Re: Apple2 disk controller questions

[dempson@actrix.gen.nz (David Empson)]

Supertimer <supertimer@aol.com> wrote:

> david@uow.edu.au wrote:
> 
> >Here is a table of floppy disk formats I have built up over the years

[Snipped table - very interesting!]

> Wow!  Ok, next question: how does the Amiga fit 880k into a DS/DD 3.5" disk?

I think the main trick is to eliminate sectors.  Each track is one long
sector (per side).  This means that the whole track has to be read or
written every time.

Going back to my earlier detailed posting on the Apple II 5.25" disk:

Assuming 50000 raw bits per track and the same 6-and-2 encoding method,
how much data could you fit onto a single track?

6-and-2 encoding requires 4 disk bytes for every 3 data bytes.

There still needs to be a header and variable length gap.    If you lop
off an arbitrary 2000 bits (250 disk bytes) for the headers (probably on
the high side), this leaves 48000 disk bits.  Divide by 8 to get 6000
disk bytes.  Divide by 4 and multiply by 3 to get 4500 data bytes.

With sectors, the track only hold 16 sectors of 256 bytes, which is 4096
bytes.

4500 / 4096 = 1.099, call it 1.1 (a 10% increase in capacity).

The same data can be (roughly) extrapolated to any format which has in
the order of 16 sectors per track (or slightly fewer sectors with longer
headers, as with Apple's 3.5" format).

There you go.  880K instead of 800K.

I don't know where the Amiga uses variable track rotation speeds, or
which data encoding format it uses.

There are encoding methods that are more effecient than MFM, such as the
RLL family (of which MFM is the simplest case), but I don't know if a
floppy is stable enough for generalised RLL.

-- 
David Empson
dempson@actrix.gen.nz
Snail mail: P.O. Box 27-103, Wellington, New Zealand