Hello All,
I am trying to install OpenDSB on on a computer with a single 20GB physical disk with XPpro sitting on a 3GB primary partition. Before rebooting (out of XP), I used Partition Magic Pro7 to create a second, but hidden, primary partition, which I chose not to format. I then popped in the floppy31.fs OpenBSD install floppy and rebooted into the Open BSD install utility.
One of the first things the install utility does if you do not want the install program to consume the entire physical disk is to drop you into fdisk(8). This is fine and I changed the second primary partition type from Unknown to A6, which is the OpenBSD partition type. I then set it to Active, and exited the fdisk(8) program, saving changes as I left. Upon leaving fdisk(8), the installer drops you into disklabel.
What I am having a difficult time is understanding the output from print command in disklabel. Am I looking at the partition geometry for just one of the primary partitions on the physical disk (and if so, which one?), or the geometry of the entire disk? I've been consulting the OpenBSD Install FAQ, but they use the geometry of a floppy disk, yeah right, and assume you only want to install OpenBSD using the entire disk. I have also read the Install.pt document, but thats just a run of the mill 101 on partitions in general.
While in disklabel I do wish to follow along with the FAQ's example, but I also do not wish to blow away my XP install. Its not that I cannot afford to lose XP as this is all on a PC that I use for experimentation anyway, but the point is to prove to myself OpenBSD can be installed onto a single physical disk with another OS on it.
Also, the inital portion of the print output states the device as being /dev/rwd0c where I know from fdisk(8) wd0 is the physical disk. Confused? I am! The OpenBSD install routine makes installing Linux look like the easiest dang thing in the world! I posted this problem on dslreports, too, in the Unix section.
I am trying to install OpenDSB on on a computer with a single 20GB physical disk with XPpro sitting on a 3GB primary partition. Before rebooting (out of XP), I used Partition Magic Pro7 to create a second, but hidden, primary partition, which I chose not to format. I then popped in the floppy31.fs OpenBSD install floppy and rebooted into the Open BSD install utility.
One of the first things the install utility does if you do not want the install program to consume the entire physical disk is to drop you into fdisk(8). This is fine and I changed the second primary partition type from Unknown to A6, which is the OpenBSD partition type. I then set it to Active, and exited the fdisk(8) program, saving changes as I left. Upon leaving fdisk(8), the installer drops you into disklabel.
What I am having a difficult time is understanding the output from print command in disklabel. Am I looking at the partition geometry for just one of the primary partitions on the physical disk (and if so, which one?), or the geometry of the entire disk? I've been consulting the OpenBSD Install FAQ, but they use the geometry of a floppy disk, yeah right, and assume you only want to install OpenBSD using the entire disk. I have also read the Install.pt document, but thats just a run of the mill 101 on partitions in general.
While in disklabel I do wish to follow along with the FAQ's example, but I also do not wish to blow away my XP install. Its not that I cannot afford to lose XP as this is all on a PC that I use for experimentation anyway, but the point is to prove to myself OpenBSD can be installed onto a single physical disk with another OS on it.
Also, the inital portion of the print output states the device as being /dev/rwd0c where I know from fdisk(8) wd0 is the physical disk. Confused? I am! The OpenBSD install routine makes installing Linux look like the easiest dang thing in the world! I posted this problem on dslreports, too, in the Unix section.