The CD - a technical question

Discussion in 'Computer Support' started by Rob K, Jan 28, 2004.

  1. Rob K

    Rob K Guest

    I'm engaged in a pointless, but funny discussion about the weight of
    media.

    (copied, not verbatim, from http://www.howstuffworks.com: )

    "A CD has a long, spiraled data track, with 1s and 0s.

    In conventional CDs, these 1s and 0s are represented by millions of
    tiny bumps and flat areas on the disc's reflective surface. The bumps
    and flats are arranged in a continuous track that measures about 0.5
    microns across and 3.5 miles (5 km) long.
    To read this information, the CD player passes a laser beam over the
    track. When the laser passes over a flat area in the track, the beam is
    reflected directly to an optical sensor on the laser assembly. The CD
    player interprets this as a 1. When the beam passes over a bump, the
    light is bounced away from the optical sensor. The CD player recognizes
    this as a 0. "

    So - theoretically of course- would a manufactured (stamped) CD which
    contained only 0s be lighter than 'the same' CD containing only 1s ?
    It's tempting to say yes.

    Would the same apply to CD-R and CD-RW ?

    --
    My E-mail address in ROT-13:
     
    Rob K, Jan 28, 2004
    #1
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  2. Rob K

    °Mike° Guest

    On Wed, 28 Jan 2004 23:58:33 +0100, in
    <>
    Rob K scrawled:

    >I'm engaged in a pointless, but funny discussion about the weight of
    >media.
    >
    >(copied, not verbatim, from http://www.howstuffworks.com: )
    >
    >"A CD has a long, spiraled data track, with 1s and 0s.
    >
    >In conventional CDs, these 1s and 0s are represented by millions of
    >tiny bumps and flat areas on the disc's reflective surface. The bumps
    >and flats are arranged in a continuous track that measures about 0.5
    >microns across and 3.5 miles (5 km) long.
    >To read this information, the CD player passes a laser beam over the
    >track. When the laser passes over a flat area in the track, the beam is
    >reflected directly to an optical sensor on the laser assembly. The CD
    >player interprets this as a 1. When the beam passes over a bump, the
    >light is bounced away from the optical sensor. The CD player recognizes
    >this as a 0. "
    >
    >So - theoretically of course- would a manufactured (stamped) CD which
    >contained only 0s be lighter than 'the same' CD containing only 1s ?
    >It's tempting to say yes.
    >
    >Would the same apply to CD-R and CD-RW ?


    No, in theory, it would be heavier -- zeros equal bumps, equals
    more mass.

    --
    Basic computer maintenance
    http://uk.geocities.com/personel44/maintenance.html
     
    °Mike°, Jan 28, 2004
    #2
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  3. Rob K

    Rob K Guest

    °Mike° wrote:

    > No, in theory, it would be heavier -- zeros equal bumps, equals
    > more mass.


    OK - see what you mean, so you're also saying that the first thing the
    CD player reads is a 1, right ?

    Any thoughts on the CD-R and CD-RW, keeping the technical diffecences
    in mind ?

    Thanks as always, °Mike°.

    --
    My E-mail address in ROT-13:
     
    Rob K, Jan 28, 2004
    #3
  4. Rob K

    °Mike° Guest

    On Thu, 29 Jan 2004 00:11:32 +0100, in
    <>
    Rob K scrawled:

    >°Mike° wrote:
    >
    >> No, in theory, it would be heavier -- zeros equal bumps, equals
    >> more mass.

    >
    >OK - see what you mean, so you're also saying that the first thing the
    >CD player reads is a 1, right ?


    No, I never said that. The player has a 50-50 chance of reading
    either, first.

    >Any thoughts on the CD-R and CD-RW, keeping the technical diffecences
    >in mind ?


    I haven't gone into how CD-R/CD-RW work, but if they follow the
    same principal, then the same applies... in theory.

    >Thanks as always, °Mike°.


    YW.

    --
    Basic computer maintenance
    http://uk.geocities.com/personel44/maintenance.html
     
    °Mike°, Jan 28, 2004
    #4
  5. Rob K

    ICee Guest

    Rob K wrote:
    > °Mike° wrote:
    >
    >> No, in theory, it would be heavier -- zeros equal bumps, equals
    >> more mass.

    >
    > OK - see what you mean, so you're also saying that the first thing the
    > CD player reads is a 1, right ?
    >
    > Any thoughts on the CD-R and CD-RW, keeping the technical diffecences
    > in mind ?


