Chapters 1 and 2 read the curve of statical stability one ordinate at a time. But the flag state, and the sea itself, care about something an ordinate cannot express: the area under the curve, which is the work the ship can absorb before she gives way. Areas under curves want integration, and the mariner’s integration is a pair of pocket sized rules published by Thomas Simpson in 1743, nearly three centuries ago. This chapter builds both rules, runs them over the GZ curve for dynamical stability and the Code’s area criteria, and then turns them loose on the hull itself: waterplane areas, the volume of displacement, and the centroids, LCF and KB, that the whole of the coming trim theory pivots on.
Heel a ship to 40° and hold her there: the moment resisting you at 40° is one ordinate’s worth of information. But to get her to 40° the sea had to work against the righting moment through every angle on the way, and work is moment times angle, summed: an area. That accumulated work is the dynamical stability, the energy a ship stores to resist external forces, and it equals the displacement times the area under the GZ curve to the angle in question. The units follow the recipe: tonnes for the displacement, metres for the lever, radians for the angle, so areas come in metre radians and dynamical stability in tonne metre radians.
One radian is 57.3°, the same angle where Chapter 1 erected the GM tangent. To convert, divide degrees by 57.3: a common interval of 10° is 10 ÷ 57.3 = 0.1745 radians. Areas taken with degree intervals are meaningless; the conversion is not optional.
Slice the area into equal intervals h and measure the ordinates, the heights from the base to the curve. Simpson’s first rule fits a parabola across each pair of intervals and integrates it exactly: for three ordinates a, b, c the area is one third of h times (a + 4b + c). Chain the sets and the multipliers add where they meet: 1,4,1 becomes 1,4,2,4,1 for five ordinates and 1,4,2,4,2,4,1 for seven. The rule serves any odd number of ordinates, and the common interval must stay common: change h midstream and the rule quietly returns rubbish.
MV Ninja’s summer departure curve (Δ 30456 t, KG 8.09 m) computed from the booklet’s KN rows as GZ = KN − KG sin θ, gives GZ = 0 at 0°, 0.402 m at 10°, 0.866 m at 20°, 1.104 m at 30° and 1.134 m at 40°. Find the area under the curve to 40° and the dynamical stability at that angle.
| Heel | Ordinate GZ (m) | Simpson’s multiplier | Product for area |
|---|---|---|---|
| 0° | 0 | 1 | 0 |
| 10° | 0.402 | 4 | 1.608 |
| 20° | 0.866 | 2 | 1.732 |
| 30° | 1.104 | 4 | 4.416 |
| 40° | 1.134 | 1 | 1.134 |
| Sum of products | 8.890 |
h = 10° = 10 ÷ 57.3 = 0.1745 radians.
Area = ⅓ × 0.1745 × 8.890 = 0.517 metre radians.
Dynamical stability = Δ × area = 30456 × 0.517 = 15746 t m r: the work, in tonne metre radians, the sea must spend to lay her over to 40°, and the energy she holds to spend back.
The second rule fits a cubic across each set of three intervals: for four ordinates the area is three eighths of h times (a + 3b + 3c + d), and the sets stack exactly as before, 1,3,3,1 becoming 1,3,3,2,3,3,1 for seven ordinates. It serves ordinate counts of 4, 7, 10 and so on. Between them the two rules cover every count from three upwards except certain even counts, six, eight, twelve and fourteen among them, which need an extra ordinate read from the fair curve. On a fair curve the rules disagree only in the third decimal, and the examiner accepts either.
Using the same curve, find the area to 30°. Four ordinates are available (0°, 10°, 20°, 30°), so the second rule applies.
| Heel | Ordinate GZ (m) | Simpson’s multiplier | Product for area |
|---|---|---|---|
| 0° | 0 | 1 | 0 |
| 10° | 0.402 | 3 | 1.206 |
| 20° | 0.866 | 3 | 2.598 |
| 30° | 1.104 | 1 | 1.104 |
| Sum of products | 4.908 |
Area = ⅜ × 0.1745 × 4.908 = 0.321 metre radians.
Note the choice was forced: four ordinates cannot feed the first rule. Had ordinates at every 5° been read from the curve, seven ordinates and either rule would have served.
The 2008 IS Code prices stability by exactly these areas: at least 0.055 metre radians under the curve to 30°; at least 0.09 metre radians to 40° or the angle of flooding, whichever is less; and at least 0.03 metre radians between 30° and 40°, or between 30° and the angle of flooding if that is less. For MV Ninja at the summer draught the angle of flooding is 52.8°, so 40° governs. The third patch comes free once the first two are known: subtract.
Assemble the areas of Worked examples 3.1 and 3.2 into the Code’s compliance table for the summer departure condition, taking GM and the peak from Chapter 1.
Area between 30° and 40° = 0.517 − 0.321 = 0.196 m r.
| Stability criterion | Requirement | Actual | Compliance |
|---|---|---|---|
| GM | 0.15 m | 2.24 m | OK |
| GZ at 30° or more | 0.20 m | 1.104 m at 30° (maximum 1.190 m) | OK |
| Angle of maximum GZ | 25° | 50° | OK |
| Area to 30° | 0.055 m r | 0.321 m r | OK |
| Area to 40° | 0.09 m r | 0.517 m r | OK |
| Area 30° to 40° | 0.03 m r | 0.196 m r | OK |
Each of the five numerical criteria clears its bar by a factor of five or more, and the peak sits at twice the required angle: a deep laden bulker with a low G is a stiff, compliant ship. Chapter 15 formalises this table, adds the angle of flooding and the weather criterion’s cousins, and Chapter 16 runs it backwards into the maximum KG table.
