Chapter 02 of 12 · GATE NM

Stability

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.

Worked examples3, fully stepped
Read time≈ 15 min
PrerequisiteHydrostatics

1. What actually rights a ship

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.

Righting moment = Δ × 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.

The key idea

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°.

2. KB, BM, KM, KG — and which of them you control

Four heights, all measured from the keel, and the distinction between them is examinable in its own right.

  • KB — height of the centre of buoyancy. Depends on hull shape and draught. Roughly 0.53 T on a normal merchant hull.
  • BM — metacentric radius, equal to IT/∇. Depends on the waterplane and the displaced volume.
  • KM = KB + BM. Still pure geometry: it is a property of the ship at a given draught, printed in the hydrostatic tables, and no amount of loading changes it.
  • KG — height of the centre of gravity. This is the loading, and it is the only one the ship's staff can change.
GM = KM − KG

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.

GM = (w · d · L_pendulum) / (Δ · deflection)

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.

3. Free surface: stability lost without moving anything

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.

FSC = i · ρ_liquid / Δ where i = l·b³/12 for a rectangular tank

Three consequences follow, and each one has been an examination question:

  • Breadth is everything. i depends on b³. A tank split down the centreline has each half at half the breadth, so each contributes one eighth — two of them give one quarter of the undivided value.
  • The quantity of liquid does not appear. A tank with 10 cm of water gives the same free surface as one half full. This is why an almost-empty tank is not almost-safe.
  • The position of the tank does not appear either. High or low, forward or aft, the correction is the same — only the surface dimensions matter.

Always apply the correction to KG, not to KM: the effective GM is KM − (KG + FSC).

4. The GZ curve and what its shape tells you

Plot GZ against heel angle and the curve reports on the whole range of stability, not just the beginning.

  • The initial slope equals GM in radians. Draw a tangent at the origin and read it at 1 radian (57.3°) — a standard construction, and a standard question.
  • The maximum is where the righting moment is greatest. Its angle matters: a maximum at 25° or beyond means the ship keeps working for you well past the point a stiff, shallow-curved hull has given up.
  • The range of stability is the angle at which GZ returns to zero — beyond it, the ship capsizes.
  • The area under the curve is energy: the work needed to heel the ship there. This is why it is called dynamical stability and why gusts and waves, which deliver energy rather than a steady moment, are assessed against area rather than against GZ.

The intact stability criteria are written almost entirely in terms of that area, in metre-radians:

Area 0–30° ≥ 0.055 m·rad Area 0–40° (or θ_f) ≥ 0.090 m·rad Area 30–40° ≥ 0.030 m·rad GZ at 30° or more ≥ 0.20 m Angle of max GZ ≥ 25° Initial GM ≥ 0.15 m
The key idea

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.

5. List and loll are not the same thing

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.

tan θ_list = (w · d)/(Δ · GM)

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.

tan θ_loll = √(2·|GM| / BM)

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.

6. Damage stability, in outline

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.

Empty hold μ ≈ 0.95 Machinery space μ ≈ 0.85 Stores μ ≈ 0.60 Sinkage = v·μ / (A_W − a·μ)

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.

7. Worked examples

Three problems that between them cover most of what GATE asks of this topic.

Worked example 1

Righting lever from a KN curve

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.

Given

Δ = 8000 t, KG = 7.20 m KN at 30° = 4.85 m, KM = 8.05 m

Required

Find GZ and the righting moment at 30°, and compare with what small-angle theory would have predicted

  1. Cross curves are computed for an assumed KG of zero.

    GZ=KN − KG·sin θ

    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.

  2. Substitute.

    GZ=4.85 − 7.20 × 0.500 =4.85 − 3.60 =1.25 m

    sin 30° = 0.500 exactly, which is why examiners like 30°.

  3. The righting moment is that lever acting on the whole displacement.

    Righting moment=Δ × GZ =8000 × 1.25 =10 000 t·m
  4. Now the comparison.

    GM=8.05 − 7.20 = 0.85 m GZ (small angle)=0.85 × 0.500 = 0.425 m

    Small-angle theory says GZ ≈ GM·sin θ, with GM = KM − KG.

  5. 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.

Worked example 2

Free surface correction and the resulting list

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?

Given

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

Required

(a) Find the effective GM
(c) What would the correction have been had the tank carried a centreline division?

  1. Solid GM first.

    GM(solid)=KM − KG =8.40 − 7.50 =0.90 m

    The value before any free surface is accounted for.

  2. The free surface correction is a virtual rise of G.

    i=l·b³/12 =12 × 10³/12 =1000 m⁴

    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.

  3. Convert that to a rise of G.

    FSC=i·ρ_l/Δ = 1000 × 1.025 / 6000 =0.171 m
  4. Effective GM is the solid value less the correction.

    GM(eff)=0.90 − 0.171 = 0.729 m
  5. Now the list.

    tan θ=(w × d)/(Δ × GM) =(40 × 9)/(6000 × 0.729) =360/4374 = 0.0823 θ=4.7°

    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.

  6. Part (c): with a centreline division the tank becomes two compartments each 5 m wide.

    i=2 × (12 × 5³/12) =2 × 125 =250 m⁴ FSC=250 × 1.025/6000 = 0.043 m One quarter of the undivided value.

    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.

Worked example 3

Dynamical stability and an IS Code check

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.

Given

Δ = 10 000 t θ: 0° 10° 20° 30° 40° GZ: 0.00 0.20 0.45 0.62 0.61 m

Required

Find the area under the GZ curve to 30° and to 40°, and check the loading against the intact stability criteria

  1. The criteria are stated in metre-radians.

    h=10° =10 × π/180 =0.1745 rad

    So the interval must be converted from degrees to radians before integrating. This is the single most-failed step in the topic.

  2. Area to 40° uses four intervals.

    Area to 40° — Simpson's first rule, four intervals
    θ (deg)GZ (m)SMProduct
    00.0010.000
    100.2040.800
    200.4520.900
    300.6242.480
    400.6110.610
    Σ4.790
    A(0–40)=(h/3) × Σ =(0.1745/3) × 4.790 =0.279 m·rad

    Even, so Simpson's first rule applies.

  3. Area to 30° is three intervals.

    Σ=0×1 + 0.20×3 + 0.45×3 + 0.62×1 =0 + 0.60 + 1.35 + 0.62 = 2.57 A(0–30)=(3h/8) × Σ = (3 × 0.1745/8) × 2.57 =0.168 m·rad

    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.

  4. The area between 30° and 40° is the difference.

    A(30–40)=0.279 − 0.168 = 0.110 m·rad
  5. Now check each criterion.

    Intact stability criteria
    CriterionRequiredActualMeets
    Area 0–30°≥ 0.055 m·rad0.168
    Area 0–40°≥ 0.090 m·rad0.279
    Area 30–40°≥ 0.030 m·rad0.110
    GZ at 30°≥ 0.20 m0.62 m

    All three area requirements are comfortably met, and GZ at 30° is well above the minimum.

  6. Dynamical stability.

    Dynamical stability to 40°=Δ × A =10 000 × 0.279 = 2790 t·m·rad

    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.

Reference sheet
60-second recall
  1. GM = KM − KG, and KM is geometry, KG is loading.
  2. GZ = KN − KG·sin θ at large angles — one correction, not two.
  3. FSC ∝ b³, and does not care how full the tank is.
  4. Convert degrees to radians before integrating a GZ curve.
  5. List: move weight. Loll: lower G first.