Is alignment to erase block size needed for modern SSD’s

alignmentssd

I now this topic has been somewhat discussed before but I haven't managed to get a clear picture.
I’m trying to figure out whether partition alignment to erase block size is necessary with recent-ish SSD’s.
I have an OCZ Vertex 3 SSD with Windows 7 and Linux (XBMCbuntu) operating systems already installed in dual boot.

Gathered my drive’s block sizes from this thread on OCZTechnologyForum

  • Page size: 8 KB
  • File alloc. unit: 4 KB (ntfs)
  • Erase block size: 2048
    KB

Windows 7 setup created the first partition with a 1MB offset (1048576 bytes).
ms32info:

Partition Disk #0, Partition #0     
Partition Size  48,83 GB (52.428.800.000 bytes) 
Partition Starting Offset 1.048.576 bytes 
Partition Disk #0, Partition #1     
Partition Size  31,23 GB (33.531.363.328 bytes) 
Partition Starting Offset 52.429.848.576 bytes

The online tool SSD Aligment Calculator says the partition is misaligned regarding the erase block size since the latter is 2MB.

Since that online tool is oldish… is this correct ?
If so, is it important and worth the effort of moving around/resizing partitions to align to a 2MB offset and multiples-of boundaries?

Best Answer

From OCZ's "The ABC Guide" to SSD's regarding alignment:

Default Vista and Win 7 create aligned partitions at an offset of 1024KiB.

[...]

Mac OSX Disk Utility aligns partitions in HFS+. As do most newer distros of GNU/Linux.

However, as you mentioned, the Vertex 3 has an erase block size of 2MB, which is not a factor of the current alignment offset; for this reason, I would indeed recommend moving the alignment to 2MB. From the OCZ forums, here's a guide on how to use gparted to align a SSD. There is also this tool for Linux to manually align a solid state drive.

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