Re: Size of DOS 3.3 800k volumes?

[Eric Smith <eric-no-spam-for-me@brouhaha.com>]

me@lazilong.com (Lazarus I. Long) writes:
> I know they split into "drive 1" and "drive 2" each 400k.
> 
> but is it really exactly 400k?  by my calculations it is: 368640
> bytes.
> 
> 256 (bytes/sect) * 16 (sect/track)  * 97 (tracks/disk) = 368640
> (bytes/disk)
> 
> do i have that wrong?  does this leave a gap between volumes or are
> they separated by the "sides" of the disk?  if the disk is separated
> by "sides" that would mean, since i must access it as a block device,
> when I read block 0, i'm getting 2 sectors, side 0 sector 0 and side 1
> sector 0 right?
> 
> anyone got any more format info on this?

The 400K and 800K formats use 512-byte sectors (called blocks).  So when
you read a particular block, it comes from only one side.

The block ordering is from 0..N-1 on track 0 side 0, N..2N-1 on
track 0 side 1, etc.  However, it gets a bit more complicated because the
number of blocks per track varies by the track number.  The outer tracks
contain more blocks.  That's the point of the variable speed.

There isn't room in the DOS 3.3 VTOC for 97 tracks.  The mapping for
800K diskettes probably uses 32 sectors per logical track, but I'm
only guessing.