Conference cookie::notes$archive:cd_v1

        >	If this should work, does anybody know how much playing time
    	>256 MB would cover?
    
    I'm gonna take an 'educated' guess at this - The CD format requires
    588 bits on the disk to encode 192 bits worth of 44,100 Hz sampled,
    stereo 16-bit words.  So 256MB gives you:
    256 Mb * (8 / (16 * 2)) * (192 / 588) = 21 MWords (stereo),
    474 seconds,
    7.9 minutes,
    minus 10% for track and time encoding,
    about 7 minutes of digital stereo?
    
    This sounds too low, what am I doing wrong here?
    I thought we had determined that a CD has 500 Mb, so this doesn't
    seem right.

    Maybe I shouldn't be using the 192/588 conversion for error-correction,
    since the erasable disc also needs error correction, and the quoted
    256Mb capacity is a formatted capacity (right?).  Going by that,
    my guess should increase to 25 or 30 minutes, so that's my real
    guess.    
    
    
    	  		Steve
                                   

Regardless of capacity, the data rate out of the disk subsystem will 
have to be:

	44100 Samples/Sec/Channel * 2 Channels * 16 bits/sample
	* 14 Recorded bits/8 bits = 2.4696 M Recorded bits/Sec

    And from all reports, you need to buy a digitizer board (ADC),
    since the built-in one is mono, and voice-grade only--8 bits.
    
    Hmm, edit the music notation on the screen, play it as MIDI data
    into a bank of synthesizers and samplers, recording the audio
    output onto the optical disk, then go back and do fancy DSP
    tricks on the audio for final production.  Hmmm.... I already
    wanted one.
    
    	Eirikur
    

    re 2 
    
    I'm not sure I understand. 256 MB and 2.24696/sec would give us
    1.7 min playing time. 
    
    I seam to remember that a CD can hold 700 MB of info; does that
    mean 4.7 mins of music, something ain't right.
    
    But either way I guess we would need considerably more than a gigabyte
    opictical disc to make it interesting. But who nows it may be
    coming....we also started off with 5MB magnetic hard discs for the
    PC's.
    
    lars
    

    re: .4
   >     I'm not sure I understand. 256 MB and 2.24696/sec would give us
   > 1.7 min playing time. 
    
    Well, I think he said 2.24696 MBITS/sec, so it's really 1.7 * 8
    = 13.6 minutes.  I'll bet that's close.
    
    		Steve

>    And from all reports, you need to buy a digitizer board (ADC),
>    since the built-in one is mono, and voice-grade only--8 bits.

    The way I understand it is that the A/D uses something called mu-level
    encoding.  It is actually a 12 bit conversion but it allocates 8
    bits for 0-1 volt and 4 bits for the higher voltages.  But as you
    say it seems that it is aimed at voice-grade applications.
    
    -Eric

    
    A full lenght CD (approx 75 minutes) = 20Giga bits
    Your optical disk = 256MegByte hence =  2.0 Giga bits
    
    Then if you want the same resolution and same Error correction
    you will have close to 7.5 minutes.   That's more than 1 or 
    2 minutes as previously said.   By the way, anybody know the
    size of those $50.00 CDs?.  3�" or 5�" or 12"?
    I beleive our RV20 is 12" isn't it?
    
    Vince.

    Re: .-1
    
    Same error correction not needed.  The reason for so much error
    correction on a CD is because of the process used to produce them.
    They are *pressed* like records from a master!

    re -.1
    
    Ok, then if we don't use the error correction we don't use the
    eight-to-fourteen modulation then we can have 13.1 minutes on
    one disk. (26.2 mins if we use an 8 bits D/A).  We're getting
    there.  But $50 for 13.1 minutes...
    
    Vince.

    re:.8
    >    Same error correction not needed.  The reason for so much error
    >correction on a CD is because of the process used to produce them.
    >They are *pressed* like records from a master!
     
    I disagree.  Most of the errors encountered are soft errors, that
    is, read errors. The hard error count, which are errors on the disc
    itself, is much smaller than the soft error count.  Also, errors
    are more tolerable for music than for storing computer data, so
    the error correction may have to be even more robust than 14-8
    decoding.  Let me know if I am totally off-base with this.
    
    			Steve

    Re: .-1
    
    Well lets put it this way.  When DEC says one of its disks has a
    capacity of 256Mbytes, DEC means *formatted* capacity, which is how
    much *real* data can be put on the disk (before you count filesystem
    overhead).  A lot of competitors quote *unformatted* capacity, which I
    think is what you are implying NeXt is doing when it says 256Mbytes (?).
    
    256Mbytes * 8bits/byte = 2Gigabits

    Opps, finished my last reply too soon.  I calculate
    
    		~ 24.18minutes (24 minutes and 10.8 seconds)
    
    (44.1Ksamples/sec * 2 channels) * 2bytes/sample = 176400bytes/sec
    
    256Mbytes / 176400bytes/sec = 1451seconds
    
    1451seconds / 60seconds/minute = 24.18
    
    This assumes the disk has a *formatted* capacity of 256Mbytes and that
    the raw disk is being used (ie. no filesystem is being used).