Learning objectives
By the end of this chapter you will be able to:
- define the transverse metacentre M and explain why the buoyancy line passes through it at small angles of heel;
- build the vertical ladder KMT = KB + BMT from the hydrostatic table and GM = KM − KG from the loading;
- classify a ship as stable, neutral or unstable from the sign of her GM;
- calculate the angle of list from tan(List) = GGH ÷ GM;
- solve full list problems by taking vertical and horizontal moments in a single table;
- allow for a suspended weight swung off the centreline;
- size the weight transfer needed to bring a listed ship upright.
Chapter 6 put G under the officer's control; Chapter 1 gave the water its say through B. This chapter stages the contest between them. The referee is a point that most people meet here for the first time: the transverse metacentre M. Once M is on the page, three of the most important questions in ship stability collapse into simple subtractions and one small triangle: is she stable, how hard will she resist heeling, and at what angle will she settle if G strays off the centreline?
7.1 The metacentre
Heel a ship a few degrees. The waterplane tilts, a wedge of hull dips in on the low side and an equal wedge emerges on the high side, and the centre of buoyancy B slides across towards the low side, to B₁. The buoyancy force still acts vertically upwards through B₁, perpendicular to the waterline. Now extend that line of action upwards: for small angles of heel it always crosses the ship's centreline at very nearly the same point, however the ship got there. That crossing point is the transverse metacentre, M, and its steadiness at small angles is what makes the whole of this chapter possible.
7.2 The ladder to GM
M has a height above the keel like every other point on the ship, and the hydrostatic table supplies it. The MCA sheet writes the relationship with a subscript T for transverse:
KB is the height of the centre of buoyancy, a little more than half the draught for a ship shaped hull; BMT is the distance from B up to M, which comes from the breadth and shape of the waterplane and the volume of displacement. Both are tabulated against draught in the booklet, so in practice KM is simply read out. At MV Ninja's summer draught the booklet gives KB 5.041 m and KM 10.330 m, so BM is their difference, 5.289 m. Then the loading takes over: Chapter 6's moments table delivered KG 8.09 m for the departure condition, and the gap left between G and M is the metacentric height:
At her summer draught MV Ninja's booklet gives KB 5.041 m and KM 10.330 m. Her loaded KG, from the moments table of Worked example 6.4, is 8.09 m. Find BMT and the metacentric height GM.
BMT = KM − KB = 10.330 − 5.041 = 5.289 m
GM = KM − KG = 10.330 − 8.09 = 2.24 m
Notice what varies and what does not. KB and KM belong to the hull and the draught: no amount of cargo planning changes them. KG belongs entirely to the loading. The officer's control over GM is exercised through one number only, and it is the one Chapter 6 taught.
7.3 Stable, neutral, unstable
Why does the gap between G and M matter so much? Heel a stable ship: the weight acts down through G, the buoyancy up through B₁, and because G lies below M these two forces form a couple that turns her back upright. Now imagine loading her so high that G climbs above M: the same construction produces a couple that pushes her further over. The sign of GM is the whole verdict:
The three states of equilibrium
- Stable: G below M, GM positive. Disturbed, she returns upright.
- Neutral: G at M, GM zero. Disturbed, she rests where she is put.
- Unstable: G above M, GM negative. Disturbed, she heels further.
A ship floats with KM 10.330 m. State her condition if her KG is (a) 8.50 m, (b) 10.330 m, (c) 10.60 m.
(a) GM = 10.330 − 8.50 = +1.83 m: stable.
(b) GM = 0: neutral equilibrium; she will rest at a small angle of heel to either side.
(c) GM = 10.330 − 10.60 = −0.27 m: unstable; she cannot stay upright.
Case (c) does not necessarily mean immediate capsize: as the ship heels, the waterplane widens and M climbs, and she may find rest at an angle of loll. That dangerous condition, and the special formula the MCA sheet holds for it, belong to a later chapter; at this stage the operational rule is simple: GM must be positive, always.
