# How much is left on a roll?

1. Aug 3, 2007

### Joe Mechanic

I am not an engineer and my math skills are not great but I have this problem that I would like to know the answer to.

We have a blister packaging process that seals labeled tyvek against formed pvc blisters that contain our product. What I would like to be able to do is figure out how many rows of labels are left on the tyvek as we switchthe rolls between product changeovers. The core of the roll is four inches. And the outer diameter of the roll is two feet. I believe each row is about four inches long.

Is there a formula that can be used to determine how many rrows are left as the diameter becomes smaller?

Joe

2. Aug 3, 2007

### Staff: Mentor

Since circumference is pi times diameter, and varies linearly, you can take the average diameter of the product minus the core diameter, squared, times the number of times it is wrapped (total thickness divided by the thickness of the sheets) to get the total length on the roll.

((Do+Dc)/2-Dc)*pi*(Do-Dc)/th = total length of product left

Do = outer diameter of the roll
Dc = core diameter
TH = sheet thickness

If I understand what you need, that answer would get divided by 4" to find the number of rows.

3. Aug 3, 2007

### K.J.Healey

Or do some weight measurements. Find out how much each piece weighs and weigh it before and after. That way for each roll you have a before weight (50lbs for say 500") and then after (5lbs). You can approximate how many are left by a simple weight reduction. The more accurately you weigh it, the better the reading. (Like the ticket-people in arcades)

4. Aug 3, 2007

### Joe Mechanic

Mr Watters,

Thank you for the quick response. I 'm envious of your math skills.

I applied the formula to a brand new roll and I was off .
In your written paragraph you said "you can take the average diameter of the product minus the core diameter, squared, times... " But the formula looks like you meant to say divided by 2 and not squared.

Here are my actual measurments from an unused roll using the formula:

Do =17.25"
Dc= 3.375"
Th=0.0085"
Space between rows 3.75"

The roll's labelling says there are 100,380 labels and I am told that's the number of rows times 12 labels per row. I may have keyed in the numbers wrong but I was off by 30,000 labels.

5. Aug 3, 2007

### LURCH

Seems like a simple counter would do the trick. Take the number of labels on a roll as a Starting number (nS), subtract the number of Blisters you produced in a run (nB) times the number of Labels on a Blister (nLB) to get the number Remaining (nR).
nR = nS - (nB*nLB)

Write the new number on the roll before you put it away, and you'll have your starting point for next time.

6. Aug 3, 2007

### Joe Mechanic

That would be easy but it doesn't work here. It's an extra step we don't need. It is easier to just calculate before returning the roll to the warehouse.

My problem is, we over estimate the number of rows left on a partial roll and install it on the machine when we should be tossing it. It is not economical to stop a production run that uses ten people for a half hour or more to replace a mostly spent roll of tyvek.

7. Aug 3, 2007

### Joe Mechanic

These rolls weigh around 100 pounds and only our floor scale can handle the load but it is located in an area that is far from its point of use and I don't think anyone would bother. That's why I'd rather rely on a formula. All of the measurements are constant except for the diameter.

8. Aug 3, 2007

### Joe Mechanic

Mr. Watters, I wrote the formula from the paragragh description and it worked out much better. Thanks for you help.

9. Aug 3, 2007

### LURCH

Perhaps I'm not understanding your process correctly; I don't see the "extra step" to which you refer. You know how many labels are on a roll before you start using it, (100,380) yes? And you must know how many packages you produce in a production run, and how many labels you put on each package?
That's what I'm suggesting. Take the number of labels you started with at the beginning of your production run, subtract the number you used during that run, and write the new number on the roll before you return it to the warehouse.
Now that I hear myself say it, it sounds a bit too simple. I am missing something, aren't I?

10. Aug 3, 2007

### K.J.Healey

They probably don't know exactly how many labels are used? Or maybe they have multiple rolls at once around the area?

11. Aug 3, 2007

### Joe Mechanic

I'll try to explain. I'm the person who has to lug these rolls back and forth so I have a vested interest in proving the economics of tossing a roll when it reaches, say an OD of 4 inches.

I want to prove to my management that it is not economical to use a leftover roll if it reaches a certain diameter by showing how little run time is left before we have to shut the line down to replace it.

Our operators do not record cycles because we're not required to. To require this info would mean a change in SOP which means a mountain of paperwork.

I want to work with indisputable givens and use them on a spread sheet. Does this make sense?

12. Aug 3, 2007

### Staff: Mentor

That was off the top of my head - I wasn't deriving with a pen and paper, so I may have done it differently each time. I may have tried to mulitply something out that didn't quite work.

Average diameter minus core diameter times pi gives you the average circumference. And total thickness divided by sheet thickness gives number of times wrapped....hmmm, I think the way I have it written in equation form is right.

Last edited: Aug 3, 2007
13. Aug 4, 2007

### NoTime

I don't understand the part in red.
The rest looks fine.

