I wrote:

> - 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).

Actually, it seems that lbnsiz includes the revectoring table, not the
replacement blocks.  With 8 copies of a 4 block table, this means that
the visible number of blocks for the OS is lbnsiz - 32, and lbnsiz is
calculated as physical size - xbnsiz - dbnsiz - rbnsiz.

-tih
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