Chapter 1 used the booklet’s KN values the way a navigator uses a lighthouse: gratefully, and without asking who keeps the lamp lit. This chapter opens the machinery. Where do the cross curves come from, why is the tabulated lever measured from the keel of all places, what exactly does the correction KG sin θ pay for, and how does the working officer move from a printed row of figures to a defensible curve for a condition the table never listed? By the end, the construction is a drill: enter, interpolate, correct, plot, verify.
The lever GZ depends on three things: the angle of heel, the displacement, and the position of G. The first two belong to the hull and the sea; the third belongs to the voyage, and it changes with every cargo plan, every bunker stem and every ballast movement. The naval architect cannot know where G will be next year, so a table of GZ itself would die the day the ship first loaded. The solution is elegant: compute the geometry once for a G that never moves, and let the mariner pay the correction for the real G on the day. The architect heels the hull form through a family of angles at a family of displacements, tracks the centre of buoyancy numerically, and measures each lever from the one point aboard that is fixed for ever: the keel, K.
Heel the ship to θ and stand in the space frame, where the buoyancy force is truly vertical. KN is the perpendicular distance from K to that buoyancy vertical: the lever the ship would have if all her weight sat on the keel. The real G sits a distance KG up the centreline, and the centreline now leans at θ to the vertical. Drop X onto the line KN directly below G, and the figure GXNZ closes into a rectangle:
In triangle KGX of Figure 2.2, show that KX = KG × sin θ, and hence find how much of MV Ninja’s summer KN of 5.149 m at 30° (the booklet rows at 29000 t and 30500 t, 5.261 m and 5.146 m, interpolated at 30456 t) is consumed by the correction when KG = 8.09 m.
The angle at K between the centreline KG and the space vertical is the heel θ, and KX is horizontal, so in the right angled triangle KGX: sin θ = KX ÷ KG, giving KX = KG × sin θ. Since XN and GZ are opposite sides of the rectangle GXNZ, KN = KX + GZ and the formula follows.
At 30°: correction = 8.09 × sin 30° = 8.09 × 0.5 = 4.045 m.
GZ = 5.149 − 4.045 = 1.104 m. Of the 5.149 m the hull offers, 4.045 m is spent hoisting G from the keel to its true height, leaving the 1.104 m righting lever of Chapter 1. The table is generous only because it assumes a ship with her entire weight in the keel plate; the correction is the honest invoice, and it is always subtracted.
Plotted, the table becomes the cross curves of stability: one curve per angle of heel, displacement along the base. The officer enters vertically with the displacement, reads KN where the entry line cuts each curve, and walks away with one row of figures. The curves for the small angles fall steeply while the ship is light and flatten as the displacement grows; the curves for the large angles run nearly level in the light condition and fall away at the deep draughts. The entering argument is the displacement, which the loading calculation gives directly; a draught would first have to be converted to a displacement, and that conversion depends on the trim and the water density.
Some booklets tabulate the levers for an assumed KG rather than for the keel. The correction is then (KG actual − KG assumed) × sin θ, subtracted when the real G is higher than assumed and added back when it is lower. The KN convention used aboard MV Ninja avoids that two way traffic: from the keel, the road only ever goes up, so the correction only ever comes off.
A professional never uses a table without shaking its hand first. At small angles the lever must obey Chapter 1’s small angle rule with G on the keel, which makes KN ≈ KM × sin θ: two numbers from different pages of the booklet that are honour bound to agree.
At the summer displacement of 30456 t the booklet gives KM = 10.330 m, and the cross curves, interpolated as in Worked example 2.1, give KN = 0.901 m at 5° and 1.807 m at 10°. Audit the table.
At 5°: KM × sin 5° = 10.330 × 0.08716 = 0.900 m, against the tabulated 0.901 m: agreement within a millimetre.
At 10°: KM × sin 10° = 10.330 × 0.17365 = 1.794 m, against the tabulated 1.807 m: the table now sits 13 mm proud, which is the emerging wedge effect that the wall sided formula of Chapter 4 will price exactly.
The audit passes: the hydrostatic page and the cross curve page describe the same hull. Had they disagreed at 5°, one of the two tables, or the officer’s reading of them, would be wrong, and the error hunt starts before the arithmetic does.
Real displacements land between printed rows, and the cure is ordinary linear interpolation, carried out angle by angle with the displacement as the entering argument.
MV Ninja arrives at a displacement of 26120 t. The booklet’s bracketing cross curve rows (Appendix A, KN in metres) read as follows; find her KN row.
| Δ (t) | 5° | 10° | 12° | 20° | 30° | 40° | 50° | 60° | 70° | 80° |
|---|---|---|---|---|---|---|---|---|---|---|
| 26000 | 0.909 | 1.822 | 2.190 | 3.684 | 5.497 | 6.859 | 7.789 | 8.253 | 8.347 | 8.131 |
| 27500 | 0.904 | 1.813 | 2.178 | 3.663 | 5.379 | 6.715 | 7.661 | 8.146 | 8.271 | 8.091 |
Fraction = (26120 − 26000) ÷ (27500 − 26000) = 120 ÷ 1500 = 0.08.
