2.8M 3.5" floppy (was: three and a quarter inch floppy?
cisin at xenosoft.com
Thu Mar 10 19:26:45 CST 2005
> On Wed, 09 Mar 2005 00:43:38 +0100, Fred Cisin <cisin at xenosoft.com> wrote:
> > BTW, in order to call it "2.88M", you would need to use3 the IBM drive
> > definition of "megabyte" of 1024000. If you define a Megabyte
> > ("Mebibyte") as 1048576, then "1.44M" has 1.40625 M, and "2.88M"
> > has 2.8125 M.
> Since this is not a binary matrix of bits, it seems most reasonable to use
> SI units, in which case a megabyte should be 1000000.
In which case, it would be 2.95M
2 sides * 80 tracks per side * 36 sectors per track * 512 bytes per sector
Likewise, a "1.44M" diskette only has 1.44M if you use the IBM "megabyte"
of 1024000. If you use an HONEST Megabyte, of 1048576, then it has
1.40625 M. If you use your 1000000 Bytes per Megabyte, then it would
2 sides * 80 tracks per side * 18 sectors per track * 512 bytes per sector
Or do you also want a Kilobyte to be 1000 bytes?
> Except that the size of a byte is undefined on a diskette, since it is not
> directly linked to a processor architecture.
Byte is 8 bits.
Word often varies, though
> And if we should talk of bytes on diskettes anyhow, what is the smallest
> individually adressable unit of information on a diskette?
That would depend on YOUR terminology, which could make it be anywhere
from a flux transition to a sector, or even a FILE!
Grumpy Ol' Fred cisin at xenosoft.com
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