Learning objectives
By the end of this chapter you will be able to:
- define the righting lever GZ and the moment of statical stability MSS = ∆ × GZ;
- calculate GZ at small angles from GZ = GM × sin θ;
- apply the wall sided formula GZ = [GM + (½ × BM × tan² θ)] × sin θ and state its limits;
- use the cross curves of stability with GZ = KN − (KG × sin θ);
- construct and read the curve of statical stability: initial slope, maximum GZ, angle of vanishing stability and range;
- verify the initial slope of the curve against GM using the 57.3° ordinate;
- explain how a rise in KG shrinks every feature of the curve.
Chapter 7 measured a ship's stability with a single number, GM, and it served well for small angles and small lists. But a ship in weather does not politely stay within five degrees, and one number cannot tell the whole story of a roll to thirty. The full story is a curve: the righting lever GZ plotted against the angle of heel. This chapter builds that curve for MV Ninja, from three formulas on the MCA sheet, and teaches you to read its every feature.
8.1 The righting lever and the moment of statical stability
Heel the ship. Her weight ∆ acts vertically down through G; her buoyancy, equal and opposite, acts vertically up through the shifted centre of buoyancy B₁. The two forces are no longer in line, and a pair of equal, opposite, parallel forces separated by a gap is a couple. The perpendicular gap between their lines of action is found by dropping a perpendicular from G onto the buoyancy line: its foot is the point Z, and the distance GZ is the righting lever. The turning moment trying to bring her upright is the lever times the force:
MV Ninja, at her summer displacement of 30456 t with GM 2.24 m, is heeled to 5° by a gust. KG is the solid value, 8.09 m. Find the righting lever and the moment of statical stability.
GZ = GM × sin θ = 2.24 × sin 5° = 2.24 × 0.087156 = 0.19523 m, say 0.195 m
MSS = ∆ × GZ = 30456 × 0.19523 = 5946 t m (the moment is formed from the unrounded lever; 30456 × 0.195 would give 5939 t m)
Nearly six thousand tonne metres from a five degree nudge: the same order of moment as several hundred tonnes of cargo shifted across the ship. A stable ship pushes back hard, and immediately.
8.2 GZ at small angles
Where does that GZ = GM × sin θ come from? Look at the small triangle formed by M, G and Z. The side MG lies along the ship's centreline; the side MZ lies along the true vertical through B₁, because at small angles that vertical passes through M by the very definition of the metacentre; and GZ closes the triangle at right angles to MZ. The angle at M between centreline and vertical is the heel θ itself, so the side opposite it is GM × sin θ:
For the same condition (∆ 30456 t, GM 2.24 m, BM 5.289 m), find GZ at 10° of heel (a) by the small angle rule and (b) by the wall sided formula of the next section. Comment.
(a) GZ = 2.24 × sin 10° = 2.24 × 0.1736 = 0.389 m
(b) GZ = [2.24 + (½ × 5.289 × tan² 10°)] × sin 10° = [2.24 + 0.082] × 0.1736 = 0.403 m
At ten degrees the two answers differ by only 14 mm of lever: the correction term is still small. It grows with tan² θ, which is why the simple rule dies quickly beyond this.
8.3 The wall sided formula
As the heel grows, the wedge of hull immersed on the low side and the wedge emerged on the high side shift real buoyant volume across the ship, and B₁ travels further than the fixed metacentre picture allows. For a hull whose sides are effectively vertical walls through the waterline, the wedge geometry can be worked exactly, and the MCA sheet carries the result:
Find MV Ninja's righting lever and moment of statical stability at 15° of heel in the summer departure condition (∆ 30456 t, GM 2.24 m, BM 5.289 m), using the wall sided formula.
tan² 15° = 0.2679² = 0.0718
GZ = [2.24 + (½ × 5.289 × 0.0718)] × sin 15° = [2.24 + 0.190] × 0.2588 = 0.6289 m, say 0.629 m
MSS = 30456 × 0.6289 = 19154 t m
The simple rule would have said 2.24 × 0.2588 = 0.580 m: the wedges are now worth 49 mm of extra lever. At 15° the deck edge (which immerses at about 18°) is still dry, so the formula is trustworthy here.
