### Eytelwein's formula

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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:

- $T\_\backslash text\{load\}\; =\; T\_\backslash text\{hold\}\backslash \; e^\{\; \backslash mu\; \backslash phi\}\; \backslash ,$

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

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

- The rope is on the verge of full sliding, i.e. $T\_\backslash text\{load\}$ is the maximum load that one can hold. Smaller loads can be held as well, resulting in a smaller
*effective*contact angle $\backslash phi$. - 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.
- 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 $e^\{\; \backslash mu\; \backslash phi\}\; \backslash ,$ 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**

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

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

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

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

Then from (5)

- $\backslash frac\{d\}\{d\backslash varphi\}\{\backslash mathbf\{s\}\}\_u\; =\; -\backslash cos\backslash varphi\backslash cdot\{\backslash mathbf\{x\}\}\_u\; -\backslash sin\backslash varphi\backslash cdot\{\backslash mathbf\{y\}\}\_u=-\{\backslash mathbf\{n\}\}\_u$ (6)

- $\backslash frac\{d\}\{ds\}=\backslash frac\{d\}\{d\backslash varphi\}\backslash cdot\backslash frac\{d\backslash varphi\}\{ds\}=\backslash frac\{d\}\{d\backslash varphi\}\backslash cdot\backslash frac\{1\}\{r\}=\backslash frac\{1\}\{r\}\backslash cdot\backslash frac\{d\}\{d\backslash varphi\}$ (7)

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

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

**2. Forces on cordage in general**

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

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

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

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

- $\backslash frac\{d\}\{ds\}\backslash mathbf\{T\}(s)=-\backslash mathbf\{q\}(s).$
**(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 $s\backslash ,$ be the length of the line from a point A where the line makes contact with the cylinder. At the point $s\backslash ,$ on the short piece $\backslash Delta\{s\}\backslash ,$ of the line acts a force from the cylinder that can be subdivided into a *tangential* component $t\backslash Delta\{s\}\backslash ,$ (friction) and a *normal* component $n\backslash Delta\{s\}\backslash ,$. That is to say that

- $\backslash mathbf\{q\}(s)=t\backslash cdot\backslash mathbf\{s\}\_u\; +\; n\backslash cdot\backslash mathbf\{n\}\_u$ (12)

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

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

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

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

Derivative of a product and (8) imply that

- $\backslash frac\{d\}\{ds\}T(s)\backslash cdot\backslash mathbf\{s\}\_\{u\}=\backslash frac\{dT(s)\}\{ds\}\backslash cdot\backslash mathbf\{s\}\_\{u\}+T(s)\backslash cdot\backslash frac\{d\}\{ds\}\backslash mathbf\{s\}\_u$

- $=\; \backslash frac\{dT(s)\}\{ds\}\backslash cdot\backslash mathbf\{s\}\_u-\backslash frac\{T(s)\}\{r\}\backslash cdot\backslash mathbf\{n\}\_u=-t\backslash cdot\backslash mathbf\{s\}\_u-n\backslash cdot\backslash mathbf\{n\}\_u$
**(15)**

Identifying components in **(15)**, we get

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

and

- $\backslash frac\{T(s)\}\{r\}=n.$ (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

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

From (18) and (19) it follows that

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

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

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

- $\backslash frac\{1\}\{T\}\backslash cdot\{dT\}=-\backslash mu\backslash cdot\{d\}\backslash varphi$
**(23)**

Integration of **(23)** yields

- $\backslash int\_\{T\_\backslash text\{load\}\}^\{T\_\backslash text\{hold\}\}\; \backslash frac\{1\}\{T\}\; \backslash cdot\{dT\}\; =\; \backslash int\_0^\backslash phi\; -\backslash mu\; \backslash cdot\; \{d\}\backslash varphi$ (24)

- $\backslash ln\; T\_\backslash text\{hold\}\; -\; \backslash ln\; T\_\backslash text\{load\}\; =\; \backslash ln\backslash frac\{T\_\backslash text\{hold\}\}\{T\_\backslash text\{load\}\}\; =\; -\backslash mu\backslash cdot\backslash phi$ (25) $\backslash Rightarrow$

- $\backslash frac\{T\_\backslash text\{hold\}\}\{T\_\backslash text\{load\}\}=\{e\}^\{-\backslash mu\backslash cdot\backslash phi\}$ (26)

Finally,

- $T\_\backslash text\{hold\}=T\_\backslash text\{load\}\backslash cdot\{e\}^\{-\backslash mu\backslash cdot\backslash phi\}\backslash quad\backslash text\{\; or\; \}\; \backslash quad\; T\_\backslash text\{load\}\; =\; T\_\backslash text\{hold\}\backslash cdot\{e\}^\{\backslash mu\backslash cdot\backslash phi\}$

## See also

- Belt friction
- Frictional contact mechanics

## References

## Further reading

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