So, I have one of these newfangled external Samsung T3 USB SSD drives here, sized at 250 GB.
Let's see what fdisk
has to say about that:
Disk /dev/sdb: 232.9 GiB, 250059350016 bytes, 488397168 sectors
Units: sectors of 1 * 512 = 512 bytes
Sector size (logical/physical): 512 bytes / 512 bytes
I/O size (minimum/optimal): 512 bytes / 33553920 bytes
Disklabel type: dos
Disk identifier: 0x7df0da81
Device Boot Start End Sectors Size Id Type
/dev/sdb1 * 64 488392128 488392065 232.9G 7 HPFS/NTFS/exFAT
Sectors are still 512 byte large (I suspect underneath the software layer, things are a bit different), the minimum I/O size is 1 sector, the optimal I/O size is 3553920 byte = 65535 sectors ≈ 31.995 MiB
We see that the factory-performed formatting consists in a single partition from block 64 to block 488392128, which is ~232.88 GiB or ~244.19 GB.
So I just want to repartition using fdisk
. fdisk
insists that the first partition start at sector 65535. Maxing it out, it ends at sector 488397167, a bit further than the factory-issued partition:
Device Boot Start End Sectors Size Id Type
/dev/sdb1 65535 488397167 488331633 232.9G 83 Linux
I know about the 1 MiB alignment boundary which manifests itself by fdisk
putting the first partition at sector 2048. The reason for that seems to be a historical accident due to a wart present in the Logical Disk Manager of Microsoft Vista.
But what about this new 65535 sector "alignment"?
Best Answer
If that
65535
refers to 512-byte sectors, it would not be aligned at all. Unfortunatelyfdisk
with dos partitions can get funny ideas about drive geometry.Ignore whatever
fdisk
is trying to do, just use1MiB
alignment anyway. If you don't need dos partitions for any particular reasons, switch to gpt. Iffdisk
doesn't work, switch toparted
,gdisk
, or whatever works.