Talk:Solid State Drives

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Revision as of 10:38, 1 July 2010 by Graysky (talk | contribs)
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@Aedit - thank you very much for the contribution to this article. Can you explain how you arrived at the numbers that relate to the head and cylinder choices? 2^8 times 49 and how does that translate into 2^8 times 512 = 128 KiB alignment?

  • Ted Tso recommends using a setting of 224*56 (=2^8*49) which results in (2^8*512=) 128 KiB alignment
  • While others advocate a setting of 32*32 (=2^10) which results in (2^10*512=) 512 KiB alignment

@Graysky - you're welcome. The alignment number is the largest power-of-two divisor of the cylinder boundary positions on the disk. The size in bytes of the cylinders is H*S*512 = (tracks per cylinder) * (sectors per track) * (sector size). So factorize H, S and sector size (512=2^9) into prime factors, and take all the 2s. In the first case above we have to ignore the non-power-of-two factor of 7^2=49.

@Aedit - interesting, so in my case, using an Intel X25-M, I used the -H 32 -S 32 to give a 512 KiB alignment which should be fine, yet Ted suggests using the values that give a 128 KiB alignment (although 128 is a factor of 512). Is there any advantage to using the smaller number, or to put it another way, is there a disadvantage to using the larger number?