|
Chip's CD Media Resource Center:
|
|
Image from Disctronics
|
| Previous | Next | |
|
|
||
Assuming that a CD-ROM was used only for digital data tracks (no digital audio), what would be its capacity? In other words, how many Mode 1 sectors, each containing 2048 bytes of data, can we put on the disc?
The first problem is in whether to include the Lead-in or Lead-out areas in addition to the Program area (or User Data area). For consistency with the CD-DA numbers I'll just use the program area. I don't actually know yet whether it makes sense to record data on the Lead-in and Lead-out areas.
Our approach is basically just a continuation of what we did in the section on CD-DA Storage Capacity. I'll work through one example and then present a table.
Track length of program area: 5378 meters
Channel bit length: 277.662 nm/bit (at 1.2 m/s)
Channel Bit capacity of program area = 5378 meters / 277.662 nm/bit
= 19,368,867,000 bits
Data capacity of program area = 19,368,867,000 bits / 588 bits/frame
= 32,940,250 frames
then 32,940,250 frames / 98 frames/sector
= 336125 sectors
and 336125 sectors * 2048 bytes/sector
= 688384000 bytes (656.49 Mbytes)
| Velocity | Bit Size | DA Time | DA Size | Data Size |
| 1.2 m/sec | 278 nm | 74 min 42 sec | 754 Mbyte | 656 Mbyte |
| 1.3 m/sec | 301 nm | 68 min 57 sec | 696 Mbyte | 606 Mbyte |
| 1.4 m/sec | 324 nm | 64 min 01 sec | 646 Mbyte | 563 Mbyte |
I do not yet know the reason for small discrepancies between these figures and published figures for the capacity of a CD-ROM. Nor do I yet know how an "80 minute" CD is made, though I suppose the linear velocity is simply reduced to about 1.1 m/s.
| Previous | Next | |
|
|
||
|
Last Updated Monday October 15, 2001 17:58:32 PDT
|