On Wed, 6 Oct 1999, Martijn van Buul wrote: > > This would fit about 149 programs per CD. > > Just curious: How did you compute this figure? 74 min * 60 s/min / (65536 bytes / 2205 bytes/s) = 149. By using both channels (a left/right selector) you will get double the amount. This assumes that there's no 2-second break between the tracks (I don't know if it actually occupies any disk surface) and that all programs occupy full 64 kilobytes. Hmm, let's compute a safety margin x for 99 tracks per CD: 74 min * 60 s/min / (65536 bytes / 2205 bytes/s + x) = 99 x = 74 min * 60 s/min / 99 - 65536 bytes / 2205 bytes/s x = 15 s I think the loader will fit in 15 seconds. > There's a minimum tracklenght too, so it might be handy to combine > multiple (small) programs into one track. And then you could use a table of programs and their starting times. But then there should be a synchronization pattern (I suggest half a second of 101010) in the beginning of each title. Half a second is 2205/2 bytes (a bit over one kilobyte), so if you have many short programs on the track, the pause should be at least one second, so that the CD can be positioned over the correct title. Marko - This message was sent through the cbm-hackers mailing list. To unsubscribe: echo unsubscribe | mail email@example.com.
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