# Label Rolls

1. Dec 12, 2009

### mlawrence2102

Not sure whether or not I have posted in the correct forum or not, anyway this is my problem.

At work we have no way of estimating how many labels are left on a roll before returning them to the warehouse. the average roll diameter is 8" with a 3.5" core diameter and the roll web has a single strip of labels per reel - label diameter is 20mm. Is there a mathematical calculation out there which could help me estimate how many labels are left on a part used roll simply by measuring the diameter?

2. Dec 12, 2009

### Danger

Welcome to PF, Mr. Lawrence.
I'm afraid that I can't help you with a formula, even though it shouldn't be too difficult. Math is not my thing.
A couple of questions come to mind, though. How much advance notice do you need of this thing running out? What is your wastage allowance for the labels?
The reason for the first question is that my first thought involves bouncing a light (maybe a laser) off of the roll to a photodetector. The core should be far less reflective than the labels, and thus will return a null signal when the last label exits.
The second question is based upon my other idea, which is to have a cantilevered probe that rides upon the surface of the roll in order to track its radius. Depending upon the accuracy, though, this could result in an 'empty' signal being recorded when there are still a couple of dozen labels on the roll.

3. Dec 12, 2009

### mlawrence2102

Running out of labels during a production run is not the problem... there is an early warning which signals this when the end of the roll is near. The problem is returning the wrong qty of labels to the warehouse at the end of a run, consequently when the labels are to be used again in the future we sometimes have insufficient to fulfill an order due to the incorrect qty showing in the system from previous wrong qty returns. Upgrading the machines is out of the question so I am hoping to find a mathematical formula which will enable me to predetermine the qty's onto a quick reference matrix.

4. Dec 12, 2009

### Danger

Ahhh...
Sorry that I misunderstood the problem. I must leave this in the hands of others. Should it occur that a mechanical contrivance of some sort is the proper way to go, I'll dive back in. Meanwhile, I'll just check this thread once in a while to see how it's going. Good luck with it. I'm sure that someone here can help you.

5. Dec 12, 2009

### cepheid

Staff Emeritus
The key problem seems to be to determine the total length of the remaining label reel, which, I'll call 'L.' If you knew L, then you would be able to determine the number of labels remaining, because if you assume that each label has a length of $\ell$, then the number of labels 'n' on a reel of length 'L' is just :

$$n = \frac{L}{\ell}$$​

or, in words:

$$\textrm{no. of labels} = \frac{\textrm{total length of reel}}{\textrm{length \textit{per} label}}$$​

The tricky part is how to estimate the length of the reel, L, when it is wound into a tight spiral around the spool. I'm taking an approach that involves approximating it as series of concentric circles of increasing radii, and adding up the circumferences of each of the circles to get an overall length. So, a key thing to determine is the number of circles, i.e. the number windings, "N" around the spool. An easy way of doing this is to notice that each winding has a thickness equal to the paper thickness, 't'. So, the total radius R of the reel should be equal to the radius of the spool plus the number of paper thicknesses added onto it by the windings:

$$R = 3.5^{\prime\prime}/2 + Nt$$​

Now, the circumference of a circle of radius R is equal to $2\pi R$. So you can add up the circumferences of all of the individual windings to get the total length of label strip. I'm going to say that the radius of the first winding is approximately equal to the radius of the spool. The radius of the second winding is approximately equal to the radius of the spool plus the one paper thickness of the first winding. Generalzing this, the radius of the "ith" winding is equal to the spool radius plus the i-1 paper thicknesses of the windings that came before it. Mathematically, we write:

$$L = \sum_{i=0}^{N-1} 2\pi (3.5^{\prime\prime}/2 + it)$$​

If this looks daunting or if you aren't familiar with the summation notation, don't worry. All it means is the following:

$$L = 2\pi (3.5^{\prime\prime}/2) + 2\pi (3.5^{\prime\prime}/2 +t) + 2\pi (3.5^{\prime\prime}/2 +2t) + 2\pi (3.5^{\prime\prime}/2 +3t) + \ldots + 2\pi (3.5^{\prime\prime}/2 +(N-1)t)$$​

