Subject: Varying Diameters for Linear FeetName: Richard
Who is asking: Other
Although I have been able to solve my equation using an Excel spreadsheet, I have yet to duplicate it in a written constant formula. Sounds odd I know, however I am very curious and would like very much to do this. I would appreciate any help.
The root of my equation is simply this, I have a roll of paper, wrapped around a corrugate core, whos diameter is 10.750 in. The outer diameter of the roll is approx. 60 in. The thickness of the paper is .014 in. I am trying to find out how much linear feet of paper is left on the roll, given only the diameter of paper remaining on the core, I'll call the remaining paper scrap.
For example, if I had a diameter of 10.750, there would be no paper on the core. If I had a diameter of 60.002 in. my scrap would be 16297.392 ft. I believe this is correct because I used a spreadsheet to get the answer. Because of the ever changing diameter of the roll as it is being consumed, is there a formula that I can use in a small Visual Basic program to identify the ramaining footage (scrap) based on 2 inputs, the thickness of the paper and the diameter of the roll.Thank you
The paper is rolled around the core in the form of a spiral called the spiral of Archimedes. Using some calculus you can find an expression for the length of the spiral.
If rO is the radius of the outside of the roll of paper and rI is the radius of the core as in the diagram and t is the thickness of the paper then the length of the scrap is given by
where ln is the natural logarithm.
If you perform this calculation with the data given in your problem, rO = 60.002/2 = 30.001 inches, rI = 10.750/2 = 5.375 inches and t = 0.014 inches then L = 195490 inches or 16290.8 feet.
If t is much smaller than rO and rI, which it is in your situation, there is an approximation of L which is much easier to calculate. In this situation
In the example above this also gives L = 195490 inches = 16290.8 feet.
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