# Capstan Equation

Posted in Mechanics and tagged on .

This note reviews the derivation of capstan equation. A capstan is a cylinder-like device that turns to wind a rope or cable around it to lift or haul things. Ropes and cables are flexible enough to be wound and work as a medium to transfer force to other bodies. The capstan’s cylindrical geometry couples rope’s tangential and radial loads as the rope is in equilibrium on the capstan. This coupling transfers a portion of the rope’s tangential tensile load into its radial contact pressure with the capstan, thus, enlarges the frictional effect. This behavior results in a distinct feature of capstan devices—the ratio of tensile forces at two ends of the rope wound on a capstan is an exponential function of the product of kinetic friction coefficient and wrap angle (slip angle to be precise) between them. Looping multiple threads of rope on the capstan would significantly increase the contact area where the friction can take place on, thus, significantly alter the ratio of two end forces of the rope, which sometimes would be nice to take advantage of. The same mechanics universally governs the problems with similar geometry and loading members, for instances, a belt transporting on pulleys or a web transporting on rollers. The capstan equation solves the tensile force developed in the rope (belt, web) undergoes kinetic friction on capstans (pulleys, rollers).

## Problem of a Belt Slipping on a Cylinder

The capstan problem is abstracted to the following planar problem (boundary conditions are marked in red while reaction loads in grey). A slender belt with the section width of $$W$$ wraps a cylinder for 90°. The belt has a tensile force $$T_0$$ applied at its lower end and is fixed at the right end, while the cylinder is rotating at a constant counterclockwise angular velocity; so the belt slips on the cylinder and is in equilibrium. Kinetic friction takes place at the contact interface between the belt and the cylinder.

We want to derive the capstan equation that solves the tensile force developed at any position in the belt, as well as the reaction force at the right end of the belt and the reaction torque on the cylinder. ## Assumptions

The belt is simplified to a 1-D structure, the meaningful geometric feature is the dimension along the belt: the belt can only have longitudinal deformation and it only depends on its axial position. Within this framework, we further assumes:

• the capstan is rigid;
• the belt
• is elastic, subjected to only the elastic constitutive law,
• is flexible, the bending rigidity is negligible, thus naturally bends without consuming any external moments,
• can only have axial direction tensile deformation, incompressible,
• accordingly, cannot have shear deformation;
• only kinetic friction is allowed at the belt-capstan contact interface and it equals to the product of kinetic friction coefficient and normal force (pressure).

The last two assumptions are paired. A 1-D belt, due to its original theoretical foundation, is unable to account for shear deformations. An important outcome, as the belt gets contact with other bodies, is the existence of static friction is also theoretically eliminated: without allowing shear deformation, any forces acted along the belt, for example, a friction force at the contact interface, would directly produce axial deformations, thus, the material points of belt there have to slip on the capstan due to the axial elongation; consequentially, the friction force becomes kinetic. The assumption of 1-D belts works very well for typical flexible ropes, belts, webs, however, fails for cables and belts made from stiff core and compliance stuffing material around, since the stiff core may hardly deform axially and the covering material usually undergoes severe shear deformation. In this case, an additional degree of freedom has to be introduced to account for the shear deformation and as shear is allowed, static friction should be allowed.

## Analysis

Derivations on the equilibrium of circular structures in polar or cylindrical coordinate system usually analyze the force balance of an infinitesimal arch of material, on which the radial pressure is intuitively uniform, thus, it has no effect on the tangential force balance. I’ve been very comfortable with this approximation, but I’m very curious about how it is made. So here I prefer to do a analysis on a finite part of material, which preserves all related quantities, then let the math tell the significance of each term.

