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 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"?