Hoffmann-Vetter, Martin wrote > > Hello, > > ... > What is if the format for the CBM-900 ist expanded in the block size to > 512 > bytes? The checksum calculation must failed! Does anyone know the mostly > used block size of the coherent system? > > Martin > According to the 8723 datasheet, this is really the case: ... GCR FORMAT The Z8000 computer will run a version of the unix operating system and will require a 512 byte sector size. This format will use the existing commodore GCR data encode scheme with variable density (see table 3.2.2). This format will allow for 1.224.704 bytes of storage per drive. The data rate for one track will vary from 41K to 33K bytes/second. for the different regions. Th multi-track transfers including step settle and latency will be 38K to 33K bytes/second. ... Table 1.7.4 GCR DATA RATES/FORMATTED CAPACITY TRACK REGION BIT TIME SECTORS/TRACK SECTORS/REGION 1-39 2.16 16 624 40-53 2.33 15 210 54-64 2.50 14 154 65-80 2.66 13 208 SECTORS/SIDE = 1 196 SECTORS/DISK = 2,392 BYTES/DISK = 1,224,704 BYTES/CONTROLLER= 2,449,408 ... If I remember correctly, this is the same feature as in the 8280, 80 tracks with sectors on both sides, whereas the 8250 (as expanded 8050) sees first the 80 tracks on one side, then the 80 tracks on the other side. On the other hand, I never tried to read a c900 disk in a 8250 or SFD1001. Regards, Simon -- View this message in context: http://nabble-com-cbm-hackers.2304266.n4.nabble.com/CBM-900-floppy-disk-format-encoding-tp4073123p4267470.html Sent from the Nabble.com: cbm-hackers mailing list archive at Nabble.com. Message was sent through the cbm-hackers mailing listReceived on 2012-01-06 00:00:11
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