"Short stroke" a hard drive

[TeeJayHoward]

I'm curious as to how exactly one would "short stroke" a drive for better access times. I believe I understand the concept behind it, but how is such a thing done? When partitioning a drive, does the partitioning utility start at the outside of a disc and move inwards until the full amount of requested space is allocated? Or vice versa? Or does partitioning just select random sectors of the hard drive and set those as a combined partition?

Or is short stroking even related to partitioning? Maybe you tell the drives firmware not to push the arm so far in?

Is it phyisical? Do you use larger platter with the inside part "disabled"?

Or do I not understand the concept at all? The outside part of a platter is the fast part because the arm moves les, right? Or would it be the inside part? And having multiple platters would be a benefit to short stroked drives, correct?

Edit: I'm just confusing myself. I'm going to try it out myself and find out if it is faster or not. I've got a spare box with a 9GB 10K SCSI drive. I'll install XP and HDTach on a 1.5GB partition, run a test three times, average the result, and then repeat with a 9GB partition.

Edit2: Server2K3 with a 1.6GB partition: 8.3ms
Edit3: Server2K3 with the full drive partitioned: 8.3ms

Thusly have I concluded that partition size has no effect. Am I doing something wrong? Are there really no benefits to "short stroking" a drive? Or is this not short stroking?

 

[mwarps]

There is no longer any point to short stroking a drive. Modern Drives have recording density zones that basically change with the distance from center. I am not sure there's been a non-zoned drive made in about a decade.

 

[unhappy_mage]

Does HDTach benchmark disks or partitions? I could write a thing to test this on Linux, but I don't know how to access partitions directly on Windows.

Essentially, the test would be to perform various-sized seeks. Read the first byte of the partition and write down the time, then the last and compare the times. Repeat with the halfway point of the partition, and the 1/4 and 3/4 marks as well.

I think this could make quite a bit of difference. Sure, there are different density zones; what does that have to do with seek times? There's still only one head for all the zones, and since seek times are related to the distance traveled while the amount of data available is related to the square of the distance travelable. In other words, if you restrict the head to only having to access half its swing, you lose less than half of the disk space.

 

[mwarps]

That's why the drives are zoned, so that the bits at the edge of the platter take up less space. Yeah, your areal density is much higher at the end of the disk, but the access time itself for a random read is still pretty much the same. the time for the disk to make on revolution is precisely the same at the zeroth track and the last track.

Edit: Ah. I see. Not angular travel time, radial travel time.. Okay, still, how long is full stroke time? Is it more or less than rotational latency? (I can't check, I'm at work, this is oddly one of the few sites not blocked.)

 

[unhappy_mage]

From the WD2500KS:
Full Stroke Seek 21.0 ms (average)
Average rotational latency for a 7200 RPM disk is 60 seconds / 7200 RPM / 2 = 4.17 ms. So average case for full stroke (doing the test I proposed) would be ~25 ms. The average read time is given as 8.9ms, but I'm not sure if that's including rotational latency or not.

So basically, full stroke time is a *lot*. You can only do 40 of those a second, compared to over 100 average seeks. So eliminating those worst-case seeks (the ones that must venture into the slow, far-away portion of the disk) could help out a lot. Next time I buy a disk I'll try it out.