One hypothesis holds the topic together: split resistance into a part that scales with Reynolds number and a part that scales with Froude number, test the model at equal Froude, and calculate the rest.
Towing a hull through water costs energy in two distinct ways, and they behave completely differently.
Friction. Water sticks to the hull. A boundary layer forms, thin at the bow and thickening aft, and dragging it along costs a tangential force over the whole wetted surface. This depends on how viscous the water is relative to the flow — that is, on Reynolds number.
Wave making. The hull pushes water aside and it does not close smoothly behind; a wave system is left astern, carrying energy away permanently. Whether the bow and stern wave systems reinforce or cancel each other depends on the ratio of ship speed to wave speed — that is, on Froude number.
Add eddy-making behind blunt terminations, appendage drag from rudder and bilge keels, and air resistance on the above-water form, and you have the full account. But the two that matter, and the two the whole method is built around, are friction and wave making.
Two different physical mechanisms, obeying two different scaling laws. Everything difficult about this topic comes from trying to model both at once.
Suppose you want to know a 200 m ship's resistance. You build a 5 m model. To make the model's flow physically similar to the ship's, you would need to match both scaling numbers at once. You cannot.
Froude's answer, and it is still the basis of every towing tank in the world, was to divide resistance in two:
Then run the model at the Froude speed, and assume that the residuary coefficient is the same for model and ship at the same Froude number. Friction is not measured at all — it is calculated for each body separately, at its own Reynolds number, from an agreed formula.
That formula is the ITTC-1957 model–ship correlation line:
It is called a correlation line rather than a friction line for a reason: it is not a pure statement of flat-plate friction, it is the curve that makes the whole extrapolation procedure come out right. Use it as prescribed and it works; interrogate it as physics and it will disappoint you.
The procedure, in the order you will use it in an examination:
Froude's split is slightly crude in one respect. A three-dimensional hull has more viscous resistance than a flat plate of the same area and length, because the flow accelerates around the curvature and the pressure distribution leaves a viscous pressure drag behind. Calling all of that difference “residuary” lumps a viscous effect in with the wave-making one — and the two scale differently.
The 1978 ITTC method introduces a form factor (1+k) to account for it:
k is found experimentally by running the model very slowly, where wave making is negligible and any excess over flat-plate friction must be form effect. Typical values run from about 0.10 for a fine hull to 0.30 for a full one.
In an examination, use whichever method the question sets up. If a form factor is given, use the 1978 form; if it is not mentioned, the two-component Froude method is intended.
The bow generates a wave system and so does the stern. Depending on speed, their crests either coincide near the stern — reinforcing, and costing extra energy — or fall out of step and partly cancel.
Plot residuary resistance against Froude number and the result is not a smooth rise but a series of humps and hollows. The important ones sit near Fn ≈ 0.30 and, far larger, near Fn ≈ 0.50.
The last hump is why displacement hulls have a practical speed ceiling. At Fn ≈ 0.4 the ship is running with a wavelength close to its own length — bow crest at the bow, trough amidships, crest at the stern — effectively climbing its own bow wave. Pushing past it costs power out of proportion to the speed gained.
This also explains a design rule of thumb: for a given speed, a longer ship has a lower Froude number and sits further from the hump, which is one reason fast ships are long and fine rather than short and full.
The extrapolation so far describes a perfectly smooth, bare hull in still air. Three additions bring it back to reality, and all three are applied to the ship side only — never to the model.
Fouling is a separate and larger matter. A season's growth can raise resistance by 20 % or more, which is why hull cleaning appears in every fuel-saving discussion and why performance monitoring trends resistance rather than measuring it once.
Once RT is known, effective power is simply the rate of doing work against it:
Everything between PE and what the engine must deliver belongs to the next chapter.
The first is the standard extrapolation, done in full. The second shows why the method has to work that way. The third is the quick comparison an examiner uses to test whether you know the limits of a shortcut.
A ship 150 m long with a wetted surface of 3600 m² is to run at 18 knots. A 6 m model is towed in fresh water and its total resistance at the corresponding speed is 38.5 N. Find the effective power of the ship. Take ν = 1.14×10⁻⁶ m²/s for the model and 1.19×10⁻⁶ m²/s for the ship, ρ_model = 1000 and ρ_ship = 1025 kg/m³, and a correlation allowance C_A = 0.0004.
L_s = 150 m, S_s = 3600 m², V_s = 18 kn L_m = 6 m, R_T(model) = 38.5 N λ = 150/6 = 25
Find the effective power of the ship
Convert the ship speed to SI, then find the model speed.
The model is towed at the same Froude number, which means speed scales with the square root of the length ratio.
Scale the wetted surface.
Areas go as λ², not λ — this is where a third of candidates lose the question.
Total resistance coefficient of the model.
From the measured force.
