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Some linux dual/multi-boot tutorials / general partitioning tutorials say that the inner side of a hard disk is faster than the outer side so partitions towards the outside tend to be slower while some say that the exact opposite is true.

Which side is actually faster?

Is the difference noticeable?

How are the partitions physically placed on the disk as compared to the partition listing order-for example gparted shows a logical ribbon layout of the partitions. Are the partitions shown to the left side of this ribbon near the outer or inner edge of the disk physically?

Are there software benchmarks like testing some disk intensive stuff for the first time on identical vanilla installs of the same OS but on different partitions, inner and outer?

EDIT: Links

http://www.dedoimedo.com/computers/dual-boot-windows-7-ubuntu.html ctrl + f for "slower" on this page

Does the order of partitions matter? Says outer side is faster

http://www.pcworld.com/article/255224/how_to_partition_your_hard_drive_to_optimize_performance.html Says inner side is faster

http://partition.radified.com/partitioning_2.htm Says outer side is faster(not about dual-booting)

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    How much faster depends on the physical disk size (inner rings are of less diameter than outer). You also generally don't get to decide where on the disk things end up. Any regular disk benchmarking software should be able to tell you if it's running faster or slower than previous benchmarks. What's your actual problem? Sep 9, 2013 at 17:59
  • My actual problem is that is it noticeable?
    – user80551
    Sep 9, 2013 at 18:02
  • @techie007 the OP is asking whether this is a significant difference and how they can test it. Perhaps by creating various partitions starting at different sectors?
    – terdon
    Sep 9, 2013 at 18:12
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    "Many linux dual/multi-boot tutorials say..." -- Please provide links to such tutorials. Sounds like a new disk myth. Outer cylinders have faster linear velocity. Without zone recording, timing would be same as inner cylinders. With zone recording, timing would be "faster" (not slower) than inner cylinders. The average rotational latency will be the same in all cases, since the angular velocity is the same in all cases.
    – sawdust
    Sep 9, 2013 at 18:34
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    Okay, you have provided two links. The first link is a tutorial, but I can find no mention of a preference for inner versus outer partition location. If there is, then please point out where that text is. The second link is not a tutorial, but a Q&A that contradicts the premise of your question. You claim that there are "many ... tutorials (that) say that the inner side of a hard disk is faster", but you have yet to provide even one. Your question is based on a faulty premise. Correct the question or I vote to close.
    – sawdust
    Sep 10, 2013 at 8:17

3 Answers 3

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Which side is actually faster?

The entire HDD platter assembly rotates at a fixed RPM, so the angular velocity is constant.
The average rotational latency will be the same in all cases, since the angular velocity is the same in all cases.

The outer cylinders have faster linear velocity.
Without zone recording, reading a sector on an outer track would be same as an inner track.
With zone recording (as used on probably all new HDDs), reading a sector on an outer track would be "faster" (not slower) than an inner track.

You seem to be misreading these tutorials. Links #1 and #3 clearly state that reads at the outer cylinders can be faster than the inner cylinders.
In link #1 the "end of the disk" refers to the innermost cylinders.
In link #3 the "early part of the test" refers to starting at cylinder 0, which is the outermost cylinder.
There are no inconsistencies or contradictions on this topic among any of the four links you have provided.

Note that optical discs (e.g. CDs, DVDs) are different from HDDs.
Optical discs use a spiral track (per usable side) that starts on the inside and spirals outward.
HDDs employ concentric circular tracks on each surface. Multiple surfaces have tracks that are organized into cylinders. The outermost cylinder is always numbered #0.

Is the difference noticeable?

That depends on what you're doing.

Since the days of 14" platters, then 8", 5.25" and 3.5", the ratio in track length between the outermost and innermost seems to never exceeded 2:1. A practical reason for not exceeding this ratio could be that more cylinders increases the maximum and average seek times.

Modern drives that employ zoned recording take advantage of the larger quantity of magnetic domains (and faster linear velocity) of the longer track lengths of the outer cylinders. By allocating sectors in each zone to a fixed number of magnetic domains, sectors use a consistent length of track per zone. Since there will be more sectors per track on the outer cylinders, the data transfer rate on these cylinders is now faster than the inner cylinders.

The data rate on the outermost cylinder could be twice as fast as the innermost cylinder. On average you might get a 50% faster data rate on an outer cylinder compared to an inner cylinder.

But this performance benefit is only on the data transfer between the R/W head and the platter. This one data transfer of the sector is only one transfer among several operations that will occur in order to satisfy a read or write request by the OS.
To read data in a random sector the steps that will occur are:

  • The OS will construct an ATAPI read request that is transmitted over the SATA bus.
  • The HDD receives the request and processes the command.
  • A seek is initiated to the the proper cylinder (this delay is called seek time and can consume 10s of milliseconds).
  • Once at the correct cylinder, the correct R/W head is selected, and a search for the proper sector is initiated.
  • On average the search for the proper sector takes about a half revolution of the platter (this delay is aka rotational latency).
  • Once the proper sector is found, the actual sector data is read into a sector buffer (usually SRAM) (this is the only operation that is affected by the outside versus inside location).
  • After the entire sector has been read, the data is validated and possibly corrected using the ECC by the onboard controller. The sector data can then be transmitted to the PC over the SATA bus.
  • The OS receives the data.

Now that is for just one sector.
For an idea of the numerous disk requests/operations to copy files, see this answer

On sequential reads that do not require a seek operation, then the time for R/W of the platter becomes a more prominent item in the total time to perform a disk access. How well you can perceive a reduction of a few microseconds is questionable.

