Offshore heavy-lift construction vessel at blue hour, lowering a subsea structure on a single taut crane wire through the sea surface

KNOWLEDGE HUB / SUBSEA LIFTING

Subsea Lifts

Q: Why is a subsea lift harder than a lift in air?

Because the water fights back four ways: buoyancy changes the effective weight and varies rapidly at the surface, slamming hits at water entry, drag resists motion with the square of velocity, and added mass makes the payload effectively heavier whenever it accelerates. Each phase of the lift has a different dominant force.

Q: How large can drag forces get on a subsea lift?

Significant: in the worked example on this page, a 15 × 10 m plate moving at a peak heave velocity of 1.57 m/s sees roughly 216 kN ≈ 22 t of drag. Because drag grows with velocity squared, halving the relative velocity cuts the drag force to a quarter; how much of that a compensator delivers on a given lift comes out of the coupled model.

Q: What is added mass and how big is it?

The surrounding water a payload must accelerate along with itself. In the page's example, a 10 × 10 × 20 m square prism in 8-second waves picks up an inertial force of about 93 t (908 kN) — on top of the static weight. Near the seabed it grows further due to confinement.

Q: Which force dominates the splash zone?

Slamming and rapidly varying buoyancy, on top of wave-particle velocity. The splash zone combines the worst of every regime in a few metres of travel, which is why it sizes the compensator more often than any other phase.

Q: How does heave compensation protect a subsea lift?

It cuts the crane-motion share of the relative velocity and acceleration between hook and payload — shrinking the velocity-squared drag and the acceleration-driven added-mass force — and holds wire tension so slack-snap has no opening; minimum tension and stroke use are verified in the coupled model. For heavy and deepwater lifts the passive compensator sized for this duty is CYGNUS; RIGEL and ANTARES cover the standard and adaptive cases.

Understand the water forces from deck to seabed.

By Peter Wang, COO · December 2025 · Reviewed August 2026 · 15 min read

START WITH THE LOAD DRIVERS

Four forces to keep separate.

Buoyancy, slamming, drag and added mass affect the lift in different ways. More than one can act during a phase; their combined response needs the lift model.

Buoyancy

Scales with: Displaced volume (ρ_wVg)
Where to consider it: From water entry; varies fast while crossing the surface
Effect / published example: Effective weight drops and fluctuates — slack-wire risk

Slamming

Scales with: Impact velocity at water entry
Where to consider it: Splash zone
Effect / published example: Short, sharp load spikes on payload and rigging

Drag

Scales with: Velocity squared (½ρ_wC_DA_⊥v²)
Where to consider it: Submerged descent through waves and current
Effect / published example: ≈ 22 t on the example plate at the prescribed relative velocity

Added mass

Scales with: Acceleration (ρ_wC_AV_R × a)
Where to consider it: Any oscillation; grows near the seabed (confinement)
Effect / published example: ≈ 93 t (908 kN) of extra inertial force in the separate prism example below

The splash zone stacks slam, fast-changing buoyancy and wave-particle velocity in a few metres of travel — see splash-zone crossing and the method reference DNV-RP-N103.

THE PUBLISHED WORKED EXAMPLES

Two bodies. Two separate calculations.

The examples give the forces a sense of scale. They do not describe two forces on the same payload.

A / DRAG

15 × 10 m rectangular plate

Prescribed relative velocity normal to the plate

Velocity: 1.57 m/s
Projected area: 150 m²
Drag coefficient: 1.14

≈216 kN≈22 tonnes-force

Read the drag calculation →

B / ADDED-MASS FORCE

10 × 10 × 20 m square prism

Prescribed acceleration along the prism’s long axis

Acceleration: 1.23 m/s²
Reference volume: 2,000 m³
Added-mass coefficient: 0.36

≈908 kN≈93 tonnes-force

Read the added-mass calculation →

Do not add these two results.

They use different bodies. For sinusoidal motion, velocity and acceleration peaks are also a quarter-period apart. The combined peak for an actual lift comes from the coupled time-domain response.

