19 Nov
2007
19 Nov
'07
11:02 p.m.
IIRC, the 2.8M is vertical recording, with a Barium
Ferrite diskette.
(also called "4M" (unformatted capacity) by NeXT)
If'n anybody wants to call it "2.88M", then multiply 2 * 80 * 36 * 512
and tell me exactly how many bytes are in your "megabyte"s.
Since that's esactly twice the capacity of a MS-DOS format HD 3.5" disk
(36 sectors/track as against 18), I assume the answer to your quesiton is
1024000
-tony