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Calculating the amount of labels left on a reel.

  1. Jun 27, 2011 #1
    I have had a look on the forums and couldn't find this question answered in a way i understood. I apologise in advance it is a repeat question.

    At work we use reels of wine labels in a production environment. Every reel has a different starting quantity (fresh out of the box) and different labels have different width,height...

    The problem we have is estimating the quantity of labels left on each reel once production has used some of them.

    I understand the VERY BASICS of what i need to do but have been unable to find a formula that works this far.

    SO as i understand it i need to know:

    diameter of the core
    diameter of the reel
    width of the label
    width of the space between each label
    thickness of the label and backing paper combined

    Any help would be greatly appreciated, my math skills are limited so simplicity would be great.
    and if the was a formula available that could be used in execl it would make my life a whole lot easier.
    Many Thanks
     
  2. jcsd
  3. Jun 27, 2011 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF.

    Are these reels tightly wound, even when partially used? Unless they are tightly wound, it may be hard to get a good estimate.

    Can you maybe use the weighing method instead? If you know the tare of the bare reels, you should be able to figure out what percentage of the paper stuff is left versus the weight of a full reel...
     
  4. Jun 27, 2011 #3
    Hi,

    Thanks for the speedy reply

    Yes these reels tend to remain tightly wound even after use.

    We have tried to use the weighing method before and found it inaccurate so were hoping for a better solution on here.
     
  5. Jun 27, 2011 #4
    Actually, I would have thought that the weighing method would be rather accurate, too...so, maybe there is just too much variation in all the parameters...thickness of label, thickness of backing paper, spacing between labels...hhhmmm, I bet they look identical from afar.

    The thing is, even for any other approach, if there are variations in the parameters...things are not going to come up very precise, either...

    So, having said that and without involving spirals...

    The first approximation to a solution would be to think of whatever is left as a bunch of concentric circles, and so:

    measure the thickness of the reel and determine the inner diameter and the outer diameter; knowing the thickness of both label and backing together, evaluate the circumference of each circle, add them up and you have your total length, divide this by the pitch and you have how many labels are left...something like this:
    • dr = thickness of both label and backing paper together
    • id = inner diameter of left over reel...this may be constant since it should probably be the outer diameter on the bare core material of reel the labels come in
    • od = outer diameter of reel
    • pitch = distance from one spot in one label to the same spot in the adjacent label; basically, the width (length?) of the label plus spacing in between...maybe make a few measurements and make sure this number is representative

    Code (Text):

    length = 0
    do loop for d from id to od every 2*dr
        perimeter = 3.1415926*d
        length = length + perimeter
    end loop
    NumOfLabels = perimeter / pitch
     
    In other words, you have N concentric circles, where N= (od - id) / (2dr)
    And just like adding a set of consecutive numbers is the same as multiplying the average so many times:

    your total length can be calculated like this:

    length = pi x ( (od + id) / 2 ) x N

    again, this is just assuming a bunch of concentric circles.

    Just an idea
     
    Last edited: Jun 27, 2011
  6. Jun 27, 2011 #5
    Carefully make precise measurements and then count the actual number of labels left on a few test reels.

    d=diameter of the core
    D=diameter of the reel
    W=width of the label
    w=width of the space between each label
    T=thickness of the label and backing paper combined

    estimated number of labels left = Pi/4*(D*D-d*d)/((W+w)*T)

    Report back when you are done with a nice table showing all the measurements and the estimated and actual label counts.

    Make sure your measurements are all in inches or are all in centimeters, just be consistent, no thickness in mills, width in tenths of an inch and diameters in centimeters or anything like that.
     
  7. Jun 28, 2011 #6
    Unless I totally misunderstood how the labels come in reels, I think Bill's equation is the wrong one...he is calculating the cross sectional area of the reel when you look at it sideways...

    Bill: care to reconsider? or correct me?
     
  8. Jun 28, 2011 #7

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, he is. and given a constant thickness for each label, that cross-section area, divided by the thickness of the labels, gives the length of all labels left. Dividing that length by the length of a single label gives the number of labels left.
     
  9. Jun 28, 2011 #8
    Well, I guess I would like to see a picture of these damn reels, then! :grumpy:
     
  10. Jun 28, 2011 #9

    SpectraCat

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    Science Advisor

    Some good mathematical solutions have been posited on here so far, but isn't the most efficient course an operational one?

    You know how many labels are on a reel to start with.
    You know how many cases of each brand were crated on any given day.

    So can't you just keep a running total of the labels left on each reel? If your labels are put on the bottles by machine, then keep a log on each machine that is updated at the end of every day (or better yet, do it in software, if your machines are computer controlled). If they are put on by people, then keep a log for each reel. This could be done very simply by just putting a sticker on the handle of a "label applicator" with the number of cases for which there are labels remaining on the reel.

    Just a thought .. certainly the mathematical solutions proposed are more elegant.
     
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