Wave theory is the whole foundation. Once the dispersion relation and the deep-water simplifications are secure, seakeeping and offshore loading follow with very little extra machinery.
Take a small-amplitude sinusoidal wave on water of constant depth. Everything about it — how fast it travels, how the water particles move, how the pressure varies with depth — follows from a single equation linking frequency to wavenumber.
The hyperbolic tangent is what makes waves interesting, because it behaves completely differently at its two extremes.
Between the two lies intermediate depth, where the full relation must be solved — usually by iterating k = (ω²/g)/tanh(kd) from the deep-water value, which converges in two or three passes.
Check d/λ before choosing a formula. Nearly every error in this topic is a deep-water shortcut applied where the wave is already feeling the bottom.
A wave carries energy — half kinetic, half potential — and the total per unit area of sea surface depends only on the height.
Note the square: doubling the wave height quadruples the energy. That single fact explains why a modest increase in sea state is so damaging, and why wave power devices are rated on H².
The energy does not travel at the speed of the individual crests. It travels at the group velocity:
Watch a group of waves moving across open water and you can see it: individual crests appear at the back of the group, travel forward through it, and vanish at the front. The crests move at c; the group, and the energy, at half that in deep water.
This is what governs how quickly a storm's energy reaches a distant coast, and it is why swell arrives sorted by period — the longest waves, travelling fastest, arrive first.
A real sea is not one wave. It is a superposition of components of many frequencies, directions and phases, and it is described statistically by an energy spectrum S(ω) — how much energy sits at each frequency.
The spectrum's moments carry the useful numbers:
Two standard spectra appear in questions. Pierson–Moskowitz describes a fully developed sea, where the wind has blown long enough over enough distance for the sea to stop growing. JONSWAP describes a fetch-limited sea and has a sharper, higher peak — it is the North Sea design spectrum, and its peak enhancement factor is what distinguishes it.
Individual wave heights within a sea state follow a Rayleigh distribution, which gives two results worth memorising:
For a three-hour storm, H_max works out at roughly 1.9 H_s — the largest wave is about twice the significant height. Design against H_s and you have designed against the average of the bigger third, not against what actually arrives.
Seakeeping treats the ship as a linear filter. Feed in waves of a given frequency and amplitude, and out comes motion at the same frequency, with an amplitude and a phase that depend on frequency. The amplitude ratio is the response amplitude operator.
That second line is the whole method: square the RAO, multiply by the wave spectrum, and integrate to get the statistics of the motion — in exactly the same way H_s came from the wave spectrum.
The RAO shapes differ by mode, and the differences are examinable:
But the frequency the ship actually feels is not the wave frequency. It depends on speed and heading:
In following seas the encounter frequency can fall to zero, at V = g/ω. The ship then sits on one part of the wave indefinitely — the condition associated with surf-riding, broaching and the loss of righting lever that comes with a crest amidships.
For a slender member — a jacket leg, a riser, a mooring chain — whose diameter is small compared with the wavelength, the wave force has two parts.
The drag term goes with velocity squared and is in phase with the velocity. The inertia term goes with acceleration and is therefore 90° out of phase with it — so the two peak at different moments in the wave cycle, and the maximum total force is not the sum of the two maxima.
Which term dominates is set by the Keulegan–Carpenter number:
The u|u| form, rather than u², is not a typographical curiosity: it keeps the drag force pointing in the direction the water is actually moving, which reverses twice per wave cycle.
Morison's equation fails when the member is not small compared with the wavelength — a large gravity-base structure or a ship-shaped hull diffracts the wave rather than sitting passively in it, and diffraction theory is needed instead. The usual dividing line is D/λ > 0.2.
One dispersion calculation done properly, one sea-state statistics problem, and the encounter frequency question that underlies most seakeeping reasoning.
A wave of period 9 s travels in water 40 m deep. Find the wavelength, the phase velocity and the group velocity. Compare with the deep-water values and comment.
T = 9 s, d = 40 m, g = 9.81 m/s² ω² = g·k·tanh(kd), k = 2π/λ
Find the wavelength, the phase velocity and the group velocity
Compare with the deep-water values and comment
Always test the water depth first.
Using the deep-water wavelength as the yardstick.
So the full dispersion relation is required.
Rearrange it into a form that can be iterated, starting from the deep-water value of k.
Three iterations are enough — it converges quickly.
| n | kₙ (rad/m) | kₙ·d | tanh(kₙd) | kₙ₊₁ (rad/m) |
|---|---|---|---|---|
| 0 | 0.04968 | 1.987 | 0.9630 | 0.05159 |
| 1 | 0.05159 | 2.063 | 0.9682 | 0.05131 |
| 2 | 0.05131 | 2.052 | 0.9676 | 0.05135 |
Wavelength and phase velocity follow.
Group velocity.
The speed at which the wave energy travels, which in general is less than the phase velocity.
