Re: Using SAN or local drives?

  • From: "Sergey Popov" <sspopov@xxxxxxxxx>
  • To: kevinc@xxxxxxxxxxxxx
  • Date: Tue, 3 Oct 2006 12:43:49 -0400

15,000 RPM HDD does a full spin in 4ms (60 seconds / 15000 rotations).
I don't have specific info for current HDD models but giving the fact
that data density is increasing I can use data for a 20GB platter in
IBM drive that is about 5-6 years old for a conservative estimate. The
source of the data is here:

Zone 0 has 624 tracks with 792 sectors per track.

Now we do simple math:
792 sectors x 512 bytes = 396Kbytes/track
A single track read plus move heads to the track nextdoor is 4ms
(single spin time) + 0.4ms (Seagate Cheetah Datasheet) = 4.4ms
Number of read a track and move the heads operations per second = 1s /
4.4ms = 227 times (we have enough tracks in this zone to read for over
2 seconds)
Amount of data read per second = 396Kbytes x 227 reads = 87.8Mbytes/s

This is using old drive data and one side of the platter. Multiply
this by number of sides and making adjustment for higher data density
(73GB on Cheetah vs 20GB on sample IBM platter with the same platter
size) and you can realistically expect newer drives to give you

This was a very simple calculation assuming the data is well organized
on the physical surface (required sectors are co-located on the drive)
and no extra rotation required after moving from track to track to
position the heads on the next sector required. A sector location
shift of about 1/10 of the circle length should fix this extra
rotation issue. Things like OS I/O buffer size (max I/O size) are not
taken into account in this calculation either.

Sergey Popov

On 10/3/06, Kevin Closson <kevinc@xxxxxxxxxxxxx> wrote:
 rage ;-)
>>>Another example of SAN overhead is connection to the server.
>>>2Gbit Fiber channel sounds fast but realistically give you
>>>200Mbyte/s throughput. Only two 15K RPM drives at 100Mbyte/s
>>>(on average) transfer rate are enough to suck life out of a

100MB/sec from a single disk? Really?



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