Chapter 03 of 13 · PE NA&ME

Ocean Engineering

A handful of standard results carry most of the marks here: get the depth regime right, keep phase and group velocity straight, and you can turn a spectrum into a design wave and a design wave into a Morison load without missing a step.

Worked examples3, fully stepped
Read time≈ 15 min
PrerequisiteNone

1. Water depth and the dispersion relation

Every wave problem on this paper starts with the same question: how does the seabed feel to this particular wave? A 12 second swell in 200 m of water behaves nothing like the same-period wave running into a 15 m approach channel, and the dispersion relation is what connects wave period, wavelength and depth so you can tell the two situations apart before reaching for a formula.

ω² = g·k·tanh(kd) links frequency ω, wavenumber k = 2π/λ, and depth d

Physically, tanh(kd) is a bottom-feeling term: it climbs toward 1 once kd is large (deep water — the bed is too far away to matter) and collapses toward kd itself once kd is small (shallow water — the whole water column moves together). Between those limits, roughly 0.05 < d/λ < 0.5, neither shortcut is valid and the full transcendental relation has to be solved, normally by iterating on k with a calculator or a short spreadsheet loop.

Deep (d/λ > 0.5): λ = 1.56T², c = 1.56T Shallow (d/λ < 0.05): c = √(gd), wave speed set only by depth

The deep-water shortcut is the one that gets misapplied most often, because it is the formula everyone memorises first. It assumes the orbital motion has died away completely before it reaches the seabed, which only holds once the wave "feels" more than about half its own wavelength of water beneath it. Applying λ = 1.56T² to a wave that is actually in intermediate depth doesn't just introduce a small arithmetic error — it quietly shifts every downstream number: wavenumber, phase speed, and especially group velocity.

The key idea

Work out d/λ (using the deep-water estimate as a first guess if you have to) before choosing a formula, not after. The regime decision is worth more marks than the arithmetic that follows it.

2. Phase speed, group velocity, and where the energy actually goes

A single sinusoidal wave train is a convenient fiction — real energy travels in groups, and the envelope of a wave group moves at the group velocity c_g, not at the phase speed c of the individual crests inside it. Watch a group of waves arrive at a breakwater and you'll see crests appear at the back of the group, run forward through it, and vanish at the front: the crests move faster than the parcel of energy carrying them.

c_g = ½c[1 + 2kd/sinh(2kd)] general form, all depths

In deep water the bracket collapses to ½, so c_g = c/2 exactly — energy takes twice as long to cross a stretch of open ocean as the individual wave crests do. In shallow water the bracket climbs to 1 and c_g → c: once the whole water column moves together, there is no longer a distinction between how fast the shape moves and how fast the energy moves. This matters directly for wave power, because the energy flux past a line is proportional to c_g, not c.

E = ⅛ρgH² energy per unit sea-surface area P = E·c_g power per unit crest length

Taking c_g as c/2 regardless of depth is the classic slip: in intermediate or shoaling water the bracket term climbs well above ½, so the deep-water shortcut understates both group velocity and wave power — sometimes by a large margin, as Example 1 shows. Group velocity is also what governs shoaling: as a wave group moves into shallower water c_g falls, and because the energy flux has to be conserved, the wave height rises to compensate.

3. Describing an irregular sea: spectra and significant wave height

A real sea surface is not one wave but a superposition of many, and the honest way to describe it is statistically, through a wave spectrum S(ω) that shows how much energy sits at each frequency. The spectral moments compress that spectrum into single numbers a designer can actually use.

m_n = ∫ ωⁿ S(ω) dω the n-th spectral moment H_s = 4√m₀, T_z = 2π√(m₀/m₂) significant height and zero-crossing period

H_s is not the height of any one particular wave — it is defined as the mean of the highest third of all waves in the record, and it turns out to work out to almost exactly four times the square root of the area under the spectrum. For a narrow-banded sea, individual wave heights follow a Rayleigh distribution, which is what lets you go one step further than H_s and estimate the largest wave you're actually likely to meet in a given storm duration, rather than quoting H_s itself as though it were a design maximum.

The key idea

H_s describes the sea state; it does not describe the worst wave in it. The number of waves in the exposure period — not H_s alone — sets how far above H_s the design wave needs to sit.