Frictional coefficient of the model.
At the model's own Reynolds number. Model and ship do not share a Reynolds number — only the Froude number is matched — so each gets its own C_F.
The residuary coefficient is what is left over.
This is the number that transfers to the ship, unchanged, because residuary resistance obeys Froude's law of comparison.
Now build the ship's coefficient.
| Coefficient | Model | Ship |
|---|---|---|
| Reynolds number R_n | 9.747 × 10⁶ | 1.167 × 10⁹ |
| C_F (ITTC 1957) | 0.003013 | 0.001502 |
| C_R (transfers) | 0.000885 | 0.000885 |
| C_A (allowance) | — | 0.000400 |
| C_T | 0.003898 | 0.002787 |
Same C_R, but the ship's own frictional coefficient at its own — far higher — Reynolds number, plus the correlation allowance for roughness and scale effect.
Convert back to a force, then to power.
Use sea water density for the ship.
AnswerP_E ≈ 4080 kW (about 4.1 MW at the towrope)
The trap: using one C_F for both. The model runs at Rn ≈ 10⁷ and the ship at Rn ≈ 10⁹ — two orders of magnitude apart, and their frictional coefficients differ by a factor of two. Froude's whole method exists because you cannot match both scaling laws at once.
A 200 m ship is to be tested using a 5 m model. Show what speed the model must run at to represent 20 knots, and show what speed would be needed to match Reynolds number instead. Comment on the result.
L_s = 200 m, L_m = 5 m ⇒ λ = 40 V_s = 20 kn = 10.29 m/s Same water for both (ν equal)
Show what speed the model must run at to represent 20 knots, and show what speed would be needed to match Reynolds number instead
Comment on the result
Froude scaling first.
Equal Froude number means V/√(gL) is the same for both, and g is the same, so V scales as √L.
Now Reynolds scaling.
Equal Reynolds number means V·L/ν is the same, and with the same water that means V·L is the same — so a shorter model needs a faster speed, by the full length ratio.
Compare.
To satisfy Froude the model must be towed at 1.63 m/s; to satisfy Reynolds it would have to be towed at over 400 m/s, which is beyond the speed of sound in air and physically absurd in a towing tank.
This is exactly why resistance is split.
The part that depends on Froude number — wave making — is measured, because Froude scaling is achievable. The part that depends on Reynolds number — friction — is calculated from a correlation line at each body's own Reynolds number, because it cannot be measured at the right scale.
AnswerV_m = 1.63 m/s by Froude; Reynolds would demand 411.6 m/s — hence the split
The trap: thinking the split into frictional and residuary resistance is an approximation someone chose for convenience. It is forced on us: no tank in the world can satisfy both scaling laws in the same run.
A ship of 14 000 t displacement requires 6000 kW at 16 knots. Estimate the power required at 18 knots at the same displacement, and say when the estimate can be trusted.
Δ = 14 000 t, P₁ = 6000 kW at V₁ = 16 kn V₂ = 18 kn, same Δ
Estimate the power required at 18 knots at the same displacement, and say when the estimate can be trusted
The Admiralty coefficient bundles displacement.
Speed and power into a figure that stays roughly constant for the same ship over a modest speed range.
With Δ unchanged.
The Δ^(2/3) term cancels between the two conditions, leaving power proportional to the cube of speed.
Substitute.
Sanity check the assumption.
A 12.5 % increase in speed costs a 42 % increase in power — steep, and that is the cube law doing its work.
Now the caveat, which is the part worth marks.
The cube law assumes the resistance coefficient is unchanged between the two speeds. That holds while both speeds sit on a smooth part of the resistance curve. If the higher speed crosses a wave-making hump — around F_n ≈ 0.30 and again near 0.50 — residuary resistance rises faster than the cube law allows and the estimate is optimistic.
AnswerP₂ ≈ 8540 kW
The trap: using the Admiralty coefficient across a large speed change or between different ships. It is an interpolation tool for one hull over a narrow range, not a law of nature.
R_T = R_F + R_RFroude's hypothesis — the basis of every tank testC = R/(½ρSV²)The same non-dimensional form for every componentC_F = 0.075/(log₁₀R_n − 2)²ITTC 1957 correlation lineR_n = VL/ν F_n = V/√(gL)ν ≈ 1.19×10⁻⁶ m²/s, sea water at 15 °CV_s = V_m·√λλ = L_s/L_m; equal Froude numberS_s = S_m·λ² ∇_s = ∇_m·λ³Areas as λ², volumes as λ³C_T = (1+k)C_F + C_W + C_A1978 form-factor methodS ≈ 1.7·L·T + ∇/TDenny–Mumford wetted surface estimateP_E = R_T·VEffective (towrope) powerA_C = Δ^(2/3)·V³/PAdmiralty coefficient — one hull, narrow speed range