Twice now this volume has left a debt unpaid. Worked example 2.2 found the booklet’s KN value at 10° sitting 13 mm proud of KM sin θ and promised that Chapter 4 would price the difference exactly. Chapter 1 admitted that GZ = GM sin θ is only honest below about 10° and offered no replacement. Both debts are settled here by one formula, derived from a single honest assumption: that over the angles in question the ship’s sides are vertical walls.
Heel a wall sided hull and a wedge of buoyancy immerses to leeward while its twin emerges to windward. The waterplane, measured along the sloping waterline, grows wider, and a wider waterplane means a larger BM: the metacentre climbs. The small angle rule fails at moderate angles precisely because it keeps M nailed to its upright position. For a wall sided hull the climb can be computed exactly: M rises by MM₁ = ½ × BM × tan² θ, so the effective metacentric height at angle θ is GM + ½ BM tan² θ, and the lever follows:
Everything in the bracket is on the booklet’s hydrostatic page: GM from KM and the day’s KG, and BM = KM − KB. The formula costs one extra term over the small angle rule and buys eight more degrees of honesty.
The derivation assumes the wedges are true triangles, which holds only while the sides at the waterline are vertical walls: the deck edge must stay dry and the turn of bilge must stay submerged. For MV Ninja at summer marks the passport therefore expires at the deck edge immersion angle of 17.95° (call it 18°) found in Volume One from tan θdei = freeboard ÷ (½ × beam) = 3.92 ÷ 12.10. Beyond it the immersed wedge is beheaded by the deck and the formula begins to flatter her, gently at first and then shamelessly.
For MV Ninja at summer marks (BM = 5.289 m), evaluate the wedge term of the wall sided formula at 5°, 10°, 15° and at the deck edge, 17.95°, and comment.
| Heel θ | ½ × BM × tan² θ × sin θ | Worth in levers |
|---|---|---|
| 5° | 2.6445 × 0.0077 × 0.0872 | about 2 mm |
| 10° | 2.6445 × 0.0311 × 0.1736 | 14 mm |
| 15° | 2.6445 × 0.0718 × 0.2588 | 49 mm |
| 17.95° | 2.6445 × 0.10495 × 0.3082 | 86 mm |
The term grows roughly as the cube of the angle: negligible noise at 5°, a centimetre and a half by 10°, and nearly nine centimetres of righting lever at the deck edge. This is the quantitative answer to Chapter 1’s rule of thumb: below about 10° the small angle rule survives on a technicality; beyond it, the invoice must be paid.
At the summer departure condition (Δ 30456 t, GM 2.24 m, BM 5.289 m, KG 8.09 m) find GZ at 10° by the small angle rule, by the wall sided formula, and by the KN route using the booklet’s 1.807 m (1.808 m at 29000 t, 1.807 m at 30500 t), and reconcile the three.
Small angle rule: GZ = 2.24 × sin 10° = 0.389 m.
Wall sided: GZ = [2.24 + (½ × 5.289 × tan² 10°)] × sin 10° = [2.24 + 0.082] × 0.1736 = 0.403 m.
KN route: GZ = 1.807 − (8.09 × sin 10°) = 1.807 − 1.405 = 0.402 m.
Reconciliation: the wall sided formula collects the 14 mm wedge term the small angle rule ignores, closing to within 1 mm of the booklet; that last millimetre is the rounding of the printed KN row to three decimal places. Worked example 2.2’s promissory note is paid in full.
Repeat the comparison at 15° against the booklet’s KN of 2.719 m (read between the 12° and 20° rows at 30456 t: 2.171 + ⅝ × (3.633 − 2.171)), then find the righting lever and the moment of statical stability at the deck edge itself.
Wall sided at 15°: [2.24 + (½ × 5.289 × 0.0718)] × sin 15° = 2.430 × 0.2588 = 0.629 m.
KN route: 2.719 − (8.09 × 0.2588) = 2.719 − 2.094 = 0.625 m: 4 mm from the wall sided lever, and most of that is the straight line between the 12° and 20° rows under-reading a curve that bends upward. At the tabulated 12° the two are 2 mm apart (0.491 m against 0.489 m), the rounding of the row.
