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Old 03-08-2011, 02:45 AM   #1
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MSDOS and GPT partitioning in 1 Hard Disk

Well I am not a newbie per se ...But I thought my question would fit in only here!

I have a BIOS affected by 1024 cylinder limit ,,(Just to explain , the BIOS will recognize only 137GB no matter how big the HDD is (160/250/320/500))
This problem doesnt affect the OS in any way --Linux can read the entire disk and so can WIN after SP1 of XP...

This problem is only upto the point of MBR handing over boot instructions to Bootloader..In that case LiLo or Grub1 or Grub2 would fail if the kernel of the OS is after the 137GB

I want to know if I can mix 2 partitioning schemes..Say use MSDOS partitioning till the limit of say 100 GB and for the rest 60 GB use GPT partitioning and install GRUB2 (as I read only GRUB2 supports GPT cleanly) ...Then , maybe I can do away with options of restricting /boot within 137 , including ata options etc...

Last edited by vikrang; 03-08-2011 at 03:02 AM.
Old 03-08-2011, 06:14 AM   #2
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Not as far as I know ... it's one or the other
Old 03-08-2011, 04:21 PM   #3
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I am still wondering how one could ever use past the ata limit. No matter your disktype the bio interface can't use enough bits to overcome the disk.

I think I played with this a long time ago and found the data was not really there but I could be wrong.
Old 03-09-2011, 09:37 AM   #4
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The problem is only with the BIOS, and since the OS doesn't use the BIOS, there is no problem accessing the entire disk with Linux (or other OS). The bootloader is somewhere in between and I don't know how much or even if it uses the BIOS (some might use it, others not, I suppose). Even if it does use the BIOS, as long as the crucial bits are within reach of the BIOS, then it should still be able to boot at least one OS. That should mean that as long as the /boot partition is within the first 1024 cylinders, it should be able to load the kernel, and as long as the kernel has the ATA driver built in or in an initrd, the kernel should be able to take over and access the entire disk.
--- rod.


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