I had two drives partitioned the same and running two RAID partitions on each.

One died and I replaced it under warranty for the same model.

While trying to partition it, the first partition can only start on sector 2048, instead of 63 that was before. Drive have different geometry as previous and remaining ones. (Fewer heads/more cylinders)

Old drive:

$ sudo fdisk -c -u -l /dev/sdb 

Disk /dev/sdb: 2000.4 GB, 2000398934016 bytes
255 heads, 63 sectors/track, 243201 cylinders, total 3907029168 sectors
Units = sectors of 1 * 512 = 512 bytes
Sector size (logical/physical): 512 bytes / 512 bytes
I/O size (minimum/optimal): 512 bytes / 512 bytes
Disk identifier: 0x000aa189

   Device Boot      Start         End      Blocks   Id  System
/dev/sdb1   *          63   174080339    87040138+  83  Linux
/dev/sdb2       174080340   182482334     4200997+  82  Linux swap / Solaris
/dev/sdb3       182482335  3907024064  1862270865   fd  Linux raid autodetect

Remanufactured drive received from warranty:

$ sudo fdisk -c -u -l /dev/sda

Disk /dev/sda: 2000.4 GB, 2000398934016 bytes
81 heads, 63 sectors/track, 765633 cylinders, total 3907029168 sectors
Units = sectors of 1 * 512 = 512 bytes
Sector size (logical/physical): 512 bytes / 512 bytes
I/O size (minimum/optimal): 512 bytes / 512 bytes
Disk identifier: 0x000d0b5d

   Device Boot      Start         End      Blocks   Id  System
/dev/sda1            2048  ...

Why is that?

  • 1
    The output does not indicate that this is an Advanced Format 4K drive: Sector size (logical/physical): 512 bytes / 512 bytes.
    – bwDraco
    Oct 31, 2011 at 23:07
  • 2
    Why was this downvoted?
    – bwDraco
    Oct 31, 2011 at 23:35
  • 1
    I disagree with the downvote, I was also not aware of the track-alignment misconception and JdeBP's info is useful and relevant to a valid question.
    – Garrett
    Oct 31, 2011 at 23:39
  • Releated questions and answers: superuser.com/q/565577/19956, unix.stackexchange.com/q/81556/3054.
    – pts
    Jan 5, 2018 at 14:48
  • 2
    Using newer disk drives, GPT is recommended which uses 2048 as start sector by default. The older disk was probably formatted with an older utility and using non-GPT partition table format. For partitioning newer "Advanced Format 4K" disks, use GPT fdisk Utilities. For RAID, be sure to use type fd00 Linux RAID and label your partitions appropriately.
    – TrinitronX
    Aug 14, 2019 at 23:27

6 Answers 6


Because your old disc was partitioned with a old utility, such as the Linux fdisk, that uselessly implemented track-alignment using the entirely fake disc geometry that you see reported, and your new disc has been or is being partitioned by a newer utility that (by default) aligns to 1MiB boundaries instead.

Further reading

  • 8
    In other words, the LBA sector number 63 corresponds to cylinder 0, head 1, sector 1 in the CHS format, which is the first sector you can use in the MBR format. However, the number 63 is not divisible by 8, which causes a problem with 4K drives, so some modern tools starts the first partition at 2048 which also provides future GPT compatibility.
    – billc.cn
    Oct 31, 2011 at 23:48
  • managed to partition the same way on fdisk by disabling DOS compatibility mode...
    – gcb
    Nov 5, 2011 at 4:11
  • 7
    fdisk is not an "old utility", it's actively maintained. Just do NOT use "fake disk geometry" via -c=dos (it's off by default!). The manpage explicitly warns against doing that as CHS is dead as disco. The link to JdBP's page is dead, but the history of the 1MiB (2048 sector) alignment can be found the Wikipedia entry Logical Disk Manager. It's a Windows Vista wart: "Using a 1-MiB alignment boundary allows safer editing of the partition table with Vista Disk Mgt." Aug 14, 2016 at 14:09
  • 1
    This is not a great answer and the linked page is only accessible via the archive: web.archive.org/web/20200214191047/http://jdebp.eu./FGA/… . Any answer that is essentially "because of historical reasons" without explaining the reasons is worthless in my opinion. At best, this answer is somewhat practical but still borderline cryptic. Dec 3, 2021 at 11:51
fdisk -c=dos

You used the old DOS partition table when creating your partition. Newer versions of fdisk do not use dos compatibility mode by default.

