Velocity Reviews > The CD - a technical question

# The CD - a technical question

Rob K
Guest
Posts: n/a

 01-28-2004
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 ?

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°Mike°
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Posts: n/a

 01-28-2004
On Wed, 28 Jan 2004 23:58:33 +0100, in
<(E-Mail Removed) id>
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.

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Rob K
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Posts: n/a

 01-28-2004
°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°.

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°Mike°
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Posts: n/a

 01-28-2004
On Thu, 29 Jan 2004 00:11:32 +0100, in
<(E-Mail Removed) id>
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.

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ICee
Guest
Posts: n/a

 01-28-2004
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°.

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So Long, and Thanks for All the Fish

Rob K
Guest
Posts: n/a

 01-28-2004
°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 ...

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Budweiser
Guest
Posts: n/a

 01-28-2004

"°Mike°" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
> On Thu, 29 Jan 2004 00:11:32 +0100, in
> <(E-Mail Removed) id>
> 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.

Harrison
Guest
Posts: n/a

 01-28-2004
On Wed, 28 Jan 2004 23:58:33 +0100, Rob K <(E-Mail Removed)>
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

°Mike°
Guest
Posts: n/a

 01-28-2004
On Thu, 29 Jan 2004 00:34:58 +0100, in
<(E-Mail Removed) id>
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 ...
>
>

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Rob K
Guest
Posts: n/a

 01-29-2004
°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 ?

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