As part of the discussion exercise on Nov. 19, you were e-mailed the following instructions and a micrograph of a CD:

Assignment: Estimate the digital storage capacity of a CD in megabytes. Hints: Assume that the smallest pit shown represents two bits of information, changing state at each edge. Remember that there are 8 bits in a byte. E-mail me your answer, with a brief discussion of your reasoning, by Friday, April 26 by 5pm.

Micrograph of Bon Jovi CD, by permission of Michael Davidson and FSU

Find the area occupied by one bit. In the image above I have delineated
several neighboring bits to emphasize the "packing" of bits on a CD.
Actually I have enclosed 2 bits in each box, a **1** bit followed by a
**0** bit
-- notice the longer stretches of pits or "missing" bits along
each track that represent several bits set (or unset) in succession.

From this I estimate that 2 bits occupies roughly a rectangular area
of 1.6 micron by 1.6 micron
(micron = micrometer = 0.001 mm = 10^{-6} m.

So the area per 2 bits is (1.6 micron) squared or 2.5 square microns
[1.3 x 10^{-6} mm^{2} per bit].
The reciprocal of this number is 800,000 bits per square mm, or
since there are 8 bits per byte:
**100,000 bytes/mm ^{2}**.

Next we must estimate the "active" area of the CD, the area where information
can be stored. By examining a disc, the following dimensions can be obtained:

-- O.D. (outside diameter) = 117 mm;
I.D. (inside diameter) = 46 mm

The formula for the area of a disk is pi times the *radius* squared.
Thus the area of the "O.D. disk" is 10,800 mm^{2} and the area of the missing
"I.D. disk" is 1660 mm^{2}.
The total active area is the difference, roughly
**9000 mm ^{2}** (around 14 sq. in. for those so inclined).

So the number of bytes/mm^{2} times the size in mm^{2} gives an estimate of the
total capacity of a CD:

100 kb/mm^{2} x 9000 mm^{2} = 900,0007nbsp;kb = **900 Mb** (megabytes).

There are some details in compact disc encoding of data that need to be incorporated
to get a more accurate estimate of a compact disc's capacity. It turns out that
the pits do not directly represent bits; an encoding system is used that converts
8 bits to 14 bit words. This 8 to 14 bit encoding insures
that a binary **1** is separated by no fewer than *two* **0**s.
There are also 3 "merging" bits for each such block of 14 bits. Our
estimate is thus too large by 18/7. In addition, there will be other contributions
to "overhead" reducing the effective capacity.

One more detail is that the
minimum length of a pit is 0.8 micron, not 1.6 micron, with
minimum land length of 0.8 micron.
Also the features "boxed" in the micrograph above represent
3 bits rather than 2...
The actual capacity of a current CD is around **650 Mb**.

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Last updated November 29, 2004.

Copyright George Watson, Univ. of Delaware, 1996.