just with partitions from all different drives? until I run out of partition sets of at least 4 different disks?
I can see why FreeNAS doesn't support the bad idea version where some disks hold more than 1 "disks" worth, but why can't it do the sensible version?
e.g.
D1 - 500gb
D2 - 500gb
D3 - 500gb
D4 - 500gb
D5 - 1000gb (split into partition A and B)
D6 - 1000gb (split into partition A and B)
D7 - 2000gb (split into partition A, B, C and D)
D8 - 2000gb (split into partition A, B, C and D)
D9 - 2000gb (split into partition A, B, C and D)
D10 - 2000gb (split into partition A, B, C and D)
so vdev1(A) would be:
(raidz2)
D1, D2, D3, D4, D5A, D6A, D7A, D8A, D9A, D10A
vdev2 (B):
(raidz2)
D5B, D6B, D7B, D8B, D9B, D10B
Vdev3 (C):
(raidz2)
D7C D8C, D9C, D10C
Vdev4 (D):
(raidz2)
D7D D8D, D9D, D10D
If any disk failed no vdev would be lost so the pool would be ok,
What I can't work out is how to add a 4tb disk :(
The other, and more simple would be:
subvdev1: (stripe)
D1, D2, D3, D4 = 2TB
subvdev2: (stripe)
D5, D6 = 2TB
vdev1:
(raidz2)
subdev1, subdev2, D7, D8, D9, D10
This is even more fault tolerant than the first example as Disks 1-6 could all fail and the vdev would be ok,
It might actually be easier to increase its size as well? though it would take some time to realise the increase, say with 4tb disks,
If the vdev "allow expansion" thing was enabled:
Replace D10 with D11 (4tb)
resilver
remove D9, make subvdev3 (stripe) D9+D10 = (4tb)
add subvdev3 to replace D9
resilver
rinse and repeat for the other positions with two more 4tb drives and ta-da! vdev1 will grow :)
Sadly there's a trade off between raidz2 storage efficiency, e.g. making vdev1 out of more smaller components resulting in a larger cost to expand,
say with the 6 x 2tb above:
66.6% space efficiency and only 3x4tb to expand,
8 x 2tb would give:
75% space efficiency but 4x4tb to expand
Having said that 6 components in RaidZ2 looks like kinda the sweet spot of that tradeoff :)
edit: Awww, tedious :( just read elsewhere that you can't nest vdevs with FreeNAS :(