The nice thing is, the manufacturer probably tells you how many labels are in a full roll. You know the length of a full roll, because you can multiply the number of labels by the length per label. You also know the radius of a full roll. Therefore, you can estimate the number of windings in a full roll. What this means is that you can sort of "reverse engineer" this calculation in order to determine the paper thickness (rather than trying to measure it, which might be subject to a lot of error). Then you can go ahead and apply that to your partial roll. In fact, with these starting parameters, it may even be possible to eliminate the paper thickness entirely get a result that depends upon the ratio of the radius of a partial roll to the radius of a full roll.

Last edited: Dec 12, 2009
6. Dec 12, 2009

### mlawrence2102

Fascinating read... although totally bewildering to someone of my mathematical ability. Would it be possible to input this formula into an excel spreadsheet as a quick calculation tool?

7. Dec 12, 2009

### eachus

Hmm... Better to take a simpler approach. You know the number of labels per roll, N and the intial and final roll diameters, let's call them I and F. Assuming the labels have a constant volume, v, we can solve for the volume using a full roll. Then you just have to measure either the diameter of the partial roll or the thickness of the labels remaining to get an answer.

Ready? We are given I = 8" and F = 4.5" I'll assume N is 1000. The volume (actually edge area since we are assuming one label wide rolls) is Pi/4*(I*I-F*F) = Pi/4*(64-20.25) = Pi/4*43.75 = 34.36 sq. inches.

v (small v) of a label (again figuring 1 label wide roll)l is 34.36/1000 = 0.03436 sq. inches. Plugging this back in, we have R the number of remaining labels = V/0.03436

What is V (big V)? Current diameter (C) squared, minus F squared times Pi/4. Since the Pi/4 can be canceled out, lets use V' = V/(Pi/4) and v' = v/(Pi/4) = 0.04375 sq inches. For a roll that is currently 5 inches in diameter (1/4") thickness of remaining labels, we have (C*C-F*F)/v' = (25-20.25) /v' = 4.75/0.04375 = ~109 labels left. Probably what you want to do is make a table up converting either the roll diameter or the thickness of the remaining labels (C = F + 2r, where r is the remaining thickness).

Then the only thing left to do is teach the operators how to measure with calipers, or better just create a template of concentric cirles labeled with the remaining number of labels. (If you have a good printer, you should be able to just print out the template, but beware of printers that scale by 97% or 102%. Don't ask me why they do that.)

8. Dec 12, 2009

### cepheid

Staff Emeritus
EDIT: went through and converted things that were supposed to be diameters into their corresponding radii.

Well, I didn't quite finish the problem, so I don't think that that formula is particularly useful to you. I worked on it some more, and I figured out that you can eliminate both the paper thickness and the number of windings, and get a result that depends only upon the ratio of the radii of a partial reel to a full reel. Here's how you do it (I've used the subscript '0' or 'naught' to represent quantites in a full reel):

Let $R_s$ = (3.5/2)" be the spool radius, $n_0$ be the number of labels in a full roll (which is given by the manufacturer to within some guaranteed error...we hope). Let $R_0$ = (8/2)" be the radius of a full reel, so that a full "ring" has a width of $(\Delta R)_0 = R_0 - R_s$ = (4.5/2)" in this case. Let $N_0$ be the number of windings in a full roll. Using the formula I derived earlier, we can actually figure out both the paper thickness and the number of windings in full roll. First, the paper thickness:

$$t = \frac{(\Delta R)_0}{N_0} = \frac{2.25^{\prime\prime}}{N_0}$$ ​

Using the formula I derived earlier, and applying it specifically to a full reel:

$$n_0 \ell = \sum_{i=0}^{N_0-1}2\pi(R_s + it) = 2\pi R_s \sum_{i=0}^{N_0-1}1 + 2\pi t \sum_{i=0}^{N_0-1} i$$ ​

