I just plugged into USB A 3.0 / C 3.1 my new external HDD to Debian Buster system.

The disk was sold as LaCie 2.5" Porsche Design P'9227 2TB USB-C.

Here is the output of fdisk -l /dev/sdc:

Disk /dev/sdc: 1.8 TiB, 2000398934016 bytes, 3907029168 sectors
Disk model: P9227 Slim      
Units: sectors of 1 * 512 = 512 bytes
Sector size (logical/physical): 512 bytes / 4096 bytes
I/O size (minimum/optimal): 4096 bytes / 33553920 bytes

I just read some articles about 4k-emulated drives (512e), this one should be the case.

I am confused as to how to format it with NTFS.

I tried to use my brain, and here is what I came with:

  • Start sector of the partition should probably start on 4096 sector (?)

    So I created a partition with gdisk like this:

    Device     Start        End    Sectors  Size Type
    /dev/sdc1   4096 3907029134 3907025039  1.8T Microsoft basic data
  • Sector size should probably be forced with the --sector-size option like I did (?) issuing:

    mkfs.ntfs --no-indexing --verbose --with-uuid --label EXTERNAL_2TB --quick --sector-size 4096 /dev/sdc1


Windows 10 fully updated did not recognize the partition and asked me to format, I used my favorite tool for that, and back to Linux here is the output of fdisk -l /dev/sdc:

Device     Start        End    Sectors  Size Type
/dev/sdc1   2048 3907028991 3907026944  1,8T Microsoft basic data

So why it must start at sector 2048, I don't understand.


I don't understand what I am doing wrong in terms of compatibility with Windows. Every time I re-partition it / re-format it and boot Windows and plug the drive in, it just offers me to Format it itself.

I am quite positive I tried everything from inside gdisk + mkfs.ntfs.

I would like to know why I am unable to do the same as Windows does from my Linux CLI.

I will answer all questions tomorrow morning as well as comments.

I am now running:

pv --progress --timer --eta --rate --average-rate --bytes -s 1953314876k < /dev/zero > /media/vlastimil/LACIE_2TB/zero

with an expected speed of 123 MiB/s.

  • 1
    The only weird thing here is the bogus "optimal io size". Nothing wrong with start sector 2048 or 4096 (MiB alignment). As to why Windows doesn't like it, who knows? Jan 27, 2019 at 19:37
  • 2
    There is no need for the partition to start on sector 4096. The only requirement for optimal alignment is that the starting sector number is divisible by eight. The physical sectors are aligned to logical sectors N*8. 4096/512 = 8. Jan 27, 2019 at 20:34
  • I fail to understand why the disk not working in Windows is a concern here, do not understand why you are finding it strange it starting at sector 2048 (it is pretty normal), and I think you are mixing up logical and physical sectors. However, you true problem is that you external adapter sees the disk in a different way than pluggin it in your Windows machine. There is another question that talks about that, but there are people here that are able to explain it better. Jan 27, 2019 at 20:34

2 Answers 2


A physical sector size of 4096 means that the data on the drive is laid out in units of 4096 bytes, i.e. disk comprised of sequential "compartments" of 4096 bytes, that have to be written atomically. For compatibility reasons, most disks with 4096 byte sectors present themselves as having traditional 512 byte "logical sectors", which means the addressing unit is a 512 byte block.

The practical implication of this emulation of a 512 sector drive with an underlying disk with 4096 byte sectors is a potential performance issue. When writing a single 512 byte sector to a 512e disk, the drive must read the whole 4096 byte sector containing the 512-byte sector, modify the sector in RAM (on the disk controller) by replacing the 512-byte sector with the new contents, and finally write the whole 4096 sector back to the disk. Things get worse if you are reading or writing a couple of consecutive 512 sectors that happen to cross a 4096 sector boundary.

File systems usually lay out their data structures well, i.e. they are aligned to multiples of at least 4096 bytes, so the bigger sector size normally does not present a problem. This all breaks down, however, if the partition containing the file system itself is not aligned properly. In the case of a 512e disk, the partitions should be aligned so that the first 512-byte logical sector number is a multiple of eight.

  • Good thinking. This is a known issue, specially when aligning VMs because all the improvement gains add up and benefit the infra-structure as a whole. +1 Jan 28, 2019 at 17:49

I did not use this external HDD for a long while, but now I needed it, so the correct command to issue (example) follows:

mkfs.ntfs --no-indexing --verbose --with-uuid --label Lacie_2TB --quick --sector-size 512 --cluster-size 4096 /dev/sdX1


  • At the end, adjust the device and partition number.

  • If you want/need indexing, leave out --no-indexing.

  • Adjust the label (Lacie_2TB) to what you fashion.

  • I switched in my head the sector size and cluster size, this was the culprit.

In case you would like to create ext4 file system on it, I would recommend something like this:

mkfs.ext4 -v -m 0 -E lazy_itable_init=0,lazy_journal_init=0,nodiscard -L Lacie_2TB /dev/sdX1

Notes on switches used:

  • -v: be verbose
  • -m 0: do not reserve any space for root
  • -E lazy_itable_init=0,lazy_journal_init=0: initialize fully, not using lazy
  • -E nodiscard: it is a HDD, not SSD, so...
  • -L Lacie_2TB: file system label for the OS

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