Question…a blank cdr contains 74 minutes or 650 mb. So WHY can you burn a music cd that contains 800 plus mb on a 650 disk? My son proved it to me by making a compi;lation cd with 19 songs (72 minutes) with 823 mb…I don’t understand!
Well, the reason is formatting. The CD was originally made to hold 74 minutes of music. I doubt that 823 MB would fit on a 74 minute CD, because usually CD quality audio is almost exactly 1MB per minute, so a 74 minute CD usually holds about 720 MB of music. However, it’s possible he used an 80 minute CD, which will fit around 800+MB of music. The reason it only holds 650 MB (74 min) or 700MB (80 min) is because the ISO9660 formatting takes up about 75MB. Essentially, a CD-ROM can only hold 650MB of data, but 720 or so MB of audio. Odd, but that’s the way it is. If the disc truly is only 72 minutes and has 823MB, then some of the original files may have been at a higher bit rate than the CD-audio standard, because usually you can’t go much over 700.
Jman
Either the burning program is mis-representing or mis-calculating the total space. 72 minutes of CD audio is not 823 megs of audio data no matter what the burning program is claiming.
It’s a format thing. Odds are, before burning the music files onto the CD, they were on your hard drive in AIFF format, which is (for reasons I don’t fully comprehend) a little bigger per file than the same music on an audio CD. Wish I could tell you more. Anyway, nothing’s gained or lost in the conversion from AIFF to audio CD or the other way 'round, so don’t sweat it.
Let’s do some basic math:
time in minutes * 60 sec/min * samples/sec * bytes/sample * number of channels
72 * 60 * 44100 * 2 * 2 = 762,048,000
add overhead.
OTOH use compression and you’ll get ten times as much music on a CD
A typical file stored on, say, an 8Gb drive, will be larger than the same file stored on a smaller storage volume like a 650Mb CD.
I’ve seen this happen plenty of times. For example, I have a file on my 8Gb drive and it measures 2Mb, but I put it on a zip disk and it only measures something like 1.5Mb. It all has to do with file allocation blocks and other techy crap that I won’t go into here.
To verify this effect, go measure the size of a file on your hard disk. Then burn it to a CD. The file on the CD should be smaller than the original on your hard drive, even though it contains exactly the same amount of data. Now copy the file back from the CD to the hard drive. It should be as big as the original file was on your hard drive (exactly as big). Compare the original file to the file put on a CD and copied back. They will be identical. The excess size is just filesystem overhead.
Think of your hard drive as a bunch of ice cube trays. If the tray has any water in it at all it is considered full. Regardless if it is entirely full or not. I don’t know off the top of my head what the “ice cube tray size” is for PC’s or Mac’s off the top of my head, but for argument sake lets say its 10 Mbs. Well if you have a file that is 8 Mbs it will take up one “tray” (or block of your hard drive). Thus it will look like a 10 mb file. So if you have 10 8 mb files, they will take up 100 mbs instead of the 80 youwould expect. This doesn’t apply to small external drives, or rather it applies somewhat differently and with CD burners it does not apply at all. There for a file that looks like a 100 MB file might actually be closer to 80 or 90 mbs.
This logic is part of the reason that defragmenting regularly is partucularly important to folks who work in audio (such as myself) or other high data mediums, you can turn your hard drive in to swiss cheese pretty quickly and having your audio files spead out over a 20 gig drive is a good way to burn a coaster instead of a CD.
And addressing the Audio compression issue for a moment… I rarely deal with MP3s or any other compressed audio because the sound quality is so attrocious, but commercial CD players (as far as I am aware) do not read compressed audio. so when you burn an audio CD from an MP3, the computer (I suspect) has to convert the MP3 to a 44.1 Khz/ 16 bit file, as this is the only format currently interpretable by consumer CD player. I bring this up because I am skeptical that an audio CD can contain more than 80 minutes of audio, reguardless of the source format.
I could certainly be wrong.
I deal with MP3 alot and personaly can’t tell the difference between a high quality MP3 and the CD track it is from but I don’t want to hijack this thread.
Most recording software that will put a MP3 onto an Audio CD decompresses it on the fly to 44.1Khz 16 bit stereo stream but that is not what the OP was about.
Depending on the brand of burner, brand of cd, and media that you are recording onto you can do what is called Overburning. It will allow you to write more than 74 minutes of audio onto a CD. I use no name generic cd’s and can get 80-82 minutes on them and they are labeled 74minute cd’s. While there are 80 minute cd’s I don’t know if they can be overburned.
Hope this helps
I love MP3s but there is a quite noticeable difference on music with a wide dynamic range (esp classical) played on a high quality home stereo system even using a MP3 recorded at the highest bit rates. That being said, the range of much popular music (esp much alternative, rock and hip hop) is compressed, even in CD format, to give it whatever tonality they are looking for in the studio, so on some stuff there may in fact not be much difference of CD vs MP3 especially if played on a PC sound system.
As a related point even the best PC based sound system (and I have both good PC and home sound systems to reference) is going to noticably worse in sonic accuracy and range than even a medium to medium-high quality home component based sound system (say 2,000 to 5,000 range).