The nice thing is that both of these sums can be evaluated without actually manually adding up each of the 'N' terms. The answers are: $\sum_{i=0}^{N_0-1}1 = N_0$ and $\sum_{i=0}^{N_0-1}i = (1/2)N_0(N_0 -1)$. Substituting in those answers, as well as the expression for t, the result is:

$$n_0 \ell = N_02\pi R_s + \pi \frac{(\Delta R)_0}{N_0} N_0(N_0 -1)$$ ​

This equation can be rearranged to solve for $N_0$, the number of windings in a full roll, with the result that:

$$N_0 = \frac{n_0 \ell + \pi (\Delta R)_0}{2\pi R_s + \pi (\Delta R)_0}$$​

So, now we know both the number of windings in a full reel, AND the paper thickness. A neat trick you can use is that it must always be true that the difference between the inner and outer radii of a label reel is equal to the number of windings times the paper thickness. This is true for both a full roll and a partial roll of radius R and number of windings N. Therefore:

$$t = \frac{(\Delta R)_0}{N_0} = \frac{\Delta R}{N}$$​

which leads to the result that:

$$\frac{\Delta R}{(\Delta R)_0} = \frac{N}{N_0}$$​

What this formula says is that if the radial width of the reel decreaes by a certain factor (compared to a full reel), then the number of windings will decrease by that same factor. For example, if the partial reel has decreased in width by a factor of 2 (so that is it now 1.125" instead of 2.25"), then the number of windings will also be half as many. Practically speaking, what this means is that you don't have to worry about the number of windings in a partial reel at all. All you have to do is measure the width. You'll see that below. We finish the problem by going back to the equation I derived for the length of a reel, only this time we apply it generally to some partial reel of radius R and number of windings N (so that the subscript 'naughts' are gone):

$$n \ell = N2\pi R_s + \pi \frac{(\Delta R)_0}{N_0} N(N -1)$$ ​

Using the relationship between the number of windings in a partial reel vs. a full reel, we can rewrite this as:

$$n \ell = \frac{\Delta R}{(\Delta R)_0}N_0 2\pi R_s + \pi \Delta R \left(\frac{\Delta R}{(\Delta R)_0}N_0 -1\right)$$

$$n = \frac{1}{\ell} \left[\frac{\Delta R}{(\Delta R)_0}N_0 2\pi R_s + \pi \Delta R \left(\frac{\Delta R}{(\Delta R)_0}N_0 -1\right)\right]$$ ​

This is the formula you would want to put into Excel.

Last edited: Dec 12, 2009
9. Dec 12, 2009

### cepheid

Staff Emeritus
eachus,

Your approach is clearly simpler, but I am really excited, because it provides an awesome double check. Using the same numbers, I got an answer remarkably close to yours! There are some conversions between your notation and mine. Most notably:

$$v = \ell t$$

Choosing a typical paper thickness of 0.1 mm, and using the v you calculated, I obtained a label length of ~ $\ell$ = 8.73". Obviously that's unrealistically large, but that's your fault for picking a total number of labels in a full roll that was unrealistically low. :tongue2: (just teasing). Note also, that, you made a minor mistake. The diameter of the spool is 3.5", not 4.5". Oh well. It doesn't matter -- I used your value to check.

Using the parameters:

$$\ell = 8.73^{\prime \prime}$$

$$(\Delta R)_0 = 1.75^{\prime \prime}$$

$$\Delta R = 0.25^{\prime \prime}$$

$$R_s = 2.25^{\prime \prime}$$

I calculated a value for the number of windings in a full roll of

$$N_0 = 445.05$$

Then, plugging this all into my final equation in my previous post, I obtained answer of:

$$n = 108.62$$