- Dec 12, 2000
- 293
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Hi folks! I have just purchased a 60 Gb drive and I was wondering whether I can or not install W2K totally above the cylinder 1024 (around 8Gb). I have the first 15 Gb of the drive filled with Linux and the rest of the space is free. Is there such "cylinder 1024 limitation" to W2K?
P.S.: I meant totally because I know that when you install W2K in a system where there is already a running W98/ME (FAT32) - usually installed below the cyl 1024 - , the W2K boot files are automatically "smartly" placed - during the W2K install - INTO THE W98/ME FAT32 partition. When this situation occurs, if you try to boot directly the W2K partition with a third party boot manager, like XOSL or another of your preference, you'll simply get an error message (what is obvious, because the boot files are not there!). So, M$ changes the things in a way that you have first to boot the W98/ME partition, get into that damn W2K boot manager and then choose between both. In my case, the entire W2K will fit totally into a NTFS partition above cylinder 1024. My question is: Is it gonna work?
Thanks in advance!
P.S.: I meant totally because I know that when you install W2K in a system where there is already a running W98/ME (FAT32) - usually installed below the cyl 1024 - , the W2K boot files are automatically "smartly" placed - during the W2K install - INTO THE W98/ME FAT32 partition. When this situation occurs, if you try to boot directly the W2K partition with a third party boot manager, like XOSL or another of your preference, you'll simply get an error message (what is obvious, because the boot files are not there!). So, M$ changes the things in a way that you have first to boot the W98/ME partition, get into that damn W2K boot manager and then choose between both. In my case, the entire W2K will fit totally into a NTFS partition above cylinder 1024. My question is: Is it gonna work?
Thanks in advance!