This is a topic I've long ignored. It has now become crucial, since I'll be backing up my music CD collection in digitally-extracted (meaning .WAV) form. Here goes:
Digital CD-quality audio takes up:
(2 channels) * (44100 samples/sec) * (16 bits per sample) / (8 bits in a byte) / (1024 bytes in a kilobyte) = 172.3 KB/sec (this is fact). To the best of my knowledge, .WAV files on your computer take 172.3KB for each second of audio.
However, 1X-speed audio CD's invariably play at 150KB/sec (this is presumably also fact). Furthermore, CD music MUST be stored at 150KB/sec in order to fit exactly 74 minutes of audio into 650MB of space.
Since CD music is of the above format, how in h3ll does it store this information? All I could find online was this info:
"A 44.1kHz rate means there are 44,100 chops every second, each one describing the waveform amplitude at that moment in time with a 16-bit number; 16-bit itself offering 65,536 steps from which to choose. With samples occupying two bytes on each of two channels - each sample yields a data transfer rate of just over 176KBps. A single-speed CD-ROM transfers data at the same rate, but a portion of the data stream is taken by error-correcting information - reducing the effective transfer rate to 150KBps. A CD can hold up to 74 minutes of encoded stereo audio data - which, when the ECC overhead is taken account of, equates to the standard CD capacity of 680MB."
Source
(They used 1000bytes in a kilobyte instead of 1024).
This only confuses me more: does CD audio omit sound information when truncating the datarate from 172 to 150KB/s? Most importantly: if WAV files take up more than raw CD audio tracks, one may be unable to fit my albums on 650MB CD's!
Does anyone understand how this business works? Please deposit your thoughts below.
--leoV
Digital CD-quality audio takes up:
(2 channels) * (44100 samples/sec) * (16 bits per sample) / (8 bits in a byte) / (1024 bytes in a kilobyte) = 172.3 KB/sec (this is fact). To the best of my knowledge, .WAV files on your computer take 172.3KB for each second of audio.
However, 1X-speed audio CD's invariably play at 150KB/sec (this is presumably also fact). Furthermore, CD music MUST be stored at 150KB/sec in order to fit exactly 74 minutes of audio into 650MB of space.
Since CD music is of the above format, how in h3ll does it store this information? All I could find online was this info:
"A 44.1kHz rate means there are 44,100 chops every second, each one describing the waveform amplitude at that moment in time with a 16-bit number; 16-bit itself offering 65,536 steps from which to choose. With samples occupying two bytes on each of two channels - each sample yields a data transfer rate of just over 176KBps. A single-speed CD-ROM transfers data at the same rate, but a portion of the data stream is taken by error-correcting information - reducing the effective transfer rate to 150KBps. A CD can hold up to 74 minutes of encoded stereo audio data - which, when the ECC overhead is taken account of, equates to the standard CD capacity of 680MB."
Source
(They used 1000bytes in a kilobyte instead of 1024).
This only confuses me more: does CD audio omit sound information when truncating the datarate from 172 to 150KB/s? Most importantly: if WAV files take up more than raw CD audio tracks, one may be unable to fit my albums on 650MB CD's!
Does anyone understand how this business works? Please deposit your thoughts below.
--leoV