I would like to know a bit about boot loaders, where they save their data, and where partitions begin

http://en.wikipedia.org/wiki/GNU_GRUB
"GRUB Stage 1.5 is located in the first 30 kilobytes of hard disk immediately following the MBR."

Does that mean that the first partition starts after the first 30.5 kilobytes of the hard disk data? (MBR has 0.5 KB)

Is this a standard for most of the operating system partitioning tools? I mean, when you create the first partition on a hard disk, in Windows or Linux, the partition will start always exactly after the first 30.5 kilobytes of the hard disk?

thanks

Yes - the first partition is considered to start on cylinder 1 (not zero). This is simply an old convention that is changing, but is true for the majority of cases.

I think the exact wording is

"stage 1.5. GRUB Stage 1.5 is located in the first 30 kilobytes of hard disk immediately following the MBR"

This could mean stage1.5 is in the first sector but not in the first partition.

Since each head has 63 sector the first sector of the first head is also reserved in which the only first 512 bytes are read by the Bios and is commonly called the MBR.

The first partition actually starts at 63x512 = 32256 bytes. 52256 - 512 (the MBR) = 31744 bytes which can accommodate the stage1.5

As a proof here is a partition table with 5 partitions all 10 cylinder each.
By definition 10 cylinders = 10x255x63x512 = 82252800 bytes. When this is divided by 1024 (unit of 1 block in LBA) to give 80325 blocks which match exactly with the size of sdb2, sdb3 and sdb4.

The sdb1 is the first partition and loses exactly 32256 bytes or 31.5 blocks therefore it has 80325-31.5 = 80293.5 blocks. In fdisk this is shown as 80293+ blocks.

It is interesting to see the logical partition sdb5 is the same size as sdb1. This is because every logical partition must have it own partition table and its first sector has been used for that purpose. The partition table of sdb1 to sdb4 are contained in the MBR.

Stage1.5 isn't used much nowadays. I never use it myself. The stage1 is normally hard-coded the hard disk address of the stage2. This enable Grub to boot an operating system virtually from any partition in a hard disk.

Each partition has a boot sector at the beginning of the partition and this is where the boot loader resides unless the user opts for installing it in the MBR.

Great answer saikee, thanks a lot !!
I would like to ask a few more questions:
- where exactly I can find the stage2 on the linux partition?
- how is Windows using the 30 KB before the first partition, does it install the NT loader there?
- if I install the boot loader (grub) on a partition, and not on the MBR, then where does it install the stage 1.5? Do partitions have also 30 KB at the beginning for the boot loaders?

thanks

zoombee,

Due to the size of stage2, of more than 120k, I believe Grub does not place stage2 in the boot sector of a partition but rather the stage1 and stage1.5. It is likely both stage1 and stage1.5 have hard coded the disk address of stage2 which is always inside either /grub (for distro having /boot partition like Red Hat or some Open Suse) and /boot/grub (for all distros that has /boot as subdirectory to /). Stage1 is also inside the boot sector when a Linux places its boot loader inside its root partition. A proof of this is when a Xp or Vista boots a Linux the pre-requisite is to copy the first 512 bytes of the Linux partition and store it in the "c" drive" of the XP or Vista. This allows Xp or Vista to hand over the PC control to Linux without leaving the NTFS partition or has the ability to read a Ext2/3 partition.

When Grub is set up to boot from the MBR it is possible for stage1 to load the stage2 directly using the hard disk location because stage1.5 can be missing will not cause a failure. This is reported every time if you use the "setup" command of Grub. Whichever Linux you use all the system files of Grub are always inside /boot/grub or /grub.

The M$ system traditionally has a common MBR that is interchangeable between Dos up to Xp. Only Vista and Win7 use a different one but still its function is the same. M$ MBR is best imaged like a blind person searching the first 4 primary partitions and boots whichever one that has the bootable flag set "on" position. In other word the M$ MBR does not know which system it is booting and I have successfully ued Xp's MBR to boot up Grub installed in a data-only Fat partition without any OS inside. Every partitioning tool exists in PC will switch off the existing bootable flag if the owner switches another primary partition as bootable. In M$ term the partition is called "active". Linux calls it bootable but Linux never uses the bootable flag.

For the above reason every M$ system even up to now must be booted from a bootable primary partition, at least in first instance as Win2k, Xp, Vista or Win7 can be installed in a logical partition as long as there is a Fat or NTFS primary partition to house its boot loader.

The above suggests a M$ system always places its boot loader inside the boot sector of the partition it resides. M$ systems have a big issue with it as the boot loader occupying the boot sector is backward compatible only. Therefore a W2k installed after Xp will not be able to boot Xp and Xp installed after Vista can't boot Vista etc.

Thanks again. I just want to make a minor observation:
According to this:
http://en.wikipedia.org/wiki/Extende...1_.28unreal.29

Every logical partition has reserved 20 sectors (1 for MBR and 19 are kept unused)

So it's 20, not 30.5

I don't know why its 30.5 in your example

Perhaps you should read the link you posted

I see, the empty space(in sectors) is equal with the number of sectors per track minus 1. Usually it's 63 sectors per track, that makes 31.5 KB

In this case, the information about GRUB on Wikipedia is not totally accurate: the space for Stage 1.5 is 31 KB, not 30 KB