Comments on: For the mathematically minded...

Question: For the mathematically minded...

tomble — Wed, 27 Apr 2005 16:39:15 -0800

Here's a question for the mathematically minded. A lot of the work I do involves rolling materials such as paper or plastic film onto rolls. Now, while I normally can guess by sight or weigh a roll to determine how much is on it, I got to wondering...

... how would one go about calculating the length of film on a roll, based on measuring from the outer edge of the core to the outer edge of the rolled film? You would need to take into account these factors - diameter of the core being wound onto, thickness of the material being wound.

Every time the film goes around the core, the core diameter is effectively increased by double the thickness of the film, and so a roll twice as thick is not twice as long.

Surely there is some elegant way of calculating such a thing? It would be nice to be able to measure a roll, enter the measurement into a spreadsheet, and know how long it is.

By: forforf

forforf — Wed, 27 Apr 2005 16:54:00 -0800

From "tee" at this webpage

A reasonably simple formula is L = pi*N*(D+d)/2 where N = (D-d)/(2*t) is the number of wraps of tape of thickness t on a roll of diameter D (when full) around a core of diameter d. The formula represents the average of two estimates using the sum of circumferences of
concentric circles (inside vs. outside diameters based on t).
The summation formula 1+2+3+...+N = N*(N+1)/2 is needed to develop each estimate.

The engineer in me would tweak this formula based on actual lengths and diameters to make it even more accurate.

By: evinrude

evinrude — Wed, 27 Apr 2005 17:00:14 -0800

I would say that the length, L, would be approximately equal to pi*(R^2-r^2)/x, where R is the "big radius" (distance from center of core to outer edge of film), r is the "little radius" (distance from outer edge of the core to outer edge of the rolled film; this would be the radious of the core being wound onto) and x is the thickness of the material. This comes from an estimate that L*x*h (the volume of the film, where h is the 'height' of the film) is approx. equal to h(pi*R^2 - pi*r^2), the volume of a cylinder with a cylinder carved out of the inside.

I could easily be wrong.

On preview, though, my formula is equivalent to the formula above.

By: OldMansHands

OldMansHands — Wed, 27 Apr 2005 17:02:23 -0800

Here's how I'd think about it -

Although the diameter changes, a roll is basically circular. So we can work out the volume taken up by the film, and then from this work out the length. (We have to assume that nothing awkward like stretching happens).

Lets say we have a roll with a central (empty) section of radius r, and overall radius (from the centre to the edge of the film) of R. Say it's w wide, and the tape is t thick.
Then the volume of tape that we have is:
(p . R^2 . w) - (p . r^2 . w)
= p . w . (R^2 - r^2)

Volume = length x width x thickness , so:

p . w. (R^2 - r^2) = l . w . t

The width cancels, and we get:

l = p . (R^2 - r^2) ) / t

If you're doing it in your head, the fact that (R^2 - r^2) = (R + r) (R - r) might be handy.

I hope that makes sense even with the formatting.

(On preview - forforf's may be easier, depending on what you can measure)

By: evinrude

evinrude — Wed, 27 Apr 2005 17:02:41 -0800

On further consideration, my formula differs from forforf's by a factor of 1/4. Er...

By: onalark

onalark — Wed, 27 Apr 2005 17:04:38 -0800

tomble, you should be able to approximate this by assuming that the film takes up the same area wrapped around a cylindrical roll as it would stacked in sheets. Does this make sense without a picture? (This also fundamentally relates to how approximations to integration in introductory Calculus work, by the way).

The formula I came up with is A[f] (area of the film) = A[D] (area of the outer edge of the rolled film) - A[d] (area of the inner edge of the rolled film). I'm assuming you know that you can calculate the area of a circle (the cylinder viewed from the side) by multiplying the constant pi against the radius squared (diameter/2)^2.

Since in a rectangular system, we know that the length of the roll * thickness equals the total area , we can use this equation to solve for the total length of the roll from the face area.

A[f]/t[f] (thickness of the film) will yield L[f] length of the film.

I'd try this equation out 'in the field' and see how accurate it is.

On preview: You can solve this system in two-dimensions without worrying about the volume, just look at it from the side.

By: onalark

onalark — Wed, 27 Apr 2005 17:06:53 -0800

My final equation is:

L = pi/t*(R^2 - r^2) where L = length, t = thickness, R = outer radius, and r = inner radius

By: gus

gus — Wed, 27 Apr 2005 17:07:57 -0800

OldMansHands seems correct to me. I think you could start off ignoring the width of the film though, - no need to calculate the whole volume, all you really need is the surface area.

