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I am aware that this is probably a common question, yet I did not find the right search terms to use. I have to buy a new hard drive. I figured out that the equation consists of: connector standard, rpm and density. My question is: Will I experience the benefits of 7200 rpm over 5400 or will SATA2 limit the speeds below the capabilities of the drive? I'm looking for a 3.5" 500 gb drive. Second question: Is data density same for all hard drives with 3.5" platters, holding 500 gb ? If not, is there any way I can tell one from another? The drive will be used for everyday multitasking and gaming - not so much large file transfers.

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"My question is: Will I experience the benefits of 7200 rpm over 5400 or will SATA2 limit the speeds below the capabilities of the drive? I'm looking for a 3.5" 500 gb drive." This part of the question, at least should be closed. No-one can possibly know what you will notice. –  Xavierjazz Dec 13 '12 at 15:06
    
well, while i think it was worded a little poorly, I think it has an objective answer: does SATA2 provide some inherent bottleneck in a 7200 RPM drive? and the answer is no? –  im so confused Dec 13 '12 at 15:12
    
The answer to that is indeed no. And you can use the edit link to rewrite if you want to. –  Hennes Dec 13 '12 at 15:13

3 Answers 3

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Will I experience the benefits of 7200 rpm over 5400 or will SATA2 limit the speeds below the capabilities of the drive?

The SATA-II (3.0 Gbit/sec) interface tops out around 270MB per second.
Top speed for current average consumer harddrives is about half that.

Top speed of 15000 RPM SAS enterprise drives is nearly 200 MB/sec. Still way below SATA-II throughput.

So no, SATA-II does not impose a limit to either 5400 RPM or 7200 RPM drives.

I'm looking for a 3½ 500 GB drive.

1 TB drives are only a fraction more expensive than 500 GB drives.

Is data density same for all hard drives with 3.5" platters, holding 500 GB ?

No, increased density means fewer platters are needed for the same capacity. So producers usually use the highest available density to cut down on cost. This means a typical more modern drive will have higher density platters, resulting in better performance and lower production costs.

If not, is there any way I can tell one from another?

Read the drives manual (usually found on-line).

The drive will be used for everyday multitasking and gaming - not so much large file transfers.

HDD only have two advantages over SSDs: Larger capacity (at a identical price), and decent large file performance. Since the latter is not important to you, consider a SSD. Performance on those is noticeably better.

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I'd state this even more strongly: SATA2 doesn't pose a limit to pretty much any non-enterprise hard drive. You never have to worry about it really. –  Shinrai Dec 13 '12 at 15:23

Based on my first-hand experience developing firmware for disk controllers, here's an answer from a different perspective. You question seems to based on a misconception that, somehow, R/W_head-to-platter transfer rates affect the drive-to-host transfer rate. The reality is that these two I/O transfers are separate operations, and do not interact on any level other than one follows the other.

There is a widespread misconception that the data bits read off the disk platter can be immediately put on the (SATA) interface. This misconception implies that the slow read rate can/will inhibit the faster rate of the interface. The problem is that the HDD does not operate in that manner.

Here's an analogy:
You have to take a trip to another city.
Half the trip is by land.
The other half of the trip is by air.
For the land portion of the trip, you can use either a car or scheduled bus.
For the air portion of the trip, you can use a jet or a prop plane.

So what combination of transportation will get you to your destination in the least amount of time?
Since each portion of the trip has to be taken in sequence, the fastest trip would obviously use the car and then jet plane.
The speed of the bus or car cannot affect the speed of any plane.
So you don't question if you use one mode for the first half, how the other portion of the trip will be affected.

A HDD operates in a similar manner, that is, two distinct and separate phases.
The data read from the platter is stored in a sector buffer.
The entire sector has to be read (from the platter) and validated (using ECC) before that sector data is ever placed on the (SATA) interface for transfer to the host.
This two-step transfer of HDD data is the basis for the above analogy.

The platter RPM is a operational specification that simply has no relation to the performance of the SATA interface, or vice versa. For data throughput, the platter characteristics will be the dominant term over the interface. But do not confuse a dominant item in a summation as an inter-dependence relationship.

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That is a great write up! Good to see we have a real expert too, who had first hand experience! +1 –  Canadian Luke Dec 13 '12 at 21:42
    
While not dependent, the question, to return to your analogy, is rather, "Can I get from South Africa to Japan in 12 hours?" Obviously if you take a scheduled bus that only runs every 4 hours, and you lose 4 hours waiting for it, it's going to "eat up" time that would otherwise be available for the air trip, and would make it that much harder to meet your deadline. Similarly, if your goal is to make "best use of" a high transfer speed within the disk (from the platters to the buffers), having a slow interface can become a bottleneck, much like waiting for the bus for 4 hours. –  ÃŁŁǫǛȉЖΦΤїҪ Dec 13 '12 at 22:16
    
Of course, now that I think of it, it'd be quite silly for a drive manufacturer to design a disk with an inherent limitation like that, wouldn't it? Imagine if we had a theoretical drive that could read from the platters at ten times the rate that a bog standard 7200rpm drive can today. Obviously, selling this drive as a supremely fast HDD that blows away the competition would be dishonest if you used a SATA 2 interface that was instantly and completely saturated by the ingress of data from the platters, and unable to keep up with it. –  ÃŁŁǫǛȉЖΦΤїҪ Dec 13 '12 at 22:19
    
That said, I do not know of any 7200rpm mechanical hard drive whose physical read rate is so high that it can saturate a 3 Gb/s bus. You can probably saturate the bus for a few milliseconds with data coming out of the buffer (some kind of NAND or SRAM?) but not over a sustained transfer where the sectors being read are different every time. So the answer to the original question is "Yes", in the sense that you are taking full advantage of your physical layer's transfer rate (the platters and read heads) and you don't have to worry about the SATA bus slowing you down. For SSDs it's different. –  ÃŁŁǫǛȉЖΦΤїҪ Dec 13 '12 at 22:28

I'd like to clarify something on the effect of rotation speed VS interface bandwidth. Rotational speed has almost no effect on transfer rates, it instead directly effects seek speed. It does have a small effect on throughput but platter density usually has a much higher effect on throughput (especially per $ spent). If your looking for maximum IOPS outside of a SSD then you can also buy a larger slower drive, say a 2TB 7200RPM instead of a 600GB 10K RPM drive and short stroke it by half and end up with more IOPS then you would with the 10K drive and still have more space in the fast section, and also have a 1TB "archive" section thats slower.

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