DVD-ROM physical sector size

Can anyone tell me what the physical sector size (in mm?) is of a typical stamped DVD-ROM?

I calculated it myself like this:

From the ECMA-267 (120mm DVD - Read-Only Disk) spec I obtain:
Channel bit length

  • 133.3 nm +/- 1.4nm (type A/B)
  • 146.7 nm +/- 1.5nm (type C/D)

Each physical sector contains 38688 channel bits.

So I get :

(38688 * 133.3nm) +/- (38688 * 1.4nm) (type A/B)
(38688 * 146.7nm) +/- (38688 * 1.5nm) (type C/D)

(5.1571104 mm) +/- (0.0541632 mm) (type A/B)
(5.6755296) +/- (0.058032) (type C/D)

Can someone tell me if I am approaching this correctly?

Thanks

I believe you are correct. It’s amazing that a complete sector can fit in only 5mm :confused:

Yes it is pretty amazing. I will have to look at the spec, but I bet the width of the track is even more amazing.

The smallest amount of data that can be put on a DVD is a “recording frame” which contains 16 sectors (619008 EFM channel bits). For each sector there are 38688 bits, but these are interleaved within the recording frame, so you end up with bits of each sector at the start all the way through to the end.

So, you’ll find the sectors data is physically spread over 16 times your figure 5.1571104 x 16 = 82.5137664mm.

Thanks for the info, you definitely need to have a good understanding of both physical and logical layout of the disc to get the whole picture.

And more detail:

A recording frame contains 16 data sectors.

To the user, a data sector is 2048 bytes; On the disc each data sector also has a 4 byte ID, a 2 byte IES, 6 bytes of CPR MAI and a 4 byte EDC.

With 16 data sectors in a recording frame we have 33024 bytes.

There are also 4832 bytes of PIPO for each recording frame used for the ECC.

So this gives 37856 total bytes in a recording frame.

On the disc, the data is stored using a pattern known as EFM. Our 37856 bytes need to be converted to EFM before being burnt to the disc. Each byte of our data becomes a 16 bit EFM symbol. For every 91 symbols there is also a start sync symbol which is 32 bits. So we have these EFM frames which are 1488 bits long.

37856 / 91 = 416 EFM frames per recording frame.
416 x 1488 = 619008 bits of EFM per recording frame.

So technically there are 38688 bits per sector (619008 / 16) but these bits are interleaved in the entire 619008 bits of the recording frame.

Note that I’ve skipped a few very important parts of encoding/decoding including scrambling, calculating DSV’s when encoding sync frames and interleaving;

See also:
http://www.pioneer.co.jp/crdl/tech/dvd/2-3-e.html

OK, so I checked out what I was saying and I just want to correct the bit about interleaving…

The interleaving doesn’t mix up the byte order of the data (like in CD), it simply interleaves the PI/PO data with the data (the data being 2064 byte sectors with the with ID,IED,etc in).

So, technically, for the first sector in the recording frame; Interleaving is done with:
the first 172 bytes of the data, followed by
10 bytes PO, followed by
182 bytes PI, followed by
172bytes of the data, followed by
10 bytes PO, followed by
182 bytes PI, followed by
172bytes of the data, followed by
etc…

So the interleaving simply mixes in the PI and PO bytes, in effect the EFM bits are written to the disc in order. So you can say your 5.1571104mm does contain one sector.

Of course, in order to decode the sector, you still need to read 619008 bits of EFM for an entire recording frame.

Phew! Hope that makes sense.

Ok, I just saw something that brought up a question for you guys. The linear density of single layer is 0.267 microns/bit. Dual layer is 0.293 microns/bit. How does this correspond to your 133.3 nm/bit? Sorry, I’m not good with math at all. Are the two completely different since one is based on 26.16Mhz bit clock and the other is based on 38688 channel bits/sector?

Also, I think the width of the track is approximately the same as 2T (2 clocks). In other words, a T2 would appear round under high magnification.

Hmmm.

0.267microns is approx (2 * 133.3nm). 0.267micron = 267nm. Is that what you were asking?

I am a little confused. I guess it would make sense if the clock speed was cut in half for your .293micron number. But that would be less dense wouldn’t it? Where did you get your numbers for bit density and clock speed from? I don’t have the ecma spec for DVD-ROM in front of me right now so I can’t review it right now.

Hey RichMan. Now that I am awake what you are saying is correct. Track length and linear bit density should be equal. If those are both 2T, your number .267microns or 267nm makes sense because I was saying “133.3nm per channel bit” (radially) is equivalent to T.

Am I making sense?

Wrong again. What is read on the disc in order is the following :
12 times (172 bytes of data + 10 bytes of PI), then 172 bytes of PO

  • 10 bytes of PI. Please read some docs before posting such wrong
    statements here (for instance, DVD do NOT use EFM encoding).

Thanks for keeping things correct. I am trying to transition to learning the DVD encoding scheme from CD and am finding it pretty interesting.

Not sure how it all ties together, but here is what I have in front of me.

Data bit length
single layer 0.267 µm
dual layer 0.293 µm

Channel bit length
single layer 0.133 µm
dual layer 0.147 µm

Minimum pit length
single layer 0.400 µm
dual layer 0.440 µm

Why is data bit length 2 x channel bit length? Is it because it needs to be 1 complete cycle of channel bit length?

Why is single layer minimum pit length 0.400 µm? (edit…Never mind on this one. It is because smallest pit is T3 (3 x 0.133 ~= 0.400)).

If I’m not making sense…just ignore me :stuck_out_tongue:

Yes, its just because the smallest symbol in EFM is 3T (3 channel bits).

I calculated length of one sector by following way:

700 MB = 5.378125 Km

therefore, 1 cm = 0.001301568855 MB
= 1364.793864032 Bytes
1 sector = 2352 Bytes

Therefore, 1 sector = 1.723337 cm
= 17.23337 mm

Please tell is it correct.

[QUOTE=maheshkanse;2012674]I calculated length of one sector by following way:

700 MB = 5.378125 Km

therefore, 1 cm = 0.001301568855 MB
= 1364.793864032 Bytes
1 sector = 2352 Bytes

Therefore, 1 sector = 1.723337 cm
= 17.23337 mm

Please tell is it correct.[/QUOTE]

Where did “700Mb = 5.378125 Km” come from?

I got this information from: http://en.wikipedia.org/wiki/Compact_Disc

Read data under “Main physical parameters” heading.

[quote=maheshkanse;2012674]I calculated length of one sector by following way:

700 MB = 5.378125 Km

therefore, 1 cm = 0.001301568855 MB
= 1364.793864032 Bytes
1 sector = 2352 Bytes

Therefore, 1 sector = 1.723337 cm
= 17.23337 mm

Please tell is it correct.[/quote]

Sorry, i forget to tell that i am talking about CDs. Please guide me.

Hi maheshkanse,

I do not think your calculations would be correct for all CDs since the track pitch and scanning velocity can vary so much on CDs. I believe that the CD sector length will have the minimum / maximum values that we see on DVD sectors but with greater distance between the min / max. So, you must use the same calculations for the CD sector lengths that are used for the DVD sector lenghts.

The experienced ones can correct me where I am wrong.

RM