### Equations of equilibrium

We set up a polar coordinate system whose positive angular coordinate is clockwise as shown below and section out a finite material arch $$AB$$ of $$R\Delta \theta$$ long from the belt with middle point $$C$$. We analyze its force balance in the tangential direction $$\vec{t_c}$$ and radial direction $$\vec{r_c}$$ at point $$C$$. $$p(\theta’)$$ and $$f(\theta’)$$ are the contact pressure and friction stress respectively on the belt applied by the cylinder, with a dimension of $$[Pa]$$. $$T(\theta)$$ and $$T(\theta+\Delta \theta)$$ are the internal tension of belt exposed at $$\theta$$ and $$\theta+\Delta \theta$$ respectively, with a dimension of $$[N]$$. List force balance in the tangential direction at $$C$$, $$\vec{t_c}$$:

$$-T(\theta)\cos \big(\frac{\Delta \theta}{2} \big) – \int_{\theta}^{\theta+\Delta \theta} f(\theta’)\cos \big( \theta + \frac{\Delta \theta}{2} -\theta’ \big) WR \, d \theta’-\int_{\theta}^{\theta +\Delta \theta}p(\theta’)\sin \big( \theta + \frac{\Delta \theta}{2} -\theta’ \big) WR \, d \theta’ + T(\theta+\Delta \theta)\cos \big(\frac{\Delta \theta}{2} \big) = 0 ,$$

and in the radial direction $$C$$, $$\vec{r_c}$$:

$$\int_{\theta} ^{\theta+\Delta \theta} p(\theta’) \cos \big( \theta + \frac{\Delta \theta}{2} – \theta’ \big) WR \, d \theta’ – T(\theta) \sin \big( \frac{\Delta \theta}{2} \big) – T(\theta +\Delta \theta) \sin \big( \frac{\Delta \theta}{2} \big)+\int_{\theta}^{\theta+\Delta \theta}f(\theta’) \sin \big( \theta + \frac{\Delta \theta}{2} – \theta’ \big) WR \, d\theta’ = 0 .$$

### Linearization

The mean value theorem for definite integrals states that for a continuous function $$f(x)$$, there exits a value of $$x=x^* \in [a, b]$$ such that the definite integral

$$\int_{a}^{b} f(x) \, dx = f(x^*)(b-a).$$

It’s reasonable to assume that $$f(\theta)$$ and $$p(\theta$$) satisfy the mean value theorem’s requirement on functions. Applying this theorem for both equations can transform the integrals into arithmetic terms,

in $$\vec{t_c}$$:

$$-T(\theta)\cos \big(\frac{\Delta \theta}{2} \big) – f(\theta^*)\cos \big( \theta + \frac{\Delta \theta}{2} -\theta^* \big) WR \, \Delta \theta -p(\theta^*)\sin \big( \theta + \frac{\Delta \theta}{2} -\theta^* \big) WR \, \Delta \theta + T(\theta+\Delta \theta)\cos \big(\frac{\Delta \theta}{2} \big) = 0,$$

and in $$\vec{r_c}$$:

$$p(\theta^*) \cos \big( \theta + \frac{\Delta \theta}{2} -\theta^*\big) WR \, \Delta \theta – T(\theta) \sin \big( \frac{\Delta \theta}{2} \big) – T(\theta +\Delta \theta) \sin \big( \frac{\Delta \theta}{2} \big) + f(\theta^*) \sin \big( \theta + \frac{\Delta \theta}{2} -\theta^* \big) WR \, \Delta \theta = 0,$$

where $$\theta^* \in [\theta, \theta+\Delta \theta]$$.

Next we consider the significance of each quantity when taking the limit of $$\Delta \theta$$. As $$\Delta \theta \rightarrow 0$$:

$$\Delta \theta \rightarrow d \theta, \, \theta^* \rightarrow \theta,$$

$$T(\theta + \Delta \theta) = T(\theta)+\frac{dT}{d\theta}d\theta + O(d\theta^2) \approx T(\theta)+\frac{dT}{d\theta}d\theta,$$

$$\cos \big( \frac{\Delta \theta}{2} \big) \rightarrow 1, \, \sin \big( \frac{\Delta \theta}{2}\big) \rightarrow \frac{\Delta \theta}{2},$$

$$\cos \big( \theta + \frac{\Delta \theta}{2} – \theta^* \big) \rightarrow 1, \, \sin \big( \theta + \frac{\Delta \theta}{2} – \theta^* \big) \rightarrow \frac{\Delta \theta}{2}.$$

Replacing with those approximations, the two equations are much simplified.