  • 6
    The accepted answer has some valuable information but this answer has the gem that I needed to fix my problem.
    – jcbwlkr
    Aug 13, 2013 at 13:05
  • fdisk wouldn't let me recreate a partition starting at sector 2048 without this argument - I got First sector (3072-314572799, default 3072):
    – tomfanning
    Apr 25, 2014 at 14:40
  • 6
    This is critical if you are trying to resize a partition, which with fdisk means deleting and recreating it; you need the partition to start at the same place.
    – mcr
    Aug 27, 2015 at 1:05
  • no longer works with fdisk from util-linux 2.28 :/ Sep 28, 2016 at 17:26
  • 1
    found it fdisk -c=dos -u=cylinders /dev/sdb in gist - gist.github.com/jkullick/febf46756435f1fa99dc56f00782de03 Sep 28, 2016 at 17:34

Maybe it will be useful to add a comment here. For LUKS partition, it is said to delete and re-create the partition at the same place, but larger before calling cryptsetup resize. But when you created your partition long time ago, it start at the sector 63. Using fdisk, the partition will be recreated at the wrong offset, resulting in a lost partition.

I've managed to recover it using fdisk -c=dos to be able to create partition from sector 63, without troubles.


The 1 MiB (2048 * 512-byte emulated block size) choice is a great catch-all for various hardware storage configurations. Since file system data structures are generally aligned with the partition start point, this can be important to maximize storage read/write speed.

For example:

RAID may use data stripes ranging from 16 to 256 KiB in size. 1 MiB is an integer multiple of this, so starting the partition at 1 MiB is compatible with the underlying RAID model.

SSDs typically have an erase block size of 128 to 256 NAND pages, which depending on the drive might be 256 KiB or 512 KiB. So here again, starting the partition at 1 MiB is compatible with the underlying SSD storage characteristics.

Advance Format 512e spinning HDDs have a 4 KiB physical block size, and again, 1 MiB is an integer multiple (albeit a rather large one) of this.

So while if you have an AF drive and are partitioning as GPT, you might be perfectly happy with your first partition starting at LBA block 40 (an integer multiple of the 8 logical blocks in each physical block of your HDD), hardly any real-world storage is lost by just starting at LBA block 2048 (1 MiB), which is just a more flexible value for partitioning software to default to since it is suitable for pretty much any hardware configuration.


While I realize that the actual question was answered, a quick fix for the 63 vs 2048 first cylinder issue is something like:

sfdisk -d /dev/sdb | sfdisk --force /dev/sda

(if you're sure you've got drives that are the same size)

You can then proceed with adding the various partitions back into the RAIDs where you got complaints about the partitions not being the same size due to the 63/2048 start cylinder difference throwing off the eventual partition sizes.

  • you should explain what it does, source destination? Apr 3, 2019 at 23:21
  • BEWARE: Unless Im doing something wrong, simply copying the same partitions 1:1 results in the same label names and even PARTUUID e.g. in case of GPT partition tables. that can be changed using FDISK in expert mode with the option u (change partition UUID). unix.stackexchange.com/a/341796/174233 Dec 9, 2021 at 16:19

Not sure why it started on 63 in the first place, but according to fdisk, your sector sizes are 512.

So if you want your partitions, and thus clusters to be aligned, your starting offset should be divisible by 512. 2048 is pretty common these days.

If you prefer misaligned sectors, then you can always get GPart to move your partition back to 63.


Ooops. Didn't see that you have a RAID. You should probably post your RAID and stripe size.

  • 2
    63 is 111111 in binary. The BIOS of PCs, sector number is/was encoded in six bits, resulting in a maximal number of 111111 (63) sectors per track. This maximum is still used for virtual CHS geometry.
    – Chris Reid
    Oct 12, 2018 at 21:29

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