How Long is the Tape?Date: 31 Jul 1995 22:32:16 -0400 From: charles dorsey Subject: math problem I need help with a problem. Jon Jones has a roll of scotch tape. He wants to figure out how long the tape is without unrolling it. He knows this about the tape: the thickness of the tape is .05 inches. The tape is wound on a core that is 1 inch in diameter. The diameter of the roll of tape, including the core dimension, is 5.5 inches. Is there a formula for determining the length of the tape? Thank you for your help. Charles Dorsey. Date: 1 Aug 1995 10:31:05 -0400 From: Dr. Ethan Subject: Re: math problem Well, I have a method for calculating it but it isn't really very pretty, so I'm not happy with it. But here it is: The first time around the tape has a half inch radius. Actually, since the tape has a thickness of 0.05 inches it has an average radius of .525 inches. You can choose whether or not you want to worry about this in your example; the difference came out to about 3 feet. Well back to the method: Since the average radius for the first one is .525 inches. We need to take 2Pi *(.525) for the length of the first time around. The next time around will be 2Pi * ( .575). Here is the pattern: 2Pi * (.475 + n * .05) as n goes from 1 to 45. (Do you see why I used 45?) So now the problem has been reduced to a standard finite sum. I used a computer to calculate this and got 459.458. You night be interested to know that if you just use the average radius, which in this case is 1.625, and calculate 2Pi * (1.625) * 45 you get the exact same answer, 459.458. Do you have any ideas for why that works? Ethan Doctor On Call Note from the archivist: Several months later, we got a note from Ujjwal Rane: I was going through your archives and found this problem with your solution. You have an interesting approach. Here is an alternative. Assuming that the tape is going to retain its volume, whether wound or straight. We can just go by the area of the spool. -Volume of the tape = L * w * t ..............L = Length (Calculated as if it w = Width were a rectangular box) t = Thickness -Volume of the tape = (PI/4)*(D^2 - d^2)*w ...D = Outer diameter (Calculated from the annular d = Inner (core) diameter shape in which it is wound) >From our assumption, the two volumes are equal, L * w * t = (PI/4)*(D^2 - d^2)*w, so L = (PI/4)*(D^2 - d^2)/t ......the unknown w cancels out L = (PI/4) * (5.5^2 - 1^2)/0.05 L = 459.5 inches _____________________________________________________ Here is another, more rigorous solution which essentialy yields the same answer. -Consider the tape as a spiral, given by polar equation, r = k * theta, where r = radius theta = angle k = tape thickness/(2*PI) = 0.0079577 -Let r1 = Inner (core) radius = 0.5" (given) r2 = Outer radius = 5.5/2 = 2.75" (given) Corresponding angles are, theta1 = r1/k = 62.83 radians theta2 = r2/k = 345.58 radians -Length of the tape will be the integral of k * sqrt(1+theta^2) with respect to theta, from limits theta1 to theta2. - Using the integral of sqrt(1+x^2)dx, which is (1/2)*[x*sqrt(x^2+1) + ln|x+sqrt(x^2+1)|] + C and substituting the limits by the Fundamental Theorem of Calculus, we get L = 459.5" as before. With best regards, Ujjwal Rane urane@asu.edu |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/