Should Know CD & DVD Data Capacity
Data Capacity of CDs [Tutorial]
Abstract
You can fit on a S/VCD without overburning:
- 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 overburning:
- approx. 650 MB of data onto a 74min/650MB disc
- approx. 703 MB of data onto an 80min/700MB disc
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 there are 75 sectors per second
1 sector
= 176400 bytes / 75
= 2352 bytes
One sector on a CD contains 2352 bytes max.
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 S/VCD
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
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Abstract
You can fit on a S/VCD without overburning:
- 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 overburning:
- approx. 650 MB of data onto a 74min/650MB disc
- approx. 703 MB of data onto an 80min/700MB disc
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 there are 75 sectors per second
1 sector
= 176400 bytes / 75
= 2352 bytes
One sector on a CD contains 2352 bytes max.
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 S/VCD
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
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