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Calculating How Much Paper Is on a Roll

Date: 04/24/2006 at 11:57:16
From: Joe
Subject: Measuring paper on a roll

I'm trying to determine how much paper is on a roll of paper.  I know
it has to do with the thickness of the paper and the size of the core
and the whole roll, but I can't figure it out.  For example, how much
paper has been used under these conditions:

  Initial diameter of roll including the core is 58"
  Final diameter of roll including the core is 48"
  Core diameter is 4"
  Thickness of paper is .014"

At the beginning, the roll had 28000 ft of paper on it.  How much
paper is still on the roll at the end?



Date: 04/24/2006 at 16:36:37
From: Doctor Jerry
Subject: Re: Measuring paper on a roll

Hello Joe,

We get this sort of question a lot.  Here's how I would approach it.

After 1 layer of paper is wrapped around the core the total length of
paper on the spool, written as L(1), is 2*pi*r, where  r is the radius
of the core, measured in feet.  This is essentially just the
circumference of the core.

After 2 layers, L(2) = 2*pi*r + 2*pi*(r+t), where t is the thickness
of the paper, measured in feet.  This is the first layer from above
plus the second layer, where the second layer is the circumference
around the core and the first layer, making the radius of the second
layer r + t.

After 3 layers, L(3) = 2*pi*r + 2*pi*(r+t) + 2*pi*(r+2t).  This is the
first 2 layers from above plus the third layer, which is now the
circumference around the core and the first two layers, making the
radius of that third layer r + 2t.

Expanding the above expressions for the first three layers we have:

  L(1) = 2*pi*r
  L(2) = 2*pi*r + 2*pi*r + 2*pi*t
  L(3) = 2*pi*r + 2*pi*r + 2*pi*t + 2*pi*r + 2*pi*2t

Factoring out 2*pi we have:

  L(1) = 2*pi*(r)                  = 2*pi*(r)
  L(2) = 2*pi*(r + r + t)          = 2*pi*(2r + t)
  L(3) = 2*pi*(r + r + t + r + 2t) = 2*pi*(3r + t + 2t)

Now imagine continuing this process for n layers, and let's find an
expression for L(n).  Note that in the factored form you will have n*r
in the parentheses.  You will also have the sum:

  t + 2t + 3t + ... + (n-1)*t

or

  t*(1 + 2 + 3 + ... + (n-1))

The parentheses contain the sum of the first (n-1) positive integers,
and by formula that is given by (1/2)(n)(n-1) or (1/2)(n^2-n).

Putting this all together, we can determine that:

  L(n) = 2*pi*( n*r + (1/2)(n^2-n)*t )

Next, the radius R(n) (in feet) of the roll after n layers, is: 

  R(n) = r + n*t

With these two formulas I can solve your problem.

  r = radius of core = 2" or 1/6 feet

  t = thickness of paper = 0.014" or 0.014/12 feet

The final radius of the roll is 24" or 24/12 = 2 feet.  So, I solve

  R(n) = r + n*t 

  2 = 1/6 + n*0.014/12

for n.  I find

  n ~ 1571.43 or 1571.

Substituting this value of n into

  L(n) = 2*pi*( n*r + (1/2)(n^2-n)*t )

I find that

  L(1571) ~ 10685 ft.

That's how much paper is still on the roll at the end of your problem.

One can do this generally.

If the final radius is Rf, then solving  Rf = r + n*t for n gives

  n = (Rf - r)/t.

Substituting this into L(n) = 2*pi*( n*r + (1/2)(n^2-n)*t ) gives the
length corresponding to the final radius:

  ( pi*( Rf - r )*(r + Rf - t) )/t 

Please write back if my comments are not clear.

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Geometry
College Higher-Dimensional Geometry
High School Geometry
High School Higher-Dimensional Geometry
High School Practical Geometry

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