[comp.sys.ibm.pc] Hard Disk Help!

dchun@aludra.usc.edu (Dale Chun) (05/31/91)

Does any one know what the hard drive type a Seagate ST-238R is? I
can't get it to work on a Leading Edge Model D2 computer. I know it is
a 32mb drive with 615 cylinders, 4 heads, Read/Write PreComp at 616
and 26 sectors. I tried drive types 36, 37, 38, 43 without success. I
get a "Hard Disk Controller Failure" during boot. (it *was* working
before the clock died). I got the clock working okay now, but the hard
disk is dead. Does anyone know what is going on? Thanks for any
information.
					...dale

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Dale C. Chun                 | * 2 + 2 = 5, for sufficiently large values of 2
PLAYMAC Technical Support    | * Hack First, Ask Questions Later.
dchun@aludra.usc.edu         | * MAC; Maybe A Computer, but probably a toy.
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Q: What is the best way to accelerate a Macintosh? | Real Programmers don't
A: 9.8 m/(sec*sec)!                                | use Macs.
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barnett@rex.cs.tulane.edu (Karey Barnett) (06/05/91)

In article <16110012@col.hp.com> ppa@col.hp.com (Paul Austgen) writes:
>A type 6 has no precomp, which I think is the same as the last
>track, essentially.  It has 20 Mb capacity unless RLL, which
>would be 26 sectors per track.
>
  I might be able to help, but I don't know what the problem is.  For
  example, what do you mean by "A type 6 has no precomp...".

ppa@col.hp.com (Paul Austgen) (06/05/91)

> / col:comp.sys.ibm.pc / barnett@rex.cs.tulane.edu (Karey Barnett) /  5:52 pm  Jun  4, 1991 /
> In article <16110012@col.hp.com> ppa@col.hp.com (Paul Austgen) writes:
> >A type 6 has no precomp, which I think is the same as the last
> >track, essentially.  It has 20 Mb capacity unless RLL, which
> >would be 26 sectors per track.
> >
>   I might be able to help, but I don't know what the problem is.  For
>   example, what do you mean by "A type 6 has no precomp...".
>

I didn't post the original basenote.  What I was trying to say
was that his disk sounds like a type 6 based on all of his
description, except that the poster gave a track precomp number
corresponding to the last track.  I believe that this is
equivalent to no compensation (or maybe it was vice-versa, but
anyway, it sounds like a type 6).

Paul