    You might want to look at the explanation at the same website for how CD
    burners work:
    http://entertainment.howstuffworks.com/cd-burner.htm

    >
    > Thanks as always, °Mike°.


    --
    "God's Final Message to His Creation:
    'We apologize for the inconvenience.'"
    So Long, and Thanks for All the Fish
     
    ICee, Jan 28, 2004
    #5
  6. Rob K

    Rob K Guest

    °Mike° wrote:
    <snipped>
    >
    > No, I never said that. The player has a 50-50 chance of reading
    > either, first.
    >

    Now I'm confused.

    I *think* the CD player starts on a given flat surface, and I don't
    doubt, as you said that a 0 is a bump.
    Doesn't it follow then that the first value read must be 1 ?

    As I said, the discussion is pointless but funny IMHO ...

    :)

    --
    My E-mail address in ROT-13:
     
    Rob K, Jan 28, 2004
    #6
  7. Rob K

    Budweiser Guest

    "°Mike°" <> wrote in message
    news:...
    > On Thu, 29 Jan 2004 00:11:32 +0100, in
    > <>
    > Rob K scrawled:
    >
    > >°Mike° wrote:
    > >
    > >> No, in theory, it would be heavier -- zeros equal bumps, equals
    > >> more mass.

    > >
    > >OK - see what you mean, so you're also saying that the first thing the
    > >CD player reads is a 1, right ?

    >
    > No, I never said that. The player has a 50-50 chance of reading
    > either, first.
    >
    > >Any thoughts on the CD-R and CD-RW, keeping the technical diffecences
    > >in mind ?

    >
    > I haven't gone into how CD-R/CD-RW work, but if they follow the
    > same principal, then the same applies... in theory.
    >
    > >Thanks as always, °Mike°.

    >
    > YW.
    >
    > --
    > Basic computer maintenance
    > http://uk.geocities.com/personel44/maintenance.html


    using your "theory" a CD-R would be the same as a "stamped" CD,after all you
    have to start with some media,and once a CD-R has been written to it cannot
    be changed.
    However a CD-RW can be re-done --as in the 0 changed to a 1--so the location
    must be like a swing gate--open or closed---the gate is still there.So no
    weight difference.
     
    Budweiser, Jan 28, 2004
    #7
  8. Rob K

    Harrison Guest

    On Wed, 28 Jan 2004 23:58:33 +0100, Rob K <>
    wrote:

    >I'm engaged in a pointless, but funny discussion about the weight of
    >media.
    >
    >(copied, not verbatim, from http://www.howstuffworks.com: )
    >
    >"A CD has a long, spiraled data track, with 1s and 0s.
    >
    >In conventional CDs, these 1s and 0s are represented by millions of
    >tiny bumps and flat areas on the disc's reflective surface. The bumps
    >and flats are arranged in a continuous track that measures about 0.5
    >microns across and 3.5 miles (5 km) long.
    >To read this information, the CD player passes a laser beam over the
    >track. When the laser passes over a flat area in the track, the beam is
    >reflected directly to an optical sensor on the laser assembly. The CD
    >player interprets this as a 1. When the beam passes over a bump, the
    >light is bounced away from the optical sensor. The CD player recognizes
    >this as a 0. "
    >
    >So - theoretically of course- would a manufactured (stamped) CD which
    >contained only 0s be lighter than 'the same' CD containing only 1s ?
    >It's tempting to say yes.
    >
    >Would the same apply to CD-R and CD-RW ?

    These use dyes which change reflectivity to emulate the raised spots
    on a conventional cd. Since the burning process neither adds nor
    removes material, the cd weight is unaffected. In essence, a full cd
    weighs the same as an empty one.

    Regarding manufactured cd's from masters, see
    http://www.groovehouse.com/assets/pdfs/How_CDs_are_Made11.pdf
    Acrobat Reader required.
     
    Harrison, Jan 28, 2004
    #8
  9. Rob K

    °Mike° Guest

    On Thu, 29 Jan 2004 00:34:58 +0100, in
    <>
    Rob K scrawled:

    >°Mike° wrote:
    ><snipped>
    >>
    >> No, I never said that. The player has a 50-50 chance of reading
    >> either, first.
    >>

    >Now I'm confused.
    >
    >I *think* the CD player starts on a given flat surface, and I don't
    >doubt, as you said that a 0 is a bump.


