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  1. Over the past couple of weeks it seems to me quite a few people have been giving out misleading information with regard to cd's.Some posts have stated that 74 min cd = 740mb , 80 min = 800mb etc and according to the Official CD Meter Sizes this is not true.

    This is the so called List:

    18 min = 158mb
    63 min = 553mb
    74 min = 650mb
    80 min = 703mb
    90 min = 823mb
    DVD-R = 3.95GB

    So its a myth after all that 800mb will fit on a 80 min cd,I WONDER WHAT THE QUALITY IS LIKE?



    PATSYM1
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    DVD-R's store 4.3GIGS by the Windows Standard (1024)

    DVD-R"s store 4.7 Gigs by the DVD Standard (1000)
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    Whatever the "Official CD Meter" is it must report "mode-1" sizes for CD's. Generally "mode-1" sizes are"stamped" on the CD or CD container. VCDs and SVCDs are recorded "mode-2" A 80 minute "mode-1" CD holds approximately 800mb when recorded "mode-2". Likewise, a 74 minute "mode-1" CD holds approcimately 740mb when recorded mode-2.
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  4. Originally Posted by patsym1
    So its a myth after all that 800mb will fit on a 80 min cd,I WONDER WHAT THE QUALITY IS LIKE?
    No it's not a myth, like hwoodwar states it depends whether you are burning using mode1 or mode2.

    Data CD is mode1 VCD/SVCD is mode2, search the forum for mode1 mode2, there are many posts on this subject.
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  5. Originally Posted by patsym1
    So its a myth after all that 800mb will fit on a 80 min cd
    Sigh...

    I cannot recall how many times I have to overturn THIS myth... Please try to research a topic fully before posting something like this. An indepth perusal of the CD-R FAQ would have made everything clear....

    ----------------------------------------------------------------

    The capacity of a CD disc
    Let us ignore for now the terms of megabyte for CD capacity and try to understand how the data is stored on a CD.

    As well all know, the data is stored digitally as binary data. This means, however the actual information is actually kept on the disc, this information is in the form of "1"s and "0"s. Physically, the information on a CD is as pits on a thin sheet of metal (aluminium).

    An a CD-R disc, the data is physically on an organic dye layer which simulates the metal layer on a real pressed CD.

    ----------------------------------------------------------------

    How is the information structured
    Now, on the CD, the information isn't just organised from beginning to end willy-nilly. Otherwise, it would be really hard to find a useful piece of information on the CD.

    Rather, the information is organised in sectors. Consider a sector as like a page in a book. Just like you are able to quickly find something in a book if you know the page number, you can quickly find something on a CD if you know the sector number.

    Now, remember that the CD was original made to hold audio data. It was decided, that the CD would would 75 sectors per second of audio. Although I cannot guess where this number comes from, it is quite appropriate for the 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 the 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 there 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 burning
    Now, audio CD was well and good, but the medium would become much more useful if you could store other data on the disc as well. This became to be know as CD-ROM of course.

    Now, the audio-CD uses the ENTIRE sector for audio data.

    However, for CD-ROMs this caused a problem. Simply, CDs and the CD reading mechanisms were not 100% faultless. That is, errors (indeed frequent errors) could be made during the reading. For audio CDs, this does not matter as much as you could simply interpolate from the adjacent audio samples. This will obviously NOT DO for data CDs. A single bit error could lead to a program being 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 the 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 burning mode is either MODE1 or MODE2 Form1.

    ----------------------------------------------------------------

    MODE2 Form2 sectors of VCDs and SVCDs
    Now, for VCDs and SVCDs, the video tracks do not necessarily require the robust error correction as normal data on a CD-ROM. However, there is still some overhead per sector that is used for something other than video data (e.g., sync headers).

    S/VCDs 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 the capacity of a disc is the total number of sectors it contains. This is similar to the total number of pages in a blank exercise book (if you recall the book analogy).

    The secondary determinant is the burning mode of the disc.

    For audio CDs, it is as if you could fill each page from top to bottom with audio data as the entire sector is used for audio data.

