Comparing buoyancy design tradeoffs

A floating hull must displace its own weight of water, $m g = \rho g V_d$, so the displacement volume it needs is $V_d = m/\rho$, where $m$ is lightship plus cargo mass.

For a box-shaped hull $V_d = C_B L B T$, so capacity can be bought with length, beam or draft — but the draft is capped by channels and berths, and the beam by locks and harbour limits.

The two quantities in competition are buoyancy capacity, which grows with volume ($\sim L^3$ for geometrically similar hulls), and frictional resistance, which grows with wetted surface $S \sim L^2$.

Since useful transport work is $W = F d$ and propulsive power is $P = F v$, the energy spent per tonne-kilometre is $e = P/(m_c v) = R/m_c$, where $m_c$ is the cargo mass, and because $S/V \sim 1/L$ a geometrically similar (hence deeper-drafting) hull becomes more efficient as it grows: doubling every dimension roughly halves $e$.

Under a fixed draft this scaling breaks down: the bottom area $V_d/T$ is fixed, so $S$ is minimised by a compact, nearly square plan ($L \approx B$), and lengthening alone no longer reduces $e$.

The designer therefore trades capacity and fuel against stability and freeboard, structural weight, draft and lock limits, speed and seakeeping, measuring success by the overall efficiency $\eta = W_{\text{useful}}/E_{\text{input}}$.

A box-shaped pontoon has $L = 20$ m, $B = 6.0$ m and a maximum permitted draft $T = 1.5$ m (above that, water reaches the deck). The lightship mass is $m_0 = 30$ t and $\rho = 1000\ \text{kg/m}^3$. What is the largest cargo mass $m_c$ it can carry?

Two hulls are geometrically similar: every dimension of hull B is twice that of hull A. At the service speed the resistance is friction-dominated, $R = k S$, with $k = 30\ \text{N/m}^2$. Hull A has wetted surface $S_A = 1500\ \text{m}^2$ and carries cargo $m_A = 2000$ t, hull B carries $m_B = 16\,000$ t. Derive the specific transport energy $e = R/m_c$, compute $e_A$ and $e_B$ in kJ per tonne-kilometre and state which design is cheaper to operate.

A canal lock is 60 m long and 10 m wide, and the water depth limits the draft to 4.0 m. Each lock-sized box barge has lightship mass 500 t. A customer needs 8000 t of cargo moved.

(a) Find the cargo per barge and the number of barges required.

(b) Compare the fleet's wetted area per tonne of cargo with a hypothetical unrestricted hull of the same total displacement at $T = 4.0$ m (for which the compact plan $L = B$ minimises $S$).

(c) Give two reasons a designer might still prefer a fleet of small barges.

Numerical design study (spreadsheet simulation). Model the barge in a spreadsheet instead of loading a real boat.

1. Inputs: $\rho = 1000\ \text{kg/m}^3$, required displacement $V_d = 9200\ \text{m}^3$, draft limit $T = 4.0$ m.
2. For $L$ from 30 m to 200 m compute $B = V_d/(L T)$, then $S = L B + 2(L+B)T$ and the metric $S/V_d$ in $\text{m}^2$ per tonne.
3. Plot $S/V_d$ against $L$, find the minimum, and shade the region forbidden by a lock of $60 \times 10$ m.
4. Add a second curve: for geometrically similar scaling (every dimension multiplied by $k$) plot $S/V_d$ and check that it falls roughly like $1/L$.
5. Write three sentences: which hull you would choose and which constraint cost you the most.

Safety. Keep the work purely numerical. If you want a physical check, float a sealed plastic box in a basin of water; keep mains-powered equipment away from the water, never add ballast to a real boat, and never enter the water yourself.

Show that buoyancy capacity grows faster than wetted surface when a hull is scaled freely, and that a fixed draft changes which plan shape minimises wetted area per tonne — the core of the design trade-off.

Two-panel sketch. Left: side and plan views of two geometrically similar box hulls with dimensions $L$ and $2L$, labelled with $V_d$ and $S$ so that $V_d$ is seen to grow as $L^3$ and $S$ as $L^2$. Right: graph of $S/V_d$ (in $\text{m}^2$ per tonne) against length $L$: curve (a) for free (geometrically similar) scaling falls like $1/L$; curve (b) for the fixed draft $T = 4.0$ m has a minimum near $L = B$ and rises for longer hulls; a shaded vertical band marks lengths forbidden by a 60 m lock.

A bulk carrier has a waterplane area of about $A = 2000\ \text{m}^2$ and floats in seawater with $\rho = 1025\ \text{kg/m}^3$. The crew wants to load $m = 6000\ \text{t} = 6.0\times10^6\ \text{kg}$ of iron ore. Each extra tonne must push aside its own volume of water: $\Delta V = m/\rho \approx 5.9\times10^3\ \text{m}^3$, so the hull sinks $\Delta d = \Delta V/A \approx 2.9\ \text{m}$ deeper. The Plimsoll line allows only about 2.5 m of further immersion before the freeboard is too small for the winter North Atlantic, so roughly 875 t of ore stays on the quay.

The floating condition $mg = \rho g V$ fixes how much water a hull of a given mass must displace; dividing by the waterplane area $A$ turns that volume into the draught $d = m/(\rho A)$. Choosing between cargo mass and reserve buoyancy (freeboard height) is exactly this tradeoff.

A racing shell for eight rowers is $L \approx 17\ \text{m}$ long and $B \approx 0.6\ \text{m}$ wide, giving a waterplane area $A \approx 7\ \text{m}^2$. Boat $96\ \text{kg}$ plus eight rowers of $75\ \text{kg}$ each is $696\ \text{kg}$, so in fresh water it must displace $V = 0.70\ \text{m}^3$ and floats only $d = V/A \approx 10\ \text{cm}$ deep. A 4 m dinghy with two people, $270\ \text{kg}$ in total, displaces $V = 0.27\ \text{m}^3$, but its short wide hull has far more wetted surface, so at the same rowing power of about $2\ \text{kW}$ the shell travels at roughly $5\ \text{m/s}$ while the dinghy does under $2\ \text{m/s}$.

Both boats obey $mg = \rho g V$, so the displaced volume is fixed by mass alone; what the designer chooses is the waterplane area $A$ and the hull shape that produce it. The wetted surface, and therefore the drag force, follow from that choice, so the same buoyancy requirement can be met at very different costs in power.

A 100 t ferry must displace $V = 1.0\times10^5\ \text{kg} / 1025\ \text{kg/m}^3 \approx 98\ \text{m}^3$. As a monohull with waterplane area $120\ \text{m}^2$ it draws $d = 98/120 \approx 0.82\ \text{m}$, too deep for a channel dredged to 0.9 m with a safety margin. Split the same volume into two hulls with a total waterplane area of $150\ \text{m}^2$ and the draught falls to $d \approx 0.65\ \text{m}$, but the extra wetted surface raises total drag at $8\ \text{m/s}$ from about 22 kN to 26 kN; with a propulsive efficiency of $\eta = 0.55$ the shaft power rises from $Fv/\eta \approx 320\ \text{kW}$ to about $378\ \text{kW}$.

Buoyancy fixes only the displaced volume $V = m/\rho$; how that volume is spread over the waterplane area sets the draught and hence which harbours the ship can enter. The added wetted surface of the twin-hull layout is paid for in drag $F$, and through efficiency $\eta = P_{\text{useful}}/P_{\text{supplied}}$ in fuel per trip.