What these screening numbers assume

Both worked examples on this page are screening calculations. They are here to give the forces a size, not to produce a design load. Specifically:

The relative velocity is prescribed, not derived. Both examples take the motion straight from a sinusoid of the stated height and period (ζω for velocity, ζω² for acceleration). A real lift gets the payload-to-water relative motion from the coupled model — vessel RAOs, crane-tip response, winch command, wire dynamics, compensator state and the water’s own particle kinematics.
Drag and added mass do not peak at the same instant. On a sinusoid the velocity peak and the acceleration peak are a quarter period apart, so the ≈22 t drag and the ≈93 t added-mass force may not simply be added to get a peak load. The combined load is a time-domain result.
Two different bodies. The drag example is a 15 × 10 m rectangular plate; the added-mass example is a 10 × 10 × 20 m square prism. Each was chosen to match a row of the coefficient table it reads from, so the two force figures are not two forces on one payload.
Coefficients are infinite-fluid values. DNV-RP-N103 table values carry no correction for the free surface or for seabed proximity. Near the seabed, added mass rises through confinement — which is exactly where landing happens.
Buoyancy and slam are excluded from both. They are covered in the four-force overview above; in the splash zone they usually govern.

For a lift you have to plan, these are the inputs a CONSTELLATION screen replaces with case-specific values — see the worked modelled case for what a fully specified run states about itself.

The full technical guide follows, with the original coefficient tables, equations, operating sequence and supporting references.

Subsea lifts are among the most complex offshore operations, requiring careful planning to account for hydrodynamic forces that act on the payload as it moves through water. The two most significant effects are drag and added mass, both of which increase the dynamic loads on the crane and lifting equipment: drag grows with velocity, added mass with acceleration.

Understanding these forces is critical for selecting the right subsea heave compensator and ensuring the crane has sufficient capacity for the operation.

How does drag affect a subsea lift?

Drag in a subsea lift is similar to drag on a car due to air resistance, and varies with the square of the speed — the speed of the payload through the water around it, not its speed over the seabed. How big the drag force will be depends on the area perpendicular to the motion as well as the drag coefficient (which we again know from cars competing to always have lower drag coefficients to increase range). The main difference from a car however is that the fluid is water rather than air, and water is about 1,000 times denser, which is another factor in the drag force.

$F_D = \tfrac{1}{2}\rho_w C_D A_\perp \dot z |\dot z|$

Where $C_D$ is the drag coefficient (more info can be found in DNV RP-N103, some examples shown below) and $\dot z$ is the vertical velocity of the payload relative to the water around it. Crane and winch motion and the water’s own particle velocity all feed into it; the worked example below prescribes that relative velocity as a screening input rather than deriving it from a vessel response model.

Drag coefficients by shape — DNV-RP-N103 reference table

Drag coefficients and projected areas by shape, for subsea lift drag calculations.
Shape	$C_D$	$A_{\perp}$	Notes
Sphere	$C_D = 0.5$	$A_{\perp} = \pi r^2$
Horizontal Cylinder	$C_D = 1.2$	$A_{\perp} = 2 r L$
Vertical Cylinder	$\frac{L}{2r}=0.5 \Rightarrow C_D=1.1$ $\frac{L}{2r}=1 \Rightarrow C_D=0.9$ $\frac{L}{2r}=2 \Rightarrow C_D=0.9$ $\frac{L}{2r}=4 \Rightarrow C_D=0.9$ $\frac{L}{2r}=8 \Rightarrow C_D=1.0$	$A_{\perp} = \pi r^2$
Cube	$C_D = 1.05$	$A_{\perp} = a^2$	Face normal to flow.
Cone	$C_D = 0.50$	$A_{\perp} = \pi r^2$	Pointed tip aligned with flow; dependent on cone angle.
Rectangular Plate	$C_D = 1.1+0.02 (\frac{L}{W}+\frac{W}{L})$	$A_{\perp} = L W$	Normal to flow.

