Wrapping a Ribbon Around a Cone

jonny_d · Nov 15, 2018

Greetings everybody. This is my first post and I am looking for help with a little math/geometry/engineering problem. This has been a real brain buster for my colleague and I the past couple days so I am hoping somebody can help. I am not sure if this is the best section for it, but it applies to an engineering design I am working on so I will put it here...I need to come up with a math model for the following problem:

A ribbon is wrapped around a cone. The cone rotates around it's axis as a machine traverses down the length of the cone (parallel to the cone axis) while applying even tension through the cross section of the ribbon. The ribbon material is of high modulus and it can be assumed to have no compliance except in the direction that it bends on to the surface of the cone. Also the ribbon must lay flat on the surface of the cone with no wrinkling or buckling of the material.

Here is a quick sketch of the problem:

Through experimentation and intuition I have determined that the wrap angle (theta) and the pitch will change as the wrap progresses. Specifically, the wrap angle will increase and the pitch will decrease as shown in the image. So for the machine to work, it will need to change the angle that it grips the material and compensate for the changing pitch as it translates along the length of the cone.

For a math model to represent the system I am thinking that it will need the following as I/O:
Inputs: cone taper angle (alpha), cone base diameter (D), initial wrap angle (theta), delta L
Outputs: pitch and wrap angle at the given delta L

I would also be interested in any advice on how to accurately model this in CAD in such a way that the ribbon is not deformed. I have access to SolidWorks and Catia.

Final note: the cone angle in my sketch is greatly exaggerated, and in actuality it will be a tapered cylinder, but I am considering it a cone here for simplicity. The actual taper is less than a degree over a length of 40 feet. This may sound insignificant but experimentation has shown that it truly matters for the application.

So does anyone want to take a crack at this? Thanks in advance!

Tom.G · Nov 15, 2018

I once worked (decades ago) at a plant that made machines to wind roving onto Helicopter rotors. So your problem isn't especially new, nor does it seem easily findable on-line. Anyhow, here is a slim lead. Hope it helps.

Excerpt from https://www.nap.edu/read/1824/chapter/7#83
Filament winding processes introduce residual fiber stresses in the fiber matrix assembly that are retained through cure. Each of these phenomena can result in local residual stresses of significance in the matrix. Procedures for computing these stresses were developed under U.S. government contract and reported by Rai and Brockman (1988); thus, the code is available on request.
(bold added)

Found with https://www.google.com/search?&q=wrapped+helicopter+blades 3 250 000 hits, this was at the bottom of the first page of results.

Cheers,
Tom

256bits · Nov 16, 2018

jonny_d said:

So does anyone want to take a crack at this?

Interesting problem, especially when one considers that a cone can be made from a flat sheet of paper.
And yet pasting flat ribbons onto the curved cone does present problems.

so what we can do is flatten out our cone and draw ribbons on that flat space.
One a sheet of paper, draw a large circle of radius R.
Draw one radius R₀.
Draw another radius R such that the area between the two R's represents the area of the cone flattened out.
Draw more R's, to represent more cones.

Now drawing parallel lines of width w across the circle will represent ribbons of width w that can be pasted flatly onto the cone.
If you draw two more circles of radii R₁ and R₂ to represent a tapered cylinder, one can see how the ribbons can be wrapped. Say for example, starting at circle R₁, the narrow end, parallel lines initially perpendicular to the radius, one can see how many wraps go around the cone to reach R₂, and the changing pitch as the ribbon moves from the smaller end to the larger. Some equation should follow.

256bits · Nov 16, 2018

here is what I mean.

Tom.G · Nov 16, 2018

Consider that you are requiring the ribbon to be parallel to the cone surface at all points. This means that the ribbon path will have a component of winding direction that is parallel to the cone axis. According to the article linked below, the winding pitch is calculable based on the cone apex angle and that it will be shown later; I haven't found "later" though.

See Problem 2.2 (about 40% down the page) at: http://pi.math.cornell.edu/~dwh/books/eg99/Ch02/Ch02.html

Cheers,
Tom

p.s. This thread may get better answers in the Mathematics forums. Here is paging @mfb to perhaps move it.