At the deck edge, 17.95°: GZ = [2.24 + (½ × 5.289 × 0.10495)] × sin 17.95° = 2.518 × 0.3082 = 0.776 m, and MSS = 30456 × 0.776 = 23634 t m.
One degree beyond, the formula starts inventing lever that the beheaded wedge no longer supplies: 20 mm too generous by 20° (0.886 m against the booklet’s 0.866 m), and 0.456 m, nearly half a metre, by 30° (1.561 m against 1.104 m). Past the edge the only honest tools are the cross curves of Chapter 2.
Moment of statical stability questions at moderate angles are a fixture of the Chief Mate and Master papers, and they are the wall sided formula’s natural habitat: compute the lever with the bracket, multiply by the displacement, and state the moment with its correct character, righting or capsizing.
Find MV Ninja’s moment of statical stability at 13° of heel at the summer departure condition, and compare it with the small angle estimate.
Wall sided: GZ = [2.24 + (½ × 5.289 × tan² 13°)] × sin 13° = [2.24 + 0.141] × 0.2250 = 0.536 m.
MSS = Δ × GZ = 30456 × 0.536 = 16324 t m.
Small angle estimate: 2.24 × 0.2250 = 0.504 m, giving 30456 × 0.504 = 15350 t m: an understatement of 974 t m, about 6%. Understating a righting moment flatters no one, but in an examination the marks follow the correct tool, and past 10° the correct tool has a tan² term in it.
If GM is negative at moderate angles, the bracket can go negative and the formula returns a negative GZ. That is not an error to be squashed with a modulus sign: it is a capsizing moment, and stating it as such is precisely what the examiner is testing. A candidate who obtains a negative lever and fails to recognise that the moment has changed sides loses the marks for that recognition.
Set GZ = 0 in the wall sided formula with a negative GM and something remarkable happens: besides the upright solution sin θ = 0, the bracket itself can vanish, at the angle where the climbing metacentre has repaid the whole GM deficit. That angle is the angle of loll, and solving the bracket gives it directly. A second manipulation prices the list from a transverse weight shift when GM is exactly zero. Both results carry the parent’s passport: valid only to the deck edge.
Suppose careless stowage lifted MV Ninja’s KG to 10.48 m at summer marks, so that GM = 10.330 − 10.48 = −0.15 m. At what angle would she loll, and what metacentric height would she hold there?
tan θloll = √(−2 × (−0.15) ÷ 5.289) = √0.0567 = 0.238, so θloll = 13.4°: safely inside the 17.95° passport, so the answer stands.
GM at the loll = (−2 × (−0.15)) ÷ cos 13.4° = 0.30 ÷ 0.973 = 0.308 m, about +0.31 m: positive, which is why a lolled ship sits at her angle instead of capsizing, and why she resists being pushed further.
Chapter 5 lives at this angle: how loll arises, how it differs from a list, and the strictly ordered drill for correcting it without rolling her over.
At summer marks with GM exactly zero, 150 t of cargo shifts 6.0 m transversely. Find the resulting list.
With GM = 0 the bracket is pure wedge: equating the wall sided lever to the heeling arm leaves ½ BM tan³ θ = w × s ÷ Δ, which rearranges to the formula above.
tan (list) = ∛(2 × 150 × 6.0 ÷ (30456 × 5.289)) = ∛0.01117 = 0.224, so the list is about 12.6°.
Notice the shape of the answer: with no GM to resist, even a modest shift produces a double figure list, and the cube root means halving the shift barely reduces it: 75 t gives 10.1°, and doubling it to 300 t gives 15.7°. Zero GM is not neutral: it is an open door, and Chapter 5 shows what walks through it.
At moderate angles the metacentre climbs by ½ BM tan² θ; add the climb to GM and the small angle rule becomes the wall sided formula.
The passport expires at the deck edge immersion angle; beyond it, use the cross curves.
The wedge term is worth 2, 14, 49 and 86 mm at 5°, 10°, 15° and 17.95° on MV Ninja: pay it above about 10°.
A negative bracket is a capsizing moment: say so.
Set GZ = 0 and the formula gifts the angle of loll; set GM = 0 and it prices the list from a shift.