Super User is a question and answer site for computer enthusiasts and power users. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have heard many claim that it doesn't matter how full a drive is until it starts cutting into temp and virtual memory space.

This doesn't make sense to me, given the nature of how the data is transacted on a hard drive. The inside of the platter presents less data per revolution than the outside of the drive does, by significant factors. The inside 40% of the radius of full size hard drive is used for the spindle, so only the outside 60% is used for data storage, but that still means that the inside track of a hard drive presents data 60% slower than the outside track.

By my calculation, a Hard drive that is only 10% full should perform about 2.25 times faster than a hard drive that is 90% full, assuming that the flow is constrained by other factors.

Am I wildly off base here? For all the drives I know, even the top speeds of the first 1% of the drive would be well within the bandwidth provided by a SATA 2 connection.

share|improve this question

closed as not constructive by Linker3000, Sathya Jun 30 '11 at 17:30

As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance.If this question can be reworded to fit the rules in the help center, please edit the question.

"For all the drives I know, even the top speeds of the first 1% of the drive would be well within the bandwidth provided by a SATA 2 connection." This is true, and the huge amount of bandwidth might seem overkill to some. The reason the SATA interface is designed with so much bandwidth, is because it is also used for connecting things like RAID arrays and other multidrive storage units. What is funny is we now actually have SSD's which are capable of fully saturating the SATA interface. Yipeee! – AaronLS Jun 29 '11 at 14:49
up vote 6 down vote accepted

You also cannot equate a linear size increase with a linear speed increase, since the circumference of the disk increases quadratically. With respect to cache saturation and real-world testing, most users find a 35% variation in transfer speed from the outer to the inner surface of the disk platter (e.g. 100 MB/s to 65 MB/s). See the bottom of this answer for sample calculations.

While generally drives do write from the outside of the platter to the inner portion, data fragmentation often skews data across the platter. With a multi-platter drive, this situation becomes increasingly complex, as data may be distributed across more then one platter.

I have heard the "don't fill your hard drive" quote in the past, but unless you are dealing with a system disk (e.g. operating system or swap/page file), you can ignore this issue. Do note that you may have troubles defragmenting a disk without enough free space.

It is worth noting that you should not fill a solid-state drive to it's maximum capacity.

To prove my speed formula, let us assume that 40% of the platter's area is taken up by the spindle, and let us also assume that the platter is 3.5" in diameter, so we have an inner radius of 1.75". That means that the inner radius of the platter is given as 1.75 times the square root of 0.4, or about 1.11" (remember, area = pi * radius ^ 2, so work backwards).

Then, we compute the inner and outer circumferences as just C = 2 * pi * radius, yielding an outer circumference of ~11" and an inner circumference of 6.95". Since the circumference dictates the linear velocity, we see that the inner circumference will have a linear velocity of only 63.2% of the outer circumference - or in other words, 36.8% slower.

If you work out the math, you can prove that the speed decrease from the outer edge to inner edge of the platter is given by 1 minus the square root of the spindle size proportion (e.g. 1 minus the square root of 0.4 in our case, which yields 1 - 0.632 = 0.368).

share|improve this answer
But the principle still holds for multiplatter drives doesn't it? The first 10% of data is just spread across the first 0.25% of writable space on 4 platters, which would mean that the speed still decreases linearly as the head approaches the center. Grant the point about fragmentation, which is part of why defragmentation helps. – Sunny Molini Jun 29 '11 at 14:59
Defragmentation helps a hard drive due to the drive head's seek time as it tries to read a single file across different locations. Hard drives sustained transfer rates are their maximum transfer rates, which is why defragmentation helps (not due to placing it closer to the edge). But you are correct, the principle does still hold for multiplatter drives (assuming the data is equally placed among them) - answer modified ;) – Breakthrough Jun 29 '11 at 15:01
It's also worth noting that by this logic, only the data actually being transacted from the last 10% moves 2.25 times slower, the overall performance would be an average of data pulled from all over the drive. That's where partitioning is handy, it lets you assign of the drive. – Sunny Molini Jun 29 '11 at 15:10
It is not 2.25 times slower - see my answer for updated calculations. The hard drive transfer speed depends on the linear velocity of the sectors being read by the drive head. – Breakthrough Jun 29 '11 at 15:39

Well, in normal usage, you're going for capacity over speed - and in general, you don't need that extra speed.

There MAY be some truth to this story however - though, its true for mostly EMPTY drives...

In some cases however, mostly in server environments, its not unheard of to 'short stroke' the drives in a raid array - to just use the outer sectors, and use multiple drives to get better speed - short stroking a single drive DOES seem to have a difference, and an array of 2-4 cheap drives can perform as well as an SSD according to tom's hardware.

I leave it as an exercise to the reader to decide if its actually worth it or not.

share|improve this answer
It appears as though short stroking drives would only provide a drastic decrease to seek times, but not a drastic increase to sustained read/write performance. – cp2141 Jun 30 '11 at 13:44
what he's talking about is pretty much short stroking, and as such entirely relevant to the topic at hand. A durian by any other name... – Journeyman Geek Jun 30 '11 at 15:23
I never said short stroking was or was not relevant to the subject at hand, I just said that it's used mostly for improved seek times (rather then transfer speed increases). – cp2141 Jul 2 '11 at 23:33

There should be no difference.

The only thing is, if the FS is nearly full, some file operations slow down. But for many reasons, it's prudent anyway to stay below 80% or so.

Some filesystems may deal better with high usage percentages, so YMMV.

share|improve this answer

I suspect that the specific file system used, along with details on fragmentation and allocation size, will have far more impact on speed than % capacity utilization. Plus, of course, the amount and type of buffering can have a huge impact.

My guess is that, for your normal modern drive which is at least a couple hundred GB in size, the only thing to worry about is staying below 90-95% full where fragmentation becomes more difficult to control.

share|improve this answer

How full is too full for mechanical hard drives?

It depends, I would be concerned when it gets to 90% full and very concerned at 95% full. But mainly because of fragmentation and concerns about installing more disks before it reaches 100%.

only 10% full should perform about 2.25 times faster than a hard drive that is 90% full

If I was concerned I'd swap a 7k RPM disk for a 10k or 15k RPM disk. If speed was more important than money I'd then swap for SSD, if not - I'd start thinking about partitioning, filesystem types, filesystem optimisation, raw vs cooked etc.

share|improve this answer

Not the answer you're looking for? Browse other questions tagged .