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Engineering & Technology · Marine, Ocean & Offshore Engineering

Buoyancy, hydrostatics and vessel stability

A floating object has enough buoyancy to balance its weight, yet it may roll away from an upright position. Vertical force balance and stability against a small rotation are different questions. Both depend on the actual loading and the submerged geometry.

Use Fluid Engineering, Civil & Structural Engineering and Materials Selection & Degradation for balances, structural response and environmental mechanisms. Physics fluid dynamics and continuum materials provide further preparation. No example supplies a marine operating procedure, stability certificate, mooring design, electrical architecture approval or environmental permission.

Back to the chapter

After reading, you can

Compute buoyant equilibrium and interpret initial metacentric stability without confusing flotation with complete vessel stability.

01

Buoyancy is a pressure resultant

In a stationary fluid of uniform density, pressure increases with depth. Integrating pressure over a closed wetted body gives an upward buoyant force equal to the weight of displaced water. For a freely floating body with no other vertical support, equilibrium requires mg=ρg∇, where ∇ is submerged volume. The centre of buoyancy B is the centroid of that volume.

Draft measures immersion relative to the vessel; freeboard describes the distance from the water surface to a specified deck or opening level. A displaced volume calculation alone does not establish freeboard, watertight integrity or what happens after damage. Water density changes also alter the immersion needed to support an unchanged mass.

02

Waterplane geometry affects the initial restoring moment

As a surface-piercing hull heels slightly, the submerged shape and B shift. The initial transverse metacentre M describes this geometric change. With a consistent vertical reference, GM=KB+BM−KG, where G is centre of gravity and BM=Iwp/∇. Here Iwp is the waterplane area’s second moment about its longitudinal centreline, not the hull material’s structural bending inertia.

Positive GM gives an initial restoring tendency in this model; negative GM gives an overturning tendency near upright. For small heel θ, a metacentric approximation uses righting arm GZ≈GM sin θ and restoring-moment magnitude W GZ. This local approximation does not supply the complete righting-arm curve, a downflooding angle or a large-angle recovery guarantee.

03

Keep the geometry and loading explicit

Take an invented rectangular floating box, length 12 m, beam 6 m and upright draft 1 m. Supply uniform water density 1025 kg/m³. Submerged volume is 72 m³ and supported mass 73800 kg. The buoyancy centroid lies 0.50 m above its flat keel. Its rectangular waterplane gives Iwp=12×6³/12=216 m⁴, so BM=3.0 m.

If the supplied KG is 2.0 m, GM=0.5+3−2=1.5 m. Raising G to 4.0 m with the same mass and upright volume gives GM=−0.5 m. Both cases satisfy the same vertical buoyancy balance but have opposite initial rotational tendencies. The figure shows only the declared small-angle approximation around upright.

Equal flotation can have opposite initial stabilityLength 12m, beam 6m, draft 1m, water density 1025 kg/m³ imply volume 72m³ and mass 73800 kg. KB 0.5m, BM 3m. KG 2m gives GM 1.5m; KG 4m gives GM−0.5m. Curves GZ≈GM sinθ for θ−5 to+5 degrees; positive GZ opposes positive heel. This local model is not a complete large-angle or damage-stability curve.Equal flotation can have opposite initial stabilityApproximate GZ / m-0.10.00.1-505Heel / degrees · approximation restricted to ±5°Solid GM=+1.5 m; dashed GM=−0.5 mPositive arm here denotes a tendency opposing positive heel
Length 12m, beam 6m, draft 1m, water density 1025 kg/m³ imply volume 72m³ and mass 73800 kg. KB 0.5m, BM 3m. KG 2m gives GM 1.5m; KG 4m gives GM−0.5m. Solid curve: GM=+1.5m; dashed curve: GM=−0.5m. Both curves use GZ≈GM sinθ for θ−5 to+5 degrees; positive GZ opposes positive heel. This local model is not a complete large-angle or damage-stability curve.

Calculate one small-angle restoring moment

With g=9.81 m/s², the original weight is 723978 N. At a supplied five-degree heel and GM=1.5 m, the approximate arm is 1.5 sin 5°≈0.13073 m, giving restoring moment about 94.65 kN·m. Its direction opposes the heel. In the negative-GM case the local tendency instead reinforces the displacement.

Actual loading changes, slack-liquid free surfaces, flooding and hull geometry can change stability. Strong initial restoring action also does not guarantee comfortable motion. The hypothetical box has no complete deck, opening, damage or wave-response assessment; the arithmetic is neither a loading instruction nor an acceptance test for a real vessel.

Try the reasoning

  1. Does unchanged total mass guarantee unchanged initial stability?

    Show the answer and reasoning

    No. Its vertical distribution can change KG, while liquid free surfaces and other changes can alter the effective restoring behavior.

  2. Which second moment appears in BM=Iwp/∇?

    Show the answer and reasoning

    The second moment of the waterplane area about the relevant horizontal rotation axis, not the material cross-section used for structural EI.

  3. Does positive GM prove large-angle or damaged stability?

    Show the answer and reasoning

    No. It describes an initial tendency near the stated equilibrium; complete stability needs the relevant geometry, loading and damage conditions.

What to carry forward

Close the buoyancy balance, then examine the loading-dependent restoring behavior and the range over which the stability model applies.

References

Teaching resources and research supporting this lesson’s explanations and examples.

  1. 2.019 Ship Hydrostatics and Structures, February 11, 2011, selected pp 1–16Chryssostomos Chryssostomidis and Yuming Liu · MIT OpenCourseWareTeaching resource · PDFocw.mit.eduRead source