Understanding Ship Stability calculations and formula

Introduction

Stability calculations depend on the position of the ship’s centre of gravity (COG). It is measured in three ways: transversely, vertically and longitudinally. The COG is the main factor that governs the list and trim of the ship.

• KG: the COG measured from the keel (vertical position).
• AG: the COG measured from the aft perpendicular (longitudinal position).
• MG: the COG measured from midship.
• LCG: the longitudinal COG, quoted from the aft perpendicular or from midship.

List is found from transverse stability, and trim from longitudinal stability. This article collects the main formulae, with worked examples.

1. List

When a weight is shifted across the ship, G moves to G₁ and the ship lists. The angle of list is found from:

tan θ = Listing moment / (W × GM) = (w × d) / (W × GM) = GG₁ / GM

• w is the weight shifted
• d is the distance it moves (from the centreline, or the total shift across)
• W is the final displacement
• GM is the metacentric height

Proof

Step 1: The weight w moves a distance d. The ship’s total weight W has its centre of gravity move in the same direction by:

GG₁ = (w × d) / W

Step 2: The ship lists until G₁ is directly below M. In the right-angled triangle formed by G, G₁ and M, the angle at M is the angle of list θ, so:

tan θ = GG₁ / GM = (w × d) / (W × GM)

2. Vertical Shift of G and Final KG

The same idea works vertically. If a weight is moved up or down through a vertical distance d, the shift of G is:

GG₁ = (w × d) / W

When weights are loaded or discharged, the final KG is found by taking moments about the keel:

Final KG = Vertical moment / W

For example, a ship of 10,000 t displacement with KG 7.0 m loads 500 t at KG 4.0 m. The final displacement is 10,500 t, so Final KG = (10,000 × 7.0 + 500 × 4.0) / 10,500 = 72,000 / 10,500 ≈ 6.86 m.

3. Free Surface Correction (FSC)

A slack tank has a free liquid surface. As the ship heels, the liquid moves to the low side and G appears to rise. This is the free surface effect. The correction is:

FSC = (i × ρ_i) / W = (L × B³ × ρ_i) / (12 × W)

• i is the moment of inertia of the free surface, L × B³ / 12 for a rectangular tank
• ρ_i is the density of the liquid in the slack tank
• W is the displacement

From the final KG and the FSC, the fluid KG is obtained: KG_f = Final KG + FSC. With KM and KG_f, the fluid metacentric height is obtained: GM_f = KM − KG_f.

For example, a slack tank 12 m long and 12 m wide holding sea water (1.025 t/m³) in a ship of 30,000 t gives FSC = 1.025 × (12 × 12³ / 12) / 30,000 = 1.025 × 1728 / 30,000 ≈ 0.06 m. Note that the breadth is cubed, so a wide tank is much worse than a narrow one.

4. BM for Simple Shapes

BM is the distance from the centre of buoyancy to the metacentre:

BM = I / V

where I is the moment of inertia of the waterplane and V is the underwater volume.

Box-shaped vessel. The waterplane is a rectangle, so I = L B³ / 12, and V = L B d. Then:

BM = L B³ / (12 L B d) = B² / (12 d)

Triangular-shaped vessel. The waterplane is still a rectangle, but V = L B d / 2. Then:

BM = (L B³ / 12) × (2 / L B d) = B² / (6 d)

5. Trim

Trim is the difference in draught between the forward end and the aft end. It is affected by the position of the COG against the COF (centre of flotation, also written LCF). The COF is the point about which the vessel trims.

tc = Trimming moment / MCTC = (w × d) / (MCTC × 100)

Here tc is the change of trim in metres, and MCTC (moment to change trim one centimetre) is in tonne-metres per cm, which is why the 100 appears. The distance d is:

• the distance between the LCF and the LCG of the weight, when loading or discharging;
• the distance between LCG₁ and LCG₂, the old and new positions, when shifting a weight.

The change of draught at the aft end is a share of the trim change:

T_A = (AF / LBP) × tc

where AF is the distance from the aft perpendicular to the LCF. The change at the forward end is T_F = tc − T_A.