Let’s do a practical example to get a feeling about the relevance. Assume you are lifting a payload that is shaped like a rectangular plate with length 15 m and width 10 m. The wave period is 8 seconds and the wave height is 4 m, assume sinusoidal waves. How big will the force be?

First, let us find the peak speed; the wave motion is given as

z = \zeta \cos(\omega t)

So we find the peak speed by finding the maximum of the derivative which is:

\dot{z}_{\text{max}} = \zeta \, \omega

Since $\omega = \frac{2 \pi}{T_p}$ we then have a peak speed of 1.57 m/s.

We can then calculate the drag coefficient as:
$C_D = 1.1 + 0.02 \left( \frac{15}{10} + \frac{10}{15} \right) = 1.14$

Our drag area is:

A_\perp = 10 \cdot 15 = 150 \,\mathrm{m^2}

We can then calculate the maximum drag force, which is:

F_D = \tfrac{1}{2}\cdot 1025\cdot 1.14 \cdot 150\cdot 1.57^2 \approx 216\,\mathrm{kN} \approx 22\,\mathrm{t}

Figure 1 — Drag on the example plate (C_D = 1.14, A⊥ = 150 m²). Drag grows with the square of velocity: at the example's peak heave velocity ζω = 1.57 m/s it reaches ≈22 t. Because the law is quadratic, halving the relative velocity quarters the drag. How much of that reduction a compensator actually delivers on a given lift is a result of the coupled model, not a fixed property of the unit.

How does added mass affect a subsea lift?

Added mass in a subsea lift is very important as it can cause big additional inertia, which in turn can cause big dynamic forces. It comes from the fact that when a payload oscillates in water due to heave motion it also has to accelerate surrounding water and mathematically it is given as:

$m_A = \rho_w C_A V_R$

Where $C_A$ is the added mass coefficient (can be found in DNV RP-N103, some examples below) and $V_R$ is the reference volume.

Added-mass coefficients by shape — DNV-RP-N103 reference table

Added-mass coefficients and reference volumes by shape, for subsea lift added-mass calculations.
Shape	$C_A$	$V_R$	Notes
Sphere	$C_A = 0.5$	$V_R = \frac{4}{3}\pi r^3$	Constant in all directions.
Cylinder	$\frac{L}{2r}=1.25 \Rightarrow C_A=0.62$ $\frac{L}{2r}=2.5 \Rightarrow C_A=0.78$ $\frac{L}{2r}=5 \Rightarrow C_A=0.90$ $\frac{L}{2r}=9 \Rightarrow C_A=0.96$ $\frac{L}{2r}=\infty \Rightarrow C_A=1.00$	$V_R = \pi r^2 L$	Vertical motion along cylinder axis in infinite fluid.
Rectangular Plate	$\frac{L}{W}=1 \Rightarrow C_A=0.58$ $\frac{L}{W}=2 \Rightarrow C_A=0.76$ $\frac{L}{W}=4 \Rightarrow C_A=0.87$ $\frac{L}{W}=8 \Rightarrow C_A=0.93$ $\frac{L}{W}=\infty \Rightarrow C_A=1.00$	$V_R = \frac{\pi}{4}W^2 L$	Motion normal to surface.
Circular Disc	$C_A = \frac{2}{\pi}$	$V_R = \frac{4\pi}{3} r^3$	Motion normal to surface.
Square Prism	$\frac{L}{W}=1 \Rightarrow C_A=0.68$ $\frac{L}{W}=2 \Rightarrow C_A=0.36$ $\frac{L}{W}=4 \Rightarrow C_A=0.19$ $\frac{L}{W}=10 \Rightarrow C_A=0.08$	$V_R = W^2 L$	Prismatic body with square base.

Let’s do another practical example. Assume you are lifting a payload shaped like a square prism — a 10 m × 10 m box, 20 m long, moving along its long axis. The wave period is 8 seconds and the wave height is 4 m, assume sinusoidal waves. How big will the force due to added mass be?

