Hi Malcolm,
The 24x512 has 512kbit worth of storage space, that's 65535 bytes.
Counting the number of bytes in your "packet":

29 09 2010 19:45 234 236 240 242

I get 14 bytes (because the last four values CAN be higher than 255, right?).
So 65535/14/60/24 = 3.25 days. So, in short, you won't be able to fit your seven days worth of data doing it that way.

Storing the date and time at startup only (and when it wraps around) will basically save you 6 bytes per record. If you know you are recording once per minute there's really no need to save save the time - every time. That's still not enough to get 7 days into a 24x512 though.

Well, just something to think about.

/Henrik.