10 Mar
2005
10 Mar
'05
5:26 p.m.
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
2949120
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
1.47M
1474560
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