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### Finding the Radial Separation of a CD Track Spiral

```
Date: 10/04/2000 at 06:25:00
From: Payten Giles
Subject: Measuring Radial Separation of CD Track Spirals

Dr. Math,

I'm having trouble with the following problem. Any help, or a nudge in
the right the direction, would be great.

Background:
The information on a CD is encoded as a series of pits in the disk. On
audio CDs the track spirals outward from a radius of 25 mm to a radius
of 58 mm.

Problem:
Devise some way of measuring the radial separation of these pits.

```

```
Date: 10/06/2000 at 20:22:09
From: Doctor Jeremiah
Subject: Re: Measuring Radial Separation of CD Track Spirals

Hi Payten,

Check out the page "How Compact Discs (CDs) Work" at:

http://www.howstuffworks.com/cd1.htm

The question you asked doesn't have enough information to solve it. At
the minimum you need to know how big a pit is, how close adjacent pits
are, and the total number of pits on the CD.

The total number of pits on the CD can be calculated easily. A normal
CD has 74 minutes of audio from 2 channels (stereo) where each channel
is recorded with 16-bit samples at 44,100 samples per second.

Each channel has:

16 bits/sample * 44,100 samples/second = 705,600 bits/second.

Since there are two channels, we have 1,411,200 bits/second. Also,

74 minutes/CD * 60 seconds/minute = 4,440 seconds/CD.

That means we have:

4,440 seconds/CD * 1,411,200 bits/second = 6,265,728,000 bits/CD.

In a digital world each bit is one piece of data and can be
represented by a pit or the absence of a pit. So that means there are
6,265,728,000 pits/CD.

Knowing how big the pits are is the next step.

- Doctor Jeremiah, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 10/06/2000 at 22:08:12
From: Payten Giles
Subject: Re: Measuring Radial Separation of CD Track Spirals

So now to get the separation between the pits it would seem we need
the length of the spiral and the length of the pits.

We know that the width of the tracks is 0.5 micrometers.

We can then divide the length of the spiral by the number of pits to
get the length per pit.

Does this sound okay? If so, is there an equation to work out the
length of the spiral knowing that the tracks are 0.5 micrometers wide
and knowing the diameter they are to fill?

Cheers,
Payten
```

```
Date: 10/07/2000 at 00:28:42
From: Doctor Jeremiah
Subject: Re: Measuring Radial Separation of CD Track Spirals

We need to know the length of the pits to calculate the length of the
spiral based on number of pits times length of a pit. Then we can take
the length of the spiral and calculate how many times around it went
and the distance between the lines (radial separation) of the spiral.

The width of the track is not really important. The length of the pits
is more important. The web site I pointed you to said that the pits
were .83 microns long. Then the length of the spiral is

6,265,728,000 pits * .00000083 meters/pit = 5200.55424 meters

There are 1609.344 meters/mile, so

5200.55424 meters / 1609.344 meters/mile = 3.23 miles

The Web site said the length would be almost 3.5 miles, and 3.23 is
pretty close.

Now that we know the length of the track, we just have to figure out
the radial separation. The formula for the spiral is the hard part. I
don't know a formula for that, so we will have to make one.

The length of a spiral from one point on the CD all the way around
back to that point is the same as the circumference of a circle with
the same average radius. So a spiral can be approximated as a set of
concentric circles, the distance between them being the radial track
separation.

The circumference of each circle is 2*pi*R and the number of pits it
accounts for is 2*pi*R / .83 microns. The smallest circle on the CD,
at R = 25 mm (R = 25,000 microns), would have this many pits:

2*pi*R / .83 microns = 2*pi*25000 / 0.83 = 189,253 pits

This seems reasonable. The number of pits an arbitrary circle would
have is

2*pi*(25000+d*n) / 0.83

where n is the number of smaller circles within it and d is the

The total number of pits is the sum of all the pits in all the circles
from R = 25000 microns to R = 58000 microns = 25000+d*N, where N is
the total number of circles on the CD.

Total number of pits = (2/0.83)*pi*(25000 + d*1) +
(2/0.83)*pi*(25000 + d*2) +
(2/0.83)*pi*(25000 + d*3) +
:
(2/0.83)*pi*(25000 + d*(N-2)) +
(2/0.83)*pi*(25000 + d*(N-1)) +
(2/0.83)*pi*(25000 + d*N)

Total number of pits = (2/0.83)*pi*25000 + (2/0.83)*pi*d*1 +
(2/0.83)*pi*25000 + (2/0.83)*pi*d*2 +
(2/0.83)*pi*25000 + (2/0.83)*pi*d*3 +
:
(2/0.83)*pi*25000 + (2/0.83)*pi*d*(N-2) +
(2/0.83)*pi*25000 + (2/0.83)*pi*d*(N-1) +
(2/0.83)*pi*25000 + (2/0.83)*pi*d*N

Here I collapse the sum. The sum of the first column of terms is just
N times the term. For the second column, we can use the fact that the
sum of all numbers from 1 to N is N(N+1)/2.

Total no. of pits  = (2/0.83)*pi*25000*N + (2/0.83)*pi*d*N(N+1)/2

We know the total number of pits from my last reply:

6,265,728,000 pits = (2/0.83)*pi*25000*N + (2/0.83)*pi*d*N(N+1)/2

And don't forget that 25000 + d*N = 58000, so d = 33000/N:

6265728000 = N*(2/0.83)*pi*25000 + (2/0.83)*pi*d*N(N+1)/2
= N*(2/0.83)*pi*25000 + (2/0.83)*pi*(33000/N)*N(N+1)/2
= N*(2/0.83)*pi*25000 + (N+1)*(2/0.83)*pi*16500
= N*(2/0.83)*pi*25000 + N*(2/0.83)*pi*16500 +
(2/0.83)*pi*16500
= (N*25000 + N*16500)*(2/0.83)*pi + (2/0.83)*pi*16500
= N*41500*(2/0.83)*pi + (2/0.83)*pi*16500

Thus:

6265728000 - (2/0.83)*pi*16500 = N*41500*(2/0.83)*pi

So:
N = (6265728000 - (2/0.83)*pi*16500) / (41500*(2/0.83)*pi)
N = 19,944

So this would mean we can approximate the track spiral with 19,944
circles. We know that d = 33000/N, so we can calculate the distance
between the circles of the spiral (radial separation):

d = 33000/N
d = 33000/19944
d = 1.65 microns

Keep in mind that this is just an approximation to a spiral. But it
will give you the idea. And since that Web site said that the distance
was 1.6 microns, we are not too far off.

Please write back if you want to discuss this some more.

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

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