A theoretical question on how ZFS works, for the experts on this board.
I am wondering about how and where ZFS puts the physical data on a mechanical 
hard drive. In the past, I have spent lots of money on 15K rpm SCSI and then 
SAS drives, which of course have great performance. However, given the increase 
in areal density in modern consumer SATA drives, similar performance can be 
reached by short-stroking the drives; that is, the outermost tracks are similar 
in performance to the average performance, and sometimes exceeding the peak, on 
the 15K drives.

My question is how ZFS lays the data out on the disk, and if there's a way to 
capture some of this effectively. It seems inefficient to do physically 
short-stroke any of the drives, but more sensible to have ZFS handle this (if 
in fact it has this capability). But if I am using mirrored pairs of 2 TB 
drives, but only have a few hundred GB of data, in effect if only the outer 
tracks are used, then the performance should be similar to if I have 
nearly-full 15 K drives, in practice. Given that ZFS can also thin provision, 
thereby disconnecting the virtual space and physical space on the drives, how 
does the data layout maximize performance?

The practical question: I have something like 600 GB of data on a mirrored pair 
of 2 TB Hitachi SATA drives, with compression and deduplication. Before, I had 
a RAID5 of four 147 GB 10K rpm Seagate Savvio 10K.2 2.5" SAS drives on a Dell 
PERC5/i caching RAID controller. The old RAID was nearly full (20-30 GB free), 
and performed substantially slower than the current setup in daily use (it had 
noticeably slower disk access, and transfer rates), because the drives were 
nearly full. I'm curious to see if I switched from these two disks to the new 
Western Digital Velociraptors (10K RPM SATA), if I could even tell the 
difference. Or because those drives would be nearly full, would the whole setup 
be slower?
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