KB, BM, KG & the Metacentre

The four vertical heights that set up every stability calculation that follows.

Marine EngineeringShip StabilityFree preview
⏱️ About 16 min

Every floating vessel has a hidden geometry of points — keel, centre of buoyancy, centre of gravity, and metacentre — whose relative heights decide whether it stays upright or capsizes.

💡
The big idea: A ship's initial stability is governed by the relative heights of four points measured from the keel: K, B, G, and M.
🎯 By the end, you'll be able to
  • Define KB, KG, BM, and KM and state what each represents physically
  • Compute BM = I/∇ for a box-shaped hull from its dimensions and draft
  • Explain why waterplane shape, not hull depth, drives the value of BM
  • Calculate KM = KB + BM for a simple geometric hull form
📎 Helpful to know first

Four Heights Above the Keel

You already know from fluid mechanics that a floating body experiences an upward buoyant force equal to the weight of fluid it displaces, acting through the centre of buoyancy B — the centroid of the submerged hull volume. Naval architecture takes this further: we care not just about whether the vessel floats, but about what happens when it heels.

Define four vertical measurements, all taken from the keel K:

  • KB — height of the centre of buoyancy above the keel. For a box-shaped hull at draft T, symmetry gives KB = T/2. For a real ship-shaped hull, KB is read from hydrostatic tables.
  • KG — height of the centre of gravity above the keel. This depends entirely on the loading condition — where cargo, fuel, ballast, and structure sit vertically. KG is the unknown in most stability problems, and it's determined experimentally (see the Inclining Experiment lesson).
  • BM — the metacentric radius: the vertical distance from B to the metacentre M. M is defined as the intersection of successive lines of action of the buoyant force as the vessel heels through infinitesimally small angles. BM is purely a function of hull geometry at the waterline.
  • KM — height of the metacentre above the keel: simply KM = KB + BM.
\[ BM = \frac{I}{\nabla}, \qquad I = \frac{L \cdot B^{3}}{12} \ \text{(rectangular waterplane)} \]
I is the waterplane's second moment of area about the centreline; ∇ is the displaced volume.
Ship cross-section showing the keel K at the bottom, the centre of buoyancy B below the waterline, the centre of gravity G above B, and the metacentre M above G, all on the ships centrelineWaterlineK — keelB — centre of buoyancyG — centre of gravityM — metacentre
The four reference points, all measured as heights above the keel K.
🔑 Why BM depends on beam so strongly

Because I depends on breadth cubed (B³) but only on length linearly, doubling a hull's beam increases I — and therefore BM — by a factor of eight. A long, narrow hull (a rowing shell) has tiny BM and tips easily; a wide, shallow-draft barge has large BM and strongly resists heeling. The waterplane is the hull's “stiffness” against rotation, the way a wide beam resists structural bending.

📝 Worked example: A box-shaped barge has length L = 40 m, breadth B = 10 m, and floats at draft T = 3 m in seawater. Find I, ∇, BM, and KM.
  1. Waterplane second moment of area: I = L·B³/12 = (40 × 10³)/12 = 40,000/12 = 3333.33 m⁴
  2. Displaced volume: ∇ = L × B × T = 40 × 10 × 3 = 1200 m³
  3. Metacentric radius: BM = I/∇ = 3333.33/1200 = 2.778 m
  4. Height of B above keel (box shape): KB = T/2 = 3/2 = 1.500 m
  5. Height of metacentre above keel: KM = KB + BM = 1.500 + 2.778 = 4.278 m
✓ KM = 4.278 m
✏️ Practice: A box-shaped pontoon has length L = 50 m, breadth B = 12 m, and floats at draft T = 2.5 m. Compute KM (in metres).
m
Solution
  1. I = L·B³/12 = (50 × 12³)/12 = 86,400/12 = 7200 m⁴
  2. ∇ = L × B × T = 50 × 12 × 2.5 = 1500 m³
  3. BM = I/∇ = 7200/1500 = 4.800 m
  4. KB = T/2 = 2.5/2 = 1.250 m
  5. KM = KB + BM = 1.250 + 4.800 = 6.050 m

Check your understanding

1. What does BM represent?
BM is the metacentric radius — the vertical distance from the centre of buoyancy B up to the metacentre M, equal to I/∇.
2. For a box-shaped hull, doubling the beam (with length and draft unchanged) changes BM by approximately what factor?
I = L·B³/12, so doubling B multiplies I — and BM — by 2³ = 8, since ∇ changes only linearly with beam at fixed draft and length.
🔎 Related lessons
Initial Transverse Stability & GM
✅ Key takeaways
  • KM = KB + BM, where KB depends on the underwater hull shape and BM on the waterplane shape
  • BM = I/∇: the cubic dependence on beam means small increases in breadth dramatically improve initial stability
  • KG is set by loading, not hull form — it's the variable the operator actually controls
➡️ Next, we combine KM with KG to get GM — the single number that tells you whether a ship is stable, unstable, or neutral.
Want to test yourself on this? Try the Marine Engineering Aptitude test →