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 — which is what heave compensation does.

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 20 × 10 m plate in 8-second waves picks up an inertial force of about 98 t — on top of the static weight. Near the seabed it grows further due to confinement.

How does heave compensation protect a subsea lift?

It cuts the relative velocity and acceleration between hook and payload — collapsing the velocity-squared drag and the acceleration-driven added-mass force — and it keeps the wire tensioned so slack-snap events cannot develop. For heavy and deepwater lifts the passive compensator sized for this duty is CYGNUS; RIGEL and ANTARES cover the standard and adaptive cases.

Subsea structure on the crane wire descending through the water column at dusk

Subsea lifting

Underwater, the water lifts with you — and against you: buoyancy, slam, drag, added mass.

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: 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.

By Peter Wang, COO · December 2025

Practical application: For practical application of this topic, see RIGEL Basic PHC and engineering studies and analysis.

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 effective weight and dynamic loads on the crane and lifting equipment.

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. 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 and not air which have 1000 times bigger mass density, 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 payload vertical velocity.

Shape	$C_D$	$A_{\perp}$	Notes
Sphere	$C_D = 0.5$	$A_{\perp} = \pi r^2$
Horiontal 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 lets find the peak speed; the wave motion is given as

z = \zeta \cos(\omega t)

So we find the peak speed by finding the max of the derivate 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. Halving the relative velocity — which is what heave compensation does — cuts the drag force to a quarter.

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.

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 that is shaped like a rectangular plate with length 20 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 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}

Next the added mass coefficient is 0.36 and our reference volume is 2000 cubic meters. We can then calculate the force using:

F=m a=1025 \cdot 0.36 \cdot 2000 \cdot 1.23=98 \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 — through the heave-resonance band; drag and added mass build → 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.

The four forces, one table

Drag and added mass are the two calculated above — buoyancy and slamming complete the picture, and each phase of the lift has a different boss:

Force	Scales with	When it dominates	What it does to the lift
Buoyancy	Displaced volume (ρ_wVg)	From water entry; varies fast while crossing the surface	Effective weight drops and fluctuates — slack-wire risk
Slamming	Impact velocity at water entry	Splash zone	Short, sharp load spikes on payload and rigging
Drag	Velocity squared (½ρ_wC_DA_⊥v²)	Submerged descent through waves and current	≈ 22 t on the example plate at peak heave velocity
Added mass	Acceleration (ρ_wC_AV_R × a)	Any oscillation; grows near the seabed (confinement)	≈ 98 t of extra inertial force in the example above

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.

Forces of this scale — a ≈98 t inertial force on top of the static weight — are why heavy subsea lifts run on a passive compensator sized for the case. It cuts the relative velocity and acceleration (collapsing the v² drag and the added-mass force) and keeps the wire tensioned so slack–snap cannot develop. 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

Impact at touchdown scales with velocity — and without compensation that velocity is set by vessel heave, which nobody controls. 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 removes the heave component (the load descends at the winch rate, not winch plus heave) and raises damping for the final approach, turning touchdown into a controlled deceleration — typically 0.1–0.5 m/s, set by the operator and largely independent of sea 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:

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	≈ 1% of gas pressure per 3 °C of cooling — roughly 5% of stroke — lost 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?

About 1% of gas pressure per 3 °C of cooling — roughly 5% of stroke — and most of it hits 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 get?

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, in practice?

The water the payload must accelerate with itself. The 20 × 10 m plate example picks up ≈ 98 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 the 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 on a lift that needs this?

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 SolutionANTARES →Adaptive passive heave compensator for subsea lifting operations.

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.