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Fabric Left on a Roll

Date: 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 

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 

- Doctor Robert   
Associated Topics:
High School Calculators, Computers
High School Euclidean/Plane Geometry
High School Geometry
High School Practical Geometry

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