Two ideas carry the topic: a lever that grows with heel, and everything on board that quietly shortens it. Free surface, a high KG and a shifted weight are the three that appear year after year.
Heel a ship and two forces stay in play: its weight acting down through G, and its buoyancy acting up through B. Upright, they are in the same vertical line and cancel. Heeled, the underwater shape is no longer symmetrical — more volume on the low side, less on the high side — so B moves towards the low side while G stays put.
The two forces are now offset. They form a couple, and the horizontal distance between their lines of action is the righting lever GZ.
That is the whole of stability. Everything else is a way of finding GZ at a given angle, or of noticing something that has made it smaller.
For small angles there is a shortcut. B moves along a circular arc whose centre is the metacentre M, and the geometry gives GZ ≈ GM·sin θ. This holds while the arc assumption holds — roughly to 10°, sometimes to 15° on a full hull. Beyond that the deck edge approaches the water, the underwater shape changes character, and the shortcut stops being true.
GM is the slope of the GZ curve at the origin. It tells you how stiff the ship feels initially, and nothing at all about what happens at 40°.
Four heights, all measured from the keel, and the distinction between them is examinable in its own right.
Because BM = I/∇ and I depends on breadth cubed, beam is the strongest lever on initial stability. Doubling the beam of a rectangular waterplane multiplies I by eight while multiplying ∇ by only two, so BM rises by a factor of four.
KG is found by taking moments of every weight on board about the keel, and its accuracy limits everything downstream. The inclining experiment is how the lightship KG is established in practice: known weights are shifted across the deck and the resulting small angle is measured with a pendulum.
The experiment is conducted with the ship as nearly empty as possible, tanks pressed up or dry, no free surface, no mooring restraint and calm conditions — every one of those precautions exists to remove something that would corrupt the measured angle.
A partly filled tank is dangerous in a way that a full tank is not. Heel the ship and the liquid runs to the low side. Its own centre of gravity shifts, and the ship's effective centre of gravity shifts with it — reducing GZ at every angle.
The convenient way to account for it is as a virtual rise of G: pretend G has moved up by an amount that produces the same loss.
Three consequences follow, and each one has been an examination question:
Always apply the correction to KG, not to KM: the effective GM is KM − (KG + FSC).
Plot GZ against heel angle and the curve reports on the whole range of stability, not just the beginning.
The intact stability criteria are written almost entirely in terms of that area, in metre-radians:
Every area criterion is in radians. If you integrate a curve plotted in degrees without converting the interval, every answer is 57.3 times too large.
Both leave the ship leaning. The causes are opposite, and so are the remedies — which is exactly why the distinction is examined.
A list is caused by an off-centre weight while GM is positive. The ship settles where the righting moment balances the heeling moment, and it will sit at the same angle steadily.
A loll is caused by negative GM. Upright, the ship has no righting lever at all and will not stay there. It heels until the increasing BM at larger angles has restored GZ to zero, and settles at that angle — on either side, and it may flop from one to the other.
The practical difference: for a list you may correct by moving weight to the high side. Do that to a ship that is lolling and you can capsize her, because the righting lever on the high side is still negative near upright. The correct action for a loll is to lower G first — press up slack tanks one at a time, starting with the low side, and only then think about transverse weight.
The examinable core is the lost buoyancy method. When a compartment floods it stops providing buoyancy; the intact part of the hull must therefore provide more, so the ship sinks and trims until the intact volume has grown by the volume of buoyancy lost.
Permeability μ is the fraction of a space that can actually fill with water — the rest is occupied by structure, machinery or cargo.
Here v is the volume of the flooded space up to the waterline and a is its waterplane area, subtracted because that part of the waterplane no longer contributes to supporting the ship.
Two further ideas complete the outline. Floodable length is the greatest length of compartment that can be flooded without the ship sinking below the margin line, and it is what sets bulkhead spacing. And the modern regulatory approach is probabilistic: rather than requiring survival of one specified damage, it compares an attained subdivision index A — a weighted sum of survival probabilities over many possible damages — against a required index R.
Three problems that between them cover most of what GATE asks of this topic.
A ship of 8000 t displacement has KG = 7.20 m. From the cross curves, KN at 30° heel is 4.85 m and KM is 8.05 m. Find GZ and the righting moment at 30°, and compare with what small-angle theory would have predicted.
Δ = 8000 t, KG = 7.20 m KN at 30° = 4.85 m, KM = 8.05 m
Find GZ and the righting moment at 30°, and compare with what small-angle theory would have predicted
Cross curves are computed for an assumed KG of zero.
Because the naval architect drawing them does not know how the ship will be loaded. To use them you subtract the real KG, resolved along the righting lever — which introduces the sine of the heel angle.
Substitute.
sin 30° = 0.500 exactly, which is why examiners like 30°.
The righting moment is that lever acting on the whole displacement.
Now the comparison.
Small-angle theory says GZ ≈ GM·sin θ, with GM = KM − KG.
The two disagree by a factor of nearly three.
And the real ship is the stiffer of the two. That is not an error: past about 10° the emerging bilge and the immersing deck edge change the shape of the underwater volume, B swings out much further than the small-angle model allows, and GZ grows faster than the sine curve. Small-angle theory is a tangent at the origin, nothing more.
