With many new hard drive disks the physical sector size is 4096. Would it be possible to make the system use a logical sector size of the same size, rather than the default logical sector size of 512?

Will it speed up bulk reads and writes? Where can it be configured?

  • See unix.stackexchange.com/a/18542/1131 for comments on alignment issues. Recent versions of mkfs.* should automatically use the optimal sector size. You can do some mkfs.* tests and inspect the result (either in the verbose output of mkfs or in a related fs utility program). Jan 13, 2015 at 14:30
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    Thanks but no alignment issues
    – Matan
    Jan 13, 2015 at 15:46

4 Answers 4


512 byte is not really the default sector size. It depends on your hardware.

You can display what physical/logical sector sizes your disk reports via the /sys pseudo filesystem, for instance:

# cat /sys/block/sda/queue/physical_block_size
# cat /sys/block/sda/queue/logical_block_size

What is the difference between those two values?

  • The physical_block_size is the minimal size of a block the drive is able to write in an atomic operation.
  • The logical_block_size is the smallest size the drive is able to write (cf. the linux kernel documentation).

Thus, if you have a 4k drive it makes sense that your storage stack (filesystem etc.) uses something equal or greater than the physical sector size.

Those values are also displayed in recent versions of fdisk, for instance:

# fdisk -l /dev/sda
Sector size (logical/physical): 512 bytes / 4096 bytes

On current linux distributions, programs (that should care about the optimal sector size) like mkfs.xfs will pick the optimal sector size by default (e.g. 4096 bytes).

But you can also explicitly specify it via an option, for instance:

# mkfs.xfs -f -s size=4096 /dev/sda


# mkfs.ext4 -F -b 4096 /dev/sda

In any case, most mkfs variants will also display the used block size during execution.

For an existing filesystem the block size can be determined with a command like:

# xfs_info /mnt
meta-data=                       sectsz=4096
data     =                       bsize=4096
naming   =version 2              bsize=4096
log      =internal               bsize=4096
         =                       sectsz=4096
realtime =none                   extsz=4096


# tune2fs -l /dev/sda
Block size:               4096
Fragment size:            4096


# btrfs inspect-internal dump-super /dev/sda | grep size
csum_size             4
sys_array_size        97
sectorsize            4096
nodesize              16384
leafsize              16384
stripesize            4096
dev_item.sector_size  4096

When creating the filesystem on a partition, another thing to check then is if the partition start address is actually aligned to the physical block size. For example, look at the fdisk -l output, convert the start addresses into bytes, divide them by the physical block size - the reminder must be zero if the partitions are aligned.

  • Thanks, your clarification of what Linux describes as logical and physical block sizes was very helpful. I recently ran into a situation where the same 8TB hard drive showed up as 512-byte blocks in one USB enclosure and 4K blocks in another, both logical and physical. I thought something was wrong because Linux didn't see the partition map when I changed enclosures, but then I found that GPT starts on the second logical block, so it was just in the wrong place for the new alignment. I used gdisk to recreate one single big partition, and all of my data was still there (ext4 with 4K blocks).
    – Raptor007
    Mar 13, 2018 at 8:08
  • Is the method you described to check the alignment reliable? I used the "parted" utility and it reported that an hdd partition is aligned, but with you approach I queried the physical block size, got 512, then took the start address of the partition from fdisk -l, divided it by 8 and then 512. The remainder wasn't 0, so the partition seems to be not aligned
    – d.k
    Mar 11, 2019 at 14:24
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    @user907860 Why do you divide by 8? A fdisk -l reports Units = sectors of 1 * 512 = 512 bytes in one of my Linux systems - thus, for a 512/4096 logical/physical drive with 2 partitions starting at 2048 and 1026048 I compute 2048*512%4096 and 1026048*512%4096 - e.g. in a python shell. Since both expressions equal zero those partitions are 4k aligned. Mar 11, 2019 at 16:49
  • thank you very much for the explanation, I didn't notice that the number was sectors, not bits. I upvoted the answer a couple of month ago, sadly, can't do it again
    – d.k
    Mar 11, 2019 at 16:58

No, it is not possible, nor would it matter if it were. IO is typically done in units of at least 4096 bytes anyhow, and usually much more.

