I have a question about the CD storage medium - if anyone knows this, I'd appreciate any info. I'm interested in knowing if there is a chart anywhere that shows where, physically, on a CD-Rom, CD-Audio, etc. A certain percentage of the track would be.

i.e. if you took a full 80 min. Audio CD, and found the exact point on the disc where 50% of the track would be, where would this be on the physical CD? I'm guessing that it would be obut 2/3 of the way towards the edge of the disc.

Hopefully what I'm talking about is fairly clear. The reason that I want to know this is so that I can diagnose CDs that are problematic when ripped with EAC. I have the capability to repair the polycarbonate side of a disc, so if I see that a disc has a problem at, say, 68%, it'd be nice to know where to look for the problem on the physical disc.

I'm not entirely sure Supacon, but I think I read that a CD player reads from inside out.

In other words, track 1 is closest to the center and the last track is further out.

How "far out" the last track is depends on the total time of the disc.

If I am correct, an error occuring during the last track of a 79min disc may be a scratch be near the outer edge(assuming an 80min disc.)

sorry I can't be more exact..

afaik, chemeye is right, cd info is written and read from the inner rings of the cd, to the outer rings. i too, dont have a specific answer to your question, but a relatively easy way to find out. just take a cd-rw or a cd-r, and burn 40 mins of music on it (assuming it is an 80 min cd) and look at the underside of it. on cd-r's and cd-rw's, you will notice a visible difference in where the data is written, and where it is blank. where the change in appearance occurs would be the 50% mark. oh, and if you are using a cd-rw, be sure you do a full erase on it before you do this test, because w/ quick erases, the visible marking of the data isn't removed. correct me if im wrong anybody...

Well the point where the problem occurs is at pi.r^2, therefore u have have the distance from the middle. Don't forget to also factor in the middle of the cd where there is no data.

VCSkier's suggestion would give you one data point per CD-R. You might get more data points by doing something like the following.

Burn a full length CD-R
Use a marker pen such as a black extra fine tip Sharpie and make a series of marks on the CD. Say about 10 of them, more or less equally spaced, about 1/4 in long, tangent to the track.
Rip the CD. I would set the retry count of the ripper to minimum to keep it from retrying forever. In this case you want to see errors.
Look at the times where the ripper reports errors. If the number of errors, or groups of errors, matches the number of marks you made, you can make a table of time vs location.

This would let you locate the defect to a relatively narrow band. If you want to locate the bad spot rotationally, I don't have a clue.

You can also burn regularly timed tracks of music, with long gaps (4 seconds?) between tracks.

Silence will show up in a different colour.

I never managed to get those rings which pressed CDs have. I tried with three writers so far in both DAO and TAO mode.

The visibility strongly depends on the dye type - in my experience, a 2 second gap is easily visible with AZO or Cyanine dyes, but very hard to spot with Phtalocyanine.

The information starts 25 millimeters from the center of the CD (+0 / -0.2 mm). According to the CD, 63 to 90 minutes can be stored between 25 and 58 millimeters from the center.

Thanks for the ideas guys. There are some ways to determine this manually, but does anyone know of a formula that actually determines the radial distance (from the inner to outer edge of a disc) and the relationship between a given radial distance to the length of linear distance at that point? I guess I'm not very geometrically sophisticated...

A CD actually is read at varying speeds at different points on the disc, right? Constant Linear Velocity, where the disc rotates faster when the inner tracks are being read than when the final, outer tracks are being read.

Given this, I'm guessing that, the first track actually takes quite a bit more radial distance on the CD than the last track on the edge... say, if you created a disc with 20 4 minute tracks, (the point being that they are all identical).

If one knows exactly the very detailed specifications of Compact Discs and the technical details of how they are read, it probably wouldn't be a complicated matter to give some kind of a calculus type equation to effectively answer my question very precisely, and generate some kind of graph.

Any math wizards around here?

Alright... I dug up some specs, and I'm trying to wrap my head around this. At this point note that I'm doing this just for the sheer nerdy academic thrill of calculating stuff for no reason at all.

A CD, obviously has 44.1KHz 16 bit Stereo audio data on it, and is divided into frames of 1/75th of a second. Do the math, and that gives you 2353 byte frames. A CD has a data transfer rate of 150kB/s (I calculated 176,400 Bytes per second to transfer the audio data at the rate that you hear it, but I have yet to verify that this is indeed the precise transfer rate.)

The track pitch (distance between tracks, from the middle of one to the middle of the next) is an average of 1.6 microns for most CDs. (This might be less for 80 min, 90 min CDs, etc).

A CD rotates at approximately 500 rpm when reading the center, and the rotation slows to 200 rpm at the outer edge of the disc.

Now, knowing the transfer rate, and track pitch and rotational speed, it should be a cinch to devise a formula that can generate the nice pretty graph I'm after

edit: The minimum pit/land length is 0.83 um. That means that a 74 minute CD has a track that is over 5 Km long, if you were to unravel it!

Damn, I think I actually found it. Now if only I hadn't slept through my integral calculus classes...

http://www.projectintermath.org/docs/4406.pdf

I need to find some fancy mathematical graphing software, and then I can actually do the graph that I had in mind. The formulas I need to do this are near the end of this PDF.

wow, im glad i stopped math at pre-calc and stats. my head hurts looking at that stuff...

