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On Sun, Mar 4, 2012 at 6:06 PM, Pandu Poluan <pandu@××××××.info> wrote: |
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> |
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> On Mar 5, 2012 4:59 AM, "Grant" <emailgrant@×××××.com> wrote: |
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>> |
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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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>> So it doesn't matter where the first partition starts? |
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>> |
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> |
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> Older BIOSes don't understand that hard disks now can have 4KiB sectors, so |
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> some of the "advanced format" hard disks report a sector size of 512B. But |
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> behind the scenes, the hard disk maps the logical sector to a subsector of |
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> the physical sector. |
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> |
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> The only sure fire way to find out if your hard disk uses 4KiB sectors is to |
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> open your computer and eyeball the hard disk. All 4KiB hard disks that I |
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> know of have statements on their surface that tell me so. |
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> |
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> Rgds, |
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I think I must be kind of late to this conversation, but as background |
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consider hdparm -i coupled with Google for the actual spec: |
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|
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c2stable ~ # hdparm -i /dev/sdg |
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/dev/sdg: |
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|
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Model=WDC WD10EARS-00Z5B1, FwRev=80.00A80, SerialNo=WD-WCAVU0415076 |
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Config={ HardSect NotMFM HdSw>15uSec SpinMotCtl Fixed DTR>5Mbs FmtGapReq } |
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RawCHS=16383/16/63, TrkSize=0, SectSize=0, ECCbytes=50 |
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BuffType=unknown, BuffSize=unknown, MaxMultSect=16, MultSect=off |
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CurCHS=16383/16/63, CurSects=16514064, LBA=yes, LBAsects=1953525168 |
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IORDY=on/off, tPIO={min:120,w/IORDY:120}, tDMA={min:120,rec:120} |
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PIO modes: pio0 pio3 pio4 |
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DMA modes: mdma0 mdma1 mdma2 |
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UDMA modes: udma0 udma1 udma2 udma3 udma4 udma5 *udma6 |
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AdvancedPM=no WriteCache=enabled |
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Drive conforms to: Unspecified: ATA/ATAPI-1,2,3,4,5,6,7 |
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|
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* signifies the current active mode |
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c2stable ~ # |
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With the model number it takes only a minute to determine that this WD |
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drive is a 4K sector drive. (Which is marked on the drive, as you |
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state, but I'd have to remove it to find that out.) |
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Now, in terms of performance, the only requirement (as I understand |
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it) is that all drive partition be aligned to sector addresses |
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divisible by 8. (512 * 8 = 4K) The reason 63 gives low performance is |
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because it's not naturally aligned by 8. With older versions of fdisk |
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if I started the first partition at 64 then the performance was fine |
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and only one sector was wasted. |
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M$, for whatever reason, decided to start at 2048, which is divisible |
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by 8, reserving the area at the front of the drive for (I think) their |
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boot loader and other M$-y things. My understanding of why fdisk now |
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enforces this is simply to be more careful about not overwriting the |
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M$ boot loaderif it's there. (But I could be very wrong about that!) |
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Remember, it's possible to make a dual boot system using M$'s loader |
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instead of grub, and important that fdisk doesn't mangle it when |
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someone is using that tool. |
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Just my views, |
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Mark |
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