    For CD-ROMs, it is as if you need to first rule a margin and then leave the bottom part of each page for footnotes (headers + ECC + EDC). The amount of text you can actually write per page is then less due to these other constraints.

    For S/VCDs, we still need to rule a margin on the page, but we don't have to worry about the 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 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

    ----------------------------------------------------------------

    Conclusions
    As you can see, the often quoted capacities of 650MB and 700MB refer to CD-ROM capacities.

    Due to the fact that S/VCDs use a different burning mode where MORE of each sector is available as user data, the relatively capacities are HIGHER.

    Now, since S/VCDs are not composed of PURELY video tracks and have some unavoidable overheads, the 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 the often quoted capacities of 740MB and 800MB come from. They are quite accurate.

    All these capacities are available BEFORE overburning. Overburning is where you burn MORE sectors than the disc is rated for. If you overburn, you can typically achieve about 1-2 minutes of additional capacity (depending on your drive and media).

    I hope this finally puts this topic to rest!

    Regards.
    Michael Tam
    w: Morsels of Evidence
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    I hate it when people get a little bit of knowledge and think they know it all.

    Dont believe stuff people tell you or stuff you read, (especially if your gonna tell other people) untill you've tried it yourself, and if your gonna read up on something, dont read half of it.

    I found out you got 795mb on an 80min CD just because i tried it and it worked, not because someone told me or i read it on some website.

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  7. Yeah Martyn and overburning erases all that hard work, eh?

    Michael - think about cut & paste into a User Guide please. Then we can link to Newbie Guides to the user guide.
    Panasonic DMR-ES45VS, keep those discs a burnin'
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  8. (Dont believe stuff people tell you or stuff you read, (especially if your gonna tell other people) untill you've tried it yourself, and if your gonna read up on something, dont read half of it. States Martyn)

    And why should we believe you Martyn? If you read the post properly you will find what I SAY IS ACCURATE its not possible to put 800mb on an 80 min cd like some people in this forum are stating and I ALSO BELIEVE the Moderators should put a stop to it especially as it is aimed at Newbies.


    patsym1
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  9. ok here's the bottom line SOMETHING TO TRY.
    create a 800mb mpg which is TO THE VCD SPEC!
    import it to Nero and try to burn it.
    You'll find that it does burn quite nicely to a 700mb CD.
    PROOF THAT 800MB does fit on a 700mb CD.
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    Originally Posted by patsym1
    you will find what I SAY IS ACCURATE its not possible to put 800mb on an 80 min cd like some people in this forum are stating and I ALSO BELIEVE the Moderators should put a stop to it especially as it is aimed at Newbies
    Ok, I don't know why Martyn is flamebaiting, but what everyone else says is correct. Your statement isn't true unless you specify that you can't fit 800MB of *data* on a disc. I burn ~800MB of SVCD onto CDR all the time. I think there is a sufficient amount of coverage of the topic to help newbies understand that they can fit different amounts of data versus audio or video on discs (when burned in the appropriate mode). I'm also pretty sure the guides explain this.

    So if anyone is saying you can fit 800MB of data on a 80 min disc, then they are newbies themselves and shouldn't be so eager to share their misinformation. The moderators and experienced users do their best to correct misinformation.
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  11. Thank you Thorn at least someone is getting the gist of what I have been saying.





    patsym1
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    Originally Posted by vitualis
    Originally Posted by patsym1
    So its a myth after all that 800mb will fit on a 80 min cd
    Sigh...

    I cannot recall how many times I have to overturn THIS myth... Please try to research a topic fully before posting something like this. An indepth perusal of the CD-R FAQ would have made everything clear....

    ----------------------------------------------------------------

    The capacity of a CD disc
    Let us ignore for now the terms of megabyte for CD capacity and try to understand how the data is stored on a CD.

    As well all know, the data is stored digitally as binary data. This means, however the actual information is actually kept on the disc, this information is in the form of "1"s and "0"s. Physically, the information on a CD is as pits on a thin sheet of metal (aluminium).

    An a CD-R disc, the data is physically on an organic dye layer which simulates the metal layer on a real pressed CD.