The zero-crossing period T_z, read straight off a wave trace as the mean time between successive upward crossings of the mean water level, is what lets you convert a storm duration into a wave count, and from there into an extreme-value estimate using the Rayleigh distribution — worked through in full in Example 2.

4. Ship motions, RAOs, and encounter frequency

A ship or floating structure has six rigid-body degrees of freedom — surge, sway, heave, roll, pitch and yaw — and in a seaway each one responds to the waves passing beneath it. The response amplitude operator, or RAO, is simply the ratio of the motion amplitude to the wave amplitude that caused it, plotted against frequency.

RAO(ω_e) = motion amplitude ÷ wave amplitude a function of encounter frequency, not wave frequency

The frequency that matters to the vessel is not the wave frequency ω an observer on the shore would measure, but the encounter frequency ω_e — the rate at which the moving vessel actually meets successive crests, which depends on heading as well as speed.

ω_e = ω − (ω²V/g)cos μ μ = 0° following seas, μ = 180° head seas

In head seas the vessel meets crests faster than a stationary observer would, so ω_e is raised above ω; running with the sea in following conditions lowers it, and at the right speed and heading ω_e can pass through zero — the vessel then sits almost stationary relative to the wave pattern, which is exactly the condition to watch for near a roll or pitch resonance. Every RAO curve peaks close to the structure's own natural frequency in that mode, and the practical design task is keeping that peak away from the encounter frequencies the vessel will actually see in service, particularly for roll, where damping is naturally low and resonant amplification can be severe.

5. Mooring and station keeping

A moored structure is, in effect, a mass-spring system with the mooring lines supplying most of the spring. In a catenary mooring, the restoring force comes from lifting line weight clear of the seabed as the structure is pushed off station — soft at small offsets and stiffening as more chain lifts. A taut mooring instead relies on the elastic stretch of the line itself, giving a stiffer, more linear response and a smaller seabed footprint, at the cost of higher peak line tension.

Either way, the mooring system has its own natural period in surge, sway and yaw, set by the platform's mass (plus added mass) and the system stiffness. The design aim is the same one that runs through this whole topic: keep that natural period well clear of the energetic band of the sea states the structure will experience, typically by pushing it out beyond the wave-period range entirely, into the tens-of-seconds region where wave-frequency energy is negligible.

Slow-drift forces — second-order, arising from the wave groups rather than the individual waves — are what excite that long-period resonance in practice, even though the first-order wave loads themselves are too fast to do it directly. A mooring check that only compares first-order loads against a stiff spring model, and never asks where the system's own natural period sits relative to the group (envelope) period of the sea state, misses the mechanism that most often governs mooring-line fatigue and peak tension.

6. Wave loading on slender members: Morison's equation

A slender vertical member in waves — a pile, a jacket leg, a riser — sees a horizontal force built from two physically distinct mechanisms, and Morison's equation keeps them as two separate terms rather than folding them into one.

F = ½ρC_dDu|u| + ρC_m(πD²/4)u̇ drag term + inertia term, per unit length

The drag term is the familiar velocity-squared resistance force, in phase with the water particle velocity u; the inertia term comes from the fluid having to be accelerated out of the way of the member, and is in phase with the particle acceleration u̇ instead. Because velocity and acceleration in a wave are 90° apart in phase, so are the two force terms — they do not peak at the same instant, and simply adding their individual maxima overstates the true peak load, sometimes substantially.

Which term matters is governed by the Keulegan–Carpenter number, KC = u_max T/D, which compares how far a fluid particle travels in one wave period against the member's diameter. At low KC the particle barely moves relative to the member and the flow never really separates, so inertia dominates almost completely; at high KC the flow separates and sheds vortices every half-cycle, and drag takes over. Most real design cases fall somewhere between the two limits, which is exactly where both terms have to be carried through the cycle together, as in Example 3.

The key idea

Morison's equation itself only applies to members that are slender compared with the wavelength. Once the diameter approaches roughly a fifth of the wavelength, the member starts to scatter the wave rather than simply being pushed by it, and diffraction theory — not Morison — is what's needed.

7. Worked examples

Three worked problems below cover the three places this topic most often costs marks: picking the right depth regime and following its consequences through to group velocity, turning a sea state's statistics into a design wave, and combining drag and inertia correctly on a loaded member.

