ÷ (0.115 + 0.079) = 3.0°. Two routes, one answer, and the examiner’s tick.
The residual curve peaks at 1.091 m at 50° (down from 1.190 m) and crosses zero between 70° and 80°, at 70 + 10 × 0.443 ÷ (0.443 + 0.008) = 79.8°: the working range now runs from the list at 3° to about 80°, and both the peak and the area under the curve, the dynamical stability of Chapter 3, have been eroded. On a ship the cure is the cure of Volume One: put G back where it belongs, low and on the centreline.
A heeling moment holds MV Ninja at a steady 30° in the summer departure condition, and later at 50°. Find the moment of statical stability resisting it in each case, and show the error the small angle rule would have made.
From the table, GZ at 30° = 1.104 m (Worked example 1.2), so MSS = 30456 × 1.104 = 33623 t m.
The small angle rule would claim GZ = 2.24 × sin 30° = 1.120 m and MSS = 30456 × 1.120 = 34111 t m, an overestimate of only 1.4 per cent. At 30° the two routes happen almost to coincide for this ship: the curve climbs above the tangent by the deck edge and drops back through it between 20° and 30°. That is luck, not law.
At 50° the curve gives 1.190 m and MSS = 30456 × 1.190 = 36243 t m, whereas the rule claims 2.24 × sin 50° = 1.716 m and 30456 × 1.716 = 52262 t m: an overestimate of 16019 t m, 44 per cent, and on the unsafe side. Below the deck edge the rule reads low (at 12° it gives 0.466 m against the booklet’s 0.489 m); above it the rule reads high, and by 60° it claims more than twice the lever the ship has. Past the deck edge the curve is the only honest witness.
The flag state judges this curve, not the ship’s good intentions. The intact criteria demand, among others: GM at least 0.15 m; a righting lever of at least 0.20 m at an angle of 30° or more; a maximum GZ occurring at an angle of not less than 25°; and minimum areas under the curve to 30°, to 40° or the angle of flooding if that is less, and between those two angles (0.055, 0.090 and 0.030 m rad). MV Ninja’s departure curve, with GM 2.24 m, a lever of 1.104 m at 30° and a maximum of 1.190 m at 50°, clears every bar with sea room to spare, and the areas, about 0.32 m rad to 30° and 0.51 m rad to 40°, are more than five times the minima; Chapter 3 supplies the area arithmetic and Chapter 15 the full judgement.
The curve is drawn for one displacement, one KG and one assumed trim: change any of them and it must be redrawn, and a vessel trimmed well off her KN trim carries levers the table never promised.
It is statical: it says nothing of the wave that arrives while she is still rolling, of water shipped on deck, of cargo that shifts at the worst moment, or of flooding through an opening immersed before the vanishing angle. The dynamical account, area rather than ordinate, begins in Chapter 3.