How are the partitions physically placed on the disk as compared to the partition listing order-for example gparted shows a logical ribbon layout of the partitions. Are the partitions shown to the left side of this ribbon near the outer or inner edge of the disk physically?

Usually the first sector (cylinder 0, head 0, sector 0) is placed on the left side of these representations. The graph or bar represents the numerical ordering of sectors, which increments the sector number fastest, then the head number (for track number), and then the cylinder number. This progression travels from the outermost cylinder to the innermost cylinder.

A partition shown on the left side would probably be actually located on the outer cylinders. GParted has property boxes that provide the actual disk addresses (by sector numbers) to verify these relationships.

Are there software benchmarks like testing some disk intensive stuff for the first time on identical vanilla installs of the same OS but on different partitions, inner and outer?

I have no idea.

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    computers are fascinating. Jan 25, 2015 at 22:24
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    Great detail! Wish I could upvote more than once.
    – culix
    Feb 11, 2015 at 4:47
  • @sawdust What happens when we have a freshly formatted 4TB disk, with say 4 platters each of 1TB? Assuming a initial data block of 100 MB data; will it get stored proportionally across all 4 platters at the outer egde, say 25MB each. Or in a single platter with 100MB. Reason I ask is to know what needs to be done if I have multiple operating systems to be installed in the same disk. Thank you for your time.
    – Rahul
    Jan 15, 2017 at 5:52
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    @Rahul - Actual sectors chosen depends on the OS/filesystem allocation scheme. Assuming contiguous allocation, the distribution of sectors among surfaces would depend on the track capacity (which you haven't even considered) and the number of R/W heads (e.g. 8). But zone-bit recording makes track and cylinder dimensions variable, and therefore unknown external to the HDD. Cylinder, not platter or surfaces, is the salient quantity; an (electro-mechanical) seek requires more time than an (electronic) R/W head switch.
    – sawdust
    Jan 17, 2017 at 1:01
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    @sawdust, I just finished mirroring a raid1 in which I was using 2 wd drives that are roughly equivalent. I noticed later in the monitoring that the throughput was decreasing as the drive was being walked, here you can find a graph of it snapshot.raintank.io/dashboard/snapshot/… as you mention, by the end of the drive you get 50% the linear performance.
    – Pablo
    May 1, 2018 at 22:34
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So here are my testing results

  • disk formatted with 4K allocation unit NTFS
  • faster sequential read/writes near track 0, the outer part of the disk
  • faster random access read/writes 512 bytes near track 0
  • 4K random access read/writes almost on par in both cases

Hope that you find it useful.


Track 0 (Outer area) ST3500320AS 500.1 GB SATA/300 7200RPM 976771055s QD32 through USB3 (Astone HDD dock)

CrystalDiskMark 3.0.3 x64 (C) 2007-2013 hiyohiyo

MB/s = 1,000,000 byte/s [SATA/300 = 300,000,000 byte/s] -- Test : 1000 MB (x5)

  • Sequential Read : 114.448 MB/s
  • Sequential Write : 113.556 MB/s
  • Random Read 512KB : 46.757 MB/s
  • Random Write 512KB : 75.253 MB/s
  • Random Read 4KB (QD=1) : 0.585 MB/s [ 142.8 IOPS]
  • Random Write 4KB (QD=1) : 1.643 MB/s [ 401.2 IOPS]
  • Random Read 4KB (QD=32) : 0.624 MB/s [ 152.3 IOPS]
  • Random Write 4KB (QD=32) : 1.594 MB/s [ 389.3 IOPS]



Track Last (inner area near spindle) ST3500320AS 500.1 GB SATA/300 7200RPM 976771055s QD32 through USB3 (Astone HDD dock)

CrystalDiskMark 3.0.3 x64 (C) 2007-2013 hiyohiyo

MB/s = 1,000,000 byte/s [SATA/300 = 300,000,000 byte/s] -- Test : 1000 MB (x5)

  • Sequential Read : 61.120 MB/s 53.40%
  • Sequential Write : 60.587 MB/s 53.35%
  • Random Read 512KB : 33.918 MB/s 72.54%
  • Random Write 512KB : 48.483 MB/s 64.42%
  • Random Read 4KB (QD=1) : 0.587 MB/s [ 143.3 IOPS] 100.34%
  • Random Write 4KB (QD=1) : 1.533 MB/s [ 374.3 IOPS] 93.30%
  • Random Read 4KB (QD=32) : 0.618 MB/s [ 150.8 IOPS] 99.04%
  • Random Write 4KB (QD=32) : 1.535 MB/s [ 374.9 IOPS] 96.30%
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See here: http://www.pythian.com/blog/hard-drive-inner-or-outer/

It depends on the bit pattern of the disk. Some disks hold the same number of bits per 'ring'. These are usually cheaper, less cutting edge platters because it's hard to make the densest disks without packing extra bits into the additional surface area on the outer potions of the ring. These disks will have faster writes at the centre of the disk where the data is more packed.

On the other hand, some drives, especially 2.5" ones, have reading heads that park closer to the center of the disk, this reduces seek times considerably for the innermost part of the drive. The expectation is that the OS will be the first thing to be written to disk and therefore parking the heads closer to the OS will result in faster OS performance in general.

Weirdly enough there are a small amount of drives that are significantly faster in the center, why and how I do not know, but they exist.

tl:dr It depends on the drive.

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