Re: Calculate 1MHz - DiskII

["RedskullDC" <redskulldc@DESPAM.yahoo.com.au>]

Hi Bryan,

"Bryan Parkoff" <bryan.nospam.parkoff@nospam.com> wrote in message
news:lIWFb.81404$HH.69258@fe1.texas.rr.com...
>     Please let me know if you think that my formula is very accurate.  It
> appears to be the most correct Hz for Apple II+, //c, and //e, but Apple
> IIgs has 2.8MHz that may be very close to my numbers below, but it is not
> really exact.
>
> 14,318,180 Hz
> 1,022,727.143 Hz    = 14,318,180 Hz / 14 Periods
> 894,886.25 Hz         = 14,318,180 Hz / 16 Periods
> 1,020,484.320 Hz    = 14,318,180 Hz / (((64 Cycle * 14 Periods) + (1 Cycle
*
> 16 Periods)) / 65 Cycle)
>
> (((64 Cycle * 14 Periods) + (1 Cycle * 16 Periods)) / 65 Cycle) =
> 14.030769230 Periods
> 1,020,484.320 Hz   = 14,318,180 Hz / 14.030769230 Periods
>
>     I suspect that Europeans use 50Hz instead of 60Hz for video so they
use
> 14.25MHz.  It is little slower than standard 14.3Mhz.  How can't 6502 MPU
be
> affected?

Speeds of operation are marginally different between NTSC and PAL apples
that
have different mainboard crystals fitted.

What about Disk II that the speed may not be the same?

Disk write timing will be slightly different between NTSC and PAL apples
with
different crystals.

Bear in mind that when the disk II is reading data from the diskette, each
byte takes
32 clock cycles (6502 at 1MHz). Therefore each bit takes 4 clock cycles.

The software re-syncs at the start of each byte (signalled by a 1bit, which
I hope answers
your previous question on GCR encoding).
Using normal DOS3.3 encoding, the way that the software it written it can
never be
more that 2 clock cycles behind the leading edge of the start bit of every
new data byte
from the disk II.
In reality, the different crystal frequencies have no impact on DiskII
operation.

Cheers,
Red


> Curious...
>
> -- 
> Bryan Parkoff
>
>