# Natural position of a string wrapping around a cone

1. Sep 22, 2014

### dawin

I've had a problem I encountered at work some time ago and took a personal interest in. I never did end up solving it, but I've recently looked at it again.

It goes like this:

You have an axisymmetric part, such as a cone, and it's positioned such that its central axis is coincident and parallel to the central axis of a ring.

A string comes from this ring and attaches to the tip. Viewing the from the side, the string forms an angle with the part's axis, $\theta$, where $90^\circ$ means the string is perpendicular to the central axis. The part then undergoes a rotation, $\phi$, that causes the string to wrap around it. Assuming the string is not slipping, it will naturally "walk up" the part until it reaches $90^\circ$. (I've attached a schematic)

I'd like to determine the function that defines the natural walk-up, but I think I'm getting hung up on some assumptions.

Assume a very small rotation $\phi$ that will produce some walk-up. For a cone, the radius is a function of the axial distance from the initial point, $z$. If I take the "traversed" area, and unravel the cone, I end up with a flattened frustum (I'm trying to assume a general case); I simplified this to a trapezoid with the string's path mapped between two corners (attachment 2--sorry, I'm drawing these in PowerPoint). Here's one area I think I might be going wrong, but don't see intuitively if this would really affect the angle.

I'm assuming that for a very small rotation, that path will form a straight line (I think this could be where I'm going wrong, but I'm not sure how else to define the path!).

If the large ring's radius is constant, $R$, and the initial distance the string attaches to from this ring is $L$, then the initial angle is:

$\theta_0 = tan^{-1}(R/L_0)$

If I look at my trapezoid, the arc produced by the rotation is $s_1 = r(z)d\phi$; the arc where the string ended after rotation is $s_2 = r(z + dz)d\phi$.

This change in position, $dz$ will change the "attachment" length to $L = L_0 - dz$, which changes the angle. I'd naturally want to calculate any new angle as:

$\theta = tan^{-1}(R/(L-z))$

But in my case, I keep ending up with $\theta$ as functions of $L-z$, and feel like I'm sticking myself in a loop. I tried approach it with the chain rule, trying to solve two separate relations $\frac{d\theta}{dz}\frac{dz}{d\phi}=\frac{d\theta}{d\phi}$ but feel like I'm just butchering it here.

If I choose a small enough $\phi$, sure, I can iterate this and get something I believe is close. But I'd like to know how to get to a closed-form solution. There are approaches out there for filament winding, but these account for a different payout eye setup. Plus, they assume the line follows a geodesic to ignore friction, which I don't necessarily need to do in this case (there are ways to keep the fiber in place).