Re: [SOLVED] Re: Partitioning And Formatting A Large Disk (2086.09GB)
On Wed, Jan 31, 2007 at 10:20:23AM -0500, Michael S. Peek wrote:
> I've solved my problem -- sort of.
>
> I have a 3ware card that creates arrays for me that show up a
> /dev/sd(b|c|d|...). My problem is that even though ext3 supports up to
> 32TB, fdisk, cfdisk, and sfdisk can't partition even my existing 2TB
> arrays -- sucks on toast. Why, I ask myself, would I want to have a
> nice, snazzy 3ware card that takes care of all my RAID needs in
> hardware, and then have to allocate tiny arrays on it only to glue them
> back together in software via mdadm or LVM?
>
> Then it hit me: If all I'm going to do is put one partition on each
> array, then why bother partitioning it at all. I'll just mkfs on
> /dev/sdb and /dev/sdc and mount them as such.
>
>
> >bkup2:~# mkdir -p /export/raid/0 /export/raid/1
> >bkup2:~# mount /dev/sdb /export/raid/0
> >bkup2:~# mount /dev/sdc /export/raid/1
> >bkup2:~# df --si
> >Filesystem Size Used Avail Use% Mounted on
> >/dev/sda1 313G 984M 296G 1% /
> >tmpfs 531M 0 531M 0% /lib/init/rw
> >udev 11M 62k 11M 1% /dev
> >tmpfs 531M 0 531M 0% /dev/shm
> >/dev/sdb 2.3T 208M 2.1T 1% /export/raid/0
> >/dev/sdc 2.3T 208M 2.1T 1% /export/raid/1
>
> I'm wondering just how unwise it is for me to turn off filesystem
> checking via tune2fs. On the one hand, the 3ware card takes care of
> managing the health of the drives, but on the other hand drive health
> and filesystem health are two separate (albeit related) things. But it
> takes about 3.5 hours to do an fsck on just one of the 2TB arrays on
> this machine. It would take all day, literally, if I had to sit through
> checking both arrays. I think instead I'll arrange for a cron job that
> unmounts and fscks them overnight on a semi-regular basis.
>
I think this is the precise reason that IBM invented JFS. Its all I
use for everything.
I'm curious why you don't use LVM on top of the hardware raid. What
happens when you need to add more drives than that one controller can
handle?
Doug.
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