AnswerGZ = 1.25 m, righting moment = 10 000 t·m
The trap: subtracting KG twice — once by reading a KN curve and again by using GM·sin θ as a correction. KN already assumes KG = 0, so exactly one correction is due.
A ship of 6000 t displacement has KM = 8.40 m and KG = 7.50 m. A rectangular double-bottom tank 12 m long and 10 m wide is slack with sea water. (a) Find the effective GM. (b) A weight of 40 t already on board is moved 9 m across the deck — find the resulting list. (c) What would the correction have been had the tank carried a centreline division?
Δ = 6000 t, KM = 8.40 m, KG = 7.50 m Tank: l = 12 m, b = 10 m, slack, ρ_l = 1.025 t/m³ w = 40 t moved d = 9 m transversely
(a) Find the effective GM
(c) What would the correction have been had the tank carried a centreline division?
Solid GM first.
The value before any free surface is accounted for.
The free surface correction is a virtual rise of G.
The liquid does not move the centre of gravity of the ship, but as the ship heels the liquid runs to the low side and its own centre of gravity shifts, which has exactly the same effect as G rising. The size of it depends on the second moment of the liquid surface about its own centreline.
Convert that to a rise of G.
Effective GM is the solid value less the correction.
Now the list.
A transverse shift of weight moves G sideways by w·d/Δ, and the ship heels until GZ balances that offset. For small angles that gives a tangent relation.
Part (c): with a centreline division the tank becomes two compartments each 5 m wide.
Free surface depends on breadth cubed, so halving the breadth cuts each compartment's contribution to one eighth — and there are two of them.
Answer(a) GM = 0.729 m (b) list = 4.7° (c) FSC = 0.043 m
The trap: assuming a nearly empty tank is nearly harmless. The correction contains no term for how much liquid is in the tank — a tank with 20 cm of water in it gives the same free surface as one that is half full. Only the surface dimensions matter.
A ship of 10 000 t displacement has righting levers of 0, 0.20, 0.45, 0.62 and 0.61 m at heel angles of 0°, 10°, 20°, 30° and 40°. Find the area under the GZ curve to 30° and to 40°, and check the loading against the intact stability criteria.
Δ = 10 000 t θ: 0° 10° 20° 30° 40° GZ: 0.00 0.20 0.45 0.62 0.61 m
Find the area under the GZ curve to 30° and to 40°, and check the loading against the intact stability criteria
The criteria are stated in metre-radians.
So the interval must be converted from degrees to radians before integrating. This is the single most-failed step in the topic.
Area to 40° uses four intervals.
| θ (deg) | GZ (m) | SM | Product |
|---|---|---|---|
| 0 | 0.00 | 1 | 0.000 |
| 10 | 0.20 | 4 | 0.800 |
| 20 | 0.45 | 2 | 0.900 |
| 30 | 0.62 | 4 | 2.480 |
| 40 | 0.61 | 1 | 0.610 |
| Σ | 4.790 |
Even, so Simpson's first rule applies.
Area to 30° is three intervals.
Odd, so the first rule will not fit. Use the second rule, whose multipliers are 1 3 3 1 with a factor of 3h/8.
The area between 30° and 40° is the difference.
Now check each criterion.
| Criterion | Required | Actual | Meets |
|---|---|---|---|
| Area 0–30° | ≥ 0.055 m·rad | 0.168 | ✓ |
| Area 0–40° | ≥ 0.090 m·rad | 0.279 | ✓ |
| Area 30–40° | ≥ 0.030 m·rad | 0.110 | ✓ |
| GZ at 30° | ≥ 0.20 m | 0.62 m | ✓ |
All three area requirements are comfortably met, and GZ at 30° is well above the minimum.
Dynamical stability.
The work needed to heel the ship to a given angle — is the displacement times the area.
AnswerA(0–30) = 0.168, A(0–40) = 0.279, A(30–40) = 0.110 m·rad — all criteria met
The trap: integrating in degrees. Do that and every area comes out 57.3 times too large, every criterion passes spectacularly, and the answer is worthless. If an area looks like 9.6 rather than 0.17, this is why.
Righting moment = Δ × GZThe whole of stability in one lineGM = KM − KG = KB + BM − KGOnly KG is under your controlGZ = GM·sin θSmall angles, to about 10°GZ = KN − KG·sin θLarge angles, from cross curvesFSC = i·ρ_l/Δ, i = l·b³/12Virtual rise of G; independent of quantity and positiontan θ_list = (w·d)/(Δ·GM)Off-centre weight, positive GMtan θ_loll = √(2|GM|/BM)Negative initial GMGM = (w·d·L)/(Δ·y)Inclining experiment; L pendulum length, y deflectionT_roll = 2π·k/√(g·GM)k ≈ 0.35B–0.40B; a stiff ship rolls quicklyAreas: 0.055 / 0.090 / 0.030 m·rad0–30°, 0–40°, 30–40°; GZ₃₀ ≥ 0.20 m, GM₀ ≥ 0.15 m