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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/