"intelligent" disk drives

[Tony Duell]

> 
> > > 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".

Especially as you've dropped an order of magnitude there (sorry, couldn't 
resist).

Assuming you meant 1024000 (1024*1000), I agree totally. I can justify 
1024^2 (1048576). I can justify 10^6 (1000000). But to mix the 2 systems 
is ridiculous

I wonder what defintions have been used for a gigabyte. 2^30 and 10^9 are 
obvious and easy to justify. Did anyone ever use 2^20*10^ or 2^10*10^6 I 
wonder?

-tony