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### 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?

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

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

-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
```
Associated Topics:
High School Calculus

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