At 30°: KN = 5.497 + 0.08 × (5.379 − 5.497) = 5.497 − 0.009 = 5.488 m, and likewise at every angle:
| θ | 5° | 10° | 12° | 20° | 30° | 40° | 50° | 60° | 70° | 80° |
|---|---|---|---|---|---|---|---|---|---|---|
| KN (m) | 0.909 | 1.821 | 2.189 | 3.682 | 5.488 | 6.847 | 7.779 | 8.244 | 8.341 | 8.128 |
In this deep part of the table the curves run nearly level, so the interpolation moves only the second or third decimal (the largest move is −0.012 m at 40°). Do it anyway, every time: low in the table the curves are steep and taking the nearer row costs decimetres of lever. At 10000 t, for instance, the 9500 t row unaltered would overstate KN by 112 mm at 20° and 68 mm at 30°.
Everything now assembles into the working officer’s drill: enter with displacement, interpolate the row, subtract KG sin θ angle by angle, plot with the Chapter 1 protocol, and verify the tangent. Worked example 2.4 runs the whole drill on the arrival condition.
After discharging in Volume One, MV Ninja has steamed on and her fluid KG now stands at 8.72 m at the same displacement of 26120 t, where the booklet gives a draught of 8.365 m and KM = 10.400 m. Using the KN row of Worked example 2.3, construct and read her curve of statical stability.
| θ | KN (m) | KG × sin θ (m) | GZ (m) |
|---|---|---|---|
| 5° | 0.909 | 0.760 | 0.149 |
| 10° | 1.821 | 1.514 | 0.307 |
| 12° | 2.189 | 1.813 | 0.376 |
| 20° | 3.682 | 2.982 | 0.700 |
| 30° | 5.488 | 4.360 | 1.128 |
| 40° | 6.847 | 5.605 | 1.242 |
| 50° | 7.779 | 6.680 | 1.099 |
| 60° | 8.244 | 7.552 | 0.693 |
| 70° | 8.341 | 8.194 | 0.147 |
| 80° | 8.128 | 8.588 | −0.460 |
(GZ is worked from the unrounded chain, so the 60° line differs by a millimetre from the difference of its printed columns.)
GM = 10.400 − 8.72 = 1.68 m: erect it at 57.3° and rule the tangent before fairing.
From the table and the fair curve: maximum GZ 1.242 m at 40° (the neighbouring 1.128 m at 30° and 1.099 m at 50° put the peak of the fair curve within a degree or two of 40°); the lever is 0.147 m at 70° and −0.460 m at 80°, so a straight line between them puts the zero at 70 + 10 × 0.147 ÷ (0.147 + 0.460) = 72.4°: angle of vanishing stability about 72°; range 0° to about 72°.
The tangent check passes (the ruled line stands at 1.68 × 5 ÷ 57.3 = 0.147 m at 5° against the plotted 0.149 m), and note the deck edge: at the shallower arrival draught of 8.365 m the freeboard has grown from 3.92 m to 3.92 + (9.600 − 8.365) = 5.155 m, so tan θdei = 5.155 ÷ 12.10 = 0.4260 and the contraflexure has moved out from the summer condition’s 17.95° (call it 18°) to 23.1°. Less displacement, more freeboard, a later deck edge: three chapters of theory shaking hands in one plot.
Draw the KN row as a continuous curve against angle, and beneath it the pure sine curve KG × sin θ. The righting lever at every angle is simply the daylight between them; where the sine curve climbs above KN, the daylight turns negative and the lever capsizes. Raising KG inflates the sine curve and squeezes the daylight shut: somewhere there is a KG at which the ship only just complies with the rules, and finding it is the business of the booklet’s maximum KG table and of Chapter 16.
At 26120 t, what is the greatest KG at which MV Ninja would still hold a righting lever of at least 0.20 m at 30°? And what is the greatest KG at which she keeps the minimum metacentric height of 0.15 m? Which governs?
Lever route: GZ = KN − KG sin θ ≥ 0.20 at 30° requires KG ≤ (5.488 − 0.20) ÷ sin 30° = 5.288 ÷ 0.5 = 10.576 m.
GM route: KG ≤ KM − 0.15 = 10.400 − 0.15 = 10.25 m.
The GM criterion bites first, so 10.25 m governs here; the booklet’s maximum KG table gives 10.251 m at this displacement, the same ceiling within a millimetre. That table plays this same game against every criterion of the 2008 IS Code at once, the area clauses and the angle of maximum GZ included, and publishes only the lowest; Chapter 16 shows how that table is built and how the governing criterion changes hands as the ship loads.
The architect computed the cross curves at one trim, usually even keel or the design trim. Trim the ship well away from it and the immersed body changes shape, B follows a different track, and the tabulated levers drift from the truth. Ships that habitually work at large trims may carry separate KN tables for separate trims; a further refinement, free trim computation, lets the model retrim itself as it heels. The working rule is modest and firm: know the trim your table assumes, and treat the levers with growing suspicion as you leave it.
KN is the lever for a G on the keel, computed once for all voyages; the mariner pays KG sin θ, always subtracted.
Enter the cross curves with displacement; interpolate between rows by displacement, angle by angle.
Audit before use: KN must match KM sin θ at small angles.
Construct with the Chapter 1 protocol and verify the tangent at 57.3°.
Respect the assumed trim, and remember the maximum KG table is this chapter played backwards.