At Fri, 30 Oct 2009 11:14:54 -0500 CentOS mailing list centos@centos.org wrote:
Cylinders are largely logical - even in magnetic disks you have no way of knowing if the logical cylinder matches up with the physical construct of a cylinder on the disk medium - in any modern(15 years ?) disk they won't. Don't worry about cylinders, just align your fs to the stripe/sector.
Cylinders only ever existed (as a *physicly mapped* thing) with MFM, RLL, and early IDE hard drives (IDE drives grew out of 'hard cards', which were MFM/RLL drives bolted to a XT (pre/early ISA) expansion card with a MFM/RLL drive controller). (And yes, floppies. Not sure about CD/DVD-ROMS, but that is a completely different can of worms.) SCSI disks *never* had any physical mapping for cylinders, heads, OR sectors. The SCSI command protocol uses LBA addressing and has from day one. At this point, CHS addressing is pretty much a BIOS 'fantasy' for any modern disks.
Obviously, the concept of a cylinder starts to go out the window with RAID, SSD's, etc.
More information about partition alignment here - http://www.ocztechnologyforum.com/forum/showpost.php?p=335049&postcount=...
--Blake
-------- Original Message -------- Subject: [CentOS] Stripe vs Cylinder alignement... From: John Doe jdmls@yahoo.com To: centos@centos.org Date: Friday, October 30, 2009 9:45:59 AM
Hi,
I modified my kickstart to do some custom partioning and formating in a pre-install script. I am trying to align the partitions on the RAID stripe (and format with a correct stride). But, sfdisk complains that it does not start/end on a cylinder boundary (used -L option to limit complaining). Since the cylinder size is not a multiple of the stripe size, I cannot align on both. I tried to align the begining on the stripe and the end on the end of a cylinder, but sfdisk still compains... Basicaly, I have a 128KB (256 sectors) stripe, and 255*32 = 8160 sectors cylinders. What I am doing is: begin = ( begin / 256 ) * 256 end = ( end / 8160 ) * 8160 -1 So, for my first partition (96MB): begin=256 size = ( ( 96 * 1024 * 2 ) / 8160 ) * 8160 = 195840 sectors end = 195840 - 1 - 256 = 195583 Any idea what I am doing wrong in my calculations or logic?
Thx, JD
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