    No, *you* said that 0 was a bump:
    " When the beam passes over a bump, the light is bounced away
    from the optical sensor. The CD player recognizes this as a 0. "

    Nowhere did you mention -- I haven't read the article -- that the
    CD player starts on a given flat surface. This postulation would
    be ridiculous, since the CD spins rapidly, and firmware/software
    then decides where the head is, and where to position it next.

    >Doesn't it follow then that the first value read must be 1 ?


    No. Where is the logic in this?

    >As I said, the discussion is pointless


    Yes.

    > but funny IMHO ...
    >
    >:)


    --
    Basic computer maintenance
    http://uk.geocities.com/personel44/maintenance.html
     
    °Mike°, Jan 28, 2004
    #9
  10. Rob K

    Rob K Guest

    °Mike° wrote:
    >
    > No, *you* said that 0 was a bump:
    > " When the beam passes over a bump, the light is bounced away
    > from the optical sensor. The CD player recognizes this as a 0. "


    True, be it that it was quoted.

    >
    > Nowhere did you mention -- I haven't read the article -- that the
    > CD player starts on a given flat surface. This postulation would
    > be ridiculous, since the CD spins rapidly, and firmware/software
    > then decides where the head is, and where to position it next.


    You're exactly pointing out why IMHO the discussion is funny. Meaning,
    there's always somehing to be learned.

    And you're absoultely right, it's *my* assumption that a CD player
    starts on a given flat surface, how else can the player detect a bump ?

    And I'm not certain that such a postulation is necessarily ridiculous
    (it may very well be ) - the CD player must have some notion of the
    cd's TOC ? How else would the software know where to move the head to
    or even, how else would the software know how to read the TOC in the
    first place ?

    :)

    --
    My E-mail address in ROT-13:
     
    Rob K, Jan 29, 2004
    #10
  11. Rob K

    Rob K Guest

    Budweiser wrote:
    <snip>
    > However a CD-RW can be re-done --as in the 0 changed to a 1--so the location
    > must be like a swing gate--open or closed---the gate is still there.So no
    > weight difference.


    Read the howstuffworks site.... :)
    I think you're referring to the technical differences mentioned there.

    A difference in dye, basically, if we're talking CD-R and CD-RW here,
    not in weight. So, yes I agree.

    --
    My E-mail address in ROT-13:
     
    Rob K, Jan 29, 2004
    #11
  12. Rob K

    Rob K Guest

    Rob K, Jan 29, 2004
    #12
  13. Rob K

    °Mike° Guest

    On Thu, 29 Jan 2004 01:17:46 +0100, in
    <>
    Rob K scrawled:

    >°Mike° wrote:
    >>
    >> No, *you* said that 0 was a bump:
    >> " When the beam passes over a bump, the light is bounced away
    >> from the optical sensor. The CD player recognizes this as a 0. "

    >
    >True, be it that it was quoted.
    >
    >>
    >> Nowhere did you mention -- I haven't read the article -- that the
    >> CD player starts on a given flat surface. This postulation would
    >> be ridiculous, since the CD spins rapidly, and firmware/software
    >> then decides where the head is, and where to position it next.

    >
    >You're exactly pointing out why IMHO the discussion is funny. Meaning,
    >there's always somehing to be learned.
    >
    >And you're absoultely right, it's *my* assumption that a CD player
    >starts on a given flat surface, how else can the player detect a bump ?


    Did you actually read the article?

    "To read this information, the CD player passes a laser beam over the
    track. When the laser passes over a flat area in the track, the beam is
    reflected directly to an optical sensor on the laser assembly. The CD
    player interprets this as a 1. When the beam passes over a bump, the
    light is bounced away from the optical sensor."

    >And I'm not certain that such a postulation is necessarily ridiculous
    >(it may very well be ) - the CD player must have some notion of the
    >cd's TOC ?


    That's not what you were saying though, is it? Obviously, there is
    a file system involved, but you were saying that the head starts on
    a flat surface, which is patently ridiculous, since it's a removable
    medium.

    > How else would the software know where to move the head to
    >or even, how else would the software know how to read the TOC in the
    >first place ?


    Like I said above, CDs use a file system (ISO 9660 / MAC HFS, etc.)

    --
    Basic computer maintenance
    http://uk.geocities.com/personel44/maintenance.html
     
    °Mike°, Jan 29, 2004
    #13
  14. Rob K

    Rob K Guest

    °Mike° wrote:
    <snip>
    >
    > Like I said above


    See what you mean °Mike°, rest assured, I know CDs have a file system.

