How Much Material on a Spool?

Date: 05/02/2007 at 07:20:43
From: Robert
Subject: length of a reel of flat industrial blade

I have a reel of blade that is 1016 mm outer diameter and has a inner 
diameter of 508 mm.  The blade width or depth is 20 mm laid flat on a 
table.  Could you tell me the length of the reel and please show me 
the formula for working the length out using the information I have?  
Thanks.

Date: 05/02/2007 at 09:53:45
From: Doctor Jerry
Subject: Re: length of a reel of flat industrial blade

Hello Robert,

Thanks for writing to Dr. Math.  Here's the way I would think about
this problem:

  r = inner radius

  n = number of layers of material

  R(n) = outer radius after n layers

  T = thickness of material being wrapped

  L(n) = total length of material after n layers

Each layer wrapped adds another thickness of the material to the outer
radius, so

  After 1 layer,  R(1) = r + 1*T;

  after 2 layers, R(2) = r + 2*T;

  ...

  after n layers, R(n) = r + n*T.

Each layer wrapped adds more material to the total length.  The amount
added is equal to the circumference of the circle (2*pi*radius) where
the radius is the outer radius, so

  L(1) = 2*pi*R(1)

  L(2) = 2*pi*R(1) + 2*pi*R(2)

  ...
 
  L(n) = 2*pi*R(1) + 2*pi*R(2) + ... + 2*pi*R(n)

       = 2*pi*(r + 1*T + r + 2*T + ... + r + n*T)

       = 2*pi*(n*r + (1+2+...+n)*T)

       = 2*pi*(n*r + n*(n+1)*T/2)

  L(n) = pi*n*(2*r + (n+1)*T)

You are given R, r, and T.  You want L(n).  From

  R(n) = r + n*T

we can work out n, namely,

  n = (R - r)/T .

So, the total length L, expressed in terms of r, R, and T is

  L = pi*(R - r)(r + R + T)/T.

For R = 1016/2, r = 508/2, and T = 20, I find

  L = 31200.4 mm

If you have further questions about this please write back.

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/