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On Mon, Mar 5, 2012 at 11:00 AM, Alex Schuster <wonko@×××××××××.org> wrote: |
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> Grant writes: |
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> |
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>> > The performance is only impacted if the sector size is something other |
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>> > than 512 bytes. The newer 4K sector size used by some higher density |
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>> > drives requires that you start partitions on a sector boundary or they |
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>> > will perform badly. There isn't an actually performance need to |
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>> > actually start on 2048 but the fdisk-type developer folks are doing |
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>> > that to be more compatible with newer Windows installations. |
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>> |
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>> All my drives says this from fdisk: |
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>> |
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>> Units = sectors of 1 * 512 = 512 bytes |
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>> Sector size (logical/physical): 512 bytes / 512 bytes |
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>> I/O size (minimum/optimal): 512 bytes / 512 bytes |
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> |
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> Neither fdisk nor hdparm seem to get the correct sector size, at least |
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> not always. That's what I read somewhere (and not only once), and it's |
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> true for my own 2TB drive which I know to have a 4K sector size. I'd say |
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> you have to look up the specs on the vendor's web size to be sure. |
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> |
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>> So it doesn't matter where the first partition starts? |
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> |
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> If you have 4K sectors (and not a Seagate drive with SmartAlign [*]), it |
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> does. |
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> |
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> BTW, here's some benchmarks I just stumbled upon: |
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> http://hothardware.com/Articles/WDs-1TB-Caviar-Green-w-Advanced-Format-Windows-XP-Users-Pay-Attention/?page=2 |
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> |
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> [*] I don't want to sound like I'm advertising for Seagate here, but at |
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> least it seems that with SmartAlign the performance impact will be |
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> much less, so it might not be worth the trouble of re-partitioning drives |
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> that are already being used. |
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> |
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> Wonko |
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|
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Also, it counts with SSDs, where alignment,or lack therof, with the |
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erase block becomes noticeable on write performance. Finding the |
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actual size of an erase block for most SSDs is rather difficult, but |
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1MB tends to be a reliable guess as a multiple of *that* as well. |
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|
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-- |
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Poison [BLX] |
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Joshua M. Murphy |