Worked example 1

Wavelength given, wrong-regime shortcut checked against the full dispersion relation

A wave rider buoy in a proposed anchorage records a wavelength of 40 m in a water depth of 10 m. Find the phase speed, period and group velocity, and check how far off a deep-water shortcut would have put the group velocity.

Given

Wavelength (measured) λ = 40 m Water depth d = 10 m g = 9.81 m/s²

Required

Find the phase speed, period and group velocity, and check how far off a deep-water shortcut would have put the group velocity

  1. First place the wave between the deep- and shallow-water limits.

    d / λ=10 / 40 =0.25 0.05<d/λ < 0.5 → intermediate depth: the full dispersion relation is required.

    That decision is what fixes which formula is safe to use.

  2. Convert wavelength to wavenumber and evaluate kd.

    k=2π / λ = 2π / 40 =0.1571 rad/m kd=0.1571 × 10 =1.571 rad tanh(kd)=tanh(1.571) = 0.9172

    Because d/λ sits in the intermediate band, tanh(kd) has to be carried through rather than approximated.

  3. Substitute into the dispersion relation to get phase speed.

    c=√[(g/k)·tanh(kd)] =√[(9.81 / 0.1571) × 0.9172] =√(62.45 × 0.9172) =√57.28 =7.568 m/s ω=ck =7.568 × 0.1571 =1.189 rad/s T=2π / ω = 5.285 s

    Then the period.

  4. Check what a deep-water shortcut alone would have given for the phase speed.

    c₀=√(g/k) =√62.45 =7.903 m/s error=(7.903 − 7.568) / 7.568 = +4.4%

    Since d/λ = 0.25 is exactly the range where it gets used out of habit.

  5. Group velocity needs the full bracket term too.

    c_g=(c/2)[1 + 2kd / sinh(2kd)] sinh(2kd)=sinh(π) = 11.55 2kd / sinh(2kd)=3.142 / 11.55 = 0.2720 c_g=(7.568/2) × 1.2720 =3.784 × 1.2720 =4.814 m/s

    And this is where the deep-water shortcut fails hardest, because the bracket is furthest from ½ at this depth.

  6. Compare with the deep-water shortcut c_g = c₀/2.

    c_g,deep=c₀ / 2 =7.903 / 2 =3.951 m/s error=(3.951 − 4.814) / 4.814 = −17.9%

Answerc = 7.57 m/s, T = 5.29 s, c_g = 4.81 m/s — the deep-water shortcut is 4% high on phase speed and 18% low on group velocity at this depth.

The trap: d/λ = 0.25 looks close enough to "deep" to a quick eye, but it sits well short of the d/λ > 0.5 threshold, and the deep-water formulas distort group velocity — and therefore wave power — far more than they distort phase speed.

Worked example 2

From significant wave height to wave power and a design extreme wave height

A structure must be designed for a sea state with H_s = 3 m and zero-crossing period T_z = 8 s, lasting 3 hours, in deep water. Find the wave power per metre of crest and the largest wave height reasonably expected during the storm.

Given

Significant wave height H_s = 3 m Zero-crossing period T_z = 8 s Storm duration = 3 hours Deep water; seawater density ρ = 1025 kg/m³

Required

Find the wave power per metre of crest and the largest wave height reasonably expected during the storm

  1. Work back from H_s to the zeroth spectral moment.

    m₀=(H_s / 4)² =(3 / 4)² =0.5625 m²

    The link between the single design number and the spectrum underneath it.

  2. Deep-water wave power per metre of crest follows from the energy flux.

    P=ρg²H_s²T_e / (64π) =1025 × 9.81² × 3² × 8 / (64π) =1025 × 96.24 × 9 × 8 / 201.1 =7,102,200 / 201.1 =35,320 W/m ≈ 35.3 kW/m

    Take the energy period T_e ≈ T_z as a reasonable working value for a wind sea.

  3. To size for the largest wave the structure will actually meet.

    N=duration / T_z =(3 × 3600) / 8 =10,800 / 8 =1350 waves

    First find how many individual waves pass during the 3-hour storm.

  4. Wave heights in a narrow-band sea state follow a Rayleigh distribution.

    H_max / H_s=√(0.5 ln N) =√(0.5 × ln 1350) =√(0.5 × 7.208) =√3.604 =1.898

    And for a sample of N waves the expected largest value scales as √(0.5 ln N) times H_s — which is where the familiar "about 1.9 H_s" figure actually comes from, rather than being a fixed constant.