7.4 The angle of list
Chapter 6 ended with G pushed off the centreline, to G₁, and the promise that the ship would not float upright. Here is what happens. She heels towards the weight, B slides across as the wedge immerses, and she settles at the angle where B₁ stands directly beneath G₁, with M, G₁ and B₁ in one vertical line. The settled angle is the list, and the small triangle M G G₁ delivers it: the side GGH lies athwartships, the side GM lies up the centreline, the angle at G is a right angle, and the angle at M between GM and the vertical MG₁ is the list itself.
MV Ninja lies at her summer displacement of 30456 t with the Chapter 6 departure condition: KG 8.09 m, GM 2.24 m. During cargo operations 250 t of grain is shifted 10.0 m across the ship. Find the resulting list.
GGH = (w × s) ÷ ∆ = (250 × 10.0) ÷ 30456 = 2500 ÷ 30456 = 0.0821 m
tan(List) = GGH ÷ GM = 0.0821 ÷ 2.24 = 0.0367
List = 2.1° towards the side the grain moved to.
Every tool here is already familiar: the shift formula is Chapter 6's, applied sideways; only the final line, turning a horizontal shift of G into an angle, is new.
MV Ninja lies upright at her winter displacement of 29751 t, KG 8.00 m. She then loads 300 t of cargo at Kg 12.00 m, stowed 7.5 m to starboard of the centreline. Find her list on completion, using the booklet KM at the final displacement.
New displacement = 29751 + 300 = 30051 t
New KG = Σ Vertical Moments ÷ ∆ = (29751 × 8.00 + 300 × 12.00) ÷ 30051 = 241608 ÷ 30051 = 8.04 m
At the final displacement of 30051 t the booklet gives, by interpolation between the 9.40 m and 9.60 m rows, a draught of 9.485 m and KM 10.329 m (at the winter draught it was 10.328 m).
GM = KM − KG = 10.329 − 8.04 = 2.29 m
GGH = Σ Horizontal Moments ÷ ∆ = (300 × 7.5) ÷ 30051 = 2250 ÷ 30051 = 0.0749 m to starboard
tan(List) = 0.0749 ÷ 2.29 = 0.0327, so List = 1.9° to starboard
Note the order of operations, which never changes: vertical moments first, because the list formula needs the final GM; horizontal moments second; the triangle last. The 300 t adds under 9 cm of draught and moves M by a millimetre, but the table is still entered at the final displacement.
7.5 List problems in full
A real loading mixes everything at once: weights on and off the centreline, weights loaded, discharged and shifted. One table handles it all. The vertical columns are exactly Chapter 6's moments about the keel; two new columns take moments about the centreline, port and starboard keeping opposite signs. The rules are unchanged: G moves towards what comes aboard, away from what leaves, parallel to what shifts, in both directions at once.
MV Ninja floats upright at a displacement of 28343 t, KG 7.80 m. She then works cargo as follows: loads 800 t at Kg 6.00 m, 4.0 m to port; loads 400 t at Kg 12.00 m, 6.0 m to starboard; discharges 300 t of ballast from Kg 2.00 m on the centreline; and shifts 200 t already on board 6.00 m downwards and 8.0 m to starboard. Find her final list. (Booklet KM at the final displacement, by interpolation: 10.331 m.)