For example, MCTC = 200 t·m/cm, LBP = 150 m and the LCF is 70 m from the aft perpendicular. Loading 400 t at 30 m forward of the LCF gives a trimming moment of 12,000 t·m. Then tc = 12,000 / (200 × 100) = 0.6 m by the head. T_A = 70 / 150 × 0.6 = 0.28 m (aft draught reduced), and T_F = 0.6 − 0.28 = 0.32 m (forward draught increased). Bodily sinkage is added to both.

6. Shifting Liquid Cargo

To find d for liquid cargo, so as to calculate GG₁ (the shift in KG), the centroid of the cargo moved is taken, not the centre of the tank. This applies only to fluid cargo.

Example. Port and starboard tanks, each 12 m × 12 m × 9 m, lie either side of the centreline. The starboard tank is full of sea water and the port tank is empty. Half of the water is transferred from starboard to port.

• Weight moved: w = 0.5 × 12 × 12 × 9 × 1.025 ≈ 664 t.
• Transverse distance: the moved water leaves the starboard tank, whose centre is 6 m from the centreline, and arrives in the port tank, also 6 m from it. So d = 12 m.
• Vertical distance: the top half of the starboard tank (centroid 6.75 m above its bottom) moves to the bottom half of the port tank (centroid 2.25 m). So it falls 4.5 m.

For a ship of 30,000 t with GM 1.5 m: GG₁ = 664 × 12 / 30,000 ≈ 0.27 m, tan θ = 0.27 / 1.5 ≈ 0.177, so the list is about 10°. G also falls by 664 × 4.5 / 30,000 ≈ 0.10 m.

7. Fresh Water Allowance, Dock Water Allowance and TPC

FWA = W / (40 × TPC) (in cm)

W is the displacement of the ship in salt water, and TPC is the tonnes per centimetre immersion in salt water. A ship in fresh water floats deeper by the FWA, so it can load to a greater draught in fresh water up to the F mark.

DWA = (Change in RD / 0.025) × FWA

In water of relative density between salt water (1.025) and fresh water (1.000), the allowance is a fraction of the FWA.

TPC = (A / 100) × density of water displaced

A is the area of the waterplane in m². For example, with A = 1950 m² in sea water, TPC = 1950 / 100 × 1.025 ≈ 20.0 t/cm. For W = 12,000 t, FWA = 12,000 / (40 × 20) = 15 cm. In dock water of RD 1.010, the change in RD is 0.015, so DWA = 0.015 / 0.025 × 15 = 9 cm.

Sinkage

When a weight w is added, the ship sinks bodily by the weight divided by TPC:

Sinkage (cm) = w / TPC

TPC is the tonnes needed to sink the ship by 1 cm, so adding 500 t at 20 t/cm gives 500 / 20 = 25 cm of sinkage.

8. Righting Lever (GZ) and Righting Moment (RM)

The righting lever is the horizontal distance between G and the vertical through the centre of buoyancy. θ is the angle of heel.

• Small angles (up to 15°): GZ = sin θ × GM
• Large angles (over 15°), wall-sided formula: GZ = sin θ × (GM + ½ BM tan²θ)

The righting moment is the weight times the righting lever:

• Small angles: RM = W × GZ = W × GM × sin θ
• Large angles: RM = W × GZ = W × (GM + ½ BM tan²θ) × sin θ

For example, with GM = 1.0 m, BM = 6.0 m and W = 20,000 t, at a heel of 30° the wall-sided formula gives GZ = 0.5 × (1 + 0.5 × 6 × 0.333) = 1.0 m, so RM = 20,000 t·m. The small angle formula would give only 0.5 m and 10,000 t·m, which shows why it must not be used at large angles.

9. Quick Reference

• List: tan θ = w × d / (W × GM)
• Final KG = vertical moment / W
• FSC = i × ρ / W = L B³ ρ / (12 W)
• BM (box) = B² / (12 d), BM (triangle) = B² / (6 d)
• Change of trim tc = w × d / (MCTC × 100) metres
• T_A = AF / LBP × tc
• FWA = W / (40 × TPC), DWA = ΔRD / 0.025 × FWA, TPC = A / 100 × ρ
• Sinkage = w / TPC
• GZ = GM sin θ (small angles), GZ = sin θ (GM + ½ BM tan²θ) (large)
• RM = W × GZ

Conclusion

These formulae let you predict the list, trim, draught and stability of a ship before and after cargo and ballast operations. Remember that every example here is a model: check the actual hydrostatic data and loading computer for your ship.

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