I think the OP forgot to include the thickness of the labels on the sheet and just took the thickness of the carrier.
Just a guess but I suspect TH should be about .0143

14. Aug 5, 2007

### LURCH

So there's no count, and you don't know how many parts you've run? How do you know when it's time for a changeover?
Ah, now I see more clearly. That is indeed a tall order; proving to any management-type that there are some times when it's just more economical to throw something away.
Maybe you could take a measurement of the next roll you grab that looks a bit light, and then record how long it took the line to shut down because you needed to change the roll. Do this the next few times, and show them the recordee results. Rather than presenting a calculation (even if you do it right, they might not believe that you did), you show them an historical fact; not that it would take x-amount of time, but that it does.
Of course, this means you don't have proof untill after it has happened several more times.

15. Aug 6, 2007

### Joe Mechanic

Oddly enough for a manufacturing facility, we stop when we run out of parts.
Then we go to the next product.

My management is usually responsive if you have data to support your theory.

Especially when it comes to saving money.

I was hoping to be able to measure a small roll and predict run time with math. If I were able to do this repeatedly, I think the folks who look at costs will take over and toss the rolls rather than return them to the warehouse.

Notime:
The thickness of the tyvek is 0.0085".

I tried to do this linearly by examining the givens. I know a full roll's diameter = 17.25" and an empty roll is 3.75". There are 8,365 cycles per roll so half a roll must have half the and a quarter roll had 25% of that etc. Then I took an average cyc time of four cycles per minute to figure out run times.

My co-worker, also a technician but some one who ciphers in his head rather than use a calculator, says it won't work because the equation is non-linear; the smaller circumferences contains less rows of print. I think he's right.

16. Aug 6, 2007

### BobG

Calculate the area of the side of the roll as if it were solid. With a diameter of 17.25, the radius is 8.625 and the area is about 233.7.

Calculate the area of the core. With a diameter of 3.375, the radius is 1.6875 and the area is 8.95

Subtract the area of the core from the area of a full solid roll. The cross-sectional area of your labels is about 225 square inches.

At the end of the day, calculate the cross sectional area of the roll. That's the outer radius (diameter/2) squared times pi. Subtract the area of the core (which you should have saved as a constant). Divide the cross sectional area of a full roll (which you should have saved as a constant).

The result is the percentage of the roll that's left. Multiply the number of labels on a full roll and you have a good approximation of how many labels are left.

Edit: You really don't even need pi, since it will eventually cancel out when you convert to a percentage. It just provides a better idea of where your numbers are coming from so it doesn't look like you're just pulling them in out nowhere.

Last edited: Aug 6, 2007
17. Aug 6, 2007

### NoTime

What is the thickness of the lable?
TH is the total lable + tyvek carrier.

18. Aug 7, 2007

### Joe Mechanic

The tyvek is a pre printed lable. Not a separate label attached to the tyvek.

BobG, It will take me some time to digest what you wrote. Thanks. I'll update after I've tried it.

19. Sep 13, 2007

This is what I did

I had to do this in a program recently (find the linear length of material remaining on a roll when all you know is the diameter of the product remaining and the core diameter).

The circumference of any layer in a roll is the linear length of that layer.
Total length on the roll is the sum of these lengths.

(It helps to draw this out - that's how I noticed the layer of material adds 2 times the thickness.)

For each layer in the roll, the circumference is given by:
pi * (core + 2 * thickness * layer number)
where the layer number is an integer from 1 to the maximum number of layers.
So, on a 50mm core, with a thickness of 1mm, layer 10 would have a diam. of 70mm, and 219.91 mm of product for that layer.

Total length in roll is the sum of these values. This works out to:
pi * ((core * N) + 2 * thickness * sum(i, 1, N))
where N is the number of layers. N is found by (diameter - core)/(2* thickness)

The sum is the hardest part to calculate. I do mine in a program. You could also do this in a spreadsheet.
I'm not really a math whiz, so I'm not sure if there is a shortcut way of generating that sum.
For example, if I have 25 layers, the sum is 1 + 2 + 3 + ... + 25, which gives us 325.

As an example, I did a roll with a 50mm core and a diameter of 100mm, with a material thickness of 1mm. This gives us 25 layers of material ((100mm-50mm)/(2*1mm)

Working it out Russ' way, I got a length of 3926.99 mm.
Working it out my way I got 5969.03 mm.
Brute forcing it (calculating each layer's diameter-- 52mm, 54mm, 56mm, on up to 100mm -- then calculating and adding the circumferences) I got the same answer.

Just my 2 cents.

20. Sep 13, 2007

It bugged me that I couldn't shortcut the sum from 1 to N without doing it iteratively. Bugged me alot.

So, I googled it. Turns out there is a famous story (well, not to me) about Gauss doing a sum from 1-100 in grade school in a few seconds.

For any series 1+2+3+...+N, the sum is equal to 1/2 * (N+1) * N.

D'oh! It just hit me, that's the integral of N (I think). And who says education is wasted?

Anyway, in the example I gave in my previous post, I said there were 25 layers. The sum of layers is then 0.5 * 26 *25, which is 325, which is what I got in my spreadsheet.

If the thickness were 0.1mm instead of 1mm, then the number of layers would be 250. The sum would go to 0.5*250*251, which is 31375. That would give a length of:

3.1416 * (250 * 50mm + 2 * 0.1 mm * 31,375) = 58983.54 mm

So it is easy to get an answer with a calculator and a pencil. Who knows, it may even be right.

Gotta go modify my program now...