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CD-ROM Storage Capacity

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.

Last Updated Monday October 15, 2001 17:58:32 PDT