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Loading for Stability

6 min read

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River Cargo Operations
Loading, waterproofing, and scheduling
Current
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Loading for Stability
Center of gravity, ballast, trim

A boat’s initial stability is its metacentric height GM = KB + BM − KG, where KB is the height of the centre of buoyancy, BM = I ÷ displaced volume the metacentric radius and KG the height of the centre of gravity, and loading changes it through KG, through the weight shifted off-centre and through loose water. For a 6 m by 2.4 m scow carrying 1,200 kg, GM is 3.27 m, rising cargo to 1.6 m cuts it to 2.74 m, a 500 kg load moved 1 m off-centre lists the boat 5.2°, and a loose layer of water 5 m by 2.2 m on deck removes 2.46 m of the 3.27 m, leaving 0.80 m.

Part of River Navigation & Ferry Systems

Loading for stability is the placing of weight, the choice of ballast and the control of loose water so that a river boat stays upright through its working range of heel.

Stiff At Small Angles Is Not The Same As Safe At Large Ones

The carrier’s rules are: heavy items low and centred, equal weight to port and starboard, slightly more weight aft than forward to raise the bow, never stack cargo higher than the gunwale, and keep freeboard of at least 15 cm. These rules are the qualitative form of a calculation, and the calculation shows which of them matters most and where a feeling of stability misleads.

For a rectangular scow of length L, beam B and draft T, floating level: KB = T ÷ 2; the waterplane second moment is I = L × B³ ÷ 12, the displaced volume is L × B × T, so BM = I ÷ (L × B × T) = B² ÷ (12 × T). Then GM = KB + BM − KG. For small heel angles the righting arm is GZ = GM × sin φ, and for a wall-sided hull until the deck edge goes under, GZ = sin φ × (GM + ½ BM tan² φ). These are standard naval-architecture relations, recalled from textbooks (Rawson & Tupper; Biran), not re-read here.

The distinction that matters is initial stability against range. A wide scow has a large BM, hence a large GM, so it resists small tilts hard. But its range is limited by the freeboard: once the deck edge enters the water the extra righting contribution stops. A boat with high initial stiffness and shallow freeboard feels fine until it does not.

What You Need

  • Dimensions: length, beam, depth, empty mass, and the height of its centre of gravity
  • A weight list with the height of each item’s centre above the bottom
  • A heavy known weight (a person, a stone) for the inclining test
  • A way to measure heel: a plumb line on a measured board, or a long pendulum
  • A calm mooring, with the boat otherwise empty of loose water

Method: Compute, Then Measure

Step 1: Compute the draft = total mass ÷ (1,000 × L × B).

Step 2: Compute KG as the mass-weighted average height of everything aboard.

Step 3: Compute GM = T ÷ 2 + B² ÷ (12 T) − KG.

Step 4: Check the range: the deck-edge angle is arctan(freeboard ÷ (B ÷ 2)); compute GZ there.

Step 5: Incline the boat. Move a known weight w across a known distance d, measure the heel φ, and GM = w × d ÷ (total mass × tan φ). Compare it with the computed GM; if they disagree, your KG is wrong.

Worked Example: The 6 m By 2.4 m Scow

Hull 600 kg with its centre at 0.3 m; cargo 1,200 kg with its centre at 0.8 m. Total 1,800 kg.

  • Draft T = 1,800 ÷ (1,000 × 6.0 × 2.4) = 0.125 m. KB = 0.0625 m. BM = 2.4² ÷ (12 × 0.125) = 3.84 m.
  • KG = (600 × 0.3 + 1,200 × 0.8) ÷ 1,800 = 0.633 m. GM = 0.0625 + 3.84 − 0.633 = 3.27 m.
  • Freeboard 0.60 − 0.125 = 0.475 m; deck edge at arctan(0.475 ÷ 1.2) = 21.6°, where GZ = sin 21.6° × (3.27 + 0.5 × 3.84 × tan² 21.6°) = 0.368 × (3.27 + 0.30) = 1.31 m.

Cargo stacked high. Cargo centre at 1.6 m, well above the 0.60 m gunwale, a stack the carrier’s rule forbids and which is computed here only to price that rule: KG = (180 + 1,920) ÷ 1,800 = 1.167 m, so GM = 0.0625 + 3.84 − 1.167 = 2.74 m. The GM falls by 0.53 m (16%) and a wide scow is still stiff; stacking harms narrow boats far more.

Off-centre load. Hull 600 kg, a cart of 1,000 kg centred at height 1.0 m, and 500 kg of animals at 0.8 m, total 2,100 kg: T = 0.146 m, BM = 3.29 m, KG = 0.752 m, GM = 2.61 m. Moving the 500 kg by 1.0 m sideways gives tan φ = 500 × 1.0 ÷ (2,100 × 2.61) = 0.0912, φ = 5.2°, against a deck-edge angle of 20.7°.

Loose water. A free surface of l × b has second moment i = l × b³ ÷ 12. For a layer 5.0 m by 2.2 m, i = 5.0 × 2.2³ ÷ 12 = 4.44 m⁴. The loss of GM is 1,000 × i ÷ total mass = 1,000 × 4.44 ÷ 1,800 = 2.46 m. GM falls from 3.27 m to 0.80 m. The effect depends on the surface dimensions, not the depth of water; the weight of the water itself is ignored here. Dividing the width into three equal strips with fore-and-aft boards cuts i to one ninth (3² = 9), 0.49 m⁴, and the loss to 0.27 m, leaving GM at 3.0 m.

Inclining test. Move 200 kg by 1.5 m on the 1,800 kg scow. Expect tan φ = 200 × 1.5 ÷ (1,800 × 3.27) = 0.051, φ = 2.9°. A reading far above this means your GM is lower than computed.

Common Mistakes

  • Confusing stiff with safe — high GM with shallow freeboard has a short range.
  • Loose water on deck — the largest loss of GM in the example comes from it.
  • Weight shifted by animals — a moving load is the off-centre case continuously.
  • Estimating KG by eye — measure heights and weights.

Safety

The dangerous case feels stable until it does not

A loaded scow in a chop can feel rock-solid at small angles, while its range of stability depends on freeboard, and loose water or animals can strip the margin in one surge. Keep loose water out of the hull with divided compartments, keep animals tied centrally, and never exceed the load line. In cold water, stay with the hull; see Safety & Capsizing Recovery.

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