Consider a standard SATA 7200 RPM Hard Disk.Assuming a single platter of size 2TB, 3.5 inches in diameter, how long will ittake to transfer the entire disk contents assuming the data are of uniformdensity throughout the entire disk? There are 1024 tracks on the disk. At whatspeed must the disk spin to achieve the maximum SATA data transfer rate of6GB/sec?
Todays green HDD's actually shift between about 5,900 rpm and just over 7,200 rpm as it permits the drive heads to seek more as the media slows down it's rotation. The end result is actually less outstanding queued I/O as they take NCQ/TCQ to new extremes.
The question is bogus for so many reasons.
- There are 1024 tracks on the disk: It doesn't make jack difference.
- Assuming a single platter: It doesn't make jack difference.
- Assuming the data are of uniformdensity throughout the entire disk: For a platter this isn't really possible unless....
It does however have an answer, it's just worded very..... almost wrongly.
Oh, I get it now.
- I'll sell you the solution to get the answer for AUD $5,000.
- Brilliant question, but poorly worded.
Uniform density IS possible. In fact it's the norm. Drives use Zone Bit Recording to maintain a fairly constant bits per square inch density across the entire surface.
The number of tracks and number of platters DOES make a difference. Less tracks and less platters means greater data density, which in turn means greater transfer rate.
Let's assume the 2TB capacity is spread across both sides of the single platter. At 1K tracks per side, that amounts to an average of 1GB per track. A 6Gbps SATA link speed can transfer data at the rate of 0.6 GB/s, so the drive would only need to rotate at 0.6 revolution per second to saturate the SATA interface. This equates to 36 RPM.
That said, a disc is typically constructed so that the radius of the innermost zone is about half the radius of the outermost zone. This means that the innermost zone has a bits-per-track value that is half that of the outermost tracks. Therefore, the transfer rate at the innermost zone (let's call it X) is about half that at the outermost zone (2X). If the average transfer rate (1.5X) occurs at the middle zone, then the drive will need to spin at 54RPM (= 36 x 1.5) in order for the innermost zone to saturate the SATA interface.