Calculating How Much Paper Is on a RollDate: 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/ |
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