I use HD Tune to measure performance of the hard drive. The tests typically take 2-3 minutes and the transfer rate of a regular hard drive decreases dramatically as the test comes to a close.

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However, the SSD drive performance stays the same over (pic below) the life of the test. This happens on all my computers. Why is that?

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    It may be be good to know what the test is actually doing behind the scenes.
    – jmreicha
    Commented Jun 22, 2012 at 18:27
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    ... and what the graphs represent. The (read) transfer rate (as indicated by the blue line) is not as significant to the (total) access time as the (average) rotational latency and (typical) seek time of a HDD. The shape of that blue line is not a performance indicator.
    – sawdust
    Commented Jun 22, 2012 at 22:22
  • i have four ssd in raid 0 (keep backups of course). I'm getting around 650-700 Mb/s. HDTune is a nice piece of software.
    – ctilley79
    Commented Jun 23, 2012 at 0:39
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    I think the title is misleading. Isn't it true that hard drives maintain performance over time but solid states degrade because of some physical degradation from too many writes?
    – mowwwalker
    Commented Jun 23, 2012 at 3:49

4 Answers 4


The mechanical HD is being scanned from the outside inward. Since the disk is spinning at a constant 7200rpm, it's covering more data per second at the outside than the inside.

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    I talked to an HDD professional recently. He said that the ratio of speed of the outside of the HDD to the inside is about 1.8.
    – Deltik
    Commented Jun 22, 2012 at 21:44
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    @Deltik: which matches the information in the graph quite nicely!
    – Dancrumb
    Commented Jun 23, 2012 at 21:43
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    Everybody addresses the hard disk, but nobody speaks of the SSD :-)
    – nanofarad
    Commented Jun 25, 2012 at 22:13
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    To add to the answer, the relationship is simply: data rate = angular velocity * radius, with angular velocity being constant for HDDs. Therefore, transfer speeds are directly proportional to the radii at the outer edge of the disk vs. along the inner sectors. That would be about 1.8 according to @Deltik.
    – user40437
    Commented Jun 29, 2012 at 17:39
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    And how quick does the SSD drive spin?
    – j_kubik
    Commented Oct 10, 2014 at 8:08

Actually what you are seeing on the X axis does not correspond to the "time", but to the "physical area" of your disk. I mean, if your disk has 250GB (100% of it's capacity), the 0-10 will mean the first 25GB of your disk, 10-20 will mean the second 25GB part of your disk, and this goes until all your 250GB (which is the 100%).

Your HDD performance does not decrease over time, but it decreases due to the physical effect caused by the 'rotational effect" of your disk(this doesn't happen on your SSD). The fist 0-10% area of your hard disk corresponds to the external area of the disk, which gives the read speed a boost because the linear speed of this area is higher compared to the internal area of your disk(the last 90-100% of your disk for example). This gives the impression that the performance of your hard disk is decreasing over the first to the last disk sectors(it actually is, as you can see on the first picture), as all SSDs are based on random access memories, all the usable "area" of your SSD have the same speed and accessing times, which corresponds to a linear performance over the entire disk. This also explains why operational systems usually use the first "area" and the first disk sections of the hard disks... For example, Windows will boot faster and make disk I/Os better than it would be if it was intalled on the last sectors.

PS: As you can see on your first picture, Hard Disks usually have a 40 to 50 percent performance loss when comparing the first sector to the last sector reading speed.


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    You guys should mention that modern HDDs use zone bit recording, where recorded data is tied to linear speed (or areal density), rather than using constant angular speed. Check out the (steady) read speed of an old HDD that does use constant angular velocity: hdtune.com/results/Conner_CP3204F.gif BTW it is not "external" and "internal" "areas of the disk", but the outer and inner tracks.
    – sawdust
    Commented Jun 22, 2012 at 19:37
  • Good explanation... except that seek times aren't reduced vastly, 450%, or at all significantly. The dominant part of seek time is rotation.
    – Ben Voigt
    Commented Jun 22, 2012 at 21:51
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    @BenVoigt "The dominant part of seek time is rotation" - You're confusing seek time with access time (which is the sum of seek time, rotational latency, data R/W time, SATA bus transfer times plus command & response processing time). Rotational latency is a random variable that the user or OS cannot control/predict. But the user/OS might be able to control or reduce seek times with optimizations like defragging/compacting the files, and/or ordering/laddering disk operations.
    – sawdust
    Commented Jun 22, 2012 at 22:02
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    @sawdust: Rotational latency can be controlled, by placing data sequentially in the order it is needed. But that's a different topic entirely than linear velocity differences between inside and outside of the platter. Merely placing data at the outside of the platter doesn't help performance anywhere near 450% as claimed.
    – Ben Voigt
    Commented Jun 22, 2012 at 22:51
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    @BenVoigt: No, that old Conner drive (it's not "mine") has fixed 5400 rpm and does not have zoned bit recording, hence the steady read transfer rate. I included that link to show that if the OP wants a flat transfer rate curve like an SSD, then he has to give up the extra capacity afforded by ZBR. (Of course there are probably no HDDs w/o ZBR in production.)
    – sawdust
    Commented Jun 23, 2012 at 5:45

Nice answers above, but there is little notion of angular size of a sector on outer cylinder vs inner cylinder.

The Answer: zoned bit recording (ZBR) is the cause. Because inner tracks have sectors with larger angular size, so they take longer time to read while disk makes a turn under the head with constant angular velocity (rpm).

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Details: at You Don't Know Jack about Disks, by Dave Anderson, June 1, 2003

... All the tracks within a given zone had the same number of sectors. A track in a zone near the outer diameter of the disk, however, might have 50 percent more sectors than a track in a zone near the inner diameter of the same disk. This would be true for a 3.5-inch drive. The advantage ZBR provides varies by media size and is a function of the relative size of the outer radius of the recording band to the inner. Drives today usually have 15 to 25 zones. ZBR added great value: 25 percent or more capacity for no additional material cost in a 5.25-inch drive, the prevailing form factor when ZBR fi rst appeared. It forced the industry to adopt a more intelligent interface—one that would hide the complexities of ZBR and, at the same time, hide the geometry and bad-block fl aw problems by pulling that functionality into the drive, as well. ...


Your hard disk spins at a constant rate, 7200RPM or whatever. the benchmark begins on the outside of the disk, where the radius is greater, and thus the linear speed is faster(one rotation in 1/120 of a second has greater distance (proportional to radius), and therefore more bits read in that time period), while inside the disk, the radius is smaller, and so less bits are read for the same angular distance(one rotation in 1/120 of a sec with smaller radius implies smaller circumference swept and thus less bits read.

Assuming an outside radius of about 2.8 in, and an inner radius of 1.6 in(due to loss for spindle, extra alignment space, landing zones), the performance loss on the inside is about a factor of 1.8.

Note that the jitter is caused by system load jitter or noise on cables, among other factors.

Also, addressing the SSD and not just the hard disk, is has an electronically network of connections set up(not mechanical) and thus the only delays are wire(in the actual memory) latency, and access "sweeps" across the data in blocks, keeping the velocity and bitrate constant, limited only by circuitry.

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