Thanks!  :-)

Here's what I've figured out about the TEST 70 parameters so far
(listing them here from your posting for reference):

>     xbnsiz :=54         [enter the number of transfer blocks]
>     dbnsiz :=48         [enter the number of diagnostic blocks]
>     lbnsiz :=83236      [enter the number of logical blocks]
>     rbnsiz :=200        [enter the number of replacement blocks]
>     surpun :=6          [enter the number of surfaces per unit]
>     cylpun :=820        [enter the number of cylinders per unit]
>     wrtprc :=820        [enter the write precompensation cylinder]
>     rctsiz :=4          [enter the size of the revectoring control table]
>     rctnbr :=8          [enter the number of copies of the RCT]
>     secitl :=1          [enter the sector interleave]
>     stsskw :=2          [enter the surface to surface skew]
>     ctcskw :=9          [enter the cylinder to cylinder skew]
>     mediai :=627327008  [enter the MSCP media ID]

The total disk size is, of course, the number of cylinders (cylpun)
times the number of surfaces (surpun) times the number of sectors per
track (17).  Then, this gets used as follows, although I don't know
what the actual order of the special allocations on disk are:

- The number of transfer blocks (xbnsiz) is always 54.
- The number of diagnostic blocks (dbnsiz) should be calculated so
  that xbnsiz + dbnsiz = one whole cylinder.
- The number of logical blocks (lbnsiz) includes the number of
  replacement blocks (rbnsiz), so that the visible number of blocks
  that can be used by the OS = lbnsiz - rbnsiz.
- The transfer blocks, diagnostic blocks and revectoring table are
  allocated outside the logical block area, so lbnsiz = physical size
  of disk - xbnsiz - dbnsiz - (rctsiz * rctnbr).

Anyone know for sure if I've deduced this correctly?

-tih
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