  5. Apply that ratio to this sea state's H_s to get the design extreme wave height.

    H_max=1.898 × 3 =5.70 m

AnswerP ≈ 35.3 kW/m of crest; about 1350 waves in the storm, with a largest expected wave of ≈5.70 m (H_max/H_s ≈ 1.90) — the standard rule of thumb, derived rather than assumed.

The trap: treating H_s as though it were the height of the worst wave you'll meet, rather than the mean of the highest third — a structure sized only to H_s carries no margin for the wave that actually loads it.

Worked example 3

Combining Morison drag and inertia that peak 90° apart in phase

A jacket leg of diameter D = 1.0 m sits in waves with maximum horizontal particle velocity u_max = 2.0 m/s and maximum horizontal acceleration a_max = 1.0 m/s², wave period T = 8 s, with C_d = 1.0 and C_m = 2.0. Find the true peak in-line force per unit length.

Given

Diameter D = 1.0 m Max particle velocity u_max = 2.0 m/s Max particle acceleration a_max = 1.0 m/s² Wave period T = 8 s C_d = 1.0, C_m = 2.0, ρ = 1025 kg/m³

Required

Find the true peak in-line force per unit length

  1. Before combining anything.

    KC=u_max T / D =2.0 × 8 / 1.0 =16

    Check which term should be expected to dominate.

  2. KC = 16 is in the range where drag is no longer small next to inertia.

    Both terms have to be carried through the cycle together, not just the larger one.

  3. Find the peak drag force and the peak inertia force separately.

    F_D,max=½ρC_dDu_max² =0.5 × 1025 × 1.0 × 1.0 × 2.0² =2050 N/m F_I,max=ρC_m(πD²/4)a_max =1025 × 2.0 × (π × 1.0² / 4) × 1.0 =1025 × 2.0 × 0.7854 × 1.0 =1610 N/m

    As if each acted alone.

  4. Drag varies as sinθ|sinθ| and peaks at θ = 90°.

    F(θ)=F_D,max sin²θ + F_I,max cosθ (0 ≤ θ ≤ π) dF/dθ=sinθ(2F_D,max cosθ − F_I,max) = 0 ⇒cosθ* = F_I,max / (2F_D,max) =1610 / (2 × 2050) =0.3927 ⇒θ* = 66.9°

    Inertia varies as cosθ and peaks at θ = 0°. The two maxima never occur at the same instant, so the true peak of the combined force is found by maximising F(θ), not by adding F_D,max and F_I,max.

  5. Substitute θ* back in to get the true peak combined force.

    sin²θ*=1 − cos²θ* = 1 − 0.3927² =0.8458 F_max=F_D,max sin²θ* + F_I,max cosθ* =2050 × 0.8458 + 1610 × 0.3927 =1734 + 632 =2366 N/m ≈ 2.37 kN/m
  6. Compare with the naive "just add the two maxima" figure.

    F_D,max + F_I,max=2050 + 1610 = 3660 N/m overstatement=(3660 − 2366) / 2366 = +55%

    To see the size of the error it would introduce.

AnswerPeak in-line force ≈ 2.37 kN/m, occurring at θ* ≈ 67° into the cycle — about 55% below the naive sum of the two individual peaks.

The trap: adding F_D,max and F_I,max together as if they happen at the same instant — they are 90° apart in phase, so that sum overstates the real peak load by roughly half in a case like this one.

Reference sheet
60-second recall
  1. d/λ decides the formula, not the water's appearance — work it out first, every time.
  2. The deep-water shortcut errs low on group velocity (and wave power) well before it errs on phase speed — check c_g separately near d/λ ≈ 0.2–0.3.
  3. H_max/H_s ≈ √(0.5 ln N): the "1.9 H_s" rule of thumb comes from Rayleigh statistics over the wave count in the exposure duration, not from a fixed multiplier.
  4. Drag and inertia peak 90° apart in phase — never add F_D,max and F_I,max directly; find θ* from cosθ* = F_I,max/(2F_D,max) instead.
  5. Diffraction, not Morison, governs once D/λ climbs to about 0.2 — check the member's size against the wavelength before trusting a KC-based force calculation.