    ----------------------------------------------------------------

    How is the information structured
    Now, on the CD, the information isn't just organised from beginning to end willy-nilly. Otherwise, it would be really hard to find a useful piece of information on the CD.

    Rather, the information is organised in sectors. Consider a sector as like a page in a book. Just like you are able to quickly find something in a book if you know the page number, you can quickly find something on a CD if you know the sector number.

    Now, remember that the CD was original made to hold audio data. It was decided, that the CD would would 75 sectors per second of audio. Although I cannot guess where this number comes from, it is quite appropriate for the 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 the 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 there 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 burning
    Now, audio CD was well and good, but the medium would become much more useful if you could store other data on the disc as well. This became to be know as CD-ROM of course.

    Now, the audio-CD uses the ENTIRE sector for audio data.

    However, for CD-ROMs this caused a problem. Simply, CDs and the CD reading mechanisms were not 100% faultless. That is, errors (indeed frequent errors) could be made during the reading. For audio CDs, this does not matter as much as you could simply interpolate from the adjacent audio samples. This will obviously NOT DO for data CDs. A single bit error could lead to a program being 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 the 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 burning mode is either MODE1 or MODE2 Form1.

    ----------------------------------------------------------------

    MODE2 Form2 sectors of VCDs and SVCDs
    Now, for VCDs and SVCDs, the video tracks do not necessarily require the robust error correction as normal data on a CD-ROM. However, there is still some overhead per sector that is used for something other than video data (e.g., sync headers).

    S/VCDs 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 the capacity of a disc is the total number of sectors it contains. This is similar to the total number of pages in a blank exercise book (if you recall the book analogy).

    The secondary determinant is the burning mode of the disc.

    For audio CDs, it is as if you could fill each page from top to bottom with audio data as the entire sector is used for audio data.

    For CD-ROMs, it is as if you need to first rule a margin and then leave the bottom part of each page for footnotes (headers + ECC + EDC). The amount of text you can actually write per page is then less due to these other constraints.

    For S/VCDs, we still need to rule a margin on the page, but we don't have to worry about the 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 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

    ----------------------------------------------------------------

    Conclusions
    As you can see, the often quoted capacities of 650MB and 700MB refer to CD-ROM capacities.

    Due to the fact that S/VCDs use a different burning mode where MORE of each sector is available as user data, the relatively capacities are HIGHER.

    Now, since S/VCDs are not composed of PURELY video tracks and have some unavoidable overheads, the 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 the often quoted capacities of 740MB and 800MB come from. They are quite accurate.

    All these capacities are available BEFORE overburning. Overburning is where you burn MORE sectors than the disc is rated for. If you overburn, you can typically achieve about 1-2 minutes of additional capacity (depending on your drive and media).

    I hope this finally puts this topic to rest!

    Regards.

    Great explanation, vitualis!
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  13. Originally Posted by patsym1
    And why should we believe you Martyn? If you read the post properly you will find what I SAY IS ACCURATE its not possible to put 800mb on an 80 min cd like some people in this forum are stating and I ALSO BELIEVE the Moderators should put a stop to it especially as it is aimed at Newbies.
    It is possible in mode2 for vcd/svcd (well 797.9MB, lets not split hairs, 800MB will fit with slight overburn)

    Where has anyone said that you can fit 800mb burnt as data mode1 onto an 80min cdr.

    Did you read vitualis' post ?
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  14. @Virtualis

    Can U or have U put this infomation into an easy to access place (not a forum thread)? And can you also explain the different modes on DVD-+R like UDF and ISO what is the difference when it comes to bits and bytes?

    By the way...best explaination I have ever seen on the subject...
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    And why should we believe you Martyn? If you read the post properly you will find what I SAY IS ACCURATE its not possible
    I never asked anyone to believe me, i asked people to find out for themselves before telling others what to do.

    Ok, I don't know why Martyn is flamebaiting
    How am i flambaiting, i only stated a point, if i came across as maybe a bit pissed off, thats because of patsym's attitute that hes right and everyone else is wrong, and he obviously hasn't tried it himself, thats all.

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