In $$\vec{t_c}$$:

$$-T(\theta) – WR \, f(\theta) \, d \theta – \frac{1}{2} WR\, p(\theta) \, d\theta^2 + T(\theta) + \frac{dT}{d\theta}d\theta = 0,$$

and in $$\vec{r_c}$$:

$$WR \, p(\theta) \, d \theta – \frac{1}{2}T(\theta) \, d\theta – \frac{1}{2}\big[ T(\theta)+\frac{dT}{d\theta}d\theta \big] d \theta + \frac{1}{2} WR\, f(\theta)\, d\theta^2 = 0.$$

Ignoring the second order terms that contain $$d\theta^2$$, the two equations can be further reduced to,

in $$\vec{t_c}$$:

$$\frac{dT}{d\theta} = WR \, f(\theta) ,$$

in $$\vec{r_c}$$:

$$T(\theta) = WR \, p(\theta).$$

### Solutions

Supplying with the friction law:

$$f(\theta) = \mu p(\theta),$$

where $$\mu$$ is the kinetic friction coefficient, the two equations finally arrive their solvable form:

$$\frac{d p }{d \theta} = \mu p,$$

$$\frac{d p}{p} = \mu d \theta,$$

which can be effortlessly solved by separating variables:

$$ln(p)=\mu \theta +C,$$

$$p(\theta)=C’ \, e^{\mu \theta},$$

$$T(\theta)=C’WR \, e^{\mu \theta},$$

where $$C$$ and $$C’$$ are constants in two forms.

Inserting the boundary conditions $$T(0)=T_0$$, we can determine the constant $$C’=\frac{T_0}{WR}$$. The solutions for the tension developed in the belt and the contact pressure at the belt-cylinder interface are

$$T(\theta) = T_0 \, e^{\mu \theta},$$

$$p(\theta) = \frac{T_0}{WR} e^{\mu \theta}.$$

The reaction force at the constraint of the right end of belt can be calculated as $$T_1=T(\frac{\pi}{2})=T_0e^{\mu\frac{\pi}{2}}$$, towards right. The reaction torque at the axel of the cylinder can be calculated as $$M=(T_1-T_0)R=T_0R(e^{\mu\frac{\pi}{2}}-1)$$, counterclockwise. The figure below plots the change of tension in the belt $$T(\theta$$) and the contact pressure $$p(\theta$$) along the belt. ## Summary

### Load approximations in curved structures

From the derivation above we can see that the contribution of friction distribution load in the tangential direction to the radial force balance is of the second order. Similarly, the contribution of contact pressure in the radial direction to the tangential force balance is of the second order. Noted, the concentrated load—the tension in the tangential direction has significant effect in both tangential and radial force balance.

We can actually generalize that in curved structures, the distributional loads over one direction has only significant contribution to the equilibrium in that corresponding direction and negligible effect in their perpendicular direction.

### Capstan equation

The capstan equation can be written as

$$\frac{T_1}{T_0}=e^{\mu \theta},$$

which is verbally stated as that the ratio of tensile forces at two ends of the rope wound on a capstan is an exponential function of the product of kinetic friction coefficient and slip angle.

In the problem above where the slip angle is $$\frac{\pi}{2}$$, if we assume the coefficient of friction $$\mu=0.3$$, $$T_1$$ is 1.6 times of $$T_0$$. If the wrap and slip angle is a full circle ($$\theta=2\Pi$$), $$T_1$$ is 6.6 times of $$T_0$$; we can see the power of exponential functions.

If we use all variables as scalars rather than vectors (by focusing on the magnitude of those variables and implying their direction by intuition), the right-hand side of the equation is always greater than 1 (when there is a nonzero slip angle). This may sometimes help us put the right tension on the right place (nominator / denominator) in calculation.

### $$\theta$$—the slip angle

We need to note that $$\theta$$ in the capstan equation is actually the slipping angle rather than the wrapping angle. Those two angles can differ significantly. We have purposefully set up our problem to make the belt completely slips on the cylinder by applying an angular velocity boundary condition, so the slipping angle equals the wrapping angle. But in other loading conditions, for example, if a torque is applied on the cylinder in this problem, the belt may only slip on a portion of the contact interface, while sticks on the other portion of the interface, depending on the magnitude of the torque. We will address this problem in the next post. 