Re: 32MB is size???

[pausch@saaf.se (Paul Schlyter)]

In article <9ryYf.64370$Jd.53108@newssvr25.news.prodigy.net>,
Bryan Parkoff <none@nospam.net> wrote:

>     I apologize when I did not explain clear.  I do not mean to decrease or 
> increase performance when 8 bits of data is read from or is written to the 
> disk.  The fact is that critical timing is always the same from DOS 3.3 and 
> ProDOS.  I suppose that 32 cycles are needed to save 8 bits or 1 byte of 
> data into RAM, but 40 cycles are used for sync detection.
>     Now, I am saying reading 512 bytes from disk is faster than 256 bytes 
> from disk.

I disagree with you here -- the more data you read from disk, the more
time it takes.... :-)

I assume you mean that it takes less time PER BYTE to read a 512 byte
sector than a 256 byte sector.  I.e. it takes less time to read one
512-byte sector than two 256-byte sectors.  That's most like true, for
reasons you explain below.  But it is also irrelevant in the case of
Apple II floppy disks, since all the Apple II OS'es used physical
sectors 256 bytes large.  Apple DOS 3.3 did so, just like Apple
Pascal, Apple CP/M and Apple ProDos.  Some other, less common, Apple
OS'es used the same 256-byte physical sectors, like OS/9 and
Fig-Forth.  The only exception from the 256-byte physical sector usage
I know of are some Apple II games, which used other sizes for physical
sectors as a method for copy protection.

Some of these OS'es then used the concept of a "logical sector" at a
somewhat higher level, where the size of the logical sector was
different from the size of the physical sector.  Apple Pascal was
first, with 512-byte logical sector, where each logical sector was
mapped onto two physical sectors.  Apple CP/M was second, with
128-byte logical sectors, where two logical sectors were mapped onto
one physical sector.  Finally, Apple ProDos used 512-byte logical
sectors, like Apple Pascal, and ProDos used the same
logical-to-physical sector mapping as Apple Pascal did.

Reading 512-byte physical sectors is somewhat faster than reading
256-byte physical sectors, true.  But reading 512-byte logical sectors
mapped onto 256-byte physical sectors is an overhead which slows down
the reading somewhat.

> After 256 bytes of data is read, one or more routines are 
> executed to cost approximately 5,000 to 10,000 cycles.  Some programmers 
> have found some ways to read 512 bytes of data at this time before routines 
> are executed.  It saves about 10,000 cycles.
>     It is not good to reuse routines twice for each sector (256 bytes) that 
> costs more cycles.  I stated earlier when I don't remember exactly what I 
> read an article.
>     Do you know what I explain well?
> 
> Bryan Parkoff 

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