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Eytelwein's formula

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Eytelwein's formula


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. For instance 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

1. Circular coordinates

\varphi=\frac{s}{r}\qquad{d}\varphi=\frac{ds}{r}\qquad\frac{d\varphi}{ds}=\frac{1}{r} (1), (2), (3)



Let \mathbf{s}_{u} and \mathbf{n}_{u} denote unit vectors;

\mathbf{s}_{u}=-\sin\varphi\cdot{\mathbf{x}}_{u}+\cos\varphi\cdot{\mathbf{y}}_{u} (4)
\mathbf{n}_{u}=\cos\varphi\cdot{\mathbf{x}}_{u}+\sin\varphi\cdot{\mathbf{y}}_{u} (5)

Then from (5)

\frac{d}{d\varphi}{\mathbf{s}}_u = -\cos\varphi\cdot{\mathbf{x}}_u -\sin\varphi\cdot{\mathbf{y}}_u=-{\mathbf{n}}_u (6)
\frac{d}{ds}=\frac{d}{d\varphi}\cdot\frac{d\varphi}{ds}=\frac{d}{d\varphi}\cdot\frac{1}{r}=\frac{1}{r}\cdot\frac{d}{d\varphi} (7)

From (6) and (7), it follows that

\frac{d}{ds}{\mathbf{s}}_u=\frac{1}{r}\cdot\frac{d}{d\varphi}{\mathbf{s}}_u=-\frac{1}{r}\cdot{\mathbf{n}}_u. (8)

2. Forces on cordage in general

Now, let's study a piece of cord in general, subject to an arbitrary force. Let s\, denote the length of the cord and let the force per unit length be \mathbf{q}(s). Consider a short piece \Delta{s}\, of the cord and introduce the cross-sectional force \mathbf{T}(s)\,. Balancing the forces, we get

\mathbf{T}(s+\Delta{s})-\mathbf{T}(s)+\mathbf{q}(s)\cdot\Delta{s}=0, (9)


\frac{\mathbf{T}(s+\Delta{s})-\mathbf{T}(s)}{\Delta{s}}=-\mathbf{q}(s). (10)

Letting \Delta{s}\rightarrow0, we conclude that

\frac{d}{ds}\mathbf{T}(s)=-\mathbf{q}(s). (11)

3. A line around a capstan

A line is wound around a cylinder(a bollard or a capstan). In this case the curvature of the line is circular which makes the problem easier. Let s\, be the length of the line from a point A where the line makes contact with the cylinder. At the point s\, on the short piece \Delta{s}\, of the line acts a force from the cylinder that can be subdivided into a tangential component t\Delta{s}\, (friction) and a normal component n\Delta{s}\,. That is to say that

\mathbf{q}(s)=t\cdot\mathbf{s}_u + n\cdot\mathbf{n}_u (12)


With the cross-sectional force \mathbf{T}(s)\, (which is tangential) we get

\mathbf{T}(s)=T(s)\cdot\mathbf{s}_u (13)

From (11), (12) and (13), it follows that

\frac{d}{ds}T(s)\cdot\mathbf{s}_u = -t\cdot\mathbf{s}_u - n\cdot\mathbf{n}_u (14)

Derivative of a product and (8) imply that

\frac{d}{ds}T(s)\cdot\mathbf{s}_{u}=\frac{dT(s)}{ds}\cdot\mathbf{s}_{u}+T(s)\cdot\frac{d}{ds}\mathbf{s}_u
= \frac{dT(s)}{ds}\cdot\mathbf{s}_u-\frac{T(s)}{r}\cdot\mathbf{n}_u=-t\cdot\mathbf{s}_u-n\cdot\mathbf{n}_u (15)

Identifying components in (15), we get

\frac{dT(s)}{ds}=-t (16)

and

\frac{T(s)}{r}=n. (17)

Dividing (16) by (17), we get

\frac{dT(s)}{ds}/\frac{T(s)}{r}=-\frac{t}{n} (18)

From (18) and reciprocal of (2), we get

\mathrm{LHS}=\frac{dT(s)}{T(s)}\cdot\frac{r}{ds}=\frac{dT(s)}{T(s)}\cdot\frac{1}{d\varphi}=\frac{1}{T(S)}\cdot\frac{dT(s)}{d\varphi}. (19)

From (18) and (19) it follows that

\frac{1}{T(s)}\cdot\frac{dT(s)}{d\varphi}=-\frac{t}{n}. (20)

Let \mu=\frac{t}{n} (21) be the coefficient of friction (no slip). Then

\frac{1}{T}\cdot\frac{dT}{d\varphi}=-\mu (22) : \Rightarrow
\frac{1}{T}\cdot{dT}=-\mu\cdot{d}\varphi (23)

Integration of (23) yields

\int_{T_\text{load}}^{T_\text{hold}} \frac{1}{T} \cdot{dT} = \int_0^\phi -\mu \cdot {d}\varphi (24)
\ln T_\text{hold} - \ln T_\text{load} = \ln\frac{T_\text{hold}}{T_\text{load}} = -\mu\cdot\phi (25) \Rightarrow
\frac{T_\text{hold}}{T_\text{load}}={e}^{-\mu\cdot\phi} (26)

Finally,

T_\text{hold}=T_\text{load}\cdot{e}^{-\mu\cdot\phi}\quad\text{ or } \quad T_\text{load} = T_\text{hold}\cdot{e}^{\mu\cdot\phi}

See also

References

Further reading

  • Arne Kihlberg, Kompendium i Mekanik för E1, del II, Göteborg 1980, 60–62.
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