8.4 The cross curves and KN
Beyond the deck edge no simple formula survives: the shape of the emerging hull takes over, and the naval architect computes the lever numerically for a family of angles and displacements. The results are published in the stability booklet as the cross curves of stability. Because KG varies from voyage to voyage, the curves are drawn for a G assumed at the keel: the tabulated lever is called KN. Correcting it to the real G costs one term:
At MV Ninja's summer displacement of 30456 t, with the departure KG of 8.09 m, tabulate the righting lever at each angle of the booklet cross curves. The booklet tabulates KN at 5°, 10°, 12°, 20°, 30°, 40°, 50°, 60°, 70° and 80° against displacement in steps of 1500 t; 30456 t lies 0.9707 of the way from the 29000 t row to the 30500 t row, so at 30°, for instance, KN = 5.261 − 0.9707 × (5.261 − 5.146) = 5.149 m. (8.09 m is the solid KG; with the free surface of the slack tanks, Chapter 9, the fluid KG is 8.113 m, which would take 0.023 × sin θ off every lever below, 23 mm at most.)
| θ | KN (m) | KG × sin θ (m) | GZ (m) |
|---|---|---|---|
| 5° | 0.901 | 0.705 | 0.196 |
| 10° | 1.807 | 1.405 | 0.402 |
| 12° | 2.171 | 1.682 | 0.489 |
| 20° | 3.633 | 2.767 | 0.866 |
| 30° | 5.149 | 4.045 | 1.104 |
| 40° | 6.334 | 5.200 | 1.134 |
| 50° | 7.387 | 6.197 | 1.190 |
| 60° | 7.920 | 7.006 | 0.914 |
| 70° | 8.115 | 7.602 | 0.513 |
| 80° | 8.011 | 7.967 | 0.044 |
Two checks. At 15°, interpolating KN between the 12° and 20° columns gives 2.719 m and a lever of 0.625 m, where the wall sided formula of Worked example 8.3 gave 0.629 m: two entirely different routes agreeing within 4 mm, a useful cross check. And at 80° the lever is still positive, but only just, 0.044 m: KN has stopped growing while KG × sin θ has not, and the ship is close to losing her righting ability altogether.
8.5 The curve of statical stability
Plot the levers of Worked example 8.4 against the angle of heel, add the origin, and draw a fair curve through the points: the curve of statical stability, which describes the ship's stability more completely than any other single diagram. Four features carry the meaning:
Reading the curve
- The initial slope is set by GM: the tangent at the origin passes through the value of GM erected at 57.3° (one radian). The stiffness of Chapter 7 is the launch angle of the curve.
- The maximum GZ and the angle where it occurs: the hardest the ship can ever fight back.
- The angle of vanishing stability, where GZ returns to zero: heel her past this and the couple works to capsize her, not right her.
- The range of stability: the span of angles, from upright to vanishing, over which she rights herself.
From the curve of Figure 8.5, read off the four features of MV Ninja's summer departure condition, and verify the initial slope against the known GM of 2.24 m.
Maximum GZ: 1.190 m at 50° of heel. The curve is flat across its top, from 1.104 m at 30° to 1.190 m at 50°; the moment there is 30456 × 1.190 = 36243 t m, six times the moment at 5° of Worked example 8.1.
Angle of vanishing stability: the lever is still 0.044 m at 80°, the last angle tabulated, and falling by about 0.47 m per ten degrees; extending the last stretch of the curve puts the zero at about 81°. Range of stability 0° to about 81°, the last degree being an extrapolation beyond the table. Long before that, at the flooding angle of 52.8° given in the booklet for this draught, openings would be under water, so the useful curve ends there.