We need to find the maximum acceleration (derivative of speed as shown in drag example), which is given by:

\ddot{z}_{\text{max}} = \zeta \, \omega^2

Thus the maximum acceleration is:

\ddot{z}_{\text{max}}=2 \cdot \left(\frac{2\pi}{8}\right)^2 = 1.23 \,\mathrm{m/s^2}

Reading the Square Prism row above at L/W = 2 gives an added-mass coefficient of 0.36, and its reference volume V_R = W²L = 10² × 20 = 2000 m³. We can then calculate the force using:

F = m_A\,\ddot z_{\text{max}} = 1025 \cdot 0.36 \cdot 2000 \cdot 1.23 \approx 908\,\mathrm{kN} \approx 93\,\mathrm{t}

Note that for subsea lifts close to seabed, the added mass may increase due to confinement.

Deck to seabed, one run

One continuous lowering: the compensator is locked on deck, unlocks just above the surface, then switches gas mode as the load submerges — stiff through the splash, soft for a compliant set-down.

1On deck — rigged and lifted off; the compensator locked out → 2Splash zone — overboard through the wave zone; slam and snap-load checked → 3Lowering — paying out wire lengthens the suspended system and shifts its natural period; drag and added mass follow the local relative motion → 4Landing — soft set-down onto the soil reaction

ANTARES adaptive passive heave compensation, deck to seabed — simulated in CONSTELLATION. For the full H_s×T_p operating-window screen of a lift like this, see operability screening.

Forces of this scale, a ≈93 t inertial force on top of the static weight, are why heavy subsea lifts run on a passive compensator sized for the case. A correctly sized unit reduces the crane-motion share of the payload’s relative velocity and acceleration, which pulls down both the v² drag and the added-mass force; the water’s own particle motion remains. Whether minimum tension stays positive (that is, whether slack–snap is actually ruled out) is checked in the coupled model for the case, not assumed. For the heavy and deepwater band that compensator is CYGNUS (12.5–10 000 t); RIGEL and ANTARES cover the standard and adaptive cases.

Landing speed: the last half-metre

Touchdown energy scales with the square of the approach velocity, and the peak contact force depends on how much of that energy the soil and the structure absorb — deceleration distance, contact stiffness, damping and soil response all enter. Uncompensated, the approach velocity carries the vessel’s heave on top of the winch rate. A hard or bouncing landing risks:

Equipment damage — precision surfaces and seals on manifolds, trees and templates Structural damage — mudmats buckle or over-penetrate under impact Alignment failure — a bounce at touchdown leaves the load out of position for later tie-ins

A compensated landing attenuates the heave component; the load follows the winch rate far more closely than it follows the vessel. Raising damping for the final approach turns touchdown into a controlled deceleration. ND screens landings against a touchdown velocity of 0.5 m/s, the acceptance limit we apply under DNV-RP-N103. Passive compensation attenuates vessel motion rather than removing it, so the velocity actually achieved stays sea-state dependent and is confirmed per case against winch command, vessel response and compensator state:

1Approach — lower to a few metres above the seabed under normal compensation → 2Final descent — damping up, winch paying out at a controlled rate → 3Touchdown — contact at 0.1–0.5 m/s; residual heave absorbed so the load does not bounce → 4Set-down — wire tension released gradually; the compensator prevents snap loads as the seabed takes the weight

Adjustable damping is the key feature for landing work: ANTARES adds an adaptive gas spring and rod locking to hold the load secure after set-down; where an exact lowering velocity must be dialled in, an active system provides that control. The motion-ratio physics behind soft landings is derived in passive heave compensation basics.

