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Re: Can't install GRUB


From: Felix Miata
Subject: Re: Can't install GRUB
Date: Mon, 05 Nov 2012 18:50:29 -0500
User-agent: Mozilla/5.0 (X11; Linux i686; rv:16.0) Gecko/20121025 SeaMonkey/2.13.2

On 2012-11-05 22:42 (GMT+0100) Paolo composed:

Just for my culture (if not too OT), My current partition schema is:

Filesystem           1K-blocks      Used Available Use% Mounted on
/dev/md1              66063436   2507340  60200176   4% /
tmpfs                  1961908         0   1961908   0% /dev/shm
/dev/md0                233321     55309    165966  25% /boot
/dev/md2             980486564    204360 930474740   1% /home

Disk /dev/sda: 2000.4 GB, 2000398934016 bytes
255 heads, 63 sectors/track, 243201 cylinders
Units = cylinders of 16065 * 512 = 8225280 bytes
Sector size (logical/physical): 512 bytes / 4096 bytes
I/O size (minimum/optimal): 4096 bytes / 4096 bytes

What this seems to say is:

 "Cylinders":243201
 Heads:255
 Sectors:63
 Bps:512
 Size:0xE8E074C1=1907726 MiB
 255*63*512=8225280 "cylinder" size
 Internal sector size 4096 bytes

     Device Boot      Start         End      Blocks   Id  System
/dev/sda1   *           1          30      240943+  fd  Linux raid
autodetect
Partition 1 does not start on physical sector boundary.
/dev/sda2              31        8386    67119569   fd  Linux raid
autodetect
Partition 2 does not start on physical sector boundary.
/dev/sda3            8386      132401   996150488   fd  Linux raid
autodetect

I manually tried various values of Start/End sector to eliminate the
"Partition N does not start on physical sector boundary." warning.

How can I calculate the right values?

and .. is this a real problem for a working system (performance or other
problems)?

Used with a logical sector size of 512, 255 logical heads and 63 logical sectors causes a data misalignment, and consequently, a performance penalty. 512 X 255 X 63 does not create an even multiple of 2048 or 4096. The question is whether your HD actually does use 2048 or 4096 byte physical sector size.

If you find out your HD is in fact using an internal sector size that is not 512 bytes, you should repartition to:

 Cylinders:1907727
 Heads:64
 Sectors:32
 Bps:512
 Size:0xE8E07800=1907727 MiB
 64*32*4096=8388608 cylinder size

to get the required alignment of logical sectors with physical sectors that avoids the penalty.

Some of the newer partitioning tools make this easy to do. Older tools weren't built with the knowledge that physical and logical sectors would not always be the same size, and that failure to align could cause a performance hit.

For more details, see: http://en.wikipedia.org/wiki/Advanced_Format
--
"The wise are known for their understanding, and pleasant
words are persuasive." Proverbs 16:21 (New Living Translation)

 Team OS/2 ** Reg. Linux User #211409 ** a11y rocks!

Felix Miata  ***  http://fm.no-ip.com/



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