Re: Size of DOS 3.3 800k volumes?

["Bryan Parkoff" <BParkoff@satx.rr.com>]

    I have an ability to put DOS 3.3 on 800K and 1600K by patching DOS
routine.

Bryan Parkoff
BParkoff@satx.rr.com

"Eric Smith" <eric-no-spam-for-me@brouhaha.com> wrote in message
qhn0nmeax3.fsf@ruckus.brouhaha.com">news:qhn0nmeax3.fsf@ruckus.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.