To do it with a program, a good approximation would be to add up the diameter of each layer:

length = 0;
for i=insideDiameter to outsideDiameter step thicknessOfFilm
length = length + 3.14159 * i;
next i;

By: evinrude

evinrude — Wed, 27 Apr 2005 17:11:05 -0800

On further further consideration, we all seem to agree; my being off by 1/4 was because I was using the radius and not the diameter; and I would just like to point out that I just spelled radius correctly, unlike in my first post.

By: onalark

onalark — Wed, 27 Apr 2005 17:15:00 -0800

Me and OldMansHands are saying the exact same thing (though I think he gave it better treatment). Give this a try in the field and report back!

By: OldMansHands

OldMansHands — Wed, 27 Apr 2005 17:15:25 -0800

We all seem to agree, so pick the explanation that makes the most sense!

Now that I think about it, I always overestimate the amount of tape left on a roll. Obviously my brain isn't too great at doing this intuitively.

By: tomble

tomble — Wed, 27 Apr 2005 17:19:23 -0800

Wow! I think I need to ask metafilter how to choose the best answer when they all seem so good!

By: orthogonality

orthogonality — Wed, 27 Apr 2005 17:20:41 -0800

I'm really really bad at math, so this is probably wrong and I look forward to correction. But figuring from first principles, I get this:

Circumference is defined as 2? * radius (the symbol is of course Pi), or ? * diameter.

So if you wrapped the roll a single time, you have an outside length of
( Core Diameter + 2 * Sheet Thickness ) * ?

Continuing another single turn around that, you have:
( Core Diameter + 2 * Sheet Thickness ) * ? + ( Core Diameter + 4 * thickness ) * ?

On the next turn,
( Core Diameter + 2 * Sheet Thickness ) * ? + ( Core Diameter + 4 * thickness ) * ? + ( Core Diameter + 6 * thickness ) * ?

Factoring gives us:
? * ( ( Core Diameter + 2 * thickness ) + (Core Diameter + 4 * thickness) + (Core Diameter + 6 * thickness) )

or abbreviating, with d for Core Diameter and t for sheet thickness,
? * ( ( d + 2t )+ ( d + 4t ) + ( d + 6t ) )

Let's call the current number of full turns of wrapping i, and the total wraps n.

So our total sheet length would be the sum as, i goes from 1 to n of ? * ( d + 2it )

Our roll diameter, call it R is the core diameter plus the number of wrappings time the thickness times two, or
R = d + 2nt

Or n = (R - d)/2t

So the length of the roll is the sum S as i goes from 1 to (R - d)/2t, of ? * ( d + 2it )

By: gus

gus — Wed, 27 Apr 2005 17:25:43 -0800

Does anyone else feel like their calculus skills are failing them....

It should be a basic line integral, but you need to use a paramtric equation to define the path, and that seems beyond me right now.

On preview:

Ortho- that is what my little program loop above does, but it should have a step of 2*thickness.

Its pretty fun to see all of the ways that people came to the same approximation.

By: OldMansHands

OldMansHands — Wed, 27 Apr 2005 17:33:32 -0800

My calculus skills are definitely failing - I'm glad it turned out to be possible with algebra!

Have you tried the program in practice, and compared the results to the equation? I doubt it would be more efficient, but it'd be interesting to see how close it gets.

By: vacapinta

vacapinta — Wed, 27 Apr 2005 17:41:58 -0800

Then of course there's...the compressibility of the material. I know none of these methods would work for calculating how much toilet paper is on a roll for example. The best you could do there would be empirical, I think.

By: Chuckles

Chuckles — Wed, 27 Apr 2005 17:47:41 -0800

I really like onalark's solution, but I'm not sure everybody will get the description. Basically it is:

Find the function that relates running length to edgewise area.

Then it doesn't matter what shape the thing is in as long as the packing is tight. Better than a volume based solution because it is the same solution no matter how wide the roll is.

By: gus

gus — Wed, 27 Apr 2005 18:11:31 -0800

Well, I tried it out for a simple case r=2, R=5, t=1

Method A)
The formula gives: l = p*(R^2 - r^2) / t = 3.14159*(25 - 4) / 1 = 65.97

Method B)
The method of adding up the length of each layer gives:

l = sum(p * 2 * r_i) for each layer i

2 * 2 * 3.14159 +
2 * 3 * 3.14159 +
2 * 4 * 3.14159 +
2 * 5 * 3.14159

= 12.56 + 18.85 + 25.13 + 31.41 = 87.96

Which is not even close.

