eyal@echo.canberra.edu.au (Eyal Lebedinsky) (11/14/90)
Hello, I was wondering how the disk is organized. Doing some number juggling I concluded that the disk is organized as: 1219 cylinders (+5 spare) 15 heads * 36 sectors = 540 - 4 spare = 536 sectors used disk capacity = 1219*536 = 653384 blocks = 319Meg. Now, from other sources I find that the disk can be had in a 1Kbyte sector configuration, 19 sectors/track. This will add up to about 335Megs available. That is 16Megs extra, blocksize = minix blocksize. The question: can the SCSI device be reformatted? easily? I mean with just some software, no special manufacturer utilities. -- Regards Eyal
dave@quick.com (David A. Wilson) (11/15/90)
This is how I do it for a cdc4702n(702MB unformatted) disk, I think it should work in general. 1. Formatting from scratch a: set block size: SCSI cmd bytes(hex) 15 Command=Mode Select 11 PF=1, SMP(save parameters)=1 00 00 0C Parameter list length 00 SCSI data bytes to write(hex) 00 Mode select hdr 00 Medium type 00 08 Block descriptor length 00 Density code 00 | 00 |- # of blocks(0=all) 00 | 00 00 | xx |- block size for format 00 | Where xx is: 10=4096, 08=2048, 04=1024, 02=512, 01=256. b. Format Drive SCSI cmd bytes(hex) 04 Command=Format unit 00 Defect list mode 00 00 |- Interleave(default[0] means 1:1) 00 | 00 # of SCSI data bytes to read(decimal) 8 I don't think read bytes are needed, but it didn't hurt. ====== I hope this is useful David A. Wilson dave@sea375.uucp ...uw-beaver!uw-entropy!quick!sigma!sea375!dave