Recovering Cassette Tape Records
Brent Hilpert
hilpert at cs.ubc.ca
Tue Jan 2 05:10:41 CST 2007
Grant Stockly wrote:
> On top of all this, I'm having a hard time understanding the format.
>
> This is a good line:
>
> 1 2 3 4 5 6 7 8 9 10 11 12 13
> S1 13 0000 0D 76 00 F3 7E 18 F9 7E 03 3C 00 00 48 38 00 2C 7E
>
> S1 means its a data record,
> 13 is the byte count,
> 0000 is the address,
> and 7E is the checksum.
>
> So what are the mystery bytes? How do you get 13 data bytes and still have
> a place for all of the rest? 13 is supposed to include the checksum! The
> checksum is the one's compliment of the sum of the all bytes except S1 and
> the checksum itself.
>
> FF-(13+0D+76+F3+7E+18+F9+7E+03+3C+48+38+2C)=7E
>
> So 7E is the checksum. I want to know where all those bytes go! :(
Can't help you with the tape read problem although the speed issue as you
suggest sounds like a good guess, but as for the
byte count issue: it's 13 hex = 19 dec, which works out exactly for the line
you show when you include the address bytes (everything after the 13).
Lesson: you're up too late at night :) ...but then so am I (!) :/
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