NATARAJAThe curve at large angles
Learning outcomes & permitted supports
- Read a righting-lever curve: the GM tangent at the origin, the maximum, the range, and the angle of vanishing stability
- Explain why the wall-sided formula fails once the deck edge is immersed — and see the two curves separate
- Use cross curves: GZ = KN − KG·sin θ, and say why KN is published without KG
- Work areas under the curve by Simpson's rules and turn them into dynamical stability
- Test a condition against the IMO intact-stability criteria and name which one fails first
- Tell an angle of loll from a list, and carry out the recovery in the right order
Supports in this lab: Calculator. Exact box-form geometry at every angle (clipped section, not the wall-sided approximation). IS Code 2008 thresholds quoted; the ship's approved stability booklet governs aboard. Engine golden-validated 840/840.
The ship at the angle she is actually at
The wall-sided formula assumes the sides stay vertical through the water — true only while the deck edge is dry and the bilge is still under. Push her past that and the geometry changes: this section is clipped exactly, at every angle.
| Displacement (per 100 m) | 14350 t |
|---|---|
| KM · KG · GM | 8.26 · 6.50 · 1.76 m |
| Regime at 30° | deck edge immersed |
| KN at 30° | 4.485 m |
| GZ exact · wall-sided | 1.235 m · 1.278 m |
| Area to 30° · to 40° | 0.285 · 0.531 m·rad |
| Dynamical stability to 40° | 7617 t·m·rad |
| Maximum GZ | 1.53 m at 41.4° |
| Angle of vanishing stability | 80.9° |
Watch the two GZ figures separate as you pass the deck-edge angle: the wall-sided formula keeps promising a lever the ship no longer has. That is the whole reason cross curves exist.
Cross curves — KN, and why they are published without KG
KN is measured from the keel, so it depends only on the SHAPE and how deep she floats — never on where the cargo sits. The yard computes these once; you get your ship's GZ from them on the day with one subtraction: GZ = KN − KG·sin θ.
| heel | KN (m) | KG·sinθ (m) | GZ (m) |
|---|---|---|---|
| 15° | 2.183 | 1.682 | 0.500 |
| 30° | 4.485 | 3.250 | 1.235 |
| 45° | 6.098 | 4.596 | 1.502 |
| 60° | 6.679 | 5.629 | 1.050 |
| 75° | 6.598 | 6.279 | 0.320 |
The intact-stability criteria, live
The IS Code does not ask "is GM positive?" — it asks for AREAS under this curve, a lever at 30°, and a maximum that does not come too early. Move KG and watch which one fails first. (Code thresholds quoted; your ship's approved booklet governs aboard.)
| criterion | required | this ship | |
|---|---|---|---|
| Area under GZ to 30° | ≥ 0.055 m·rad | 0.285 m·rad | pass |
| Area under GZ to 40° | ≥ 0.090 m·rad | 0.531 m·rad | pass |
| Area between 30° and 40° | ≥ 0.030 m·rad | 0.246 m·rad | pass |
| GZ at 30° or beyond | ≥ 0.20 m | 1.235 m | pass |
| Angle of maximum GZ | ≥ 25° | 41.4 ° | pass |
| Initial metacentric height | ≥ 0.15 m | 1.762 m | pass |
All criteria met
Angle of loll — and the drill that gets her back
Put KG above KM and the ship has no righting arm at all upright. She falls over until the wall-sided term rebuilds the lever, and sits there: an angle of loll is an equilibrium, a list is a moment. They look identical from the bridge and are cured by opposite actions.
| GM (negative) | -0.250 m |
|---|---|
| Angle of loll — √(−2GM/BM) | 17.95° |
| Angle of loll — exact curve | 17.95° |
Two derivations, one answer — the formula from the wall-sided algebra, the other from the exact curve's zero crossing. They agree to 10⁻¹³ of a degree in the golden suite.
Put the drill in order: