hp.c (4.1bsd)

[expert at lbl-csam.arpa]

Can somebody help me, please?

We run a VAX 750 with 4.1bsd and SI disk drives.  We are planning on
installing a 300Mbyte (pseudo-RM05) drive and I've been looking through
hp.c.  There seem to be some magic numbers in the "size" structure
(partitioning info.) for the drives, which I don't understand.

"Size" has two parts: 'daddr_t nblocks' (which turns out to be type 'int')
and 'int cyloff'.  It appears that 'cyloff' is the cyl. offset to the
start of the partition.  But where do the numbers for 'nblocks' come from?
There are 8 partitions set up in my hp.c:
	nblocks		cyloff		comment
	-------		------		-------
	15884		0		"A" = cyl 0 thru 26
	33440		27		"B" = cyl 27 thru 81
	500384		0		"C" = cyl 0 thru 822
	15884		562		"D" = cyl 562 thru 588
	55936		589		"E" = cyl 589 thru 680
	86240		681		"F" = cyl 681 thru 822
	158592		562		"G" = cyl 562 thru 822
	291346		82		"H" = cyl 82 thru 561
According to my calculations, this means that the number of 'nblocks'
per cly. is anywhwere from 588.3 to 608.0.

(I can also see that the final partition scheme I use will make a big
difference in how much space I can use:
	Using "A"+"B"+"H"+"D"+"E"+"F" gets 498730 blocks
	Using "A"+"B"+"H"+"G"         gets 499262 blocks
	Using "C"                     gets 500384 blocks;
I can believe this, due to wasted space in partitioning.)

Further, it appears that the numbers to use in /etc/mkfs are 1/2 of
the 'nblocks' number (for "size") along with the standard '3 304' (for
"m" and "n").  Is this correct?

Help, please.

					Richard Collins
					(trwspf!expert at lbl-csam)