On Fri, 9 Mar 2001, Steffen Grunewald wrote:
> On Fri 2001-03-09 (00:03), [EMAIL PROTECTED]
>wrote:
>
> [Hi Björn, why do I get such a strange sender address???]
??
>
> >
> >
> > 75 sectors/second
> > audio sector = 2352 bytes
> > data sector = 2048 bytes
> > 75*60*74 = 333000 sectors in 74 min
>
> And that is what the specs require. In reality you will find CD-Rs with
> 335000 sectors (or even more than 336000) but you can rely only on 333000.
> (for 74' CDs).
> 80' CDs must be able to hold 80 minutes *minus* some seconds only.
> cdrecord will happily display the number of sectors on a disc..
>
Yes, I was only giving the formula for calculating the size
of the data on the cd per unit of time. In this example
I used 74 min. Since each sector contain 2352 bytes of audio samples,
and there are 75 sectors/second it's easy
to calculate the size/time. In case it's data(mode1) only 2048
bytes of the sector is used for "real" data.
> > 2352*333000 = 783216000 bytes (audio) = 747 MB
> > 2048*333000 = 681984000 bytes (data with error correction) = 650 MB
>
> Yes. If you make an ISO file(system) your mileage may vary since ISO9660
> is organized in an other way than say ext2 or ufs, so 'du -s' might
> provide wrong numbers. If in doubt, perform a dummy mkisofs run to
> get the final fs size
Of course the data area available for "real" data will be less
when using a filesystem. The numbers I wrote are just the "raw"
size.
/Björn
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