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Re: assistance with disk ][ and modern 1.2M floppies



Supertimer <supertimer@aol.com> wrote:

> DuoDisk manual page 14 paragraph 4:
> 
> "A single Diskware disk holds 143,360 (formatted) bytes
> of information."
> 
> That's 280 blocks times 512 bytes per block = 143,360
> bytes.  I'm thinking about ProDOS.

It doesn't matter what operating system you are talking about.  A
formatted 5.25" disk on the Apple II (assuming 35 tracks) holds 140 KB:

35 tracks x 16 sectors x 256 bytes per sector

= 35 x 4 x 1024 bytes

= 140 x 1024 bytes

= 140 KB

A "kilobyte" is 1024 bytes (2^10), not 1000 bytes.

The sometimes quoted "143 kilobyte" figure is only valid if you regard a
kilobyte as being 1000 bytes.  I don't.

If you claim a 5.25" disk is 143 KB, then you must be consistent.  Using
this principle, a double density 3.5" disk on the Apple II (or Mac)
would be 819 KB instead of 800 KB; the same disk on a PC would be 737 KB
instead of 720 KB; a high density disk would be 1474 KB instead of 1440
KB; the maximum size of a ProDOS partition is 33.5 MB instead of 32 MB.

Extending this to memory capacities: an Apple IIe with extended
80-column card now has 131.072 KB of RAM, while a fully populated IIgs
has 8.51968 MB (8519.68 KB), instead of 8.125MB.

Hard drive manufacturers use this technique to give an inflated figure
for the capacity of a drive.  A so-called "850 MB" hard drive is really
850 million bytes, which is 830 MB (1 MB = 1048576 bytes).

The Zip disk is another example.  It claims to hold 100 megabytes, but
it is really 96 MB (about 100 million bytes).


There is a growing suggestion of inventing new terms for the "binary"
kilobyte and megabyte, to reduce this sort of confusion and misleading
information.  I believe the suggested terms are "kibibyte" and
"mebibyte" (short for "kilo-binary-byte" and "mega-binary-byte", as far
as I can tell).  I'm not sure what is being suggested for the
abbreviations.

-- 
David Empson
dempson@actrix.gen.nz
Snail mail: P O Box 27-103, Wellington, New Zealand