6 Mar
2005
6 Mar
'05
1:55 a.m.
On Sat, 5 Mar 2005, Eric Smith wrote:
Sellam wrote about 8-inch drives:
Ok, makes sense. So the new calculation is:
107520 / 512 = 210 / 15 = 14. The second time I tried to format a disk
under DOS (to demonstrate to Mr. Johnny Lawson) I heard way more
recalibrations than I did the first time. 14 tracks worth sounds about
right.
Aren't there supposed to be 26 sectors per
track?
That would be the standard format for DD (MFM) 256-byte sectors,
or SD (FM) 128-byte sectors. But if you set the PC BIOS to
believe it is a 5.25" HD disk, it should think it's PC 1.2M format,
15 sectors of 512 bytes. It will also think that there are 80
tracks/side, while the drive only supports 77, so it's not surprising
that there were problems on the last tracks. It never would have occurred to me that you'd try
to use MS-DOS to
format 8-inch disks. I thought you were just trying to read disks.
Well, I needed some way to test whether the setup was working. It
basically is. Now, on to reading those CP/M disks.
It was fun booting my PC into DOS 6.22 off an 8" drive though.
Anyhow, older versions of MS-DOS allowed you to tell
FORMAT
how many tracks to use, though I'm not sure it was actually
willing to accept arbitrary values. And I'm not sure what
DOS 6.22 allows.
I experimented very briefly with this the last time I tried getting this
to work but not enough to come up with anything conclusive. DOS does seem
to expect a certain combination of "sane" values.
--
Sellam Ismail Vintage Computer Festival
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