You should be able to figure defect radius as a function of time just by area without delving into distance along the track etc. Just figure out the seconds/sq mm ratio, which you can get from measuring the starting and ending radii of the pressed area of a commercial CD (if you can't see the difference between burned and unburned on your CDR's) and knowing the total play time.

You don't really need calculus, just simple geometry.

OT: You mentioned CLV vs CAV. Laserdisks (predecessor of DVD) were available as either CLV or CAV. CLV could store 2x as much video on a side, but CAV was set up so
a video frame took exactly 1 turn of the disk. So for freeze-frame the head could just stay still and read the same frame over and over.

DonP, you're probably right, but what fun would that be?

I probably lack the mathematical sophistication to do either, to be honest, but if I plug away at this for a while, I can probably generate some kind of a graph. It's kind of interesting for me... I always have wondered about how CDs work, and now I know very intimately. This equation could allow me also to calculate curiousities such as "How much bigger would a CD have to be to get 99 minutes of audio with the same pit/land length and track pitch?".

I have yet to verify this, but I believe that they could have made CDs another 5mm bigger in diameter and easily made them fit 99.9Minutes (Almost 880 MB) of audio with the same technology, which seems like a much nicer, more round number than 74 min.. Why they decided on the size they did might be interesting to me...

You don't need all that formulas for that. As DonP said simple geometry would do:

From the pdf you linked:
R1=58mm (disc outer radius)
R2=22,5 mm (disc inner radius)

The discs usable surface would then be:
S = pi*R1^2-pi*R2^2 = 2857,75*pi mm^2

This surface area would correspond to 74 mins of audio data or 80 mins of audio data, depending on the CD type. Now if for example you have an 74 minute CD with 60 mins of audio data and you would want to find out in what distance from the center of the disc 60% of the data is then the corresponding surface up to that point would be: (60/74)*0,6*2857,75*pi = 1390,26*pi mm^2.
To find out where that is you just have to solve this equation:

pi*Rx^2-pi*R2^2 = 1390,26*pi => pi*Rx^2-pi*22,5^2 = 1390,26*pi => Rx = 43,55 mm

So to get to 60% of the 60 minutes you would need to measure 43,55mm from the centre of the disc or (43,55-22,5=) 21,05mm for the start of the data region. I don't think it is possible to locate the exact point on the disc but this would give you the radius of where the problem is.

Note: I use commas as decimal separators.

One criterion was to fit in a DIN size car player. I don't know if fitting a standard computer HD bay was a consideration. Legend has it that the play time was chosen to fit the President of Sony's favorite symphony.

Ballpark to get an extra 25% area (compared to an 80 minute CD) you would need 12% increase in diameter, which would be close to 15 mm. A bit less since you don't have to scale up the center region, but not all that much different.

I am not really sure what the difference is between 74 and 80 minute CD's.. do they burn closer to the edges, or a tighter track pitch? If the technology initially wasn't there to do 80 minutes then to have 100 minutes as the available time on first release of the format, you would need 33% extra data area.

The answer to that would be:
pi*Rx^2 - pi*R2^2=(99,9/74)*2857,75*pi => Rx = 66,06 mm
if you consider the same track pitch as a 74 minute CD. I don't know for sure but I think 80 minute CDs have a tighter track pitch, not burning closer to the edges (AFAIK that is what overburning is). In that case if you would use the same track pitch as an 80 minute CD that would be:
pi*Rx^2 - pi*R2^2=(99,9/80)*2857,75*pi => Rx = 63,84 mm

Hmm... so basically, you'd need a disc that is 13cm wide instead of 12cm wide. I would think that such a size would still work in computers and car stereos (but it would be a wee bit tight, perhaps).

For Random Access applications, a bigger disc would probably mean somewhat slower access, in a CLV environment. (BTW, interesting comment about Laser Disc, don. I used to have one here but had nothing to play it in! Out of curiousity, does anyone know what the data capacity of a laserdisc is?)

I guess we're pretty much stuck with that disc diameter anyways, for any future optical formats that come out. Can't wait until I have a BD-ROM player in my car!

Digital audio was added a few years into the format, but the video signal is analog, so it it is a bit like asking the data capacity of a vinyl record.

So if you wanted to push it maybe you could format the audio track for data (sort of as they did when they added DTS), then store data on the video signal as with PCM-1 (digital recording on beta video tape) then I think you would end up with a couple of gigabytes.

I heard once that laserdiscs were analog, but I never really understood how you could have analogue video... er... how could an optical storage medium be analogue? Instead of pits and lands, you have grooves with varying heights?

I don't know what the physical profie of the groove is, but the signal picked up by the
laser reflection is basically a ~9 MHz wide signal with frequency modulated (FM) cariers for video, audio, and digital audio. If all carriers are FM, then changes in signal amplitude should not be critical.

laserdisc


While looking around I found this. There was a short lived video disc format hat actually used a stylus to read the groove:

CED

But legend is probably wrong : www.snopes.com/music/media/cdlength.htm

A bit OT, but I don't really trust anything on snopes.com. Too many of their articles are on the side of "this sounds like a myth, so it can't be true." They have to denounce nearly everything as false to make their website interesting.

I recall that the user data starts at 25 mm from the center, not 22.5.

80 minutes CD have a tighter groove, and maybe also run beyond 58 mm. It seems that the ThinkXtra CDR 80 are in fact 99 minutes CDR with a faster linear velocity. 1.5 m/s instead of 1.2.