256bits · Nov 16, 2018

Tom.G said:

it will be shown later

We could start with a conical helix.
http://www.mathcurve.com/courbes3d.gb/heliceconic/heliceconic.shtml
Or a conical spiral, but that would work only for laying a string upon the cone with a particular constant pitch.
http://www.mathcurve.com/courbes3d.gb/spiraleconic/pappus.shtml

For laying a ribbon, the geodesic has to be involved for the ribbon to lay flat.
For a general surface,
http://www.mathcurve.com/courbes3d.gb/lignes/geodesic.shtml
which does not give any intuitive sense of what the required straight lines would look like if applied for a cone.

Note that the geodesics of a cylinder or a cone, and more generally of developable surfaces are the curves that transform into lines when the surface is applied onto a plane

, which is exactly what Post #3 and #4 are explaining.

Instead we go to the cone of revolution,
http://www.mathcurve.com/surfaces.gb/conederevolution/conederevolution.shtml
which gives an equation for the geodesics.

mfb · Nov 17, 2018

Tom.G said:

p.s. This thread may get better answers in the Mathematics forums. Here is paging @mfb to perhaps move it.

The mathematics part looks nicely covered already, the engineering questions are still open.
I put a redirect in the mathematics section, I think the main thread is fine here.

Tom.G · Nov 17, 2018

Although you may have already addressed it, another fine point just came to mind.
Even with the traveler moving parallel to the cone axis, ideally the throat of the traveler feeding the ribbon should be parallel to the cone surface. Note that this is a rotation in two dimensions, albeit by less than one degree.

Baluncore · Nov 19, 2018

I do not believe you can wind a straight tape or ribbon onto a rigid cone without distorting the ribbon. Consider a small square of ribbon glued onto the surface near the thick end of the cone. Extrapolate two opposite sides of that square as lines along the conical surface. They will diverge which means the ribbon cannot fit the conical surface.

256bits · Nov 19, 2018

Baluncore said:

I do not believe you can wind a straight tape or ribbon onto a rigid cone without distorting the ribbon. Consider a small square of ribbon glued onto the surface near the thick end of the cone. Extrapolate two opposite sides of that square as lines along the conical surface. They will diverge which means the ribbon cannot fit the conical surface.

That is not correct.
A sphere, a 3d surface will not accept squares lain upon its surface.
Since the cone is a 2d curved flat surface it can accept.

There can be parallel lines marked upon the surface of a cone.
Only method is using the geodesics for the ribbon to lay flat, in which case the ribbon runs out of cone quite quickly.
Or, a strip of ribbon lain longitudinally, in which case the strips will have to be tapered for no overlap.

If one uses overlap while wrapping the cone, then the conical spiral might suffice if the thickness of the warp is sufficient.

Baluncore · Nov 19, 2018

256bits said:

There can be parallel lines marked upon the surface of a cone.

But the two parallel lines will be of different lengths, which must distort or skew the tape.

OmCheeto · Nov 19, 2018

Baluncore said:

But the two parallel lines will be of different lengths, which must distort or skew the tape.

You can do what I did, and actually make one, and see that 256bits is correct.
It just won't look anything like the doodle in the OP.
I didn't understand posts 3 & 4 until I doodled my own flat cone.

Quite boring, as the ribbons don't even make it around the cone even once!
But then I realized that if I doodled in extra equally spaced radial lines, I could get the ribbon to circle the cone/layers of cone quite as often as I like.

In other words, in post #4, the cones labeled 1 thru 6 aren't separate cones, they are the same cone. Layers, if you will. The radial lines represent where the cones/layers line up.