Method b (revised))
The problem with what I described above it is taking the inner edge of the layer, and using it find the length. In other words, the r is wrong. Also, it is adding 1 too many layers.

If you think about layer one being between r=2 and r=3, than you can use the radius r=2.5 to describe that layer. Then the next layer goes from r=3 to r=4, so you can use 4=3.5. Doing this you get:

l = (2*2.5 + 2*3.5 + 2*4.5) * 3.14159 = 65.97

which aggrees with method a.

I'm still waiting for a physics mefite to come by and tell us the calculus way to do it....

By: Zurishaddai

Zurishaddai — Wed, 27 Apr 2005 18:43:32 -0800

I'm still waiting for a physics mefite to come by and tell us the calculus way to do it....

It is a one-dimensional (plane curve) problem. Here's the one-dimensional curve that describes the film winding around in the xy plane:

? = ( (r + t?/2p) cos ?, (r + t?/2p) sin ?)

where ? is our parameter (=angle), r is the radius of the spool, and t is the thickness of the material.

Now, what tomble can measure is how many LAYERS he has wound up = ?/2p.

He wants to know the arc length of ? from t=0 to t=?. I'll work out the integral (of d?/dt) and post it in a minute.

By: Zurishaddai

Zurishaddai — Wed, 27 Apr 2005 18:45:39 -0800

Whoops, HTML character trouble there. I meant:

I'm still waiting for a physics mefite to come by and tell us the calculus way to do it....

It is a one-dimensional (plane curve) problem. Here's the one-dimensional curve that describes the film winding around in the xy plane:

γ = ( (r + tθ/2π) cos θ, (r + tθ/2π) sin θ)

where θ is our parameter (=angle), r is the radius of the spool, and t is the thickness of the material.

Now, what tomble can measure is how many LAYERS he has wound up = θ/2π.

He wants to know the arc length of ? from t=0 to t=?. I'll work out the integral (of dγ/dt) and post it in a minute.

By: Zurishaddai

Zurishaddai — Wed, 27 Apr 2005 19:17:28 -0800

OK guys, I gotta finish watching Halloween in a minute here, but I've solved the problem analytically (according to the assumptions mentioned before). I get the same answer from my integral as gus above (66.04 length units of stuff, when spool r=2, thickness=1, and you wind it around 3 times).

Renaming theta/2*pi as just N (=number of times wound around), the arc length integral was

integral from 0 to N of the square root of —
t^2 + 4pi^2(r+nt)^2
dn

The integral solution itself is too messy to post in this format. Try this in Maple or some such program:
int(sqrt(t^2+4*Pi^2*(N*t+r)^2),N);

By: plinth

plinth — Wed, 27 Apr 2005 19:36:39 -0800

Here's what I remember from doing this in Calculus: you do a messy integral and it reduces to the algebra answer. IIRC if you work in polar coordinates for the definition of a spiral, you get r = a * theta. The length of this curve is a intgeral(sqrt(1+ theta^2) d theta). If you solve the definite integral, you want to run theta from 0 to 2pi * number of turns, which is 2pi * (R - r)/t
so, that integral is 1/2 ln(theta + sqrt(1 + theta^2)) evaluated from 0 to 2pi * (R-r)/t so that's:
1/2 ln(2pi (R-r)/t + sqrt(1 + 4pi^2 (R-r)^2/t^2) - 1/2ln(1)
Now, if I haven't slopped anything up here, this should reduce down fairly nicely to exactly what we had before. I hope.

By: Galvatron

Galvatron — Wed, 27 Apr 2005 20:41:39 -0800

You slopped something up there, plinth. The equation for the spiral is r = t / (2pi) theta, for thickness t. The differential arclength is r * d theta, so this is what you want to integrate. If the inner radius is r_inner and the outer radius is r_outer, then the initial and final angles would be (2pi/t) * r_inner and (2pi/t) * r_outer. So the total arclength would be (in LaTeX notation) \int_{(2\pi / t) r_inner}^{2\pi / t) r_outer} (t/(2\pi)) \theta d\theta . This is integrated trivially to agree with the solution that various people found up above. Except it's better because it has fancy symbols.

By: Justinian

Justinian — Wed, 27 Apr 2005 20:49:04 -0800

I've devised an elegant solution to this problem but the proof is too long to fit in this margin.

By: exceptinsects

exceptinsects — Thu, 28 Apr 2005 14:11:40 -0800

Hm. Someone else was just wondering this very same thing.