Bocephus, what cd program are you using? DiskJuggler is great.
Also, DiskJuggler can burn a cd of 1 gig from a regular cdrom. You can write it but can’t read it but a 1 gig cdrom reader can.
There is a program on the net that can burn 1.2gig of data to a regular cdrom with some compression.
Per Handy’s Diskjuggler suggestion they have an excellent overview of CD recording concepts, media format types and CDRW technology explanation(s).
See http://www.padus.com/CdjHtml/cdjAdvanced_Concepts.htm
Sample below
Storage Capacity
Very similar to the “old” vinyl, the compact disc is formatted in a contiguous spiral of blocks (CD sectors) of 2352 bytes each, going from the center hole to the outer diameter. The block at logical address 0 (beginning of the disc) is located near the center of the disc; the last addressable block (end of the disc) is located near the outer edge of the disc.
The common unit of measurement for CD capacity is time. The following are the formulas to convert time in blocks:
minute = 60 seconds
second = 75 blocks (frames)
block = 2352 bytes
Blank discs are usually available in three formats:
21 minutes = 94500 blocks
63 minutes = 283500 blocks
74 minutes = 333000 blocks
The size of the block is a direct consequence of the way the analog audio signal is converted in digital samples.
The audio data is sampled at 44.1 kHz, 16 bits, 2 channels (stereo):
(44.1 x 1000) x 16 x 2 = 1,411,200 bits/sec
441,200 / 8 = 176,400 bytes/sec
176,400 / 75 = 2352 bytes
Duplicating Digital Audio Discs
The audio CD standard was designed from the beginning for sequential access (audio streaming) only; the digital audio was intended to be read in real time, converted to an analog signal and sent immediately to a stereo amplifier.
Reading audio with random access and moving digital data over the data bus without converting it to an analog signal, is a relatively new feature added only recently to CD-ROM drives.
The main problem with reading CD audio data over the data bus is the low degree of seek accuracy provided by CD-ROM drives when accessing CD audio blocks. This is caused by the fact that CD audio data are stored in a different format that CD computer data.
In a computer data block there are 2,048 bytes of user data plus header and error correction information. The header information in a computer data block contains the precise address of the block allowing the drive to precisely seek the correct block before reading.
In an audio data block, the 2,352 bytes of the physical block are entirely filled up by audio data. There is no header containing the block address, no synch codes and no error correction. This means that the drive must use the Q sub-code information to find an individual block. Unfortunately, the Q sub-code information was only designed to allow consumer audio CD players to provide audio positioning and position display within an accuracy of ± 1 second. As a result, Q sub-code addressing is only approximate.
When searching for a specific audio data block, a CD-ROM drive moves the laser assembly to a position near where the block should be located, starts reading, and compares the Q sub-code information to the desired block address. When a Q sub-code address close to the desired block address is located, the drive begins transferring data (or playing). Most CD-ROM drive specifications state that “the actual starting audio address will be within ± four (4) Q sub-code addresses of the requested starting audio address”, in other words ± 4 audio blocks or ± 4/75th of a second. As a result, given the address of a single block, a read request might return any one of 9 blocks (according to drive manufacturer specifications). Some drives may be even less accurate than ± four (4) Q sub-code addresses.
A second problem with CD audio data capture occurs when the computer cannot accept audio data from the drive fast enough. This is referred as a “buffer overflow”, because the CD-ROM drive must write data into its internal buffer before the computer has finished retrieving data already in the buffer. During a CD duplication process, a buffer overflow condition is usually a direct result of the CD-ROM drive reading faster than the CD Recorder writes. It is common to see 32X or 36X CD-ROM drives while the fastest CD Recorders are usually 4X-6X.
When a buffer overflow occurs, the read operation must be restarted. The next read may not begin transferring data at exactly the same sample (a side effect of the, ± 4 audio blocks accuracy problem) resulting in a few lost or repeated samples. This lost or extra data can create audible artifacts in the resulting sound file.
Because of the previous problems, it is not possible to reliably read accurately an audio stream without a re-matching or re-synchronization procedure. DiscJuggler employs three different re-matching methods to read audio reliably. See the “Audio Resynchronization” section in the “Copying Compact Discs” chapter for a detailed description.
Handy, I think you are missing the point that we are talking about CD audio format. Of course if you use compression you can get much more in. It has already been mentioned. But CDaudio is CDaudio.
Bocephus–a clarification, please.
Was the ORIGINAL material 823 MB on your hard drive, or did the CD say it was 823 MB?
If you can pop the CD into your computer and read how much is allegedly stored on the disc, we’ll be able to know whether the computer somehow squeezed the original bulky data or if the CD inexplicably holds more than its advertised capacity.
I thank all who responded to my question! Now a clarification…the CD writer is a HP CD-Writer plus 8200 series, using NERO software. The CD burnt was MP3’s coverted to wave format writing to a maxell CDR-74. NERO indicated 72 minutes with 751 MB not the 800 plus I previously mentioned.