What depth does to the compensator

The four forces above act on the payload. Three more effects act on the compensator itself as it descends, each one shifting the unit off its working point, and a fourth is kept out by engineering discipline:

Depth and temperature effects on a passive heave compensator — what each does to the compensator and how it is answered.
Effect	What it does to the compensator	The answer
Buoyancy	Net payload weight drops in water → the equilibrium position walks inward, in the worst case to full retraction	Adaptive: gas pressure down, equilibrium recentred — and a longer natural period as a bonus
Temperature	Gas pressure falls with cooling on descent — the stroke cost depends on gas volume, preload and the descent profile — and most arrives fast through the thermocline	Adaptive: onboard high-pressure gas tops it back up
Water pressure	≈ 1 bar per 10 m of depth acts on the rod area and pushes the rod in — a 250 t lift with a 180 mm rod at 1,000 m sees ≈ 26 t, shifting equilibrium from mid-stroke to ≈⅙ of stroke	Adaptive: pressure adjusted down with depth
Leakage	Water ingress — historically anything from lost performance to corrosion to explosions	Design + testing + maintenance: double seals, external-pressure testing at FAT, compartmentalisation, scheduled seal replacement

The first three are equilibrium problems an adaptive unit corrects on the fly; the fourth is engineering discipline.

How much does temperature matter?

Gas pressure falls with cooling on the way down — the stroke cost depends on gas volume, preload and the descent profile — and most of the change arrives quickly through the thermocline. Adaptive units inject onboard high-pressure gas to hold pressure.

What does depth pressure do to the compensator?

About 1 bar per 10 m acts on the rod area and pushes the rod in. Example: a 250 t lift with a 180 mm rod at 1,000 m picks up ≈ 26 t of retraction force, moving the equilibrium from mid-stroke to about one-sixth. Adaptive pressure adjustment removes the shift.

How is water leakage into the compensator prevented?

Double seals on two different sealing surfaces on all volumes, external pressure testing at FAT, compartmentalisation to contain any leak, offshore-capable back-seal testing in regular maintenance, scheduled seal replacement — and redundancy options for mission-critical work.

Why choose an adaptive compensator for deep work?

Because buoyancy, cooling and depth pressure all shift the working point, and an adaptive unit corrects all three automatically — staying centred and soft down the whole water column instead of drifting off its design point.

Subsea lifting — frequently asked

Why is a subsea lift harder than a lift in air?

The water fights back four ways: buoyancy changes the effective weight, slam hits at water entry, drag resists with the square of velocity, and added mass makes the payload effectively heavier whenever it accelerates. Each phase has a different dominant force.

How large can drag forces get on a subsea lift?

In the worked example above, a 15 × 10 m plate at 1.57 m/s peak heave velocity sees ≈ 216 kN ≈ 22 t. Drag grows with v² — halve the relative velocity and the force drops to a quarter.

What is added mass and how big is it?

The water the payload must accelerate with itself. The 10 × 10 × 20 m square-prism example picks up ≈ 93 t of inertial force in 8-second waves — and the effect grows near the seabed through confinement, which matters during landing.

Which force dominates the splash zone?

Slamming plus rapidly varying buoyancy, on top of wave-particle velocity — the worst combination of the whole lift, packed into a few metres. It sizes the compensator more often than any other phase.

How does heave compensation protect a subsea lift?

By cutting relative velocity and acceleration between hook and payload — collapsing both the v² drag and the acceleration-driven added-mass force — and by keeping tension in the wire so slack–snap events cannot develop. Product-wise: CYGNUS for heavy/deepwater, RIGEL for standard passive cases, ANTARES when automatic damping control pays.

Working the numbers on a subsea lift?

We size compensators against drag, added mass and snap-load — send the subsea lift case for a recommended product and dimensions (for heavy lifts, typically the CYGNUS passive heave compensator).

ND SolutionCYGNUS →Passive heave compensator for heavy and deepwater subsea lifts, 12.5–10,000 t SWL.

ND SolutionRIGEL →Passive heave compensator for standard subsea lifts.

Subsea ChallengesOverview of the unique challenges faced during subsea lifting operations.

Splash Zone CrossingThe splash zone is the most critical phase — learn how to manage it safely.

Heave Compensator SelectionChoosing the right compensator for subsea lift requirements.