| Item | w (t) | Kg (m) | Vertical moment (t m) | Distance | Horizontal moment (t m) |
|---|---|---|---|---|---|
| Ship as she lies | 28343 | 7.80 | 221075 | — | — |
| Load tween deck | +800 | 6.00 | +4800 | 4.0 P | 3200 P |
| Load deck cargo | +400 | 12.00 | +4800 | 6.0 S | 2400 S |
| Discharge DB ballast | −300 | 2.00 | −600 | — | — |
| Shift 200 t down, to starboard | (200) | −6.00 | −1200 | 8.0 S | 1600 S |
| Totals | 29243 | 228875 | 800 S |
KG = 228875 ÷ 29243 = 7.83 m, so GM = 10.331 − 7.83 = 2.50 m
GGH = 800 ÷ 29243 = 0.0274 m to starboard
tan(List) = 0.0274 ÷ 2.50 = 0.0110, so List = 0.6° to starboard
The port and starboard moments very nearly cancel: 3200 port against 4000 starboard. Good stowage planning aims at exactly this near cancellation, and the table shows at a glance which side is winning.
MV Ninja, displacement 26000 t, KG 7.30 m, uses her crane to lift 60 t from the bottom of a hold (Kg 3.00 m, on the centreline) and swings it outboard until it hangs 9.0 m to port of the centreline. The crane head is at Kg 21.50 m, and the booklet KM at this displacement is 10.406 m. Find the list while the load hangs outboard.
From the instant of lift off the weight acts at the crane head (Chapter 6), an effective vertical shift of 21.50 − 3.00 = 18.50 m, so during the lift:
KG = 7.300 + (60 × 18.50) ÷ 26000 = 7.300 + 0.043 = 7.343 m
GM = 10.406 − 7.343 = 3.06 m
Swung 9.0 m to port, the suspended weight pulls G off the centreline: GGH = (60 × 9.0) ÷ 26000 = 0.0208 m to port
tan(List) = 0.0208 ÷ 3.06 = 0.0068, so List = 0.4° to port
Both effects of the suspended weight act together: the virtual rise of G reduces GM at the very moment the outreach creates the listing moment. Heavy lift calculations always use the lifted condition for both.
7.6 Correcting a list, and onwards
Run the list formula backwards and it becomes a tool for putting things right. A listed ship needs G back on the centreline; the transfer moment w × d that does it must equal the listing moment GGH × ∆. From tan(List) the required GGH follows, and the transfer is sized in one line.
At her summer displacement of 30456 t with GM 2.24 m, MV Ninja is found listing 3° to starboard. Ballast can be transferred from a starboard tank to its port pair across 14.0 m. How much must be moved to bring her upright?
The list corresponds to GGH = tan(3°) × GM = 0.0524 × 2.24 = 0.1174 m off the centreline.
Listing moment = GGH × ∆ = 0.1174 × 30456 = 3575 t m
w = 3575 ÷ 14.0 = 255 t transferred to port.
Always transfer towards the high side, and always check the vertical consequence too: a transfer between tanks at the same height leaves KG untouched, which is exactly why paired wing tanks are the tool of choice.
One more idea completes the picture and opens the next chapter. In the heeled ship, weight down through G and buoyancy up through B₁ form a couple; the horizontal distance between their lines of action is the righting lever, GZ, and at small angles GZ = GM × sin θ. Everything this chapter did with one number, GM, Chapter 8 does properly with GZ across the whole range of heel.
Interactive: the list simulator
Set the ship's condition and apply an off centreline weight shift. The ship heels to the calculated list, live.
Interactive: the equilibrium explorer
Slide KG up and down against a fixed KM of 10.330 m and watch the verdict change.
Chapter summary
- The transverse metacentre M is where the heeled buoyancy line cuts the centreline; at small angles it is effectively fixed.
- KMT = KB + BMT comes from the hydrostatic table; GM = KM − KG marries it to the loading.
- GM positive, zero or negative means stable, neutral or unstable.
- tan(List) = GGH ÷ GM, with GGH = Σ Horizontal Moments ÷ ∆.
- Vertical moments first for GM, horizontal moments second for GGH, the triangle last.
- A suspended weight swung outboard reduces GM and creates the listing moment at the same time.
- To correct a list: w × d = GGH × ∆, transferring towards the high side.
Self test questions
Work each question with pencil and paper first. Your score appears in the bar below.