Nothing in Simpson cares that the ordinates were righting levers. Feed the rules half breadths of a waterplane at equally spaced stations and out comes the waterplane area, doubled for the two sides. From the area, TPC follows at once, which hands us a beautiful audit: the area computed from the ordinates must reproduce the TPC the booklet prints.
At her ballast draught of 5.00 m, MV Ninja’s waterplane has the half ordinates below at eleven stations spaced h = 14.8 m over the 148 m between perpendiculars. Find the waterplane area and audit it against the booklet.
| Station | 0 (AP) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 (FP) |
|---|---|---|---|---|---|---|---|---|---|---|---|
| ½ ordinate (m) | 1.6 | 7.1 | 11.9 | 11.9 | 12.1 | 12.1 | 12.1 | 12.1 | 11.9 | 11.6 | 2.4 |
| Station | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Multiplier | 1 | 4 | 2 | 4 | 2 | 4 | 2 | 4 | 2 | 4 | 1 |
| Product | 1.6 | 28.4 | 23.8 | 47.6 | 24.2 | 48.4 | 24.2 | 48.4 | 23.8 | 46.4 | 2.4 |
Sum of products = 319.2.
Area = 2 × ⅓ × 14.8 × 319.2 = 3149 m² (the factor of two restores the port side).
Audit: TPC = (Aw × 1.025) ÷ 100 = (3149 × 1.025) ÷ 100 = 32.28 t, and the booklet’s hydrostatic row at 5.00 m prints 32.28: the ordinates and the table describe the same waterplane, to the last figure.
To find where an area balances, weight each product by its distance from a chosen axis, sum the moments, and divide by the sum of the area products. Levers are conveniently counted in whole intervals from midships, forward positive; multiplying the final quotient by h converts intervals back into metres. For a waterplane the balance point is the centre of flotation, the pivot about which the ship trims, and Chapter 6 is built on it.
Using the ordinates of Worked example 3.4, find the position of the centre of flotation of the 5.00 m waterplane.
| Station | Product for area | Lever | Product for moment |
|---|---|---|---|
| 0 (AP) | 1.6 | −5 | −8.0 |
| 1 | 28.4 | −4 | −113.6 |
| 2 | 23.8 | −3 | −71.4 |
| 3 | 47.6 | −2 | −95.2 |
| 4 | 24.2 | −1 | −24.2 |
| 5 | 48.4 | 0 | 0 |
| 6 | 24.2 | +1 | +24.2 |
| 7 | 48.4 | +2 | +96.8 |
| 8 | 23.8 | +3 | +71.4 |
| 9 | 46.4 | +4 | +185.6 |
| 10 (FP) | 2.4 | +5 | +12.0 |
| Sums | 319.2 | +77.6 |
Centroid from midships = h × (Σ moment products) ÷ (Σ area products) = 14.8 × 77.6 ÷ 319.2 = 3.60 m forward of midships, that is 74.0 + 3.6 = 77.6 m foap.
The booklet’s hydrostatic row prints LCF = 77.51 m: agreement within a decimetre, which is the honest price of eleven ordinates read to one decimal place. At this light draught the full bow waterplane drags the pivot well forward of midships; by the summer draught it has migrated abaft. Where that pivot sits decides how every loaded tonne changes the draughts, which is exactly Chapter 6’s business.
Waterplane areas at a ladder of draughts are themselves ordinates: integrate them vertically and the answer is the volume of displacement; add draught levers and the second pass finds the height of its centroid, KB. This is precisely how the architect built the hydrostatic table, so the results must land back on it.
The architect’s calculation sheet gives MV Ninja’s waterplane areas at five draughts from the keel to the summer marks, at h = 2.4 m. Find the volume of displacement, the salt water displacement and KB at 9.60 m, and audit all three against the booklet.
| Draught (m) | 0 | 2.4 | 4.8 | 7.2 | 9.6 |
|---|---|---|---|---|---|
| Waterplane area (m²) | 2400 | 2939 | 3132 | 3319 | 3442 |
| Draught (m) | Area (m²) | Multiplier | Product for volume | Lever (m) | Product for moment |
|---|---|---|---|---|---|
| 0 | 2400 | 1 | 2400 | 0 | 0 |
| 2.4 | 2939 | 4 | 11756 | 2.4 | 28214.4 |
| 4.8 | 3132 | 2 | 6264 | 4.8 | 30067.2 |
| 7.2 | 3319 | 4 | 13276 | 7.2 | 95587.2 |
| 9.6 | 3442 | 1 | 3442 | 9.6 | 33043.2 |
| Sums | 37138 | 186912.0 |
Volume = ⅓ × 2.4 × 37138 = 29710 m³; displacement = 29710 × 1.025 = 30453 t against the booklet’s 30456 t: agreement within the rounding of the areas. Note the volume itself is, to the tonne for tonne, the fresh water displacement column of the same row.
KB = (Σ moment products) ÷ (Σ volume products) = 186912.0 ÷ 37138 = 5.03 m against the booklet’s 5.041 m: within a centimetre.
The volume, the displacement and KB recovered from five areas and an eighteenth century mathematician’s rule. When the booklet, the cross curves and Simpson all tell the same story, the officer can trust the arithmetic with a ship on top of it.
The common interval h must stay common; if the curve ends awkwardly, read an extra ordinate from the fair curve rather than stretch an interval.
Odd ordinate counts (3, 5, 7, 9) feed the first rule; counts of 4, 7, 10 feed the second; seven suits both, and either answer earns the marks.
Levers in whole intervals keep the moment table clean; one multiplication by h at the end restores metres.
Every Simpson answer offers an audit: waterplane area against TPC, volume against displacement, moments against LCF and KB. Take the audit; it is free.