I will have to work out the mathematical solution for myself though, as the references 256bits has provided are a bit noisy.
Doesn't really look that hard. And I won't have to learn a bunch of fancy shmancy math words and phrases: directrices? "principal radii of curvature"? [from the "cone of revolution" link]

Fun problem!

votingmachine · Nov 20, 2018

It might be productive to not use a circular cone but to use a pyramid shape as the basis for calculating the angles. I can wind a ribbon around a triangle. each time I flip across an edge, I fold with a fairly defined symmetry around that edge. My approaching two edges hit the triangle edge and wrap, forming a "V" shape. The "V" gets smaller with each successive fold. (I believe the change in pitch is half the internal triangle end angle ... but I'm playing around not being careful).

It seems straightforward for the triangle to calculate the change in the pitch angle, relative the the angle presented by the triangle. I have no idea if the calculation gets worse with a triangular pyramid, or a square-based pyramid, or a higher number of sides one. I doubt you need a very high number of sides to compensate for the angle.

I'm probably simplifying too much, but the problem of wrapping a ribbon around a cylinder is not much different from wrapping a ribbon around a long rectangle. And you say the angle is small between your "cone" and a cylinder. You might be able to just calculate the angular drift from the case of the triangle, and be close enough.

And of course if you use a very large number of sides for the base of the pyramid, the accuracy can be arbitrarily high.

256bits · Nov 21, 2018

Baluncore said:

But the two parallel lines will be of different lengths, which must distort or skew the tape.

It is tricky to visualize.
But take a sheet of paper and make a cone out of it - that you agree should be possible.

Undo the paper and cut out, or mark, a square.
Make a cone out of that paper with the cut out square showing.
The lines must continue to be parallel.

Maybe even better is to use some squared graphing paper.

But, you really are right in a way, ( ie I did say incorrect but I should have said with reservation because )
If one draws a two circles around the cone with the centre being at the axis, those two lines should be parallel, but curved, ( ias one circle is longer than the other. )
( Parallel curved lines ?? - equidistant along the length is probably a better description )
Take a paper coffee cup and unwrap it.
Even though the cup when on the table sits nice and sturdy, one can see that the top and bottom are actually curved when flattening it out.
So a ribbon, since it cannot follow those curved lines has to work its way up the cone ( narrow end to larger end ) to remain flat on the surface.

256bits · Nov 21, 2018

OmCheeto said:

Quite boring, as the ribbons don't even make it around the cone even once!
But then I realized that if I doodled in extra equally spaced radial lines, I could get the ribbon to circle the cone/layers of cone quite as often as I like.

In other words, in post #4, the cones labeled 1 thru 6 aren't separate cones, they are the same cone. Layers, if you will. The radial lines represent where the cones/layers line up.

I will have to work out the mathematical solution for myself though, as the references 256bits has provided are a bit noisy.

Fun problem!

I admit, my picture is not self explanatory - thanks for the better description.
Quite boring?? And surprising too.
The ancient Greeks must have had a whole lot of time on there hands, as they just loved geometry, angles, and all this fun stuff to the max.
It really is a mind bender of sorts for cones and their spirals.

256bits · Nov 21, 2018

Tom.G said:

Although you may have already addressed it, another fine point just came to mind.
Even with the traveler moving parallel to the cone axis, ideally the throat of the traveler feeding the ribbon should be parallel to the cone surface. Note that this is a rotation in two dimensions, albeit by less than one degree.

One engineering problem for the OP to consider.

OmCheeto · Nov 21, 2018

256bits said:

It really is a mind bender of sorts for cones and their spirals.

I've gotten started on the problem, and have run into a few realizations/problems/constraints already.
This is not a trivial problem.

Tom.G said:

Even with the traveler moving parallel to the cone axis

This makes absolutely no sense to me in three dimensions. But I think I understand what you meant, in the context of the discussion.
File it under constraints and realizations.

Constraint: The ribbon cannot intersect with the apex of the cone.
Realization: ~~The problem is trivial if the ribbon runs parallel to the 2D cone axis. But if not, it can be rectified by rotating the 2D cone.~~
I'm not sure how to describe this properly, without a bunch of doodles and maths.

ps. Someone remind me in about 4 hours to take a break from this problem, and start working on cooking my turkey.

Baluncore · Nov 21, 2018

I reduced the problem to a long thin cone, from the apex, infinitely extended. Included apical angle = alpha.
That conical surface can be mapped to 2D, drawn with the apex at the origin. One side of the seam lies along the x-axis, the other side of the seam is a diagonal line from the origin at an angle beta, to the x-axis. beta = Pi * alpha.

Now draw the ribbon at an angle (heading) gamma from the x-axis, to contact the diagonal seam line.
Transfer the seam contact points back to the x-axis, using arcs centred on the origin (= distance from apex).
Change plane heading of the ribbon by adding beta to gamma, then draw the next turn on the new heading.

Repeat until gamma directs the ribbon onto a heading that remains forever between the seams;
Or until the ribbon covers the apex at the origin, when the two sides of the ribbon can take quite different paths.

A double cone must then be considered, the path of the ribbon from one cone side to the other must pass through the apex point. The ribbon section must then be a tightly rolled spiral, with the sides of the spiral separated by zero in space but by the width of the ribbon in the spiralized flat surface world.

jrmichler · Nov 22, 2018

Baluncore said:

I reduced the problem to a long thin cone, from the apex, infinitely extended. Included apical angle = alpha.
That conical surface can be mapped to 2D, drawn with the apex at the origin. One side of the seam lies along the x-axis, the other side of the seam is a diagonal line from the origin at an angle beta, to the x-axis. beta = Pi * alpha.

Now draw the ribbon at an angle (heading) gamma from the x-axis, to contact the diagonal seam line.
Transfer the seam contact points back to the x-axis, using arcs centred on the origin (= distance from apex).
Change plane heading of the ribbon by adding beta to gamma, then draw the next turn on the new heading.

Repeat until gamma directs the ribbon onto a heading that remains forever between the seams;
Or until the ribbon covers the apex at the origin, when the two sides of the ribbon can take quite different paths.

AHA! So simple after it's solved. So I made a model cone 10" along the side (diagonal measurement) and 2" diameter at the base. The included angle of the flat pattern is 36 degrees. I wrapped a ribbon around it starting 2" from the tip.

ci

I then cut it open:

then c

It looks like the CAD model, with some allowance for not getting the ribbon started straight, nor wrapping it perfectly flat:

The next step is to develop the math model requested in the OP. It looks like the wrapping head angle changes by a fixed amount per revolution, which leaves the winding pitch per revolution to be calculated. Or is the ribbon wide enough that the wrapping head could be mounted on a swivel so that it tracks automatically?

Baluncore · Nov 22, 2018

Yes, each line does change direction by the same angle, my beta. With the cone surface seam along the x-axis, a simple mechanical ribbon feed will come from somewhere on the x-y plane. By drawing those ribbon centre lines on the surface to infinity we can see where they all meet, which will not be a point but more probably a caustic of some sort, if that is the right word for the silhouette of a line of tangents.

The ribbon can then be fed past a circular spool, mounted with axis vertical parallel to z-axis, centred at a position in the x-y plane), with a diameter and circumference that approximates the caustic.

There will be a 90° twist in the ribbon between the spool and the cone, a twist that may help or hinder the winding. Such a mechanical system will not handle the complexity of the spiral counter-twist near the apex, nor the situation where the ribbon is close to parallel with the seam.

As a general problem it is understood. But for a specific engineering solution we need to know the range of cone dimensions, the ribbon width and the range of the cone coverage required. Only then can the optimum linkage or mechanism be synthesised.

votingmachine · Nov 22, 2018

Baluncore said:

Yes, each line does change direction by the same angle, my beta. With the cone surface seam along the x-axis, a simple mechanical ribbon feed will come from somewhere on the x-y plane. By drawing those ribbon centre lines on the surface to infinity we can see where they all meet, which will not be a point but more probably a caustic of some sort, if that is the right word for the silhouette of a line of tangents.

The ribbon can then be fed past a circular spool, mounted with axis vertical parallel to z-axis, centred at a position in the x-y plane), with a diameter and circumference that approximates the caustic.

There will be a 90° twist in the ribbon between the spool and the cone, a twist that may help or hinder the winding. Such a mechanical system will not handle the complexity of the spiral counter-twist near the apex, nor the situation where the ribbon is close to parallel with the seam.

As a general problem it is understood. But for a specific engineering solution we need to know the range of cone dimensions, the ribbon width and the range of the cone coverage required. Only then can the optimum linkage or mechanism be synthesised.

I believe that twist can be used to take out the angular shift mentioned in post #8:

Tom.G said:

Although you may have already addressed it, another fine point just came to mind.
Even with the traveler moving parallel to the cone axis, ideally the throat of the traveler feeding the ribbon should be parallel to the cone surface. Note that this is a rotation in two dimensions, albeit by less than one degree.

If you can make the 45-degree idler roll that the ribbon wraps around on a floating pivot, tension should make it compensate for the 2nd rotation.

jonny_d · Nov 23, 2018

Excellent discussion guys. Thank you all for diving into this fun problem.

We did come up with a solution that I would like to share:

After researching how to model the problem we stumbled across an image that showed a cone rolling over a line on a flat plane. It also showed how this line would transfer to the surface of the cone. This made us realize that we could imagine our system with the ribbon laid out on the floor and the tapered cylinder rolling over top of it.

To start, let's see how we can calculate the pitch of a ribbon wrapped around a normal cylinder. It is quite simple:

Now to frame up the problem as a tapered cylinder rolling over the ribbon:

You will see that I have extended the surface of the tapered cylinder to an apex. Knowing that as it rolls it will "roll around" the apex, you can get l2 (arc length) and then theta. Also to explain l2: it is the arc that is imprinted on the floor by the small diameter of the cylinder as it rolls. So it can be calculated from the circumference of the small end and the number of revolutions. If you actually want to do this for a cone it makes this even simpler.Now to derive theta (the change in wrap angle). Sorry for being too lazy to type this up:

Finally, we can form an equation for pitch on a tapered cylinder and summarize the model:

I didn't type out the derivation for pitch, but it is pretty simple. Just form an equation for diameter at any point along a tapered cylinder given the starting (large) diameter, distance traveled down the cylinder, and taper angle. Plug this in for diameter in the basic pitch equation. Also, knowing that the wrap angle changes, we need to use gamma (which equates to the initial plus the change in wrap angle) for the angle in the denominator. One thing to note is that the angles developed here are relative to the surface of the tapered cylinder. It still remains that the machine (or as someone called the "traveler") will be traversing parallel to the cylinder axis. However the taper angles I am dealing with are so slight that any error I get in these calculations might be negligible. Either way if it does matter it would be simple to take the projection of the angles onto a plane that passes through the axis and modify the equations.

That is pretty much it. Any feedback would be appreciated!

Baluncore · Nov 27, 2018

Ignoring special cases, the centre line of a flat ribbon wound towards the apex around a cone will continually turn until it winds back away from the apex. The closest point reached to the apex is critical to designing a simple mechanism that would feed ribbon to the cone. It appears that the ribbon can be wound onto a rotating cone, towards the apex, by using a fixed circular guide centred on the apex. That does not require directional control of the ribbon once the winding has begun.

PF will triple the data size and blur it beyond reading but if you can see the attached cone_caustic_circle.png

The yellow lines show the ribbon centre line on the flattened cone surface as it moves towards and then away from the apex. The blue extensions to the yellow form a caustic that is close to the circle drawn in red. I have not proved that the caustic is a circle in all cases. The blue lines are each separated by an angle beta, so the points tangent to the circle that form the caustic must also be separated by beta.

If the ribbon is not guided by the circle, but is placed by a swept radius arm, that arm must be swept at a linear rate as the cone turns. Because the ribbon direction gamma changes by a fixed angle beta for each turn, the sweep of the radius arm can be driven by rotation of the cone. The drive ratio will need to be alpha * Pi for each turn of the cone.

I find the system easiest to consider with the cone apex at the origin, with a directrix or the surface seam along the x-axis. The axis of the cone is then alpha/2 below the x-axis, in the x-z plane.

The minimum generating equations I used are;
Parameters;
alpha = included apical angle of cone = aperture.
beta = Pi * alpha = unrolled surface area, angle at apex.
gamma = direction of ribbon centre line relative to x-axis direction.
r = distance from apex to ribbon centre line on the surface seam.

Given r and gamma at the start of a turn, for each turn around the cone;
Solve for point x,y at end of this turn using;
y = r * tan(gamma) * tan(beta) / ( tan(gamma) - tan(beta) )
x = y / tan(beta)
Compute seam crossing distance at end of this turn = start of next turn, using;
r = Sqrt( x*x + y*y )
Compute direction heading of next turn from r relative to x-axis using;
gamma = gamma - beta

iakc · Aug 10, 2023

I came across this post in exploring what it might take to wrap a baseball bat from end-to-end without overlap.

Out of curiousity, does this help?

Assumptions:

The baseball bat's handle and barrel are represented as a cylindrical section.
The ribbon starts at the bottom of the handle and wraps around the cylindrical section.

Parametric Equations:

x(u, v) = R * cos(u)
y(u, v) = R * sin(u)
z(u, v) = v * H / (2 * pi)

(u, v) parameters.
R, radius of the cylindrical section (varying along the length of the bat).
H, height of the cylindrical section.

Presume the (u, v) parameters define a point (x, y, z) on the ribbon's path. As u varies from 0 to 2 * pi, the ribbon wraps around the cylindrical section. As v varies, it controls the height of the ribbon above the base of the cylindrical section, creating the helical wrap.

WDYT?

Baluncore · Aug 10, 2023

iakc said:

WDYT?

Welcome to PF.

The length of tape required will be the surface area of the bat, divided by the width of the tape.
Since the section changes along the bat, to avoid gaps or overlap, the tape will need to be slightly elastic.

Why do you define the helix of the tape?

iakc · Aug 10, 2023

Baluncore said:

Welcome to PF.

Thank you! :)

Baluncore said:

The length of tape required will be the surface area of the bat, divided by the width of the tape.
Since the section changes along the bat, to avoid gaps or overlap, the tape will need to be slightly elastic.

Agreed; might work with electrical tape. But unfortunately, in my case, I was considering very thin paper and decoupage medium.

Baluncore said:

Why do you define the helix of the tape?

Partially inspired by @256bits from their earlier post, plus wanting to create something parametric. Instead of wrapping the object with a single continuous object, you'd use a series of separate ribbon-like shapes along the path defined by the parametric equations.

Alas, I don't think my approach was correct. This was as far as I got:

SCAD:

// Parameters
R = 10;        // Radius
H = 100;       // Heigh
resolution = 100; // Smoothness
ribbon_width = 0.5;

// Parametric equations
module ribbon(u, v) {
    x = R * cos(u);
    y = R * sin(u);
    z = v * H / (2 * PI);
    
    translate([x, y, z]) {
        // Placeholder ribbon geometry approximating a baseball bat shape
        difference() {
            cylinder(h=2, r1=0.5, r2=0.5);
            translate([0, 0, -0.25]) cylinder(h=2.5, r1=0.3, r2=0.3);
        }
    }
}

// Generate ribbon path
module ribbon_path() {
    for (i = [0 : resolution - 1]) {
        u = 2 * PI * i / resolution;
        v = i / resolution;
        ribbon(u, v);
    }
}

// Display ribbon path
ribbon_path();

So, clearly not what we were going for. Apologies. :)

Wrapping a Ribbon Around a Cone

Attachments

Attachments

Attachments

Attachments

Attachments

Attachments

What is the purpose of wrapping a ribbon around a cone?

What materials do I need to wrap a ribbon around a cone?

How do I wrap a ribbon around a cone?

What are some tips for wrapping a ribbon around a cone evenly?

Are there any other creative ways to wrap a ribbon around a cone?

Similar threads

Hot Threads

Recent Insights