Fabric Left on a RollDate: 03/15/2001 at 08:49:23 From: Rob Spear Subject: Geometric progression? I am going to write a program for my TI-83+ (I can also use a TI89 if necessary) to calculate how many feet of material are left on a roll. I know the outside diameter of the core (this is the tube around which the film is wrapped, I know the outside diameter of the roll, and I know the thickness (mil thickness) of the material. What I don't know is the formula to use to calculate the feet of material left on the roll. What I thought I could do is start out with the distance around the core. In this particular case, that would be pi*3.75 (the outside diameter). Then each additional layer of material would be the previous outside diameter plus twice the thickness (1 mil = 0.001"). I'd have a field in which I'd keep a running total of these inches. I'd set up another field to keep track of the total number of "wraps" of material, which would be a calculated field (outside diameter of the roll/2 - 3.75/2 [this removes the core's dimension] divided by the mil thickness of the film). I'd put a loop in the program to keep adding to the running total of inches until the number of wraps = 0. This would seem to be the hard way, but as I said, I don't know what formula to use. I know there has to be a formula I can use for this calculation to make things a lot easier. Does anybody out there know the formula I'm looking for? I'd welcome any suggestions or ideas. Thanks. Date: 03/15/2001 at 15:04:46 From: Doctor Robert Subject: Re: Geometric progression? I'll take a whack at this one. No guarantee of success. Let me define the diameter of the tube as Do and the diameter of the roll as D. Let t be the thickness of the material. Then I figure that the number of wraps, n, on the tube, is: n = (D-Do)/(2*t) and we need to find the total length of these n wraps. I'll figure the length of a wrap to be the circumference of a circle whose radius is from the center of the tube to the center of the wrap I'm dealing with. In other words, the radius of the first wrap is Do/2+(1/2)t; the radius of the second wrap is Do/2+ (3/2)t; the radius of the third wrap is Do/2 + (5/2)t. Now the circumference is 2pi times the radius and we need to sum a bunch of circumferences. Generalizing on the formula above, the radius of the kth wrap is: radius of kth wrap = Do/2 + t(2k-1)/2 where k= 1,2,3,... Length of material = sum from k=1 to n of 2*pi*[Do/2 + t*(2k-1)/2] = n*2*pi*Do/2 + sum from k=1 to n of 2*pi*t(2k-1)/2) = n*pi*Do + pi*t times sum from k=1 to n of (2k-1) Now the sum of 1+3+5 + ...+2n-1 = n^2 So, length = n*pi*Do + pi*t*n^2 But length = n*pi(Do + n*t) But n = (D-Do)/(2t) so the formula is Length = (D-Do)*pi/(2t)*[Do + (D-Do)/2] = (D-Do)*pi/(4t)[Do+D] = (D^2-Do^2)*pi/(4t) I hope that this formula is correct and serves you well. There is no guarantee. - Doctor Robert http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/