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## Main Question or Discussion Point

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, [itex]\theta[/itex], where [itex]90^\circ[/itex] means the string is perpendicular to the central axis. The part then undergoes a rotation, [itex]\phi[/itex], that causes the string to wrap around it. Assuming the string is not slipping, it will naturally "walk up" the part until it reaches [itex]90^\circ[/itex]. (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 [itex]\phi[/itex] that will produce some walk-up. For a cone, the radius is a function of the axial distance from the initial point, [itex]z[/itex]. 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, [itex]R[/itex], and the initial distance the string attaches to

[itex] \theta_0 = tan^{-1}(R/L_0) [/itex]

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

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

[itex]

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

[/itex]

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

If I choose a small enough [itex]\phi[/itex], 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).

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, [itex]\theta[/itex], where [itex]90^\circ[/itex] means the string is perpendicular to the central axis. The part then undergoes a rotation, [itex]\phi[/itex], that causes the string to wrap around it. Assuming the string is not slipping, it will naturally "walk up" the part until it reaches [itex]90^\circ[/itex]. (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 [itex]\phi[/itex] that will produce some walk-up. For a cone, the radius is a function of the axial distance from the initial point, [itex]z[/itex]. 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, [itex]R[/itex], and the initial distance the string attaches to

*from*this ring is [itex]L[/itex], then the initial angle is:[itex] \theta_0 = tan^{-1}(R/L_0) [/itex]

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

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

[itex]

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

[/itex]

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

If I choose a small enough [itex]\phi[/itex], 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).