Initial slope: the tangent drawn at the origin, extended to the 57.3° ordinate, reaches a height of 2.24 m there: exactly the GM found in Chapter 7. The same tangent stands at 2.24 × 5/57.3 = 0.195 m at 5° against 0.196 m on the curve, and at 0.391 m at 10° against 0.402 m. If a plotted curve fails this check, either the curve or the GM is wrong, and the error must be found before the examination answer is written down.
8.6 What raising G does to the curve
Every term in GZ = KN − (KG × sin θ) is fixed by the hull except KG. Raise G and the subtraction bites harder at every single angle: the whole curve settles downwards, the peak drops, and the vanishing angle walks in towards the upright. This is Chapter 6's lesson drawn as a picture: control of KG is control of this curve.
Repeat Worked example 8.4 with KG raised to 9.00 m, and compare the two conditions.
| θ | KN (m) | KG × sin θ (m) | GZ (m) |
|---|---|---|---|
| 5° | 0.901 | 0.784 | 0.117 |
| 10° | 1.807 | 1.563 | 0.244 |
| 12° | 2.171 | 1.871 | 0.300 |
| 20° | 3.633 | 3.078 | 0.554 |
| 30° | 5.149 | 4.500 | 0.649 |
| 40° | 6.334 | 5.785 | 0.549 |
| 50° | 7.387 | 6.894 | 0.493 |
| 60° | 7.920 | 7.794 | 0.126 |
| 70° | 8.115 | 8.457 | −0.342 |
| 80° | 8.011 | 8.863 | −0.852 |
GM falls from 2.24 m to 10.330 − 9.00 = 1.33 m. The KN column is unchanged; only the subtraction is new (each column is rounded separately, so at 20° the printed columns subtract to 0.555 m where the unrounded values give 0.554 m). The maximum lever falls from 1.190 m at 50° to 0.649 m at 30°, a loss of 45 per cent. The lever crosses zero between 60° and 70°: a straight line between the two tabulated levers puts the zero at 60 + 10 × 0.126/(0.126 + 0.342) = 62.7°, and interpolating KN itself between 60° and 70° gives 62.2°; call it 62°. The range has shrunk from about 81° to about 62°. The hull, the displacement and the sea are unchanged; a rise of 0.91 m in KG accounts for the whole difference.
One idea remains before the volume moves on. The moment resists the heel at every angle, so the work the sea must do to heel the ship is the moment summed across the angles: geometrically, the area under the GZ curve, in metre radians, multiplied by the displacement. That product is the ship's dynamical stability, in tonne metre radians; for MV Ninja in the departure condition the area to 30° is 0.317 m rad, so the dynamical stability to 30° is 30456 × 0.317 = 9655 t m rad. The intact stability criteria set legal minimum areas, and Simpson's rules, later in the series, are the tool that measures them.
Interactive: the curve plotter
Slide KG and watch MV Ninja's curve grow or shrink. The plotter recomputes GZ = KN − (KG × sin θ) from the booklet cross curves at 30456 t (KN at the ten tabulated angles, 5° to 80°, taken along straight chords between them), live, and reports the features. The maximum is reported at the nearest whole degree, so it may differ by a few millimetres from the tabulated peak.
Interactive: the MSS calculator
Small angle rule against the wall sided formula, side by side, with the moment of statical stability for each.
The slider stops at 18°: MV Ninja's deck edge immerses there at the summer draught, and beyond it the wall sided formula is no longer trustworthy.
Chapter summary
- GZ is the righting lever between weight through G and buoyancy through B₁; MSS = ∆ × GZ is the righting moment.
- GZ = GM × sin θ at small angles; the wall sided formula adds ½ × BM × tan² θ while the deck edge stays dry.
- Beyond that, the cross curves take over: GZ = KN − (KG × sin θ).
- The GZ curve is read by four features: initial slope (checked against GM at 57.3°), maximum GZ, vanishing angle, range.
- Raising KG shrinks every feature of the curve at once.
- The area under the curve is dynamical stability, measured later in the series with Simpson's rules.
Self test questions
Work each question with pencil and paper first. Your score appears in the bar below.