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: http://www.storagereview.com/guide2000/ref/hdd/geom/tracksZBR.html
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 100Mbytes/s.
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.
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?