19 Nov
2007
19 Nov
'07
11:37 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.
On Mon,
19 Nov 2007, Tony Duell wrote:
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
I don't think that there is any question that 102400 is a totally
indefensible, irresponsible, and ridiculous number for defining a
"Megabyte".