Now compare with what the deep-water formulas would have given.
| Quantity | Deep-water value | Actual | Error |
|---|---|---|---|
| Wavelength λ (m) | 126.4 | 122.4 | 3.2 % |
| Celerity c (m/s) | 14.04 | 13.60 | 3.2 % |
| Group velocity c_g (m/s) | 7.02 | 7.72 | 10.0 % |
The lesson is in the last row.
Wavelength and celerity are only a few per cent out, so a candidate using the deep-water shortcut might not notice. Group velocity — and therefore energy flux, and therefore any wave power or shoaling calculation — is nearly 10 % wrong.
Answerλ = 122.4 m, c = 13.60 m/s, c_g = 7.72 m/s
The trap: reaching for λ = 1.56T² without checking d/λ. It is a deep-water result, and it stops being true the moment the wave begins to feel the bottom.
A wave spectrum has zeroth moment m₀ = 1.20 m² and second moment m₂ = 0.740 m²/s². Find the significant wave height and the mean zero-crossing period. In a three-hour storm at this sea state, how many waves would be expected to exceed 6 m, and what is the most probable largest wave?
m₀ = 1.20 m², m₂ = 0.740 m²/s² Storm duration 3 hours Rayleigh distribution: P(H > h) = exp(−2(h/H_s)²)
Find the significant wave height and the mean zero-crossing period
Significant wave height.
The factor of four comes from the Rayleigh distribution: H_s is defined as the mean of the highest one third of waves, and that works out to four times the RMS surface elevation.
Mean zero-crossing period.
From the ratio of the zeroth and second moments.
Number of waves in the storm.
Probability that any one wave exceeds 6 m.
Expected number in the storm.
| Quantity | Value |
|---|---|
| Significant wave height H_s | 4.38 m |
| Zero-crossing period T_z | 8.00 s |
| Waves in three hours N | 1350 |
| P(H > 6 m) | 0.0234 |
| Waves over 6 m | ≈ 32 |
Most probable maximum wave in N waves.
The standard extreme-value result for a Rayleigh distribution.
Note the ratio.
The largest wave in a three-hour storm is close to twice the significant wave height. That is the number to carry — it is why a design wave is not the same thing as a sea state, and why deck heights and green-water assessments are set from H_max rather than from H_s.
AnswerH_s = 4.38 m, T_z = 8.00 s, about 32 waves over 6 m, H_max ≈ 8.3 m
The trap: reading H_s as an amplitude. It is a height — crest to trough. Halving it somewhere in the working turns a survivable sea state into a benign one.
A ship steams at 18 knots in a sea of 10 s period. Find the encounter period in head seas and in following seas. Then find the speed at which, in following seas, the encounter frequency falls to zero, and say why that condition matters.
V = 18 kn, T = 10 s ω_e = ω − (ω²V/g)·cos μ μ = 180° head seas, μ = 0° following seas
Find the encounter period in head seas and in following seas
Wave frequency and ship speed in SI.
The speed term, which is the same magnitude in both cases.
Only its sign changes with heading.
Head seas: μ = 180°, so cos μ = −1 and the term adds.
The ship meets waves more often than a stationary observer would.
Following seas: μ = 0°, cos μ = +1 and the term subtracts.
The ship runs with the waves and meets them far less often.
Now the zero-encounter condition.
It occurs when the ship's speed component matches the wave celerity — the ship is riding along with a wave, neither overtaking it nor being overtaken.
Why it matters.
Near zero encounter frequency the ship sits on the same part of the wave for a long time — typically on the forward face of a following sea, with the bow buried and the stern lifted. Directional stability degrades, the ship can slew broadside (broaching), and stability is at its worst because a wave crest amidships reduces the waterplane at both ends and therefore reduces GZ.
This is also why roll is worst in beam or quartering seas rather than head seas.
At μ = 90° the speed term vanishes entirely, ω_e = ω, and a long-period swell can sit right on the roll natural frequency for hours.
AnswerHead seas T_e = 6.28 s; following seas T_e = 24.6 s; zero encounter at 30.4 knots
The trap: the sign of the cosine. Getting it backwards makes head seas look gentle and following seas look violent — the exact opposite of the truth, and it turns a seakeeping answer inside out.
ω² = g·k·tanh(kd)The dispersion relation; check d/λ before simplifyingDeep: λ = 1.56T², c = 1.56TValid for d > λ/2Shallow: c = √(gd)Valid for d < λ/20; no dispersionc_g = ½c[1 + 2kd/sinh 2kd]c/2 deep, c shallow — energy travels at c_gE = ⅛ρgH²Energy per unit surface area; goes as H²H_s = 4√m₀ T_z = 2π√(m₀/m₂)From the spectral momentsP(H > h) = exp(−2(h/H_s)²)Rayleigh distribution of wave heightsH_max ≈ H_s√(ln N/2)≈ 1.9 H_s in a three-hour stormω_e = ω − (ω²V/g)cos μ180° head, 90° beam, 0° followingF = ½ρC_d D u|u| + ρC_m(πD²/4)u̇Morison; KC decides which term dominates