Catenary mooring: chain length, tension and anchor load

A floating offshore wind platform is held in position by mooring lines, and the first decision about them is whether they hang or are pulled tight. The distinction is not aesthetic: it determines what kind of load reaches the anchor, and therefore what kind of anchor is buried at the other end.

The catenary case has a closed-form solution, and it is worth having in full because every number in a mooring layout follows from three inputs.

Every number in this article is recomputed from its inputs by a script that runs on each build of this site.

The shapePermalink to “The shape”#

A perfectly flexible line of constant submerged weight $w$ per unit length, hanging in still water, satisfies the catenary equation. The useful form starts from the fact that the horizontal component of tension is constant along the line — nothing acts horizontally on it — so with $H$ that constant and $z$ the height above the touchdown point:

$$T = H + w z$$ $$s = \sqrt{z\left(z + \frac{2H}{w}\right)}$$ $$x = \frac{H}{w}\,\operatorname{arccosh}\!\left(1 + \frac{z w}{H}\right)$$

where $s$ is the suspended length and $x$ the horizontal distance from touchdown to the fairlead. Three formulas, three inputs, everything else follows.

A worked linePermalink to “A worked line”#

Take a studless chain with a submerged weight of 1.5 kN/m in 100 m of water, at a horizontal tension of 1200 kN. Then $2H/w = 1600.0000$ m and

$$s = \sqrt{100 \times (100 + 1600)} = \sqrt{170000} = 412.3\ \text{m}$$ $$x = 800 \times \operatorname{arccosh}(1.1250) = 800 \times 0.494933 = 395.9\ \text{m}$$ $$T = 1200 + 1.5 \times 100 = 1350.0\ \text{kN}$$

Three things are worth reading off that.

The vertical weight of the line is the entire difference between top and bottom tension. $T - H = wz = 150$ kN, which is 12.5% of the horizontal tension. In deep water this term dominates: the chain is largely carrying itself, which is why chain is displaced by wire and then by synthetic rope as depth increases.

The line is nearly straight. The suspended length exceeds the straight-line distance between its ends by 16.4 m out of 412.3 — 4%. At working tension a catenary is a shallow curve, and the intuition of a deeply sagging cable is wrong for a mooring line that is doing its job.

The angle at the fairlead is $\arccos(H/T) = 27.3°$. That, not the tension, is what sizes the fairlead and decides the vertical load the platform must carry.

The point of the whole arrangementPermalink to “The point of the whole arrangement”#

At the anchor end, provided some chain still lies on the seabed, the line arrives horizontally. The anchor sees 1200 kN of horizontal load and zero uplift.

That is the entire engineering case for catenary mooring. A drag embedment anchor develops its capacity by ploughing further into the seabed as it is pulled, and it does that only under horizontal load; apply uplift and it walks out. Keeping the touchdown point on the seabed under every load case is therefore a design requirement, not an outcome, and it is what sets the total deployed line length.

Raise the horizontal tension by 20%, to 1440 kN, and

$$s = 449.4\ \text{m}, \qquad x = 434.5\ \text{m}, \qquad T = 1590.0\ \text{kN}$$

The suspended length has grown by 9.01% and the span by 9.73%, both drawn from the chain that was lying on the bottom. The line has to carry enough spare length to give that up under the extreme load case and still keep the anchor loaded horizontally — and the extra chain is weight the platform must support and cost that scales with water depth.

What a taut line does insteadPermalink to “What a taut line does instead”#

Run a straight line between the same two points and the geometry changes completely. The straight distance is 408.4 m, arriving at the anchor at 14.2° above horizontal. For the same horizontal component the anchor load becomes

$$T = \frac{1200}{\cos 14.2°} = 1238\ \text{kN}, \qquad V = 1200 \tan 14.2° = 303\ \text{kN}$$

303 kN of uplift, which a drag embedment anchor cannot take. A taut system requires an anchor that resists vertical load — a suction caisson, a driven pile, a vertically loaded plate anchor — and in exchange it gives a much smaller footprint, less line to buy and carry, and a stiffer restoring response.

Neither is better in general. The catenary buys a cheap anchor with an expensive footprint; the taut system buys a small footprint with an expensive anchor. What decides it is water depth, seabed competence, and how much of the seabed the array is allowed to occupy.

Where the closed form stops being enoughPermalink to “Where the closed form stops being enough”#

The three formulas above assume a perfectly flexible, inextensible line of constant weight in still water, hanging in a single plane. Real design departs from all of it: chain stretches elastically and the segments differ, current drags the line sideways and out of plane, the platform moves so the problem is dynamic, and the portion on the seabed develops friction that changes where the touchdown point actually sits.

Which is the familiar pattern. The closed form does not survive contact with the real load case — and that is precisely what makes it valuable, because it is the thing a dynamic mooring analysis can be checked against in the still-water limit. A numerical model that cannot reproduce 412.3 m and 1350 kN with the current switched off has a defect that will not be any smaller once the current is switched on. That is the whole argument for benchmarking against closed forms, and it applies as much to a mooring solver as to a finite element engine.