# The capstan equation

15 Nov 2021 - tsp
Last update 15 Nov 2021
3 mins

So since I had encountered this equation again I thought it was a good exercise to see if I still know the derivation of the equation and summarize it here.

## What is the capstan equation?

The capstan equation describes the friction of a rope wound around a bollard, winch or a capstan. It’s also known as Eytelweins equation. To use the equation though there are a few assumptions that one has to take:

• The rope is on the verge of sliding so the hold force really is exactly the required force to hold the load.
• The rope is not rigid - if this would be the case forces would be required to bend the rope.
• The rope is not elastic.

In this case the capstan equation simply relates the hold force $F_H$ to the load force $F_L$:

[ T_L = T_H * e^{\mu \phi} ]

## Derivation

The derivation is pretty simple. First one assumes the cable is running along the surface of a cylinder with an contact area for $\alpha$ radians. On the left side the force $F_1$ describes the load, on the right side the force $F_2$ describes the hold force

Then one looks into a differential (infinitesimally small) section of the contact area. The opening angle is $d\phi$. The force of the rope can be split into a tangential part $dT$ and a radial part $dR$. The contact length of the section along the surface of the cylinder is $ds$. On the left side a force $S + dS$ is applied to the rope, on the right side the force is $S$ - so in this model the force decreases from left to right. As one can see from the tangentially extended triangles the forces are acting under an angle of $\beta = \frac{d\phi}{2}$

Now one can use the standard trick for statics - applying force equilibrium along the tangential and radial direction:

• Tangential: $dT - (S + dS) * cos(\frac{d\phi}{2}) + S cos(\frac{d\phi}{2}) = 0$
• Radial: $dR - (S + dS) * sin(\frac{d \phi}{2}) - S * sin(\frac{d \phi}{2}) = 0$

If one assumes that angles are small ($\phi « 1$) one can assume that:

• $cos(\frac{d\phi}{2}) \approx 1$
• $sin(\frac{d\phi}{2}) \approx \frac{d\phi}{2}$

This massively reduces the complexity of the equations in both directions:

[ dT - (S + dS) + S = 0 \\ \to dT - dS = 0 \\ \to dT = dS \\ dR - (S + dS) \frac{d\phi}{2} - S \frac{d\phi}{2} = 0 \\ \to dR - 2 S \frac{d\phi}{2} - \underbrace{dS \frac{d \phi}{2}}_{neglecting} = 0 \\ \to dR - S d\phi = 0 \\ \to dR = S d\phi ]

In the radial equation second order terms have been neglected. Now one has to assume the special boundary case on the verge of slipping: $T = \mu R$

[ T = \mu R \\ dT = \mu dR \\ dS = \mu S d\phi \\ \frac{dS}{S} = \mu d\phi ]

Integrating over the whole contact area (and thus forces $S_1$ to $S_2$ or angles from $0$ to $\alpha$):

[ \int_{S_1}^{S_2} \frac{dS}{S} = \mu \int_0^\beta d\phi \\ ln(S_2) - ln(S_1) = \mu (\beta - 0) \\ ln(\frac{S_2}{S_1}) = \mu \beta \\ e^{ln(\frac{S_2}{S_1})} = e^{\mu \beta} \\ \frac{S_1}{S_2} = e^{\mu \beta} \\ S_1 = S_2 * e^{\mu \beta} ]

Dipl.-Ing. Thomas Spielauer, Wien (webcomplains389t48957@tspi.at)

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