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. Itβs 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}
]
This article is tagged: Math, Tutorial, Physics, Mechanics