# Capstan equation

An example of when knowledge of the capstan equation might have been useful. The bent white tube contains a cord to raise and lower a curtain. The tube is bent some 40 degrees in two places. The blue line indicates a more efficient design.
An example of holding capstans and a powered capstan used to raise sails on a tall ship.

The capstan equation or belt friction equation, also known as Eytelwein's formula,[1][2] relates the hold-force to the load-force if a flexible line is wound around a cylinder (a bollard, a winch or a capstan).[3] [4]

Because of the interaction of frictional forces and tension, the tension on a line wrapped around a capstan may be different on either side of the capstan. A small holding force exerted on one side can carry a much larger loading force on the other side; this is the principle by which a capstan-type device operates.

A holding capstan is a ratchet device that can turn only in the counter clockwise direction so that once a load is pulled into place it can be held with a much smaller force. A powered capstan, also called a winch, rotates so that the applied tension is multiplied by the friction between rope and capstan. On a tall ship a holding capstan and a powered capstan are used in tandem so that a small force can be used to raise a heavy sail and then the rope can be easily removed from the powered capstan and tied off.

In rock climbing with so-called top-roping, a lighter person can hold (belay) a heavier person due to this effect.

The formula is

$T_\text{load} = T_\text{hold}\ e^{ \mu \phi} ~,$

where $T_\text{load}$ is the applied tension on the line, $T_\text{hold}$ is the resulting force exerted at the other side of the capstan, $\mu$ is the coefficient of friction between the rope and capstan materials, and $\phi$ is the total angle swept by all turns of the rope, measured in radians (i.e., with one full turn the angle $\phi =2\pi\,$).

Several assumptions must be true for the formula to be valid:

1. The rope is on the verge of full sliding, i.e. $T_\text{load}$ is the maximum load that one can hold. Smaller loads can be held as well, resulting in a smaller effective contact angle $\phi$.
2. It is important that the line is not rigid, in which case significant force would be lost in the bending of the line tightly around the cylinder. (The equation must be modified for this case.) For instance a Bowden cable is to some extent rigid and doesn't obey the principles of the Capstan equation.
3. The line is non-elastic.

It can be observed that the force gain grows exponentially with the coefficient of friction, the number of turns around the cylinder, and the angle of contact. Note that the radius of the cylinder has no influence on the force gain.

The table below lists values of the factor $e^{ \mu \phi} \,$ based on the number of turns and coefficient of friction μ.

Number
of turns
Coefficient of friction μ
0.1 0.2 0.3 0.4 0.5 0.6 0.7
1 1.9 3.5 6.6 12 23 43 81
2 3.5 12 43 152 535 1881 6661
3 6.6 43 286 1881 12392 81612 437503
4 12 152 1881 23228 286751 3540026 43702631
5 23 535 12392 286751 6635624 153552935 3553321281

From the table it is evident why one seldom sees a sheet (a rope to the loose side of a sail) wound more than three turns around a winch. The force gain would be extreme besides being counter-productive since there is risk of a riding turn, result being that the sheet will foul, form a knot and not run out when eased (by slacking grip on the tail (free end), or in land talk, one lets go of the hold end.

It is both ancient and modern practice for anchor capstans and jib winches to be slightly flared out at the base, rather than cylindrical, to prevent the rope (anchor warp or sail sheet) from sliding down. The rope wound several times around the winch can slip upwards gradually, with little risk of a riding turn, provided it is tailed (loose end is pulled clear), by hand or a self-tailer.

For instance, the factor 153552935 means, in theory, that a newborn baby would be capable of holding the weight of two USS Nimitz supercarriers (97 000 ton each, but for the baby it would be only a little more than 1 kg).

## Proof of the capstan equation

The first step is to relate the radial or normal force $F$ (Newtons/radian) at any point of the rope wrapped around a capstan to the tension $T$ (Newtons) in the rope. This can be done with a simple energy argument as follows: Assume that the rope is inelastic (nonstretching) and the capstan is frictionless and compressible. Then the energy required to squeeze down the capstan radius $R$ by an amount $\Delta R$ with one turn of rope equals the energy supplied by pulling the rope (Force) x Distance = Force x (Distance).

$( 2\pi {F}) \Delta {R} =T(2 \pi \Delta {R})$

Cancel $2\pi \Delta {R}$, on each side leaving $F=T$

So the frictional force over a wrap angle $d\varphi$ is

$\mu {F} d\varphi = \mu {T} d\varphi$ where $\mu$ is the coefficient of friction (nonslip).

The increase in rope tension $dT$ over a wrap angle $d\varphi$ is the frictional force over that angle so

$dT=\mu {T} d\varphi$
$\frac{1}{T}dT=\mu d\varphi$

Integration of both sides yields

$\int_{T_\text{hold}}^{T_\text{load}} \frac{1}{T} \;{dT} = \int_0^\phi \mu \; {d}\varphi$
$\ln T_\text{load} - \ln T_\text{hold} = \ln\frac{T_\text{load}}{T_\text{hold}} = \mu \phi$

and exponentiating both sides,

$\frac{T_\text{load}}{T_\text{hold}}={e}^{\mu \phi}$

Finally,

$T_\text{load} = T_\text{hold}{e}^{\mu \phi}$