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    Not sure how this statement relates to my question.. logical sector size may imply smaller chunks in some part of the IO pipeline, plus needless emulation on the HDD firmware. Care to clarify?
    – Matan
    Jan 13, 2015 at 15:48
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    @Matan, I can't make sense out of your comment at all. I explained that IO is not performed 512 bytes at a time, so the fact that the disk is addressed in 512 byte sectors doesn't matter. The only time the drive has to do any emulating is if you try to do a write that is not 4k aligned, and since IO is normally done in multiples of 4k anyhow, and modern partitioning tools make sure the partition starts on a 4k boundary, that won't happen.
    – psusi
    Jan 13, 2015 at 18:36
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    This "answer" is entirely wrong, both in its claims and its conclusions. Logical size is set when a volume is formatted. Mismatches between logical and physical size have a definite cost in technologies such as flash-based storage. Jul 28, 2018 at 17:47
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    Incorrect. The problem exists any time you have software trying to operate on units smaller than the storage system's physical size. This is especially true with logic units smaller than a flash erase block size - software ends up having to duplicate the untouched portions to a new physical block, and that costs erase life, regardless if it's done by the operating system file system code or by code internal to the drive. Aug 21, 2018 at 13:06
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    Please actually read my comment. If you format the drive with a logical size smaller than the physical size, then you have a problem. Your claim only applies if you format with a logical size and alignment equaling or exceeding the physical size. Your answer was defective because it did not include that critical qualifier which you have now belated added, but only as a comment rather than to the text of the defective answer itself. Aug 21, 2018 at 13:15

Yes it is possible, however doing so would cause the drive to fill much quicker than it should. For files less than 512K, each file would then take up a full 4096K (4MB) and fill the rest of the sector with 0's due to the inability for most filesystems (NTFS and the like) to allow files to share sectors. The best option for a filesystem would be to allow variable sector sizes, however this increases the size of the MFT (master file table) and increases the risk of data corruption while reducing the ability to recover data easily. In other words, the boundaries would be not entirely known by the recovery software. So, while a 4096K logical sector size is awesome for large files, for a normal everyday use PC, it's just a bunch of 0's. Now, with that said, there is the option of storing data in the MFT itself when it comes to data smaller than the logical sector size. This, however, means that your MFT becomes huge and data would be written twice (there are two copies of the MFT on your HDD). You would also have to specify the maximum size of the MFT which can cause for problems when either you reach its maximum or the drive usage exceeds what would be free for the MFT to use. All of this is based on the usage of an NTFS file system. On the brighter side of things, NTFS does allow you to use native compression for files at the block level for any logical sector size of 4MB or less. This limitation is applied due to the way in which the NTFS compression works. 4MB blocks are read and compressed regardless of the logical sector size. This, of course, cannot occur for anything larger than 4MB in sector size due to it then crossing boundaries and losing data.

So, does this clear things up for you a bit?

  • I think this answer would be better if it didn't use NTFS as the example filesystem, as (unless things have changed in the past few years) it isn't supported well or at all under Unix and friends
    – Fox
    Feb 24, 2017 at 3:16
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    The K stands for Kilo which is a metric unit prefix meaning one thousand. 4 MiB is 1024 sectors and not 1 as you suggest. 4096 bytes is 4 KiB or 0,00390625 MiB.
    – Daniel
    Feb 15, 2018 at 4:21
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    This is not an answer. OP's talking about sector size of HDD, not block size of filesystem. That's different layer. Dec 18, 2018 at 6:07

1) Logical Sector: Called Native Sector.

Manufacture default setting. user cannot change.

Before 2010 year: 512b/sector

After 2010 year: 4k/sector.

Few manufacture provide HDD tool to change native sector.

2) Physical Sector: Called Cluster(or allocation unit - FAT windows) or Block(Linux/Unix)

User can change physical sector size 512b,1k,2k,4k,... by format or partition tool. Physical sector contains one or few more native sectors.

(example1: if you have HDD 512b/native sector: user can set 4K/Physical sector. this mean 1 cluster = 4 native sector)

(example2: if you have HDD 4K/native sector: user can set 4K/Physical sector. this mead 1 cluster = 1 native sector)

3) File system deal with Physical sector(or block or Cluster) only.

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