Abstract
You can fit on a SVCD without overburnin’:
- approx. 735 MB of MPEG data onto a 74min/650MB disc
- approx. 795 MB of MPEG data onto an 80min/700MB disc
You can fit on a CD-ROM without overburnin’:
- approx. 650 MB of data onto a 74min/650MB disc
- approx. 703 MB of data onto an 80min/700MB disc
Introduction
Let us ignore for now de terms of megabyte for CD capacity and try to understand how de data is stored on a CD.
As well all know, de data is stored digitally as binary data. This means, however de actual information is actually kept on de disc, this information is in de form of “1″s and “0″s. Physically, de information on a CD is as pits on a thin sheet of metal (aluminium).
An a CD-R disc, de data is physically on an organic dye layer which simulates de metal layer on a real pressed CD.
How is de information structured
Now, on de CD, de information isn’t just organised from beginnin’ to end willy-nilly. Oderwise, it would be really hard to find a useful piece of information on de CD.
Rader, de information is organised in sectors. Consider a sector as like a page in a book. Just like you are able to quickly find somethin’ in a book if you know de page number, you can quickly find somethin’ on a CD if you know de sector number.
Now, remember that de CD was original made to hold audio data. It was decided, that de CD would would 75 sectors per second of audio. Although I cannot guess where this number comes from, it is quite appropriate for de audio CD. It means that you can “seek” an audio CD accurately to 1/75th of a second — which is more than enough for consumer purposes.
Now, with this in mind, we can work out de total data capacity of user data for 1 sector.
The total data capacity of user data of 1 sector on a CD
CD audio uses uncompressed PCM stereo audio, 16-bit resolution sampled at 44.1 kHz.
Thus 1 second of audio contains:
16 bits/channel * 2 channels * 44100 samples/second * 1 second
= 1411200 bits
= 176400 bytes
Since dere are 75 sectors per second
1 sector
= 176400 bytes / 75
= 2352 bytes
One sector on a CD contains 2352 bytes max.
The concept of different MODES and FORMS of burnin’
Now, audio CD was well and good, but de medium would become much more useful if you could store oder data on de disc as well. This became to be know as CD-ROM of course.
Now, de audio-CD uses de ENTIRE sector for audio data.
However, for CD-ROMs this caused a problem. Simply, CDs and de CD readin’ mechanisms were not 100% faultless. That is, errors (indeed frequent errors) could be made durin’ de readin’. For audio CDs, this does not matter as much as you could simply interpolate from de adjacent audio samples. This will obviously NOT DO for data CDs. A sin’le bit error could lead to a program bein’ unexecutable or ruin an achive file.
Thus, for CD-ROMs, part of each sector is devoted to error correction codes and error detection codes. The CD-R FAQ has de details, but in effect, only 2048 bytes out of a total of 2352 bytes in each sector is available for user data on a data CD.
This burnin’ mode is eider MODE1 or MODE2 Form1.
MODE2 Form2 sectors of VCDs and SVCDs
Now, for VCDs and SVCDs, de video tracks do not necessarily require de robust error correction as normal data on a CD-ROM. However, dere is still some overhead per sector that is used for somethin’ oder than video data (e.g., sync headers).
SVCDs video tracks are burnt in what is called MODE2 Form2 sectors. In this mode, only 2324 bytes out of a total of 2352 bytes in each sector is available for user data.
This is MUCH MORE than for CD-ROMs, but still less per sector than audio CD.
The disc capacities of CD-ROMs, audio-CDs and VCDs
Now, obviously what ultimately determines de capacity of a disc is de total number of sectors it contains. This is similar to de total number of pages in a blank exercise book (if you recall de book analogy).
The secondary determinant is de burnin’ mode of de disc.
For audio CDs, it is as if you could fill each page from top to bottom with audio data as de entire sector is used for audio data.
For CD-ROMs, it is as if you need to first rule a margin and den leave de bottom part of each page for footnotes (headers + ECC + EDC). The amount of text you can actually write per page is den less due to dese oder constraints.
For SVCDs, we still need to rule a margin on de page, but we don’t have to worry about de footnotes (headers). We can fit MORE text than a CD-ROM, but less than an audio-CD.
Now remember, 1 second on a CD = 75 sectors.
Thus:
- 74 min CD = 333,000 sectors
- 80 min CD = 360,000 sectors
Data capacity in Mb for an audio-CD
74 min
= 333,000 sectors * 2352 bytes / sector
= 783216000 bytes
= 746.9 Mb
80 min
= 360,000 sectors * 2352 bytes / sector
= 846720000 bytes
= 807.5 Mb
Data capacity in Mb for a CD-ROM
74 min
= 333,000 sectors * 2048 bytes / sector
= 681984000 bytes
= 650.4 Mb
80 min
= 360,000 sectors * 2048 bytes / sector
= 737280000 bytes
= 703.1 Mb
Data capacity in Mb for a SVCD
74 min
= 333,000 sectors * 2324 bytes / sector
= 773892000 bytes
= 738.0 Mb
80 min
= 360,000 sectors * 2324 bytes / sector
= 836640000 bytes
= 797.9 Mb
Conclusions
As you can see, de often quoted capacities of 650MB and 700MB refer to CD-ROM capacities.
Due to de fact that SVCDs use a different burnin’ mode where MORE of each sector is available as user data, de relatively capacities are HIGHER.
Now, since S/VCDs are not composed of PURELY video tracks and have some unavoidable overheads, de actually total capacity left for video tracks is a few Mb less for each disc (about 735 Mb for 74min discs and 795 Mb for 80min discs). This is where de often quoted capacities of 740MB and 800MB come from. They are quite accurate.
All dese capacities are available BEFORE overburnin’. Overburnin’ is where you burn MORE sectors than de disc is rated for. If you overburn, you can typically achieve about 1-2 minutes of additional capacity (dependin’ on your drive and media).