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### Determining the Length of a Coil of Ribbon

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Date: 08/31/2001 at 15:28:35
From: George
Subject: Using Calculus to determine the length of a coil of ribbon

I have a coil of ribbon. The ribbon spirals around 34 times. The
solved this problem just by using the average circumference for one
loop. I found this by averaging 1.12 and 2.22, 1.67" multiplied by
2pi. By division, I found the width of the ribbon was about .0333".

So how will calculus give me the same answer I found with algebra and
geometry (29'9")?
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```
Date: 08/31/2001 at 16:51:32
From: Doctor Fwg
Subject: Re: Using Calculus to determine the length of a coil of
ribbon

Dear George,

Herein and below are a few possibilities. Only the third method
involves calculus:

TOTAL LENGTH OF SPIRAL GROOVE OR ROLL OF TAPE

METHOD I (Area Method):

There are several ways to solve this problem. However, a few
dimensional measurements are required first. These measurements must
provide the outermost radius of the groove (Rmax), the innermost
radius of the groove (Rmin), and the distance between successive turns
in the spiral groove (D).

One must also assume that D is a constant. If not, the following
method will not be very accurate. In any case, measuring Rmax and Rmin
is pretty straightforward. To estimate D for a spiral groove, mark
out the distance between about 11 (or 101) successive groove turns
and measure the distance between them with a good ruler (one with
fine divisions). If the distance between 11 successive groove turns
is X cm, divide by 10 to get the distance between two successive
groove turns. If the distance between 101 groove turns is measured,
divide by 100 to get the average distance between any two
successive groove turns. For better accuracy, use a magnifying lens or
other low power optical device to count the number of groove turns.

Next, calculate the area (A) of the annulus between Rmax and Rmin,
where:

A = Pi [(Rmax)2 - (Rmin)2]

(Note: Pi is approx. 3.14159...)

Next, imagine that it is possible to completely "unwind" the spiral
groove so that the thickness of the resulting strip of material is
exactly equal to D. The total length (L) of this spiral is the unknown
quantity. However, the top edge area (L x D) of this long strip of
imaginary material is equal (almost exactly) to A (see above). So:

L x D = Pi [(Rmax)2 - (Rmin)2]

The only unknown here is L. One can solve for L pretty easily. This is
also a pretty neat way to find the total length of material on a long
coiled roll of tape. In that case, a micrometer may be used to measure
the thickness (D) of a small section of the tape instead of counting a
large number of "coils" or turns.

METHOD II (Summation Method):

1) Measure distance between the center of a disk and the first turn of
its spiral groove. Let this radius be Ro.

2) Measure the distance between 11 (or 101) successive turns in the
spiral groove. Let this value be X. Divide this distance X by 10 (or
100). This ratio (i.e., X/10 or X/100) gives the average distance
between any two successive turns in the spiral groove. Let this
average distance between any two successive turns in the groove be D.

3) Measure the distance between the innermost turn of the spiral
groove and the outermost turn. Let this distance be H.

4) Multiply H by X/10 (or X/100 if the distance between 101 successive
groove turns was measured). This gives a good estimate for the total
number of groove turns (N), so one doesn't have to count them all.
Accuracy here will depend on the accuracy of the dimensional
measurements. If one can ignore mistakes in counting, there is
absolutely no error in counting an integer number of anything.

5) If it can be assumed that each spiral turn is approximately
circular, it is possible to calculate the circumference of each
"ring," then add them all together to get the total length. Even
though this may seem a little tedious, there is a fast way to do this.

6) Let C be the total length of the spiral groove so:

C = 2(Pi)Ro + 2(Pi)(Ro + D) + 2(Pi)(Ro + 2D) + 2(Pi)(Ro + 3D) + ...+
2(Pi)(Ro + [N -1]D).

7) Notice that the total number of 2(Pi)Ro terms is N, so factoring
out the 2(Pi)Ro terms and multiplying by N yields:

C = 2N(Pi)Ro + 2(Pi)[D + 2D + 3D + ...+ (N -1)D].

8) Now factoring out the D terms yields:

C = 2N(Pi)Ro + 2(Pi)D[1 + 2 + 3 + ...+ [N -1]].

9) Now if it is noticed that the sum (S) of a series of successive
integers (starting from 1 and increasing by 1 unit at a time up to the
Nth integer), is:

S = (N)(N + 1)/2

or in the case of (N - 1) terms:

S = (N -1)(N - 1 + 1)/2 = N(N - 1)/2.

10) So:  C = 2N(Pi)Ro + 2(Pi)D[N(N - 1)/2].

11) Factoring out the 2N(Pi) terms, one gets:
C = 2N(Pi)[Ro + D(N - 1)/2].

So, by estimating values for D and N, one can find a good estimate for
the Total Length of the Spiral Groove.

METHOD III (Calculus Method):

1) Measure the distance between the center of a record and the first
turn in the spiral groove. Let this radius be Ro.

2) Measure the distance between 11 (or 101) successive turns in the
spiral groove. Let this value be X. Divide this distance X by 10 (or
100). This ratio (i.e., X/10 or X/100) gives the average distance
between any two successive turns in the spiral groove. Let this
average distance between any two successive groove turns be D.

3) Measure the total time needed to "play" the record. Let this time
be T (in minutes).

4) Note that the radius (R) at any point in time is:

R = Ro + [D/(2{Pi})]beta,

Where beta is the angle (in radians) that the record has turned
through at any intermediate time between 0 and T (minutes). Note: when
beta equals 2(Pi) radians, the record has turned around once, so the
value of R at that point in time is [Ro + D], and so on. Note: it is
assumed here that the record is running backward because the play time
is independent of rotational direction. In any case, it doesn't make
any difference in the answer provided that the distance to the
innermost part of the spiral groove can be measured just as accurately
as the distance to the outermost turn in the spiral groove.

5) Since we now have R as a function of beta, it is possible to find
the total distance (S) that the record needle travels through for any
value of beta. The differential form looks like this:

dS = R dbeta.

6) The integral limits on S are between 0 and S and on beta between 0
and beta(max).

7) Note that beta(max) = [2(Pi)(33.3)T], and the units here are in
radians. It is also assumed here that the angular turning velocity of
the record is 33.3... rev/min.

8) Now, rewriting the differential form:

dS = [Ro + (Dbeta)/(2{Pi})] dbeta.

9) Now, integrating between the limits one should end up with the
following:

S = Ro(beta) + [D/(2{Pi})][beta2/2].

10) Use the value: beta = beta(max) = [2(Pi)(33.3)T] in the equation
above to find S.

11) The expression for S (above) can be simplified somewhat, if
desired, and it will produce the quantity of interest but one still
has to measure or calculate values for certain necessary variables. In
this case, one needs to measure or calculate values for Ro, D, and T.
However, if the angular velocity is known to be 33.3...rev/min, this
value can be used as is shown above. One should also keep in mind that
the relative velocity of the needle, with respect to any particular
point within the spiral groove, is not constant in this problem even
though the angular velocity is constant.The needle can cover a much
greater distance per unit time near the outermost part of the spiral
groove versus the innermost part of the spiral groove because the
angular rotational velocity is constant.

I hope this has been helpful.

- Doctor Fwg, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Calculus
High School Euclidean/Plane Geometry
High School Geometry

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