    Methinks it's time to end the discussion, for now, at least.

    Here in Dutchland we've got a proverb saying something like "one fool
    can ask more questions than a thousand wise men can answer", and I'm
    sure you lot over there on the island to the west of us have got
    something similar.

    Thanks °Mike°, I'm off to you know where.

    --
    My E-mail address in ROT-13:
     
    Rob K, Jan 29, 2004
    #14
  15. Rob K

    Sultan Guest

    Harrison is absolutely correct, nothing is actually "added" to the cd mass
    wise so an empty cd weighs as much as a full one.

    Sultan



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    Sultan, Jan 29, 2004
    #15
  16. Rob K

    Jerry G. Guest

    When burning a CD, the laser etches in a pattern representing 0's and 1's.
    You can say that the 1's are more reflective than the 0's. This is detected
    as a 0 and 1 condition.

    There are a lot of very complex operations that go in to burning, and
    reading CD disks. The engineering behind this is very advanced.


    Read the link below. This has an excellent explanation. Just keep reading
    this page!
    http://www.repairfaq.org/sam/cdfaq.htm#cdtek


    A very good site that explains things very simply.
    http://music.concordia.ca/Programs/Electroacoustics/Compact_Disk_Intro.html


    There are also a number of standards for CD disks. Most consumer CD units
    are multicompatable.

    --

    Greetings,

    Jerry Greenberg GLG Technologies GLG
    =========================================
    WebPage http://www.zoom-one.com
    Electronics http://www.zoom-one.com/electron.htm
    =========================================


    "Rob K" <> wrote in message
    news:...
    I'm engaged in a pointless, but funny discussion about the weight of
    media.

    (copied, not verbatim, from http://www.howstuffworks.com: )

    "A CD has a long, spiraled data track, with 1s and 0s.

    In conventional CDs, these 1s and 0s are represented by millions of
    tiny bumps and flat areas on the disc's reflective surface. The bumps
    and flats are arranged in a continuous track that measures about 0.5
    microns across and 3.5 miles (5 km) long.
    To read this information, the CD player passes a laser beam over the
    track. When the laser passes over a flat area in the track, the beam is
    reflected directly to an optical sensor on the laser assembly. The CD
    player interprets this as a 1. When the beam passes over a bump, the
    light is bounced away from the optical sensor. The CD player recognizes
    this as a 0. "

    So - theoretically of course- would a manufactured (stamped) CD which
    contained only 0s be lighter than 'the same' CD containing only 1s ?
    It's tempting to say yes.

    Would the same apply to CD-R and CD-RW ?

    --
    My E-mail address in ROT-13:
     
    Jerry G., Jan 29, 2004
    #16
  17. Rob K

    nemo Guest

    Rob K <> wrote in message
    news:...
    > I'm engaged in a pointless, but funny discussion about the weight of
    > media.
    >
    > (copied, not verbatim, from http://www.howstuffworks.com: )
    >
    > "A CD has a long, spiraled data track, with 1s and 0s.


    "Spiraled" is incorrect, unless you're saying something like, "The trach was
    straight, but now it's been spiralled."

    Spiral is OK here.
    >
    > In conventional CDs, these 1s and 0s are represented by millions of
    > tiny bumps and flat areas on the disc's reflective surface. The bumps
    > and flats are arranged in a continuous track that measures about 0.5
    > microns across and 3.5 miles (5 km) long.
    > To read this information, the CD player passes a laser beam over the
    > track. When the laser passes over a flat area in the track, the beam is
    > reflected directly to an optical sensor on the laser assembly. The CD
    > player interprets this as a 1. When the beam passes over a bump, the
    > light is bounced away from the optical sensor. The CD player recognizes
    > this as a 0. "


    Odd way to describe it.

    They're not bumps, they're pits in a flat surface. A pit is a 1, and the
    absence of a pit is a 0. At least I think it's that way round.

    The track is read by means of a fluctuation in the reflected light caused by
    the bottoms of the pits being further away from the LASER assembly and
    therefore out of focus compared with the surface.

    > So - theoretically of course- would a manufactured (stamped) CD which
    > contained only 0s be lighter than 'the same' CD containing only 1s ?
    > It's tempting to say yes.
    >
    > Would the same apply to CD-R and CD-RW ?
    >

    I'm not sure, but I think with these, the reflected light from the
    equivalent of a pit is of a different polarity.
     
    nemo, Jan 30, 2004
    #17
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